Resonances, waves and fields

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19. References used for the mathematics of relativity

Classical Electrodynamics by J. D. Jackson, a widely used graduate physics level textbook.
Engineering Electromagnetics by Hayt and Buck, a junior level engineering electromagnetic textbook. No relativity, but electric and magnetic field equations are clearly explained.
Classical Electricity and Magnetism by Panofsky and Phillips. A senior level textbook on electric and magnetic fields. Has a good section on relativity, including that applied to electric and magnetic fields, probably more complete than Jackson.
Field and Wave Electromagnetics by Cheng, a junior level engineering electromagnetic textbook somewhat more sophisticated and complete than Hayt and Buck.
Feynman Lectures in Physics, a senior level physics text that has electromagnetics, relativity and general physics. Intended as a text for freshmen at Cal Tech but makes good reading in general.
Fields and waves in communication electronics by Ramo, Whinnery and Van Duzer, a graduate level engineering electromagnetic textbook. Theoretically less sophisticated than Jackson, but more complete on applications. Does not cover relativity.
Lorrian and Corsan, a junior/senior physics textbook. Follows a unique approach to building up magnetic fields as moving electric fields using relativistic concepts similar to those presented here.
Berkely physics course - volume 2 - electricity and magnetism, a sophomore level physics textbook that treats things in its own special way. Makes good reading.
Wikipedia, many relativity topics are covered at all sorts of levels.
There are many other very good sites on the internet both in text form and in video form.

Separating the transformed Maxwell equations

2011-07-17T10:54:00.000-07:00

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18. Separating the transformed Maxwell equations

In Fig. 18.1 below, we restate the transformed Maxwell Equations as derived in the previous two chapters:

	The first Maxwell equation as transformed: Chapter 16, Fig. 16.5
	The second Maxwell equation (a vector equation) as transformed: Chapter 17
	The third Maxwell equation as transformed: Chapter 17
= 0	The fourth Maxwell equation (a vector equation) as transformed: last equation in Chapter 17
Fig. 18.1. Transformed Maxwell Equations as derived in the previous two chapters.

Unfortunately Maxwell's equations transformed as mixes of Maxwell's equations in the prime reference frame. We now need to separate the mixes. In order make the equations more manageable, we will use the abbreviations shown below in Fig. 18.1a. Note that each of the newly defined quantities are the left sides of the wished for Maxwell equations in the primed frame. We hope to show that the above equations (in Fig. 18.1 above) imply that all these quantities listed in Fig. 18.1a are zero.

	Definition of m1, equal to the left side of the first transformed Maxwell equation.
	Definition of the vector m2, equal to the left side of the second Maxwell equation (a vector equation).
	Definition of m3, equal to the left side of the third Maxwell equation.
	Definition of the vector m4, equal to the left side of the fourth Maxwell equation (a vector equation).
Fig. 18.1a. Definitions of variables used to separate out and isolate the various Maxwell equations.

Using the definitions in Fig. 18.1a, we can be expressed the equations in Fig. 18.1 as:

(C1)	The first Maxwell equation as transformed written in terms of m1 and m4_x.
(C2x) (C2y) (C2z)	The second Maxwell equation (a vector equation) as transformed and written in terms of the variables m2 and m3.
(C3)	The third Maxwell equation as transformed and written in terms of the variables m3 and the x-component of m2.
(C4x) (C4y) (C4z)	The fourth Maxwell equation (a vector equation) as transformed and written in terms of the variables m1 and m4.
Fig. 18.2. Maxwell's equations as transformed by the above work, using the Lorentz transforms and written in terms of the variables defined in Fig. 18.1a .

We need to solve the above equations for the variables m1, m2_x, m2_y, etc:

Inspecting the equations, we see that four of the variables are equal to zero, i.e. (C2y), (C2z), (C4y), and (C4z). Thus ⇒	m2_y = m2_z = m4_y = m4_z = 0
Multiplying (C1) by V and subtracting it from (C4x) yields m4_x − V²·m4_x = (1 − V²)m4_x = 0 which implies that ⇒	m4_x = 0 .
Substituting m4_x = 0 into (C1) yields ⇒	m1 = 0 .
Similarly, we can multiply (C2x) by V and add this to (C3) to yield: m3 − V²m3 = (1 − V²)m3 = 0 which implies that ⇒	m3 = 0.
Inserting m3 = 0 into (C2x) gives ⇒	m2_x = 0 .
Fig. 18.3. Solving for m1, m2_x, m2_y, etc.

In summary, all eight of the the variables m1, m2_x, m2_y, ... , m4_z are zero. Using the definitions of these variables as listed above in Fig. 18.1a gives:


Fig. 18.4. Final result of the above work to transform Maxwell's equations using the Lorentz transforms. This shows that in the moving reference frame the laws governing electric and magnetic fields appear to be the same as they were in the stationary reference frame.

Thus, we have successfully shown that the Lorentz transforms when operating on Maxwell's equations, produce the same equations in the new reference frame.

Most text books would do this derivation much more efficiently using more advanced methods, briefly mentioned in Chapter 14. We have presented here this vector algebra derivation as an alternative, in part to illustrate that most physics can be derived in a number of ways and we can learn from each way.

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Transforming the second, third and fourth Maxwell equation

2011-07-17T08:22:00.000-07:00

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17. Transforming the second, third and fourth Maxwell equations

Transforming the second Maxwell equation

We start transforming the second Maxwell equation. This is a vector equation:

Because of its complexity, we now finish the transformation component-by-component:

Note that 1/γ² = 1 − β² where β = V/c (see the definition of γ ).

Transformation of the third Maxwell equation

The third Maxwell equation transforms similarly to the first one above:

Transformation of the fourth Maxwell equation

The fourth Maxwell equation is a vector equation and will transform in a similar manner to the second Maxwell equation above. We start by transforming the left side of the fourth Maxwell equation:

Now we transform this component-by-component:

The right side of the fourth Maxwell equation is transformed as:

where we have used the relation: c² = 1/(ε₀μ₀ ) which can be solved for μ₀, i.e.: μ₀ = 1/(c²ε₀) .

Combining the above equations, we get:

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Transforming Maxwell equations - intro and 1st equation

2011-07-16T09:12:00.000-07:00

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16. Transforming Maxwell equations-intro and 1st equation

Maxwell's four equations:

. (16.1)

. (16.2)

. (16.3)

. (16.4)

We would like to consider how the equations need to be changed so that they apply in a lab aboard a space ship moving at velocity, V, in the x direction. The most obvious effect is that a stationary charge is now seen as moving in the negative x direction and thus being seen as a current density as well as a charge density.

We are trying to verify that the Lorentz transformations result in the same set of Maxwell's equations. Thus we will use the Lorentz transformations, as well as the transformed electric and magnetic field equations which resulted from the Lorentz transforms.

Transformation of partial derivatives

One tricky aspect of transforming Maxwell's equations from the stationary reference frame to the moving one is the need to transform the partial derivatives in those equations from using x, y, z, and t as variables into using x', y', z', and t'. In Fig. 16.1 we show the standard inverse Lorentz transform of these variables and some useful partial derivatives of these. We don't show the relationships of y and z to y' and z' since these are trivial, i.e. y' = y and z' = z and ∂y'/∂x=0, ∂y'/∂y=1, ∂y'/∂z=0, ∂y'/∂t=0, ∂z'/∂x=0, ∂z'/∂y=0, ∂z'/∂z=1, and ∂z'/∂t=0.



Fig. 16.1a. Inverse Lorentz transforms of x and t.	Fig. 16.1b. Partial derivatives of the Lorentz transforms at the left.	Fig. 16.1c. More partial derivatives of the Lorentz transforms at the left.

In Fig. 16.2 below we show the chain rule for partial derivatives applied to a typical electromagnetic field quantity, E_x in this example. We calculate the differential of E_x in terms of differential changes of the new variables x', y', z' and t'. This equation involves taking partial derivatives of E_x with respect to each of these new variables while the other new variable is held constant. One important aspect to note is which variables are held constant in the various terms.

Fig. 16.2. Calculation of the differential of a typical electromagnetic field quantity E_x in terms of differentials of the variables x' and y' of the moving reference frame.

In the next step, in Fig. 16.3, we form the partial derivative of E_x with respect to x, in the form it is found in the original Maxwell's equations, but now in terms of the partial derivatives in Fig. 16.2 above. This is one step in transforming Maxwell's equations from being written in terms of x, y, z, and t to being written in terms of x', y', z', and t'.

Fig. 16.3. Calculation of the partial derivative with respect to x of a typical field quantity (E_x shown here as an example). This operation is one step in changing the variables of E_x from x, y, z, and t into x', y', z', and t', i.e. into the moving reference frame. The second equation above (boxed) is simply the first one rewritten separating out the operations from E_x.

In Fig. 16.3, we have dropped the terms that involve partial derivatives with respect to y' and z' because:

The whole equation (see the left hand side) is directed towards calculating the partial of E_x with respect to x while keeping y, z, and t fixed.

But y' = y and z' = z.

Thus y' and z' are to be kept fixed and are not variables for this equation.

Thus we are only interested in the x' and t' dependence for this equation.

Fig. 16.4. Calculation of the partial with respect to t.

Transformation of the first Maxwell equation

We start out transforming the left side of the first Maxwell Equation (as shown in (16.1) above):

We next write out the transformation of the right side of the first Maxwell equation (see (16.1) above).

Finally, we use these in the first Maxwell equation (written with all terms moved to the left):

Fig. 16.5. First Maxwell equation transformed to moving reference frame.

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Transformation of electric and magnetic fields

2011-07-16T08:19:00.000-07:00

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15. Transformation of electric and magnetic fields

Computational strategy

We cannot feel or see electric and magnetic fields. We hypothesize them to account for forces and trajectories of charged particles and currents. Electric and magnetic fields are detected or "felt" with charges and currents. Along these lines, in order to figure out how electric and magnetic fields transform under change in reference frame, we must see how the Lorentz force changes upon change of reference frames:

. (15.1)

Box 15.1. Writing the force F in terms of forces in the primed reference frame
To start, we need the transformation for force. We define force by the effect it has on the motion of objects using Newton's second law: F = dp/dt . (15.2) We need the momentum transform from Chapter 11, which to repeat is: which means , (15.3) where ε is energy as discussed in Chapter 11. Script "E" is used to distinguish energy from electric field. We've also assumed that the velocity V of the reference frame does not change, which also means γ does not change either. We also need the inverse Lorentz transform for time from Chapter 4: which means that . Dividing dp by dt gives us the force in Equation (15.2): . (15.4) Dividing both the numerator and denominator by dt' and replacing the components of dp'/dt' with the corresponding components of F', we have: . (15.5) We can relate the time derivative of energy in the above equation to force using: , (15.6) This makes (15.5) become: . (15.7) Continuing to simply: (15.8)

Box 15.1. Writing the force F in terms of forces in the primed reference frame

To start, we need the transformation for force. We define force by the effect it has on the motion of objects using Newton's second law:

F = dp/dt . (15.2)

We need the momentum transform from Chapter 11, which to repeat is:

which means , (15.3)

where ε is energy as discussed in Chapter 11. Script "E" is used to distinguish energy from electric field. We've also assumed that the velocity V of the reference frame does not change, which also means γ does not change either.

We also need the inverse Lorentz transform for time from Chapter 4:

which means that .

Dividing dp by dt gives us the force in Equation (15.2):

. (15.4)

Dividing both the numerator and denominator by dt' and replacing the components of dp'/dt' with the corresponding components of F', we have:

. (15.5)

We can relate the time derivative of energy in the above equation to force using:

, (15.6)

This makes (15.5) become:

. (15.7)

Continuing to simply:

(15.8)

Next we need to transform the components of u' into corresponding components of u. This may seem like a step backwards, but in the end we need primed variables times unprimed velocities, u, to match the form of (15.1) above where E and B will be functions of E', B' and V (the relative velocity of the primed and unprimed reference frames).

Box 15.2. Converting the velocities in the primed reference frame to the unprimed frame.
We start by repeating the transform for velocity from Chapter 9: (15.9) The messiest factor in this transform will be converting the u'_x in the 1 + u'_xV/c² factor occurring in the denominators of (15.8) to u_x. We first transform and simplify this factor: We use this to transform the u' in (15.8) into corresponding components of u as promised: . We cancel factors as appropriate to write: (15.10)

Box 15.2. Converting the velocities in the primed reference frame to the unprimed frame.

We start by repeating the transform for velocity from Chapter 9:

(15.9)

The messiest factor in this transform will be converting the u'_x in the 1 + u'_xV/c² factor occurring in the denominators of (15.8) to u_x. We first transform and simplify this factor:

We use this to transform the u' in (15.8) into corresponding components of u as promised:

We cancel factors as appropriate to write:

(15.10)

Box 15.3. Introducing electric and magnetic fields.
We next need to substitute the Lorentz force (containing electric and magnetic fields) in for the forces. Just to remind you, the Lorentz force in the primed reference frame is: . (15.11) When we substitute the various force components of (15.11) into (15.10), F becomes: . (15.12) This last substitution has reintroduced some primed velocities. We transform these into velocities in the unprimed reference frame using (15.9): . (15.13) With some rearrangement, this becomes (15.14) where a number of terms in the first line have been collected and designated as T given by: . (15.15) T can be further simplified by the following steps: Substituting this into (15.14) we have: and (15.16)

Box 15.3. Introducing electric and magnetic fields.

We next need to substitute the Lorentz force (containing electric and magnetic fields) in for the forces. Just to remind you, the Lorentz force in the primed reference frame is:

. (15.11)

When we substitute the various force components of (15.11) into (15.10), F becomes:

. (15.12)

This last substitution has reintroduced some primed velocities. We transform these into velocities in the unprimed reference frame using (15.9):

. (15.13)

With some rearrangement, this becomes

(15.14)

where a number of terms in the first line have been collected and designated as T given by:

. (15.15)

T can be further simplified by the following steps:

Substituting this into (15.14) we have:

and

(15.16)

We need to interpret the above in terms of the standard Lorentz force:

. (15.17)

Matching up (15.16) with (15.17), looking at the multipliers of the various velocities, we see that (15.16) will become (15.17) provided that

. (15.18)

Summary

Below we have repeated (15.18) as well as a vector notation of the results (in terms of the parallel and perpendicular components). These are the transforms of the electric and magnetic fields to be used when changing from one reference frame to another.

. (15.19)

. (15.20)

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Four-vectors

2011-07-16T08:05:00.000-07:00

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Fighting four dragons at the same time.

14. Four-vectors

An astute observer will notice that (13.11) and (13.12) for transforming current and charge densities are operationally the same transforms as we presented earlier for transforming x and t (time). To further emphasize this, we present these two sets of transforms side by side in Table 14.1 below.

Pairings of four quantities that relativistically transform the same as x and t are called four-vectors. The math of four-vectors is well developed and offers a very elegant, compact way to express Maxwell's equations, as well as work with the math of relativity. Their use is often coupled with concepts of matrices, covariant vectors, contravariant vectors, and Einstein's summation convention of repeated indices.

Table 14.1. Four-vectors
Transform of current and charge	Transform of position and time	Commonly used four-vectors
		position and time: (x, y, z, ct) Velocity four-vector: (γv_x, γv_y, γv_z, γc) Momentum and energy: (p_x, p_y, p_z, E/c) = (γm₀v_x, γm₀v_y, γm₀v_z, γm₀c) Current and charge density: (J_x, J_y, J_z, cρ) Vector (magnetic) and scalar (electric) potentials: (A_x, A_y, A_z, φ/c) Four-vector del operator, the d'Alembertian:
More on four-vectors and how to use them can be found at many web sites and other references. There is a wide variety of conventions used for four-vectors. Some references use the above convention, while others divide the above by a factor of c. Some put the scalar item (last one in each line of the above list) first. Some add the imaginary constant i in front of the scalar item to make the magnitude calculation more "natural". The list in Table 14.1 follows the convention of Lorrain, and Corson. The trick is to be consistent once you start a calculation using four-vectors. One of the most useful four-vector facts is that the "magnitude" of a four-vector (e.g. x² + y² + z² − c²t²) is invariant under change of reference frame. This magnitude is defined as the normal magnitude of the 3-vector part minus the square of the scalar item, i.e. the last item. A very common use of this trick is to apply it to the momentum/energy four-vector: (p_x, p_y, p_z, E/c). It allows us to equate this magnitude in the frame (the proper frame) where the particle is stationary to another frame in which the particle is moving. In the proper frame the magnitude is -E₀²/c² and in the other frame the magnitude equals p² − E²/c² . Equating these and multiplying through by c² yields −E₀² = p²c² − E² . This equation is useful for relating the momentum p = γm₀v of a particle to its total moving energy E = γm₀c² and its rest mass energy E₀ = m₀c² .

Table 14.1. Four-vectors

Transform of current and charge

Transform of position and time

Commonly used four-vectors

position and time: (x, y, z, ct)

Velocity four-vector: (γv_x, γv_y, γv_z, γc)

Momentum and energy:
(p_x, p_y, p_z, E/c) = (γm₀v_x, γm₀v_y, γm₀v_z, γm₀c)

Current and charge density: (J_x, J_y, J_z, cρ)

Vector (magnetic) and scalar (electric) potentials:
(A_x, A_y, A_z, φ/c)

Four-vector del operator, the d'Alembertian:

More on four-vectors and how to use them can be found at many web sites and other references.

There is a wide variety of conventions used for four-vectors. Some references use the above convention, while others divide the above by a factor of c. Some put the scalar item (last one in each line of the above list) first. Some add the imaginary constant i in front of the scalar item to make the magnitude calculation more "natural". The list in Table 14.1 follows the convention of Lorrain, and Corson. The trick is to be consistent once you start a calculation using four-vectors.

One of the most useful four-vector facts is that the "magnitude" of a four-vector (e.g. x² + y² + z² − c²t²) is invariant under change of reference frame. This magnitude is defined as the normal magnitude of the 3-vector part minus the square of the scalar item, i.e. the last item.

A very common use of this trick is to apply it to the momentum/energy four-vector: (p_x, p_y, p_z, E/c). It allows us to equate this magnitude in the frame (the proper frame) where the particle is stationary to another frame in which the particle is moving. In the proper frame the magnitude is -E₀²/c² and in the other frame the magnitude equals p² − E²/c² . Equating these and multiplying through by c² yields

−E₀² = p²c² − E² .

This equation is useful for relating the momentum p = γm₀v of a particle to its total moving energy E = γm₀c² and its rest mass energy E₀ = m₀c² .

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Transformation of charge and current densities

2011-07-13T09:17:00.000-07:00

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⇐Fields of a stationary charge. The fields created are electric fields.

Fields of a moving charge. The motion adds magnetic field and transforms the charge into a mix of charge and electrical current. The pure electric field is transformed into a mix of electric and magnetic fields.⇒

13. Transformation of charge and current densities

Assumptions:

The charge on an object will be the same in all reference frames, e.g. the charge of an electron will be the same independent of how fast it is moving.
Space is subject to Lorentz contractions due to speed.
The velocity transforms derived above from Lorentz transforms are valid.

Results:

Transformation equations for charge density ρ and current density J.

The object of this chapter is to derive transformations for charge density and current density that allow us to calculate these in one reference frame using values of these in another reference frame. We will assume that we have moving charges from the points of view of both reference frames. It is convenient to use a third reference frame, one in which the charges are stationary. Thus we have three reference frames to keep track of in this chapter. These are summarized in Figure 13.2 below.

The charge on an object is usually assumed to be independent of velocity. This means the charge of an electron will be the same no matter how fast it is moving. Certainly this is consistent with Einstein's idea that physics should look the same in all reference frames. That is, the charge of one electron will be the same and there will be the same number of electrons making up a charge.

Charge density is another matter. Because for charge density, the charge in a tiny volume is divided by the volume and volume is subject to Lorentz contraction; charge density undergoes a transformation when changing reference frames. We might think that is all there is to it, but it turns out that the transformation also involves current density.

To understand this, let's consider a general case involving both charge density and current density, that is we have charges moving at velocity u in the unprimed reference frame. Note that the velocity u has components in all three directions. We call this reference frame, "frame 1" shown in Fig. 13.2b.

For this derivation, we will assume all charges have the same velocity. It would be easy to generalize for the case where charges have various velocities by splitting the charges up into species, each with their own velocity, transforming each separately and then recombining them. Because the transform we will derive next is linear and very simple, it is easy to see that the result will be the same as in the single velocity case.

To start, we need to go back to the reference frame where the charges are stationary. This is possible because we are treating the case where all charges have the same velocity. We call this frame 0 shown in Fig. 13.2a. Another accepted title for this frame is the "proper frame". The charge density in frame 0 is labeled ρ₀ . In the following box, we derive the transform between this frame and frame 1.

Box 13.1. Transform between stationary charge and a frame in which the charge is moving

Because these charges are stationary in frame 0, the current density due to these charges will be zero. Thus if we transform from this frame to another frame there will be no effect in the second frame due to current density in the frame 0. When we switch to the frame 1, the tiny volume elements in the denominator of the charge density will shrink due to Lorentz contraction by the γ factor making the charge density increase by this same factor. For this box (and this box only) we assume we have rotated the coordinates so the x is in the direction of the relative velocity between frames 0 and 1, i.e. in the direction of u.

, (13.1)

Fig. 13.1. Setup for charge density calculation.

where γ₁ is associated with the velocity u and is given by:

. (13.2)

The current density in frame 1 is simply the above charge density times its velocity:

. (13.3)

Box 13.2. Charge as viewed from the three reference frames and the related γ's

Fig. 13.2a. View of charge density from the point of view of frame 0, the frame in which the charge is stationary.	Fig. 13.2b. View of moving charge density from the point of view of frame 1.	Fig. 13.2c. View of moving charge density from the point of view of frame 2.
	γ concerned with the relative velocity between frames 0 and 1.	γ concerned with the relative velocity between frames 0 and 2.
	γ concerned with the relative velocity, V, between frames 1 and 2.

Box 13.3. Velocity and γ in a frame moving with respect to frame 1

Now we would like to calculate the charge density in a third reference frame, frame 2 shown in Fig. 13.2b. The charges are moving at velocity u' relative to frame 1. Reference frame 2 shown in Fig. 13.2c is moving at velocity V in the x direction relative to frame 1. From equation (9.4) of chapter 9, the velocity of the charges in reference frame 2 in terms of their velocities in frame 1 is given by:

. (13.4)

where γ (without subscripts) is associated with the relative velocity, V, between reference frames 1 and 2, is shown at the bottom of box 12.2:

. (13.5)

In our derivation we will need to use u' to calculate γ₂ as shown in Box 13.2 above. We first calculate the hardest part of γ₂:

. (13.6)

Using (13.6) to calculate γ₂ , the relativity factor associated with velocity u', we have:

, (13.7)

where we have used (13.4) to change from velocity u' of these charges in frame 2 (the primed reference frame) to velocity u of the charges in frame 1.

Box 13.4. Calculating the charge density in frame 2

Equation (13.1) above relates the charge density (labelled ρ₀ in the frame where it is stationary) to the same charge density (labelled ρ' in this reference frame) as observed in a frame in which this charge is moving. We can write this equation in a more general way as:

ρ_moving = γρ₀

where γ is the relativistic factor written in terms of the relative velocity between these two frames. Equation (13.1) is the application of this equation to the relative motion between frames 0 and 1. We can also applied it to the relative motion between frames 0 and 2, in which case it would be written as:

ρ' = γ₂ρ₀

In the next equation we first use the equation immediately above to relate the charge density as seen in frame 2 to that in frame 0. We then apply (13.1) to relate the charge density in frame 0 to that in frame 1:

, (13.8)

where γ₁ and γ₂ are as given in (13.2) and (13.7), respectively.

Substituting (13.2) and (13.7) to into (13.8) we get an expression for ρ' (the charge density in reference frame 2) in terms of the parameters of reference frame 1:

, (13.9)

where γ (without subscripts) is defined by (13.5) above and relates to the relative velocity, V, between reference frames 1 and 2.

Box 13.5. Calculating the current density in frame 2

In a similar way using (13.4) and (13.9), we can calculate the current density J':

. (13.10)

Box 13.6. Summary - current density/charge density transform

Summarizing what we have derived above (equations (13.9) and (13.10)), we have:

. (13.11)

. (13.12)

We might have guessed the transform for current density would be that shown in (13.12). That is, the γ is simply due to the length contraction which concentrates the charge when it is moving. Since current is just charge times velocity, this concentration will have the same multiplying effect on current density. The second, or γVρ term is simply the extra current due to the fact that in this reference frame, the charge is moving and becomes a current density in addition to its role as a charge density.

The first term in (13.11) is also easy to explain. It is simply the same charge multiplied by the γ factor which is again due to length contraction. Harder to explain on an intuitive basis is the second term in (13.11). That is, "how does a current density become a charge density?" I don't know an easy way to explain it. To get the formula for it you need to go through the above math.

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Transforming electromagnetic fields and Maxwell's equations

2011-07-12T17:38:00.000-07:00

previous: relativistic kinetic energy

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12. Transforming electromagnetic fields and Maxwell's equations

Relativity has its roots in a problem associated with Maxwell's electromagnetic theory: that there was no obvious way to address the question of reference frames in Maxwell's equations. It was further prompted by the Michelson-Morley experiment which indicated that the speed of light as measured is independent of the velocity of the reference frame from which it is measured.

"Relativity" was created to make a world in which all objects, when moving at very great velocities, would behave in a manner consistent with their internal structures being determined by Maxwell's equations and the constant speed of light. Part of this world was a set of "Lorentz" transforms which allowed us to convert observations taken in one reference frame into observations in another reference frame.

Lorentz transforms have the property that they preserve the form of the operators in both Maxwell's equations and the electromagnetic wave equations. At the same time, this does not mean that the electromagetic variables, the electric and magnetic fields, and the charge and current densities, stay the same when changing reference frames. In fact, we shall find, in the next few chapters, that all these variables do change.

We shall find, that when we change reference frames a pure electric field becomes a mix of electric and magnetic fields. The same is true of a pure magnetic field. Also a pure charge density becomes a mix of charge density and current density. A pure current density becomes a mix of current and charge density.

In a sense, it is obvious. After all in a reference frame where a charge is stationary, it is a pure charge (not a current) and it creates a pure electric field. However, when we change reference frames so that the charge appears to be moving, we have a moving charge which is not only a charge, but also is a moving charge, i.e. an electrical current. In this case we would expect not only an electric field (from the charge) but also a magnetic field (from the electrical current). The pure electric field is changed into a mix of electric field plus magnetic field with the addition of motion.

We shall derive the equations for these transformations. In the final four chapters we shall come full circle and transform not only the operator part of Maxwell's equations, but also the electric and magnetic fields and also the current and charge densities. With quite a lot of brute force work, we shall show that the set of equations in fact do transform into an identical looking set of Maxwell's equations in the new reference frame.

James Clerk Maxwell	Hendrik Lorentz	Albert Einstein

The great men of relativity

Chapters on transforming electromagnetic fields

13. Transformation of charge and current densities

14. Four-vectors

15. Transformation of electric and magnetic fields

16. Actual transformation of Maxwell's equations

17. Transformation of the second Maxwell equation

18. Separating the transformed Maxwell equations

19. References used for the mathematics of relativity

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Relativistic kinetic energy of a particle

2011-07-12T17:14:00.000-07:00

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Einstein's equating of mass and energy through his famous equation E = mc² was later verified by the invention of the atomic bomb. The atomic bomb converts a small amount of mass into heat and various forms of electromagnetic radiation. It is also thought that the sun derives its energy from the continual conversion of large amounts of mass into heat and light. Many other nuclear reactions similarly convert mass into energy.

Most physicists today consider mass and energy to be attributes of the same property. That is to say, that mass and energy are always proportional to each other as given by the above equation. Thus a hot object will be slightly more massive than when it is cold, although the difference in mass is so slight as to be very hard to detect.

11. Relativistic kinetic energy of a particle

If the mass increases with velocity as discussed in the previous chapter, what is the relationship between kinetic energy and velocity?

Classical kinetic energy

In classical (low speed) physics we are taught that the kinetic energy of a particle is given by:

KE = ½mv² . (11.1)

We can derive this as

. (11.2)

where all the vectors are in the forward direction of motion (in the x direction) so we just show these as scalars. Finishing the derivation, we have:

.
(11.3)

Derivation of relativistic kinetic energy

In the relativistic case, we repeat the same steps as above except that now the mass is a function of velocity as given by:

. (11.4)

We now repeat the steps in (11.2) and (11.3) with this new non-constant mass that varies with velocity v. This calculation will be more laborious because of this new addition and will take a few lines of equations:

, (11.5)

, (11.6)

where we have used integration by parts for the second step. Also, please excuse the monster integral signs. This is one of the few flaws in the equation editor I use, MathCast, which is otherwise extremely good, especially considering that it is free, very easy to install, and makes entering equations very fast.

, (11.7)

where u is defined as:

. (11.8)

Continuing with (11.7), we have:

, (11.9)

and:

. (11.10)

Interpretation of relativitic energy

It is standard to call the left-most term of the final expression in (11.10) the "total relativistic energy" or E_total and the right-most term the "relativistic rest mass energy" or E₀:

, (11.11)

. (11.12)

Making this substitution, we write (11.10) as:

. (11.13)

Interpreted this way, the total energy of a particle (or object or mass) has two components, the kinetic energy and the rest mass energy. Identifying E₀ as an energy was quite a leap of faith for Einstein, because at the time there was no conceivable way to extract this energy for use or even for experimental verification that this term indeed was an "energy" i.e. an ability to do work. He made the leap primarily because it made a nice equation for the energy as shown in (11.11) above. The ensuing years revealed the annihilation of matter and antimatter, nuclear fission, and nuclear fusion, all of which converted this rest mass energy to the more normal forms of energy (light and heat) verifying his bold assumption.

Incorporating the rest mass into the energy also allowed the energy and the three momentum components (p_x, p_y, and p_z) to make up a "four vector", a concept we shall briefly explain in Chapter 14.

Momentum/energy transform

We can use our new mass/energy relationship (11.11) to rewrite the momentum/mass transformation, equation (10.25), at the end of the previous chapter as:

, (11.14)

where in this equation we represent the total energy of the object (rest energy plus kinetic energy) as E . We have also multiplied the bottom items in (10.25) on either side of the equal sign by c² .

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Relativistic mass

2011-07-11T15:41:00.000-07:00

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A small mass requires a small force to get it to accelerate at a certain rate.	A large mass requires a large force to get it to accelerate at the same rate.

A small mass moving at a certain velocity will have a smaller momentum than a large mass moving at the same velocity. In a collision, the object with the smaller mass will undergo a more abrupt change in its velocity.

A large mass moving at the same velocity will have a larger momentum. In a collision, the object with the larger mass will experience less change in its velocity.

10. Relativistic mass

Mass has to do with momentum. Indeed the desire to maintain the principle of momentum conservation in all reference frames led Einstein to decide that mass must vary with velocity (at extremely high speeds).

Incidently, Lorentz's view would be that the mass of a particle would actually increase at very high speeds (relative to "the" absolute reference frame), i.e., that this is not just a reference frame thing, but instead an actual physical phenomenon.

The following derivation follows that in Feynman [Lectures in Physics, Vol 1, 1963 ch16 p6+], also found on the web here. It leads us through a thought experiment involving the collision of two identical particles (red and blue) whose trajectories are shown in Fig. 10 below. The lengths and directions of the trajectory lines are also the velocity vectors before and after the collision of the particles. The derivation below ends up with a formula for the increase in mass due to high speed motion.

The setup

We consider an elastic collision between two identical particles. "Elastic" means that no energy is lost in the collision. In any collision, we can always change to another reference frame such that the center of mass of the objects has no velocity. For our collision this means that in this reference frame the particles will have the same speed, but are heading in the opposite direction, initially heading straight towards each other, then after the collision, directly away from each other.

We will also rotate the coordinate system so that the collision lies in the x-y plane and is symmetric with respect to the x and y axes as shown in Fig. 10.1a.

We consider yet another change in reference frame. Suppose we pick a frame that is moving in the x direction at the same x-speed as the lower blue particle. In that reference frame our collision looks like Fig. 10.1b where the blue particle is seen to travel vertically upwards before the collision and vertically downwards after the collision. We label the speeds as shown in Fig. 10.1b, i.e. w is the initial (and final) vertical speed of the lower blue particle, while v is the initial and final speed of the upper red particle. v_x is the horizontal (or x component, shown in green) of the velocity v. The trajectory of the red particle makes an angle θ relative to the x-axis. The vertical component of the red particle's velocity (y-component, also shown in green) is given by v_y = v_x tanθ .

We also look at the collision in Fig. 10.1a in a reference frame going along with the red particle. In that reference frame the collision looks like that shown in Fig. 10.1c. Because the two particles are identical, Figs. 10.1b and 10.1c are symmetric images of each other.


Fig. 10.1a.	Fig. 10.1b.	Fig. 10.1c.

Transforming velocities

To transform from the reference frame of Fig. 10.1b to that of Fig. 10.1c we need to match the horizontal velocity of the red particle in Fig. 10.1b. In other words, we need to add a horizontal velocity of V = v_x (in the negative x direction) to the reference frame to get from Fig. 10.1b to Fig. 10.1c. Thus for this experiment, the magnitude of the quantity v_x is both the x velocity of one of the particles in Figs. 10.1b and 10.1c, AND the relative velocity between the two reference frames in these illustrations.

We can use this fact and the equation (9.1) of the previous chapter (equation for transforming velocities) to arrive at a relationship between w in Fig. 10.1b and the vertical component of v in Fig. 10.1c:

. (10.1)

In this case γ is given by because v_x is the relative speed of the two reference frames of Figs. 10.1b and 10.1c. (We previously have used V for the relative speed of reference frames.)

So now we have an equation (i.e. equation (10.1) ) relating the variables v_y and w in the Figs. 10.1b and 10.1c to each other.

Momentum conservation?

We will now focus on the collision as shown in Fig. 10.1b (reproduced at the right). We write the equation for momentum conservation in the y direction for the collision shown in this figure:

Fig. 10.1b

Δp_y(red particle) = Δp_y(blue particle)

Written in terms of the above variables, we have:

2m_redv_y = 2m_bluew

Substituting (10.1) in for v_y , we have:

m_redw /γ = m_bluew

or:

m_red = γm_blue , (10.2)

where gamma in this case is given by:

. (10.3)

But the two particles are supposed to be identical, so their masses are supposed to be equal. But they cannot be equal if the y component of momentum is to be conserved in Fig. 10.1b.

The fix

Momentum conservation can be restored if we let mass vary with velocity. After all, if an object's length is changing with velocity, why not let the mass vary? Thus, the "fix up" that would allow the red and blue masses to have a ratio such that (10.2) was true. Expressed in an alterate way, their ratio needs to be given by:

. (10.4)

So now we have the ratio of two relativistic masses.

We next wish to relate a relativistic mass (one that is moving at a speed close to that of light) to a non-relativistic mass (one that is stationary or moving very slowly compared to the speed of light). We call this slower mass, the rest mass of the particle, designated as m₀.

Fig. 10.2. Collision with very small vertical velocities, i.e. a glancing blow to an almost stationary blue particle.

To do this (relate the relativistic ratio of (10.4) to the low speed mass) we consider the limiting case where the vertical velocity w is very small, so that the blue particle is moving at non-relativistic speeds as illustrated in Fig. 10.2, at the right. The red particle is still moving extremely fast (at relativistic speeds) with a trajectory that is almost parallel to the x axis. Since the blue particle is moving slowly, the mass is simply the regular low speed mass, the rest mass, i.e. m_blue ≈ m₀ . Using (10.3) the mass of the high speed particle (the red particle) is given by:

In general, this gives the mass of a moving particle (or any object) and is usually expressed as:

, (10.5)

where v is the velocity of the object relative to the observer and m₀ is the rest mass of the object (mass when it is at rest or moving slowly). γ in this case is calculated using the total speed of the object relative to the observer, not just its horizontal component.

Even though we derived the above using a rather special collision, (10.5) is generally true, verified by many, many experiments. At the bottom of this chapter we shall check it using a more general expression. In the next section we do a little check to see if (10.4) and (10.5) are consistent with each other.

Checking the relativistic mass equation

So far we only have shown that equation (10.5) holds for the limiting case where one particle is moving relatively slowly. We next show that (10.5) insures that the ratio in equation (10.4) holds in the more general case where both particles move at relativistic speeds.

Fig. 10.1b

We focus again on the collision of Fig. 10.1b (again copied at the right). In this case, the mass of the blue particle is easy, since its velocity is simply equal to w in the vertical direction and (10.5) yields:

, (10.6)

To calculate the relativistic mass of the red particle in Fig. 10.1b we need the magnitude of its velocity. Its x velocity is v_x and its y velocity is given by (10.1). Using Pythagoras' theorem to calculate the magnitude of the red particle's velocity from these two components, we have:

. (10.7)

Substituting (10.7) into (10.5) we get:

, (10.8) which can be continued as:

. (10.9)

Now to use (10.6) and (10.9) to calculate the ratio of masses:

. (10.10)

which is the result we were aiming for, the same as equation (10.4)

Thus we conclude that use of (10.5) is sufficient to insure the ratio of masses agrees with (10.4), even in the case where both particles are moving at relativistic velocities.

Interpretation of relativistic mass

Now that we have an expression for how the mass varies with extremely high velocities, what does it mean? How can a one gram mass become more massive by just moving faster?

Note first that a 1 gram mass would still seem to be 1 gram to a person moving along with the mass. The increase would only be apparent to a person who is stationary while the 1 gram mass is zipping past. This is consistent with Einstein's wish that inside a laboratory, even if it were moving very rapidly, all would seem normal. It is just when you compare things in the lab with things outside (going at a different velocity) that the weirdness begins.

As mentioned before, this increase in mass might seem in keeping with the fact that the same stationary observer would also report that the object is shorter. Lorentz would say that an object moving with respect to the absolute reference frame would really be more massive, just as it would really be shorter while in motion. He would also say that if an object is at rest with respect to the absolute reference frame then it is really not more massive or shorter even though observers in a moving space ship might think it is. In this case their judgment would be impaired by the warping of their rulers and clocks. Einstein would say that we will never know which frame is absolute and all this talk of what is "really" happening is silly.

This increase in mass is directly linked to the impossibility of an object other than light itself (and perhaps some really weird subatomic particles like neutrinos) going as fast as light. As a particle approaches the speed of light, its mass increases making it more and more difficult to accelerate.

Physicists use particle accelerators to study subatomic particles and their interactions. These accelerators verify the above mass increase formula every time they are turned on. One such accelerator, located at Stanford University in California, has accelerated electrons so that they become 40 times more massive than protons, 80,000 times more massive than electrons normally are. Note that normally electrons have a mass of one two thousandth (1/2000) the mass of a proton and are easily stopped by a piece of paper. The very fast moving electrons from this accelerator can pass through meters of concrete pushing many, many protons out of the way in the process. This feat would be impossible if mass did not increase with very high velocity. Incidentally, these electrons were traveling at 99.97% the speed of light when they emerged from the accelerator.

Fig. 10.3. Aerial photograph of the 2 mile long electron accelerator at Stanford University.

Parallel and transverse force and mass

Occasionally we use the words parallel and transverse mass to emphasize the fact that it is harder to accelerator a high speed particle in the direction of its motion than it is to accelerate it perpendicular to its motion.

We can understand this by considering the relativistic form of Newton's second law of motion:

. (10.11)

Note that in the above equation, (10.11), the bold faced variables are vectors, a shorthand notation for all the components of a vector to be treated together: all to be multiplied by m or all to have their time derivative taken.

The time derivate of γ is done as follows:

, (10.12)

where we have used the following:

. (10.13)

In the above equation, v⋅v represents the dot product between the vector v and itself (which yields the magnitude of v squared.)

Substituting (10.12) into (10.11) yields:

. (10.14)

Case I, transverse force

In the case of force applied perpendicular to the direction of motion (perpendicular to the trajectory of the particle to bend its trajectory), the first term in (10.14) is zero because v is perpendicular to dv/dt making the dot product zero. Or alternately, in this case the magnitude of velocity is not changing, making dγ/dt zero. With either argument, we are left with only the second term in (10.14), giving us:

. (10.15)

Case II, parallel force

In the other case, when the force is applied in the same direction as the particle's velocity to speed up the particle, then both terms of (10.11) are present. The dot product in the first term is equal to the magnitude of the two vectors. Since all three vectors in that term are in the same direction, we can use any one of them to indicate the vectorial direction of the term. We choose dv/dt to be the vector and the other two vectors (the v's) to be scalars. The force equation (10.14) written out becomes:

. (10.16)

We see that the parallel force increases with the cube of γ whereas the transverse force increases linearly with γ. Some put the γ³ dependence into an effective mass as:

. (10.17)

This allows use of the simple second law of motion with an enhanced mass:

. (10.18)

Transforming mass

We have derived the equation (10.5) in the rather special case shown in Fig. 10.1 above involving two identical particles. Equation (10.5), however, is very general and preserves momentum conservation in collisions of a slowly moving objects and rapidly moving ones, with equal and with unequal masses. This we will demonstrate at the end of this chapter.

In this section, we will derive a transform that relates the mass of an object moving at one relativistic velocity, u , to its mass when it is moving at another relativistic velocity, u' . We will work in the unprimed and the primed reference frames. In both frames the object is moving, but with different velocities. Using (10.5) we can relate the mass in each reference frame to the object's proper mass (mass when it is stationary):

and .

Dividing one equation by the other and solving for the unprimed mass yields:

. (10.19)

We now wish to simplify this a little. Most of the difficult calculation for this is the calculation of the denominator which we wish to express in terms of u instead of u'. We can do the velocity transforms using the equations from Chapter 9.

, (10.20)

where γ is dependent on the relative velocity V between the reference frames themselves (not between the objects in the two frames) and is given by:

. (10.21)

Substituting (10.20) back into (10.19) yields:

. (10.22)

This can be alternately expressed in terms of the x component of the momentum p_x as:

. (10.23)

Comparing this with the setup (Fig. 10.1) used to derive (10.4), we see that to go from Fig. 10.1b (the unprimed frame) to Fig. 10.1c (the primed frame), that for the blue particle the unprimed x momentum is zero, making (10.23) become in this case m' = γm. This is consistent with (10.4) because the red mass of Fig. 10.1b should be the same as the blue mass of Fig. 10.1c due to symmetry.

Transforming momentum

We actually used momentum conservation to derive the mass increase with speed in the first part of this chapter. We now want to formalize the transformation equations for momentum itself.

Momentum p is simply the mass of an object times its velocity, i.e. p = mu , where m is the object's mass. We use the velocity transformations (9.4) along with the mass equation (10.23) to arrive at:

. (10.24)

Summary - momentum/mass transforms

We put (10.23) and (10.24) together, as is common:

. (10.25)

We see that the momentum and mass transforms fit together in special relativity much the same as space coordinates and time do. In fact many graduate level textbooks on this subject really just assume (10.5), i.e. m = γm₀ , in order that momentum and mass transforms result in the nice arrangement shown in (10.25), similar to the Lorentz transform for space and time. This allows the elegant treatment of special relativity using four-vectors, a concept that we will very briefly summarize in Chapter 14. We might emphasize that the derivation of (10.5) done at the beginning of this chapter is not technically rigorous. It uses a special collision to suggest that (10.5) would be a nice way to restore momentum conservation for such collisions.

Below we check whether the mass transform we "derived" above in fact does restore momentum conservation in the case of a general collision involving objects of differing masses and with some energy loss (partially elastic).

Checking for momentum conservation in the case of a general collision

	Galilean velocity transforms: v'_x = v_x − V v_y' = v_y v_z' = v_z m' = m	Lorentz velocity transform:
Fig. 10.4a. A general two body collision with arbitrary angles and velocities. The two masses, m₁ and m₂ , are of unequal size. The collision is partially elastic (some energy is lost). The initial velocity of m₁ is v_1i and its final velocity (after the collision) is v_1f. The initial velocity of m₂ is v_2i and its final velocity (after the collision) is v_2f.	Fig. 10.4b. These are the transforms for velocity for everyday life. The only change in an object's apparent velocity (in changing reference frames) is the addition or subtraction of the reference frame's velocity.	Fig. 10.4c. These are the transforms for very high speed motion, near the speed of light. These transforms are considerably more complex than the Galilean velocity transforms shown to the left. γ is defined as:

In this section we check if momentum conservation is upheld upon change of reference frame in the case of a general collision of two objects, objects that have different masses and lose some of their energy during the collision. A diagram of such a collision is shown in Figure 10.4a. The equations for the momentums before and after the collision are:

p_initial = m_1iv_1i + m_2iv_2i p_final = m_1fv_1f + m_2fv_2f . (10.26a)

We will assume that momentum is conserved before collision, i.e. that p_initial = p_final:

m_1iv_1i + m_2iv_2i = m_1fv_1f + m_2fv_2f (assumed). (10.26b)

and endever to show, with this assumption, that momentum is still conserved after we change reference frames (to the primed frame), i.e. that p'_initial = p'_final , or to show:

m'_1iv'_1i + m'_2iv'_2i = ? m'_1fv'_1f + m'_2fv'_2f (to be shown). (10.26c)

We first do the checking using Galilean transformations, shown in Figure 10.4b (reproduced here just below).

Galilean velocity
transforms:

v'_x = v_x − V
v_y' = v_y
v_z' = v_z
m' = m

Fig. 10.4b

I. Using Galilean transforms:

In low speed velocity transforms, the masses of the colliding objects do not change as a result of the collision, so:

m_1i = m_1f = m₁ and m_2i = m_2f = m₂ (10.26d) .

We'll only transform the x components of the velocities because the y and z components of velocity don't change as indicated in Fig. 10.4b. The new x components (before and after collision) of momentum in the transformed reference frame are:

Substituting in for the primed velocities from the Galilean transforms of Fig. 10.4b, we get:

. (10.27a)

Using (10.26b) above, we can change the last line of (10.27a) into:

. (10.27b)

This shows that the final momentum still equals the initial momentum when we change reference frames using Galilean transforms.

Note that the above mathematics uses the Galilean velocity transformations which are not consistent with the speed of light being invariant under changes in moving reference frames.

We next see if momentum is conserved in a new reference frame when we use Lorentz velocity transformations. Lorentz velocity transformations are shown in Fig 10.4c and reproduced just below.

Lorentz velocity
transforms:

Fig. 10.4c

II. Using Lorentz transforms:

We start again with (10.26a). We then transform only the velocities, naively hoping the masses can stay the same:

. (10.28)

With all the complicated denominators in (10.28) there is no way to factor the two lines to make them the same.

So we see that if mass does not change with reference frame, i.e. speed, then the transformed momentum equations become a tangle that will not preserve momentum conservation.

We now add the relativistic mass transform (10.22):

m' = γm(1 − v_xV/c²) . (10.29)

We also need to distinguish between the mass of each object before collision and after collision, since the masses will change along with the change in velocities before and after a collision.

Substituting (10.29) into(10.28) yields:

. (10.30)

If we examine the factors in (10.30) we see that we can cancel out all the troublesome terms to yield the simple results:

(10.31a)

(10.31b)

Using (10.26a), (reproduced here)

p_initial = m_1iv_1i + m_2iv_2i p_final = m_1fv_1f + m_2fv_2f (10.26a) ,

allows (10.31a) and (10.31b) to be further reduced to:

(10.31c)

(10.31d)

The the last terms in the x components of (10.31c) and (10.31d) are linked to relativistic energy conservation. In the next chapter, we derive the famous relation that energy E = mc² , i.e. that energy and mass are proportional. If we assume that the total relativistic energy before the collision equals the energy after the collision then

m_1i + m_2i = m_1f + m_2f (10.32) ,

If we substitute momentum conservation in the initial frame (i.e. Equation (10.26b) or p'_initial = p'_final ) and also (10.32) both into (10.31d), we get:

(10.33)

Thus momentum is conserved in the second (i.e. the primed) reference frame, as long as we assume relativistic energy conservation, i.e. (10.32).

Note that relativistic energy will be conserved even when kinetic energy is not. A collision can lose kinetic energy to heat, but if the heat is contained inside the colliding objects, it will cause them to be heavier than they otherwise would be as given by E = mc² where the energy E also include the kinetic energy due to the thermal vibrational motion of the atoms in each colliding object. Thus relativistic energy (and momentum conservation) will be maintained, even for partially (or totally) inelastic collisions. On the other hand, if electromagnetic radiation is given off during the collision, then the energy and momentum of this radiation needs to be factored into the energy and momentum conservation equations so that energy and momentum conservation is maintained.

previous: transforming velocities

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Transforming velocities

2011-07-11T09:52:00.000-07:00

previous: Lorentz transforms from length contraction and time dilation

next: relativistic mass

9. Transforming velocities

The idea behind this section can be a little hard to wrap your mind around, because it involves two different velocities at the same time. In the stuff above, we have worried about just one velocity. In the previous chapters in one reference frame, the object in question was moving, and in the other frame, the frame moving with the object, the object is at rest or stationary. Now we are considering the situation where the object is moving with respect to both our reference frame, moving with respect to the "stationary" frame and moving with respect to the "moving" frame. The question that we address here is: if the object is moving at velocity U with respect to the stationary frame, then how fast will it be traveling with respect to our moving reference frame, moving with velocity V in the x direction?

We will consider two cases: when the object is moving perpendicular to the motion of the moving reference frame and secondly, when it is moving parallel to the motion of the moving reference frame. You can use these two transforms for transforming the perpendicular and parallel components of a velocity at any arbitrary angle. You can read more about the general case here.

Velocity perpendicular to the motion of the reference frame

Here we see how an object moving perpendicular to the direction of the motion of the reference frame changes its y' coordinate (in the moving reference frame), i.e. Δy' for a given time period, Δt'. We assume that the object starts at y' = y₁' and ends the time period with y' = y₂'. The beginning and ending times are t' = t₁' and t' = t₂' , respectively. We use the Lorentz transforms to convert these quantities to the stationary reference frame (the unprimed frame).

. (9.1)

Above we've used the fact that the motion is perpendicular to mean that Δx is equal to zero. The final result shows that the y motion is reduced by the γ factor, which you might attribute to time dilation (time appears to be slowed down in another reference frame).

Velocity parallel to the motion of the reference frame

In this section we repeat the calculation for a velocity parallel to the motion of the moving reference frame, i.e. in the x direction. The calculations are made a little more complicated by the fact that the x transformation is more involved than the y transformation, and also by the fact that we can't declare the Δx term to be zero as we did above. In the end we have a little more complicated expression than we did in the previous derivation.

. (9.2)

We continue by dividing both the numerator and denominator by Δt :

. (9.3)

Summary of velocity transform

Table 9.1. Transformation of velocities
Component of velocity parallel to the relative motion of the reference frame	(9.4a)
Component of velocity perpendicular to the relative motion of the reference frame	(9.4b)
Other component of velocity perpendicular to the relative motion of the reference frame	(9.4c)

Transforming the speed of light in an arbitrary direction

If we change from one reference frame to another, we would expect the direction of any light pulse to change, however the speed of light should stay constant. We next check this out using the above equations:

In the last step we made use of the fact that the magnitude of U equals the speed of light:

The above algebra shows that the velocity transforms (9.1) and (9.3) are consistent with the speed of light being invariant, i.e. anything traveling at the speed of light in any direction will be traveling at the speed of light in a Lorentz transformed frame.

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Derivation of the Lorentz transforms starting with length contraction and time dilation

2011-07-11T09:31:00.000-07:00

previous: length contraction and time dilation from a constant speed of light

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8. Derivation of the Lorentz transforms starting with length contraction and time dilation

We begin by explaining the Galilean relativity motion transforms and then go on to derive the Lorentz transforms, both shown in Fig. 8.1 below. Both sets of transforms assume that the unprimed reference frame (with coordinates x, y, z, and t) is stationary and the primed frame (coordinates x', y', z', and t') is moving in the x-direction at constant velocity V.

Galilean transforms			Lorentz transforms


			where
Fig. 8.1. Table of Galilean and Lorentz transforms.

The length contraction and time dilation formulas are:

, (8.1) and

Δt = γ Δt' , (8.2)

where the relativistic factor γ is given by:

(8.3)

The x transform:

The first Galilean transform is easy to understand. It says that a stationary point in the stationary reference frame having x-coordinate value of x will appear to be sliding to the left from the point of view of someone in the moving frame. In that reference frame, the point will appear to shift an amount −Vt, i.e. its x' coordinate value will be given by the initial x value plus this shifting amount:

x' = x − Vt (8.4) .

In the relativistic case we need to account for the length contraction in the x-direction. That is, the x' scale will be shrunk and the same value in the x coordinates will translate to larger x' values because of this shrinking. The relativistic factor

(8.5)

is needed here to multiply the x' coordinate in (8.4) and complete the transform:

x' = γ (x − Vt) (8.6) .

The y and z transforms:

The y and z transformations are really not transformations at all, because there is no relative motion in these directions and also no length contraction in these directions. So in both the Galilean and relativistic cases:

y' = y and z' = z (8.7) .

The time transform:

In the Galilean case, the time transform is trivial, i.e. there is no transform and t' = t.

On the other hand, in the relativistic case, deriving the time transformation takes some work. The first term in the Lorentz time transform is easy: it is just the time dilation factor γ applied to the original time t.

t' = γ t (8.8) (incomplete!).

So what about the extra −Vx/c² term in the last Lorentz transform equation in Fig. 8.1? It might be called the synchronization term. It insures that time at the various points in the moving frame will appear to be synchronized to a person in that frame, as you might expect.

We would expect our time to be synchronization, that time did not seem to vary with the coordinate x . However we saw in the previous chapter that when we change reference frame, previously simultaneous events are no longer simultaneous, if they have different x values.

In the previous chapter, we found that two events that are simultaneous in the stationary frame will have a time difference between them of Δt = γΔxV/c² (see Equation (7.11) ). If we wish to correct for this time difference we need to subtract it from t in Equation (8.8):

t' = γ (t − xV/c²) . (8.9)

Equation (8.9) is the time part of the Lorentz transformations. Without this last term, a person on a moving frame would sense that clocks located in different x-positions in his frame would not be synchronized, assuming they were synchronized the stationary frame.

Simultaneity in high speed reference frames:

Because of the last term in (8.15), simultaneity is not longer a simple issue at relativistic speeds, i.e. at speeds close to that of light. At low speeds, if two events occur at the same time, i.e. simultaneously, then they occur at the same time in all moving reference frames, even if the two events are at two different locations. Thus if lightning strikes two different trees at the same time as witnessed by a stationary observer, then an observer flying at low speeds overhead will also see the two strikes occurring at the same time assuming the observers are about the same distance from the two strikes. On the other hand, if an airborn observer is traveling at relativistic speeds, even if he is equidistant from the two strikes (that are simultaneous in a stationary frame), the two strikes will not appear occur at the same time to this flying observer.

The last term in (8.15), i.e. the synchronization term, means that for high speed motion, the time in a high speed moving reference frame depends of both the x position and on the velocity of the frame, V. Thus, two events occurring at the same in the stationary frame, so that t₁ = t₂, will not have the same time in the moving frame if they have different x coordinates which make the last terms in (8.15) differ and in turn make t₁' ≠ t₂'. In general, for relativistic speeds, simultaneity of two events that have different x coordinates, only holds in a particular reference frame.

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Length contraction and time dilation from a constant speed of light

2011-07-10T10:45:00.000-07:00

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This posting includes flash animations showing the physics discussed. Most computers have a flash player already installed, but if yours does not, download the free Adobe flash player here.

7. Length contraction and time dilation from a constant speed of light

This chapter follows Einstein's logic. Einstein starts with the assumption that physics will be the same in all the uniformly moving reference frames (also called inertial reference frames) and also with the corollary that the speed of light is constant in all these reference frames. From these concepts he derives the laws of high speed motion. This particular chapter uses these two concepts to derive length contraction and time dilation, as well as constancy of length perpendicular to motion and changes in simultaneity. From these, a person can derive the Lorentz transforms (see Chapter 8) and the many other equations of special relativity (some of which are derived in Chapters 9 - 18 of this posting).

A novice to the concepts of relativity might find these derivations a bit strange. How can little timing differences in light beams mean that time and distance is warped? Well, a constant speed of light is truly odd and Einstein's thought experiments are just sticking our faces into the logical consequences of the speed of light really being the same for all uniformly moving reference frames. In order that simple experiments concerning light beams show that light is moving at the same speed to both a stationary and a moving observer, we end up concluding strange things: that either space and time warp, as Einstein held, or that objects distort and slow their internal processes, as Lorentz held. If you find this chapter confusing and haven't read my earlier blog on understanding special relativity this might be a good time to go back and read it. It explains Einstein's view of these oddities, as well as Lorentz's. Many readers might find Lorentz's view of these oddities more understandable. Both views are mathematically equivalent.

Assumptions and conclusions of this chapter

Assumptions

1. Physics obeys the same rules and equations in all uniformly moving reference frames.

2. Light moves at the same constant value in all uniformly moving reference frames (this is implied by #1 above, i.e. it's a corollary to #1).

Conclusions of this section

1. Lengths perpendicular to the relative motion of the moving reference frame are unchanged by the motion.

2. Temporal processes are slowed in moving objects. This is called time dilation.

3. Lengths in the direction of the relative motion are shrunk by the motion. This is call length contraction.

4. Events at different x coordinates (parallel to the direction of motion) which occur simultaneously in a particular reference frame will not be simultaneous in other reference frames moving relative to the first one in the x direction.

Relativity and reference frames

Relativity is all about high speed motion, i.e. the properties of particles and masses when they are moving at velocities close to the speed of light. The physics of relativity is commonly presented in terms of changing reference frames. This allows the physics of low speed objects to be transformed into the physics for very rapidly moving objects.

Often we speak of two reference frames, one stationary and one moving. The moving reference frame is taken to be moving along with a speeding particle that we happen to be studying, so that to someone in that reference frame the particle would be at rest and its physics well understood. In the stationary frame the particle would be moving. We make transforms that let us adapt the physics in the moving frame (in which the particle is at rest) so that it is valid in the stationary frame (in which the particle is moving at very high speeds).

It is common to designate properties in the two reference frames as primed and unprimed, such as x' and x. We will take the primed frame to be the moving frame, in which the particle is often at rest. The stationary frame will have its properties not primed, i.e. such as x. Note that it is common to call the properties as measured in a reference frame in which a particle is at rest the "proper" properties, such as the "proper length" or "proper mass".

Perpendicular to motion, no change in length

In this section we will analyze a thought experiment which shows that assumptions above require that lengths perpendicular to motion not change when changing reference frame. Thus, assuming that the relative motion is in the x direction, then this means that y = y' and z = z' .

Fig. 7.1a. An animation illustrating the difficulty of measuring the length of a moving object. The viewer can compare the "measured length" when the end positions are determined (or "marked") simultaneously and when they are not. This animation shows the error in failing to mark the two end positions at the same time.

Mouse over the animation to start it and mouse off to suspend it. Click on it to restart it, or if it fails to start.

To "measure" the length of a moving ruler, we first determine the positions of its two ends, both at the same instant in time; then we measure the distance between these positions. It is critical that the positions be determined at the same instant in time, i.e. simultaneously, or the motion of the moving ruler will add an error. The animation in fig. 7.1a illustrates this. Mouse over the animation to start it.

The proof of the lack of perpendicular length change is based on the next animation, Fig. 7.1b. In this figure we see two identical rulers (except for their color), identical when both are brought to rest beside each other, both oriented perpendicular to the relative motion of the reference frames. In this thought experiment, one ruler is made to move along with the moving frame while the other is fixed to the stationary frame. We show one ruler longer just to illustrate that perhaps one ruler's length is changed by the motion. We will end up concluding that this is not true, that in fact they both remain equal in length despite their relative motion.

Fig. 7.1b. An animation illustrating that in the perpendicular direction, lengths are not changed by relative velocity. If they were to be changed by motion, then one of the lengths would be measurably longer than the other in both reference frames (as incorrectly shown in this animation). This would mean that to an observer in one reference frame, motion would seem to shorten a ruler while in the other reference frame motion would appear to lengthen an identical ruler. This difference would violate the first assumption above.

The core of the animation is really just the comparing of the rulers' lengths when they pass and seeing that if they differ in length, observers in both reference frames would agree on which ruler was longer. The light pulses resolve a secondary issue, simultaneity. These pulses are shown in red and blue and demonstrate that the upper ends of the rulers pass each other at the same instant in time as the lower ends, as seen in both reference frames, so that we do not have simultaneity issues which were addressed in Fig. 7.1a.

Mouse over (or click) the figure to start the animation. Mouse off to suspend it and click on it to restart it. Read the text at the left to further understand it.

The rulers are positioned so that their centers go past each other. We adjust the coordinate system so the the rulers' centers are on the x axis as shown. At the instant that the rulers go past each other, observers in both reference frames can tell if they are the same length, i.e. if their ends coincide at the instant they pass each other.

Much of the animation is directed towards demonstrating that the two ends of each ruler pass the other ruler at the same instant in time, that there is no time offset between the passing of the upper and lower ends which could confuse the issue of simultaneity as illustrated in Fig. 7.1a. The demonstration of simultaneity is done by emitting light pulses at the instant the rulers pass each other and showing that the upper light pulses (from the upper ends of the rulers) arrive at the centers of the rulers at the same times as do the lower light pulses.

Because the setup is symmetrical around the x axis, the pathlengths the light pulses travel from the top sources are the same as the paths that the bottom pulses travel. This means that the light from the top sources will arrive at the detectors (i.e. centers of the rulers) at the same times as those from the bottom sources. This is true when we are considering light propagating to the stationary ruler's detector or we are considering the moving ruler's detector. Click on Fig. 7.1b to see this in animation. With equal pathlengths, coincident arrival times mean that the top end of the rulers pass over each other at the same time that the bottom ends do, both from the point of view of the moving ruler and from the point of view of the stationary ruler.

To further facilitate the ease of comparing the two lengths, we arrange marking mechanisms at all the ends of the rulers. These put marks on the the opposite ruler's space when the rulers pass by each other. You can see the yellow and blue x's being made in the animation at the instant one ruler passes the other. As was pointed out in the above discussion concerning light pulses, both reference frames agree that the two marks are made at the same time.

The marks and the rulers ends are physical entities that can be examined in either reference frame. It is also true that their relative positions will be the same in either reference frame. We have purposely (and incorrectly) drawn the moving ruler longer than the stationary one so you can understand this point. You can see that after the rulers pass, the marks on either ruler clearly show that the blue (moving) ruler is longer than the yellow (stationary) ruler. (Temporarily mouse off the animation to suspend it part way through so you can examine the markings more easily.)

Two results are possible: either the marks coincide with the other ruler's ends or they don't. If they don't, then one party's ruler appears shorter in both reference frames. In other words: to a stationary observer, the moving blue ruler appears longer and the motion of the blue ruler has increased its length. To an observer moving along with the blue ruler, the yellow ruler (which will appear to be moving relative to this observer) appears shorter: the relative motion of the yellow ruler appears to have shunk the yellow ruler's length. This means that the laws of physics of motion would be different in the two reference frames: in one case motion increased the moving ruler's length, in the other case it shrunk the moving ruler's length. We could use this to determine which reference frame had greater velocity relative to some absolute stationary frame, a concept that Einstein resolutely did not believe and which clearly violates assumption 1 above.

Thus, if we hold our assumptions to be valid, motion cannot result in perpendicular lengths being altered and we are incorrect in drawing our two rulers in Fig. 7.1b as different in length, since they are supposed to be identical when stationary. In conclusion: two rulers with the same proper lengths will stay equal in length when one of them is set into motion relative to the other provided that the rulers are orientated perpendicular to the direction of their relative motion.

Derivation of time dilation

For this section, we use a very special clock which is based on the propagation of light, shown in Fig. 7.2a, appropriate because of the special place light had in Einstein's theory. The nice feature of this clock is that it is particularly easy to understand the effect that relative motion has on it if we keep to our assumption about the constant speed of light. The result will be same as we would get with another clock design based on electromagnetic interactions. This special clock involves a light source that emits a light pulse that travels upwards to a mirror and back down to a detector located right beside the source. At this point another pulse is sent upwards and the process repeats again and again, presumably with a counter to count off these "ticks" of the clock. However, here we will concern ourselves with just one tick of this clock.

Fig. 7.2a. Animation showing time dilation. The illustration shows two of Einstein's light clocks, one stationary (at the far left) and one moving across the screen. The light beams in both clocks travel at the standard speed of light. However, because the diagonal distance that the light has to travel in the moving clock is greater, the clock cycle in that clock takes longer than in the stationary clock, i.e. that clock is slowed down. In the text below we calculate the amount of this slowing.

We show the light pulses leaving trails to further highlight the path lengths of the two pulses. At the end of one clock cycle, we freeze the motion to allow you to better study the paths.

Mouse over (or click) the figure to start the animation. Mouse off to suspend it and click on it to restart it. Read the text below to understand it.

Fig. 7.2a animates the two clocks while Fig. 7.2b is a snap shot at one instant of time, highlighting the paths in the animation.

As we have just seen in the previous section, the vertical lengths of the paths will be unaffected by the motion and will equal h, where h is the non-moving height of the clock. The horizontal length of the half path under the diagonal (shown in Fig. 7.2b) is VΔt/2 where V is the speed of the clock and Δt is the time of one "tick" of the moving clock. The diagonal distance is cΔt/2 because the pulse travels at c, the speed of light. Writing out the Pythagorean theorem for this right triangle, we have:

. (7.1a)

Fig. 7.2b. Snapshot of the animation above highlighting the trajectory of the light pulse. We have also labeled the three sides of the right triangle in the above Pythagorean theorem equation.

Equation (7.1a) can be solved for 2h which we will need in a moment:

. (7.1b)

We next slip into the moving reference frame. In that reference frame, the clock will be at rest. Also, according to Einstein (and a wealth of experimental evidence) the speed of light will be the same standard value in this reference frame. Thus, in the moving frame, the light pulse will be vertical (or nearly so) as illustrated by the stationary "clock" in the animation, Fig. 7.2a. In this reference frame (the moving reference frame) we use primes on values, so one clock tick in this reference frame will be designated as Δt'.

In one tick, the light travels upwards the height of the clock, then downwards the same distance again, this height h being the same value in both reference frames, as we saw in the section above. Thus we write 2h = cΔt'. Substituting 2h from (7.1b), we have:

This can be solved for Δt:

where γ is defined as:

. (7.2)

In summary, assuming that light travels at the same speed in all uniformly moving reference frames, we have shown that the time interval Δt (as measured by our "light clock") will be increased by motion as given by:

. (7.3)

Note that γ equals 1 for zero or low velocities (V << c) and becomes larger at higher velocities. It trends to infinity as the velocity, V, approaches c. This means that at low velocities, the time intervals is not affected by the motion and Δt equals Δt'. On the other hand at velocities very near the speed of light, Δt can be much greater than Δt' depending on the speed and the time interval can be greatly lengthened.

So far we have been considering time durations in our experiment shown in Fig. 7.2a, but this result is much more general than that. If we take light to be the essence of all electromagnetic interactions, we can surmise that if the apparent speed to light is really the same for all uniform reference frames, all electromagnetic processes must slow down as per (7.3) and that Δt could refer to the time length of all sorts of physical process inside speeding objects. For example Δt can refer to the length of one cycle of a clock on board an orbiting satellite, or it can refer to the average time it takes for the subatomic particle, the muon, to decay when it is in high speed motion. And indeed, clocks aboard an orbiting satellite do in fact run slow as compared with their counterparts on the ground, consistent with (7.3). Also, muons traveling at great speeds inside a particle accelerator do in fact have longer decay times that those that are stationary, consistent with (7.3).

The slowing of temporal events and physical processes due to speed as per (7.3) is called time dilation and has been experimentally verified many, many times in various ways.

Derivation of length contraction

The approach to this derivation is to calculate the time it takes for a light pulse to travel the length of a moving ruler in the direction of motion. See figure 7.3 for the setup. We correct for the distance the ruler has moved in this time. We then assume that the light pulse traveled at a speed equal to that of light, i.e. c and calculate the length of the ruler based on this.

Fig. 7.3. Experiment to determine the transformation of lengths in changing from a stationary to a moving reference frame. The setup here is basically one of Einstein's "clocks" shown in the previous animation, but here it is tipped on its side so the light pulse travels in the direction of the motion of the whole clock. We have the light pulse leave a trail in the stationary reference frame, so you can better see its path in this frame.

Mouse over (or click) the figure to start the animation. Mouse off to suspend it and click on it to restart it. Read the text below to understand it.

We start by examining the first part of the experiment, the part when the light travels from the source to the mirror. We assume it takes time Δt₁ to travel from the source to the mirror. During this time, the ruler has moved a distance:

Δx_ruler = VΔt₁ . (7.4)

During the same time the light pulse has traveled the distance:

d = cΔt₁ . (7.5)

This includes the length ℓ of the ruler plus the distance the ruler has moved, i.e.:

d = ℓ + Δx_ruler . (7.6)

We can substitute (7.4) and (7.5):

cΔt₁ = ℓ + VΔt₁ .

This can be solved for Δt₁ :

. (7.7)

There is a mirror at the far end of the ruler, so that the light is reflected back to the source (and detector). This second part is assumed to take a time Δt₂. Now the light pulse is going in the opposite direction as it was before so now in place of (7.6) we get:

d = ℓ − Δx_ruler , (7.8)

i.e. the ruler's motion reduces the length the light travels. Substituting as before for d and Δx_ruler (i.e. using (7.4) and (7.5) ) and using Δt₂ in place of Δt₁, we have:

cΔt₂ = ℓ − VΔt₂

This can be solved as:

The total time for the light to go from the source to the mirror and back to the source is given by:

. (7.9)

Next, we switch into the other reference frame, the moving one, in which the ruler is at rest. Einstein says that the speed of light is the same value, and in this reference frame there is no motion other than that of the light pulse. So we can write Δt' = 2ℓ'/c where Δt' is the time in the moving frame of one clock cycle, and ℓ' is the length of the ruler in this frame. Using the above derived time dilation expression (7.3), this becomes Δt/γ = 2ℓ'/c or Δt = 2γℓ'/c .

Substituting this into the above expression for Δt , i.e. into (7.9), we get 2γℓ'/c = (2γ²ℓ/c) which can be solved for ℓ to yield:

. (7.10)

This last expression captures the essence of length contraction, that an extremely rapidly moving object appears shorter in the direction of motion than it is when at rest or moving slowly.

Simultaneity

One of the oddities of relativity is that two events that occur at the same time in one reference frame will not be simultaneous in another reference frame. To understand this, consider the experiment shown in Fig. 7.4. In this experiment we have a moving ruler with a light pulse source in the middle. The light source sends pulses to detectors at the two ends of the ruler. We consider the pulses reaching their respective detectors to be the two spacially separated events in this experiment.

In the frame moving along with the ruler, the two light pulses would reach the two detectors simultaneously, meaning that in this, the moving reference frame, the events are simultaneous. In the stationary frame, in which the ruler is moving, the situation is quite different and the motion will shorten the transit time of the left pulse and lengthen that of the right pulse. Thus, in the stationary frame, the two events are not simultaneous.

We next calculate the error in synchronization (i.e. in simultaneity).

Fig. 7.4. Animation of a gedanken or thought experiment on relativistic simultaneity described above in the text. In this animation, we show the paths of both light pulses to help with the understanding. It may look like the two paths are too different to be real, but if you examine the animation carefully (stopping it when needed) you will see that the difference in the path lengths is purely due to the motion of the detectors. The difference is due to one beam having to catch up to a detector moving away from it while the other beam only has to run into a detector coming towards it. The animated velocity of the ruler is V = 0.7c .

Mouse over (or click) the figure to start the animation. Mouse off to suspend it and click on it to restart it.

The distance that the right light pulse must travel in the stationary frame is given by

where we have applied length contraction (i.e. the 1/γ or Eqn. (7.10) ) to the half ruler length, ℓ'/2, as measured in the moving frame. Also Δt₁ is the propagation time (measured in the stationary frame) and VΔt₁ is the distance the moving assembly moves in that time. We can solve this equation for Δt₁ as:

We repeat this for the left light pulse. The distance this pulse travels is:

which can be solved for the transit time of the left pulse:

The difference in transit times (measured in the stationary frame) for the synchronization pulses to reach the two clocks is:

Rewritten, using Δt as the difference in time intervals on the left side of the above equation, we have:

. (7.11)

In summary, in the line above we calculate the error in synchronization between two clocks that are synchronized in a moving reference frame. We will use this as one of the terms in the Lorentz transformations.

Summary

Item	Effect	Equation
Lengths perpendicular to reference frame motion	No change	Δy = Δy' Δz = Δz'
Lengths parallel to reference frame motion	Length contraction	Δx = Δx'/γ
Time intervals	Time dilation	Δt = γΔt'
Two simultaneous events separated in the x direction	Change in simultaneity upon change of reference frames	Δt = γΔxV/c²

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Derivation of constant speed of light from Lorentz transforms

2011-07-09T09:37:00.000-07:00

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6. Derivation of the constant speed of light from the Lorentz transforms

Velocity of light in the x direction in the moving frame:

We consider a light pulse traveling in the x direction, crossing the point x₁ then the point x₂. The distance traveled is related to the time it takes by:

x₂ − x₁ = c (t₂ − t₁) .

We use this and the Lorentz transforms (4.8) to calculate the velocity of light in the primed or moving reference frame.

(6.1)

Equation (6.1) shows that in the x direction, the velocity of light is the same value in the moving frame as in the stationary frame, i.e. c' = c .

Velocity of light in the y or z direction in the moving frame:

We next derive a similar result in a direction perpendicular to the relative motion. We will use the y direction for convenience, but we could equally well use any perpendicular direction.

We start by noting that to get light to appear to go vertically in the moving frame, it must be angled in the stationary frame as illustrated in the sketch at the right. This is similar to rain in a windless day coming straight down, but appearing to be coming down at an angle directed towards you when viewed from a moving automobile or bicycle.

As noted on the sketch, we immediately have two relationships:

Δx = VΔt and diagonal = cΔt .

Using Pythagoras's theorem, we solve for the remaining side:

This can be solved for Δt: Δt = γb/c .

Now to calculate the perpendicular velocity of light in the moving (primed frame):

Note that we also used the fact that the y coordinate is unchanged during the Lorentz transformation, making Δy' = Δy = b .

Thus, c' = c . As it was in the x direction, the velocity of light in the y direction is not changed by relative motion.

Summary

The velocity of light is unchanged by Lorentz transformations in both the parallel and perpendicular directions. We will address this question again for the arbitrary direction case after we have derived transformations for velocity (in Chapter 9).

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Derivation of length contraction and time dilation from Lorentz transformations

2011-07-09T09:29:00.000-07:00

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5. Derivation of length contraction and time dilation from Lorentz transformations

In this chapter we start with the Lorentz transformations and show that they can be used to derive length contraction and time dilation.

Derivation of length contraction:

Length contraction refers to a moving object appearing to look shorter when viewed in the stationary reference frame. (Lorentz would say that it really does become structurally shorter, while Einstein would say the metric has changed making it shorter.) We consider measuring the length of our object as taking measurements of the positions of the two ends x₁ and x₂ at the same time in the stationary reference frame. That is to say t₁ = t₂.

We need a transform that would let us use this fact, as well as relating x₁ and x₂ to x₁' and x₂'. That is to say we need a transform that involves t₁, t₂, x₁, x₂, x₁', and x₂'. Looking through the list of Lorentz transforms and inverse Lorentz transforms, we decide that the first Lorentz transform involves these variables:

x' = γ (x − Vt) (5.1)

where (5.2)

Applying this transform to the measurement of the two ends of the object, we have:

x₁' = γ (x₁ − Vt₁) (5.3)

x₂' = γ (x₂ − Vt₂) (5.4)

Subtracting the first equation from the second to get the length, we have:

length' = x₂' − x₁' = γ [(x₂ − x₁) − V(t₂ − t₁)] = γ (x₂ − x₁) = γ × length (5.5)

where we have used the fact that t₁ = t₂ makes the time difference factor zero. Inverting the equation we get:

length = length'/γ . (5.6)

Since γ is always equal to or greater than 1, we see that the length in the stationary frame (the unprimed frame) will be equal to or smaller than the length in the moving frame (the primed frame, inside of which the object appears stationary). The faster the object is moving, the larger γ will be and the greater the effect of length contraction will be.

Derivation of time dilation:

Time dilation refers to the time duration of a physical process in a moving object appearing to take longer when viewed from the stationary reference frame (which the object is moving relative to). Similar to the derivation of length contraction, we consider measuring the time duration of a process in the object which is stationary in the moving frame, but we wish to transform the starting and ending times of the process to the stationary frame. Since the object is stationary in the moving frame, x₁' = x₂' where x₁' and x₂' are respectively the starting and ending positions of the object in the moving reference frame.

We need a transform that will let us use the fact that x₁' = x₂' and also relates t₁'and t₂' to t₁ and t₂. That is, we need the Lorentz transform that involves x₁', x₂', t₁', t₂', t₁, and t₂. Looking through all the Lorentz transforms and inverse Lorentz transforms, we decide that the last inverse Lorentz transform involves these variables:

t = γ (t' + Vx'/c²) . (5.7)

Applying this transform to the start and finish times of our process, we have:

t₁ = γ (t'₁ + Vx₁'/c²) . (5.8)

t₂ = γ (t'₂ + Vx₂'/c²) . (5.9)

Subtracting the first equation from the second to get the time duration of the process, we have:

Δt = t₂ − t₁ = γ [t₂' − t₁' + (x₂' − x₁')V/c²] = γ (t₂' − t₁') = γ Δt' . (5.10)

Or in summary (equating the first and last):

Δt = γ Δt' . (5.11)

Since γ is always equal to or greater than 1, we see that the time duration Δt of a moving process observed in the stationary frame is equal to or larger than the time duration, Δt', observed in the moving frame, moving along with the process. The faster the object is moving, the larger γ will be and the greater the effect of time dilation will be. Another way to say this is to say that moving processes seem to slow down when viewed from the stationary reference frame.

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Inverting the Lorentz transforms

2011-07-09T08:54:00.000-07:00

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4. Inverting the Lorentz transforms

We need to use inverse Lorentz transforms in several of the derivations in the coming chapters, so we derive these here. The Lorentz transforms from the previous chapter are:

(4.1a)	(4.1b)
(4.1c)	(4.1d)

where .

These transforms give the primed or moving coordinates as functions of the unprimed or stationary coordinates. We will invert these to give the stationary coordinates as functions of the moving coordinates.

Inverting y and z transforms:

The y and z coordinates, (4.1b) and (4.1c), are trivial:

y' = y inverts to give y = y' and

z' = z inverts to give z = z' .

Inverting x and t transforms:

The transforms for x and t are intertwined and must be handled together. We repeat the original x and t transforms:

x' = γ (x − Vt) (4.2a)

t' = γ (t − Vx/c²) (4.2b)

We solve (4.2a) for x:

x = x'/γ + Vt (4.3)

and substitute this into (4.2b):

. (4.4)

Multiplying (4.4) by γ yields:

This can be solved for t:

. (4.5)

Substituting (4.5) into (4.3) yields:

. (4.6)

The factor in parenthesis in (4.6) can be reduced as follows:

Using this in (4.6) gives:

. (4.7)

Summary

Lorentz transforms	Inverse Lorentz transforms
x' = γ(x − V t) (4.8a)	x = γ(x' + V t') (4.9a)
y' = y (4.8b)	y = y' (4.9b)
z' = z (4.8c)	z = z' (4.9c)
(4.8d)	(4.9d)

Equations (4.9a) through (4.9d) are the inverse Lorentz transforms. They give the stationary (unprimed) x and t coordinates in terms of the moving coordinates (the primed ones). They are also interesting because except for the minus signs in front of the second terms in each equation, they are identical to the original Lorentz transforms, Equations (4.8a) through (4.8d). They can be had from the original transforms by simply substituting −V in place of V. It means that to the moving reference frame, except for the stationary frame appearing to be moving in the opposite direction, it has the same length contraction and time dilation that people in the stationary frame see when looking at the moving frame.

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Invariance of the electromagnetic wave equation to Lorentz transforms

2011-07-08T08:32:00.000-07:00

previous: wave equation from Maxwell's equations

next: inverting the Lorentz transforms

3. Invariance of the electromagnetic wave equation to Lorentz transforms

In this chapter, we show that the Lorentz transformations preserve the form of the wave equation, and that the older Galilean transformations do not.

To repeat what we stated in the last posting, the wave equation is of the form:

, (3.1)

where the ψ can be replaced by many different variables, depending on the type of wave involved. This equation is used in many areas of physics, wherever simple wave motion occurs.

Using Galilean transforms on the wave equation

We start with the Galilean transforms. These are listed below. They essentially say that in a uniformly moving reference frame, all is the same, except for in the x direction (the direction of motion of the moving frame). The x coordinate is offset by an increasingly negative value which represents the motion of the moving frame.



Fig. 3.1. Table of Galilean transforms. These are the transforms used to change to the coordinates of a uniformly moving reference frame moving at a speed V in the positive x direction. These are valid for speeds that are considerably less than the speed of light, and are thus valid for most everyday experiences. An example might be to change from a ground based coordinate system to a coordinate system aboard (or fixed to) a moving train, moving boat, or airplane.

We now use the above Galilean transforms to convert (3.1) from the x and t variables into those of the moving reference frame, x' and t'. We use the chain rule for partial derivatives. You may wonder if we should be transforming ψ also. First of all, we, as yet, don't have such a transformation in this discussion and don't know exactly what we are representing by ψ (we want to keep it general). Secondly, we only are interested in transforming the operator part of the equation, not the variable to which this operator is applied. If we show that this operator is the same, then the transformed equation will have the same solutions and the laws of physics (the waves in this case) will be same in the transformed reference frame. The "laws" are represented by the operator part of the equation.

So we start applying the Galilean transforms to the wave equation, i.e. equation (3.1).

Fig. 3.2. Galilean transformations applied to the wave equation.

We see the result in the last line of the above table. This is not the wave equation (at least not the simple wave equation), nor can it be converted into the wave equation by any allowable easy change. The difference represents the fact that in a Galilean world, waves travel at a certain velocity relative to a given fixed reference frame, the reference frame moving along with the wave media. In a second reference frame moving relative to the wave media, the wave equation will not be the standard simple form, and will reflect the fact that in this reference frame the media is moving. An example of this is the propagation of sound on a windy day. A stationary person will not be at rest relative to the wave media (the moving air) and therefore sound will appear to travel faster in some directions and slower in others.

Using Lorentz transforms on the wave equation

In 1887 Hendrik Lorentz came up with a set of space transformations that he claimed would be identical to Galilean transformations at low speed, but would allow Maxwell's equations to keep their same form upon transform. This was to be useful for explaining the experimental determination by Michelson and Morley that the speed of light is constant. In this section, we repeat the transformation process that we did above, but here we use Mr. Lorentz's transforms in place of the Galilean transformations.

We start by listing the Lorentz transformations.



where
Fig. 3.3. Table of Lorentz transforms.

Next we use these transformations to transform the wave equation (3.1) written in terms of the x and t coordinates into an equation written in terms of the x' and t' coordinates. We use the chain rule, as we did above for the Galilean transformation.

Fig. 3.4. Lorentz transformations applied to the wave equation.

We have used the following fact to cancel terms in the next to last line above:

The last line is just the normal simple wave equation written in terms of the primed coordinates, those of the moving reference frame.

We can reverse the above logic to show that if the wave equation is to stay the same (because of the Michelson-Morley result), then the Lorentz transforms must be the correct transformations to be used in changing between initial reference frames.

Einstein concluded from this that the Lorentz transforms therefore indicate how time and space must change in a high speed reference frame. Lorentz concluded that these transforms indicated how the apparent time and space must change.

Conclusions

1. The above derivations shows that the operator part of the wave equation is preserved by Lorentz transformations, but not by Galilean transformations.

2. This means that if the Lorentz transforms are valid then the same electromagnetic waves will occur in all inertial reference frames and that these waves travel at the same speed as determined by the coefficient of the time derivative term.

3. Furthermore, because the wave equations are so tightly coupled to Maxwell's equations (as shown in the previous chapter), this means that Maxwell's equations will hold in all inertial reference frames. We have not addressed the source side of the wave equations which contain the charge density and current density. We shall see in future chapters that the charge and current densities are changed by changing the reference frames; however, this does not affect our conclusion that Maxwell's equations stay the same in all inertia reference frames. Instead, the values of the charge and current density must change.

Einstein interpreted the Lorentz transformations as meaning that time and space were altered by high speed motion, while Lorentz held that these transformations were more mathematical in nature, allowing us to understand what the warping of our physical lengths and temporal events would do to our perception of the time and space. For more on their disagreement, see my introduction to relativity.