Covariant formulation of Coulomb's Law

Click For Summary
The discussion explores the covariant formulation of Coulomb's Law using 4-vectors in the context of Maxwell's Equations. It presents the potential and fields of a point charge in terms of 4-velocity, revealing that the potential decreases as 1/r and the fields as 1/r² when the charge is at rest, consistent with classical expectations. The author notes an interesting symmetry: flipping the charge's sign and its 4-velocity yields identical potential and fields, supporting the interpretation of positrons as electrons moving backward in time. These formulas provide a neat and tidy representation, contrasting with traditional vector notation. The insights contribute to a deeper understanding of electromagnetic theory and special relativity.
Rootboy
Messages
1
Reaction score
0
I was noodling about with 4-vectors and Maxwell's Equations when I took it upon myself to figure out how to write the field of a point charge in terms of 4-vectors, since I'd never seen it before (if anyone knows a book that has this I'd be happy to see it). It's actually relatively simple, which I thought was pretty cool.

So let's take a point particle with charge q and four-velocity U^{\mu} that passes through the origin at t = 0. Let X^{\mu} be the coordinate where we want to evaluate the potential \Phi^{\mu} and fields F^{\mu\nu}. The speed of light and the Coulomb constant are both 1. I'm using metric signature (+, -, -, -), because I like timelike vectors to have positive norms, as God intended.

So after mucking around a bit I got this:

\Phi^{\mu} = \frac{qU^{\mu}}{\sqrt{(X^{\alpha}U_{\alpha})^{2} - X^{\alpha}X_{\alpha}}}

F^{\mu\nu} = \frac{q(X^{\mu}U^{\nu} - X^{\nu}U^{\mu})}{((X^{\alpha}U_{\alpha})^{2} - X^{\alpha}X_{\alpha})^{3/2}}

The (X^{\alpha}U_{\alpha})^{2} - X^{\alpha}X_{\alpha} expression in the denominator of both equations works out to just r^{2} if we're in a frame where the particle is at rest (U^{\mu} \rightarrow (1, 0, 0, 0)), so you get that the potential drops off as 1/r and the fields as 1/r^{2} as they should. If the particle is moving that expression gives the "length contracted" r^{2} that we expect from Special Relativity. Note that this expression is also (up to a constant and a sign) equal to the norm of the X^{\mu}U^{\nu} - X^{\nu}U^{\mu} expression in the numerator of F^{\mu\nu}. I'm not sure if that means anything but I thought it was kind of interesting.

Okay, so these are pretty neat and tidy formulas. But then I noticed that they kind of give you CT symmetry! If we flip the sign of the particle's charge and it's four-velocity, the potential and fields are exactly the same. So here we have a purely classical justification for the whole "a positron is an electron moving backwards in time" idea in QED. Which was new to me.

Thought this was cool and felt like sharing.
 
Physics news on Phys.org
F^{\mu\nu} for a charge moving with constant velocity is done in most EM textbooks by Lorentz transforming F^{\mu\nu} from rest.
You have given the covariant form of these expressions, which are usually given in vector notation in terms of \phi, A, E, and B.
 
Last edited:
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

Similar threads

  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
939
  • · Replies 19 ·
Replies
19
Views
3K
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 1 ·
Replies
1
Views
610