Definitions

# Relativistic electromagnetism

Relativistic electromagnetism is the idea of explaining electromagnetism based on relativistic (Albert Einstein 1905) arguments. It was first put forward in 1963 by Edward M. Purcell who wrote an innovative electromagnetism text in which he used special relativity to derive the existence of magnetism and radiation. His method illustrates the physical reason behind many important electrodynamic phenomena. This approach is mathematically easier than the more usual treatment based on the Biot-Savart law, Ampère's law, and Maxwell's equations.

Most of Purcell's explanation is based on using the Lorentz contraction factor:

$sqrt\left\{1 - v^2/c^2\right\}$

## Introduction

Purcell argued that the question of an electric field in one reference frame, and how it looks from a different reference frame moving with respect to the first, is crucial to understand fields created by moving sources. In the special case, the sources that create the field are at rest with respect to one of the reference frames. Given the electric field in the frame where the sources are at rest, Purcell asked: what is the electric field in some other frame?

He stated that the fundamental assumption is that, knowing the electric field at some point (in space and time) in the rest frame of the sources, and knowing the relative velocity of the two frames provided all the information needed to calculate the electric field at the same point in the other frame. In other words, the electric field in the other frame does not depend on the particular distribution of the source charges, only on the local value of the electric field in the first frame at that point. He assumed that the electric field is a complete representation of the influence of the far-away charges.

## Uniform electric field — simple analysis

Consider the very simple situation of a charged parallel-plate capacitor, whose electric field (in its rest frame) is uniform (neglecting edge effects) between the plates and zero outside.

To calculate the electric field of this charge distribution in a reference frame where it is in motion, suppose that the motion is in a direction parallel to the plates as shown in figure 1. The plates will then be shorter by a factor of:

$sqrt\left\{1 - v^2/c^2\right\}$

than they are in their rest frame, but the distance between them will be the same. Since charge is independent of the frame in which it is measured, the total charge on each plate is also the same. So the charge per unit area on the plates is therefore larger than in the rest frame by a factor of:

$1oversqrt\left\{1 - v^2/c^2\right\}$

The field between the plates is therefore stronger by this factor.

## More rigorous analysis

Consider the electric field of a single, infinite plate of positive charge, moving parallel to itself. The field must be uniform both above and below the plate, since it is uniform in its rest frame. We also assume that knowing the field in one frame is sufficient for calculating it in the other frame.

The plate however could have a non zero component of electric field in the direction of motion as in Fig 2a. Even in this case, the field of the infinite plane of negative charge must be equal and opposite to that of the positive plate (as in Fig 2b), since the combination of plates is neutral and cannot therefore produce any net fields. When the plates are separated, the horizontal components still cancel, and the resultant is a uniform vertical field as shown in Fig 1.

If Gauss's law is applied to pillbox as shown in Fig 1, it can be shown that the magnitude of the electric field between the plates is given by:

$|E\text{'}|= \left\{sigma\text{'} overepsilon_0\right\}$

where the prime (') indicates the value measured in the frame in which the plates are moving. $sigma$ represents the surface charge density of the positive plate. Since the plates are contracted in length by the factor

$sqrt\left\{1 - v^2/c^2\right\}$
then the surface charge density in the primed frame is related to the value in the rest frame of the plates by:

$sigma\text{'} = \left\{sigmaoversqrt\left\{1 - v^2/c^2\right\}\right\}$

But the electric field in the rest frame has value σ / ε0 and the field points in the same direction on both of the frames, so

$E\text{'} = \left\{Eoversqrt\left\{1 - v^2/c^2\right\}\right\}$

The E field in the primed frame is therefore stronger than in the unprimed frame. If the direction of motion is perpendicular to the plates, length contraction of the plates does not occur, but the distance between them is reduced. This closer spacing however does not affect the strength of the electric field. So for motion parallel to the electric field E,

$E\text{'} = E$

In the general case where motion is in a diagonal direction relative to the field the field is merely a superposition of the perpendicular and parallel fields., each generated by a set of plates at right angles to each other as shown in Fig 3. Since both sets of plates are length contracted, the two components of the E field are

$E\text{'}_y = \left\{E_yoversqrt\left\{1 - v^2/c^2\right\}\right\}$

and

$E\text{'}_x = E_x$

where the y subscript denotes perpendicular, and the x subscript, parallel.

These transformation equations only apply if the source of the field is at rest in the unprimed frame.

## The field of a moving point charge

A very important application of the electric field transformation equations is to the field of a single point charge moving with constant velocity. In its rest frame, the electric field of a positive point charge has the same strength in all directions and points directly away from the charge. In some other reference frame the field will appear differently. In applying the transformation equations to a nonuniform electric field, it is important to record not only of the value of the field, but also at what point in space it has this value.

In the rest frame of the particle, the point charge can be imagined to be surrounded by a spherical shell which is also at rest. In our reference frame, however, both the particle and its sphere are moving. Length contraction therefore states that the sphere is deformed into a spheroid, as shown in cross section in Fig 3. Consider the value of the electric field at any point on the surface of the sphere. Let x and y be the components of the displacement (in the rest frame of the charge), from the charge to a point on the sphere, measured parallel and perpendicular to the direction of motion as shown in the figure. Because the field in the rest frame of the charge points directly away from the charge, its components are in the same ratio as the components of the displacement:

$\left\{E_y over E_x\right\} = \left\{y over x\right\}$

In our reference frame, where the charge is moving, the displacement x' in the direction of motion is length-contracted:

$x\text{'} = xsqrt\left\{1 - v^2/c^2\right\}$

The electric field at any point on the sphere points directly away from the charge. (b) In a reference frame where the charge and the sphere are moving to the right, the sphere is length-contracted but the vertical component of the field is stronger. These two effects combine to make the field again point directly away from the current location of the charge. (While the y component of the displacement is the same in both frames).

However, according to the above results, the y component of the field is enhanced by a similar factor:

$E\text{'}_y = \left\{Eyoversqrt\left\{1 - v^2/c^2\right\}\right\}$

whilst the x component of the field is the same in both frames. The ratio of the field components is therefore

$\left\{E\text{'}_y over E\text{'}_x\right\} = \left\{E_y over E_xsqrt\left\{1 - v^2/c^2\right\}\right\}$

So, the field in the primed frame points directly away from the charge, just as in the unprimed frame. A view of the electric field of a point charge moving at constant velocity is shown in figure 4. The faster the charge is moving, the more noticeable the enhancement of the perpendicular component of the field becomes. If the speed of the charge is much less than the speed of light, this enhancement is often negligible. But under certain circumstances, it is crucially important even at low velocities.

## The origin of magnetic forces

In the simple model of events in a wire stretched out horizontally, a current can be represented by the evenly spaced positive charges, moving to the right, whilst an equal number of negative charges remain at rest. If the wire is electrostatically neutral, the distance between adjacent positive charges must be the same as the distance between adjacent negative charges.

Assume a positively charged particle, with charge Q, outside the wire and traveling (initially) in a direction parallel to the current (Figure 5). This particle is the test charge. Assume the speed of the test charge to be v, the same as the speed of the moving charges in the wire. The test charge should experience a magnetic force as can be easily confirmed by experiment.

In this situation where both the source charges and the test charge are moving, we consider what happens in a different reference frame, for simplicity. Consider a reference frame where the test charge is at rest, at least initially (Figure 6). The reference frame is the 'test charge frame' and the original reference frame was the 'lab frame'.

In the 'test charge frame' the only possible force is the electrostatic force QE, since the electric field is defined as the force exerted on a unit positive test charge that is at rest. The same situation as in the fig, but viewed from the reference frame in which the test charge is initially at rest. Here the positive charges in the wire are at rest while the negative charges in the wire are moving to the left. The distance between the negative charges is 'length-contracted' relative to the lab frame, while the distance between the positive charges is not 'length-contracted', so the wire carries a net negative charge. The situation in the test charge frame is shown in the figure. The positive charges in the wire are also at rest, but the negative charges in the wire are moving to the left with speed v.

If we now consider length contraction, the negative charges are at rest in the lab frame but moving in the frame of the test charge, so the distance between these negative charges is smaller in the test charge frame than in the lab frame. Their average separation is given by:

$l\left(-\right) = \left\{lsqrt\left\{1-v^2/c^2\right\}\right\}$

Conversely, the positive charges are moving in the lab frame but at rest in the test charge frame. So the distance between the positive charges is larger in the test charge frame than in the lab frame. Their separation in the test charge frame is:

$l\left(+\right) = \left\{loversqrt\left\{1-v^2/c^2\right\}\right\}$

Both of these effects combine to give the wire a net negative charge in the frame of test charge. Now it is known that a negatively charged wire exerts an attractive force on a positively charged particle. The test charge will therefore be attracted and will move toward the wire.

If the currents are in opposite directions, consider the charge moving to the left. No charges are now at rest in the reference frame of the test charge. The negative charges are moving with speed v in the test charge frame so their spacing is again:

$l\left(-\right) = \left\{lsqrt\left\{1-v^2/c^2\right\}\right\}$

The distance between positive charges is more difficult to calculate, but should be less than 2v due to special relativity. For simplicity, assume the spread is 2v. The positive charge spacing contraction is then:

$l\left(+\right) = \left\{sqrt\left\{1-\left(2v/c\right)^2\right\}\right\}$

relative to its value in their rest frame. Now its value in their rest frame was found to be

$l\left(+\right) = \left\{loversqrt\left\{1-v^2/c^2\right\}\right\}$

So the final spacing of positive charges is:

$l\left(+\right) = \left\{loversqrt\left\{1-v^2/c^2\right\}\right\}\left\{sqrt\left\{1-\left(2v/c\right)^2\right\}\right\}$

To determine whether l(+) or l(-) is larger we assume that v<

$\left(1.x\right)^p approxeq 1+pxmbox\left\{ when \right\}x<<1$

Applying this approximation to both equations above gives:

$l\left(-\right) = \left\{loversqrt\left\{1-v^2/c^2\right\}\right\}$

After some algebraic calculation it is found that l+ < l- so the wire is positively charged in the frame of the test charge.

## Calculating the magnetic field

### The Lorentz force law

A test charge near a wire carrying current will experience a magnetic force dependent on the velocity of the moving charges in the wire. If the current is flowing to the right, and a positive test charge is below the wire, then there is a force in a direction 90 deg CCW from the direction of motion.

### The magnetic field of a wire

Calculation of the magnitude of the force exerted by a current-carrying wire on a moving charge is equivalent to calculating the magnetic field produced by the wire. Consider again the situation shown in figures. The latter figure, showing the situation in the reference frame of the test charge, is reproduced in the figure. The positive charges in the wire, each with charge q, are at rest in this frame, while the negative charges, each with charge −q, are moving to the left with speed v. The average distance between the negative charges in this frame is length-contracted to:

$sqrt\left\{1 - v^2/c^2\right\}$

where is the distance between them in the lab frame. Similarly, the distance between the positive charges is not length-contracted:

$sqrt\left\{1 - v^2/c^2\right\}$

Both of these effects give the wire a net negative charge in the test charge frame, so that it exerts an attractive force on the test charge.

## References

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