Undergrad Why is Displacement Current Excluded from Four-Current?

[particlezoo]

To put this in another way, is there some reason from first-principles as to why we have j as the spatial component of the four-current rather than the total current density which includes the displacement current? Has anyone tried to see what the experimental consequences of this would be?
Kevin M.

 

[Dale]

I am pretty sure that it wouldn’t transform as a four vector any more.

 

[particlezoo]

To put this in another way, is there some reason from first-principles as to why we have j as the spatial component of the four-current rather than the total current density which includes the displacement current? Has anyone tried to see what the experimental consequences of this would be?
Kevin M.

I am pretty sure that it wouldn’t transform as a four vector any more.

Hmmm... in the limiting case where there is just an (approximate) plane wave traveling at c, there is a "displacement current density" orthogonal to the wave vector. So the displacement current doesn't necessarily "propagate".

But in the real world, we can have (approximate) plane waves produced by moving charges, such as in dipole radiation in the far field, though the field behaves differently in the near vs. far field. Now, since the behavior of the charges would look different in one inertial frame vs another, the emitted electromagnetic field should also look different in one inertial frame vs. another. The emitted electromagnetic field should transform in such a way that it "agrees" with the transformed motion of the source charges.

If the motion of source charges (i.e. electric current) would transform as the spatial part of a four-vector, why wouldn't the displacement currents, which essentially have an equal and opposite divergence as that of the source charges, be capable of the same?

 

[Dale]

why wouldn't the displacement currents, which essentially have an equal and opposite divergence as that of the source charges, be capable of the same?

Because components of a rank 2 tensor don’t transform like a four vector. And tensors are linear so if you add two tensors you get a tensor but if you add a tensor and a non tensor then you get a non tensor.

 

[Dale]

in the limiting case where there is just an (approximate) plane wave traveling at c, there is a "displacement current density" orthogonal to the wave vector.

I just realized that this provides physical justification for not incorporating the displacement current with the four-current. The timelike component of the four-current is charge density and the displacement current does not turn into a charge density in other reference frames.

 

[PeterDonis]

is there some reason from first-principles as to why we have j as the spatial component of the four-current rather than the total current density which includes the displacement current?

The terminology you are using is misleading you (unfortunately it is common terminology, even though it is misleading). The term "displacement current" should not be taken to imply that it is an actual current composed of moving charges. It isn't. The "displacement current" is the time derivative of the electric field, which belongs on the LHS of Maxwell's Equations (the "field" side), not the RHS (the "source" side). Writing Maxwell's Equations in 4-tensor form (instead of the more common "scalar and 3-vector" form) makes this more obvious, since there is then no way to move the displacement current term to the RHS without "breaking" the EM field tensor apart, which you can't do in tensor notation.

Another way of looking at it is to note that, with Maxwell's Equations written in tensor form, the divergence of both sides is zero. For the LHS, this is an identity; for the RHS, it expresses charge-current conservation. But all of that only works if the displacement current is part of the LHS (which, as noted above, it has to be in tensor form). If you insist on breaking apart the EM field tensor and moving the displacement current to the RHS, the divergence no longer vanishes and you no longer have charge-current conservation. I would say that is the best physical justification for keeping the displacement current where it belongs on the LHS.