# A Electromagnetic tensor and time reversal

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1. Dec 5, 2018

### Andrea B DG

Consider equation (2.7.8) page 42 in the book Gravitation and Cosmology by Weinberg
F' αβ = Λαγ Λβδ Fγδ
Now consider the time reversal Lorenz transformation
Λμν = 0 if μ ≠ ν, 1 if μ = ν = 1..3 and -1 if μ = ν = 0
then
F' 00 = 0
F' 0i = -F 0i
F' ij = F ij
Using equation (2.7.5) of the same book this is equivalent to
E'i = -Ei
B'i = Bi
but the result should be
E'i = Ei
B'i = -Bi
since time reversal reverse the sign of the electric charge current.
It look like for time reversal the equation (2.7.8) needs a minus sign.
How to justify this exception?

2. Dec 5, 2018

### Paul Colby

Have you accounted for the differences between $F^{\alpha\beta}$ and $F_{\alpha\beta}$? There are some sine flips involved.

 yeah, sines those cancel. The other suggestion is one is looking at the dual tensor but I don't have this particular book.

3. Dec 6, 2018

### Andrea B DG

Yes I have considered the signs flips you mentioned.

Note that when considering space inversion (parity) Lorenz transformation
Pμν = 0 if μ ≠ ν, -1 if μ = ν = 1..3 and 1 if μ = ν = 0
I obtains the same result of time reversal (but with the correct physical meaning).

4. Dec 6, 2018

### Paul Colby

I found this paper that might help. I think things are more complicated than just applying time reversal Lorentz transformation as you have done.

Time reversal in classical electromagnetism - Philsci-Archivephilsci-archive.pitt.edu/3280/1/arntzenius_greaves_TRCE.pdf

 Also, how does the 4-vector potential transform under time reversal? Just applying the T from the Lorentz group yields, $\phi \rightarrow -\phi$ and $A\rightarrow A$. I think the opposite signs are what happen by the usual conventions. So, I'm as in the dark as you on this one.

5. Dec 7, 2018

### Andrea B DG

In the reference you mentioned I find: “Next let us consider the electric and magnetic fields. How do they transform under time reversal? Well, the standard procedure is simply to assume that classical electromagnetism is invariant under time reversal.”
I think that time invariance is contained in Maxwell's equations and must not be assumed.

Concerning 4-vector potential transformation under time reversal I get your own result using
Fμν = ∂μAν - ∂νAμ

In this particular case it look like for all Lorenz transformation Λ
F'αβ = Λαγ Λβδ Fγδ
except for time reversal T where
F'αβ = - Tαγ Tβδ Fγδ
I do not see how to get out of this.

6. Dec 7, 2018

### Paul Colby

I finding the paper very illuminating. One must choose a convention for how some basic things change under time reversal. I'm far from understanding all the fine points, but here is my 2 cents. A better staring point for me is the 4-current, $J^\mu$. Classically this is,

$J^t = \rho(x)$ for the time component
$\bar{J} = \rho(x) \bar{v}(x)$ for the space components

The charge density, $\rho(x)$, clearly shouldn't change under time reversal by the usual arguments and conventions while the velocity, $\bar{v}(x)$, clearly does. This leads directly to the usual time reversal behavior of EM. The paper goes on to discuss the Feynman view in which charge is reversed under time reversal (given I'm reading things correctly, always suspect). This view leads to the transformation rules given by the improper Lorentz transformation that you are doing. I'm guessing since the Feynman view intersects with charge conjugation, the preference in physics is to stick with the usual time reversal conventions. Hope this helps.

 Looking at your last post again I suggest the main difference boils down to how charge is -assumed- to change under time reversal. Your transformation (pure improper LT) changes the sign of the -scalar- charge while the usual/standard one does not.

7. Dec 8, 2018

### Andrea B DG

I have re-read our discussion, for which I thank you, and it seems to me that Maxwell's equations impose the following covariant formalism:

For all Lorenz transformation Λ except for time reversal T
J'μ = Λμν Jν
F'μν = Λμγ Λνδ Fγδ
→ A'μ = Λμν Aν

For time reversal T
J'μ = -Tμν Jν
F'μν = -Tμγ Tνδ Fγδ
→ A'μ = -Tμν Aν