**Required math: algebra**

**Required physics: special relativity**

Reference: Moore, Thomas A., *A General Relativity Workbook*, University Science Books (2013) – Chapter 7; Problems 7.5.

We’ve been using the electromagnetic (EM) field tensor* * in several problems without saying where it comes from, so it’s time to fill in the gap with a look at how the form of this tensor was deduced.

To say this is a derivation of the EM tensor is probably stretching things a bit; it’s more of a plausibility argument. As always in relativity, the idea is to generalize the equations of classical physics by putting them in tensor form.

We start with the charge density and current. As we’ve already seen, densities (of either mass or charge) are not invariant under a Lorentz transformation because of length contraction. If we start with a charge density at rest and transform to an inertial frame moving at velocity , the density becomes . However, we also generate a current due to the motion of the charge, which is .

The four-current is defined as a four-vector whose component is the charge density and whose spatial components are the current. That is, in a frame moving with velocity relative to the charge, we have

In the charge’s rest frame, this is

so the invariant square of the four-current is

The next step is to look at Gauss’s law in differential form. This has the form

Moore uses the alternative form

where . If we generalize the RHS to the four-current (which is a four-vector), then we need to make the LHS a four-vector as well. In its current form, is a scalar, so we need to find some four-vector of which this is the 0 component (since is the 0 component of ). Since we need to take the derivative on the LHS, we can try the form

where is a rank-2 tensor to be determined.

Note that there is one slight snag in this argument. In general, the derivative of a tensor is *not* another tensor, so the quantity on the LHS is not a tensor in a general coordinate system. However, if the transformation between coordinate systems involves partial derivatives that are constants (that is, ), then the derivative of a tensor *is* a tensor. For Lorentz transformations, this condition is true, since the transformation depends only on the (constant) relative velocity of the inertial frames. Thus in special relativity, the above equation is valid.

To make this equation consistent with Gauss’s law, therefore, we need

where the dashed entries are to be determined.

Next we look at the electrostatic force, which in Newtonian terms is

If we generalize the LHS to , we again need the RHS to be a four-vector. Since is a rank-2 tensor, we need to contract it with something to give us a four-vector. Taking partial derivatives won’t work here, since the force equation involves on its own, not its derivatives. At this point, we make a leap of logic and propose that we contract with the four-velocity . This certainly isn’t a derivation; its only justification is that it works. So we get the relativistic form of the force law:

Note that is the covariant version of the four-velocity, so its time component , while the other components are the same in both forms.

From here, we can use the following identity:

In line 3 we used the fact that in flat space is a constant. In line 4, we used the symmetry of the flat space metric: to swap the indices on the second term.

Now the invariant is a constant, so . Therefore

In order for this to be true in general, that is for all four-velocities, we need to impose a condition on . We *could* just require , but that wouldn’t get us anywhere, since that would mean the electric field would have to be zero. The trick lies in the summation; if we require then

where in the last line we swapped the two dummy indices. Thus if is anti-symmetric, the condition is automatically satisfied. This allows us to fill out the tensor a bit more:

For a particle at rest, for and , so for from above we have

which is the same as 8, since for a particle at rest . So far, so good. However, we can now use 9 to see what happens when the particle is not at rest. We will fill in the tensor with the (admittedly suggestive) symbols shown:

At this point, we don’t know what the s are; we’re just using them as placeholders in the tensor. Then from 9 we get for the component

This looks a lot like the Lorentz (this guy gets around) force law. If we use , and for we get

Finally, we note that

The final form is therefore

with similar forms for the and components. This really is the Lorentz force law, and as we’ve seen from the derivation, it is valid in relativity as well as Newtonian physics, since we used relativistic four-vectors throughout, and never made the approximation of small velocities.

In one sense, we can take this as a *prediction* of the magnetic field and of the Lorentz force law, since these came out of our generalization of the equations for electrostatics (without any reference to magnetic fields).

There is a lot of ‘try it and see’ in this derivation, but it seems that’s the way a lot of physics works. It’s not like mathematics where we specify a minimal set of axioms and then rigorously derive every other result from them.

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