# Prove that 4 vector potential is really a 4 vector?

1. May 2, 2008

### LHS1

Prove that 4 vector potential does really a 4 vector? Most of the textbooks I found only mention that divergence of 4 vector potential equals to zero and the d'Alembertian of it is a four vector current and therefore it 'should be' a four vector. However I do not see there is any tensor theorem to get this conclusion. Could anyone prove it for me, thank you.

Last edited: May 2, 2008
2. May 2, 2008

### pmb_phy

The covariance of the Lorenz gauge, i.e. div A +(1/c)@(phi)/@t = 0 (where @ = partial sign), implies that Av = (phi,A) is a (contravariant) 4-vector. This follows from the quotient theorem which states that if O is a 4-operator and OA = scalar then A must be a 4-vector.

Pete

3. May 2, 2008

### LHS1

Reply to pmb phy, to the best of my knowledge, quotient theorem does not apply to operator acting on a tensor. Are you sure you are right about this?

4. May 2, 2008

### pam

If you already believe that $$j^\mu$$ is a 4-vector (from the invariance of charge), then $$\partial_\nu\partial^\nu A_\mu=4\pi j^\mu$$ proves that $$A^\mu$$ is a 4-vector.

Last edited: May 2, 2008
5. May 2, 2008

### pam

I can't seem to get the tex right. It should be 4\pi j^\mu.

6. May 3, 2008

### LHS1

Reply to pam, I do not think you are right. What you said is just like x^2=1, then x must be equal to 1. But it is not true, x can be -1. This is just A-level logic tells us that a proposition is true does not mean its converse is true.

7. May 4, 2008

### pmb_phy

Yes. If you recall the derivation of the quotient theorem you will note that the derivation only requires the transformation properties of a quantity and since the vector and operator transform as a covariant vector and a contravariant vector it follows that the quotient theorem also applies to operators.

Pete

8. May 4, 2008

### LHS1

Reply to pmb phy, I am still skeptic about this. It is because when I refer to the derivation of quotient theorem, I found that even though it only requires the transformation properties of the quantity acting on the tensor, but that quantity must be arbitrary. Am I wrong about this? Anyway, I found a way to prove that 4-vector potential is indeed a 4-vector. Thank you for your help sincerely.

9. May 5, 2008

### LHS1

Here I have to thanks those who have attempted to help me solve my problems, expecially to pmb phy and pam.

10. May 6, 2008

### pmb_phy

Your interpretation about arbitray is a bit wrong. What is arbitrary is that the 4-vector (in this case 4-potential) is an arbitraty 4-vector and the derivative of that 4-vector is not zero in general. With these facts it follows that the 4-potential is a 4-vector. Jackson's EM text explains this in the same exact way, but my explanation is clearer (Jackson does not mention the quotient theorem but does state that the invariance implies that the 4-potential is a 4-vector).

Pete

11. May 6, 2008

### pam

That's not what I said. In my post, the LHS is a scalar times a vector. I am saying that is a vector. My upper and lower \mu got a bit mixed up, but I couldn't correct it.

12. May 7, 2008

### Hans de Vries

1) charge is invariant under Lorentz transform then.

2) The charge/current vector must transform like $(1,\beta_x,\beta_y,\beta_z)$

3) The charge/current density $j^\mu$ must transform like $(\gamma,~\beta_x\gamma,~\beta_y\gamma,~\beta_z\gamma)$ because of Lorentz contraction

4) $A^\mu$ must also transform like $(\gamma,~\beta_x\gamma,~\beta_y\gamma,~\beta_z\gamma)$ because of $\Box~A^\mu=j^\mu$

The d'Alembertian $\partial_\mu\partial^\mu$ is a Lorentz invariant contraction, it has one raised and
one lowered index. The result of the operator transforms the same as the operand.
See for instance Jackson at the end of section 11.6

Regards, Hans.

(Edit, pmb_phy is right, but it is because of the d'Alembertian)

Last edited: May 7, 2008
13. May 9, 2008

### samalkhaiat

Last edited: May 9, 2008
14. May 10, 2008

### Phrak

Last edited: May 10, 2008
15. May 16, 2008

### Hans de Vries

Last edited: May 16, 2008
16. May 17, 2008

### samalkhaiat

17. May 17, 2008

### pmb_phy

Shhh! No yelling please. Some of us may have a hangover.
Please show me the source of this expression. I don't recall ever seeing it before.

Pete

18. May 17, 2008

### samalkhaiat

19. May 17, 2008

### samalkhaiat

Last edited: May 17, 2008
20. May 17, 2008

### Hans de Vries

In the Coulomb gauge the electric potential propagates with infinite speed while the
magnetic vector potential propagates with c. The final results of Bjorken & Drell are
only Lorentz invariant simply because V is set to zero and only $\vec{A}$ is used...

The Coulomb gauge does violate Special Relativity. Changes in V which propagate
instantaneously (dV/dx) give observable violations. Therefor it is not physically
meaningful.

True, the Lorentz gauge does not determine the potentials uniquely. The Lorentz
gauge does contain the Liènard Wiechert potentials which are uniquely determined
and the Liènard-Wiechert potential $A^{\mu}$ does transform like a 4-vector.

Using other potentials as the Liènard-Wiechert potentials can give rise to all kinds
of problems like Lorentz violations, space being not "simply-connected" and erroneous
results for the spin density of the electromagnetic field in vacuum:

$${\cal S}^\mu\ \ =\ \ \frac{e^2}{4\pi^2}\ \mbox{\large \varepsilon}^{\mu\alpha\beta\gamma} A_\alpha\partial_\beta A_\gamma\ \ =\ -\frac{e^2}{4\pi^2}\ \mbox{\large \varepsilon}^{\mu\alpha\beta\gamma}\ F^{\mu\alpha}A_\alpha$$

Which depends explicitly on the absolute value of the potentials.

Regards, Hans

Last edited: May 17, 2008