Prove that 4 vector potential is really a 4 vector?

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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.
 
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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
 
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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
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?
 

pam

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

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I can't seem to get the tex right. It should be 4\pi j^\mu.
 
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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.
 
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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?
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
 
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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.
 
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Here I have to thanks those who have attempted to help me solve my problems, expecially to pmb phy and pam.
 
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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?
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
 

pam

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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.
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.
 

Hans de Vries

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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.


If you start with the assumption that:

1) charge is invariant under Lorentz transform then.

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

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

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


The d'Alembertian [itex]\partial_\mu\partial^\mu[/itex] 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)
 
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samalkhaiat

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Prove that 4 vector potential does really a 4 vector?

The vector potential is not a 4-vector!

Under Lorentz transformation, the vector potential transforms according to

[tex]a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu} + \partial_{\mu}\Omega[/tex]

This means that [itex]a_{\mu}[/itex] is a 4-vector, if and only if;

[tex]\partial_{\mu}\Omega = 0[/tex]

Since this is not compatible with the arbitrary nature of the gauge function, [itex]\Omega[/itex], we conclude that [itex]a_{\mu}[/itex] is not a 4-vector.

Deriving the transformation law of [itex]a_{\mu}[/itex] from the gauge-fixed Maxwell equation

[tex]\partial_{\mu}\partial^{\mu} a_{\nu} = J_{\nu}[/tex]

is a wrong practice. The gauge-invariant Maxwell equation is

[tex]\partial^{\nu}f_{\mu \nu} = J_{\mu}[/tex]

Lorentz covariance(of this gauge invariant equation) requires

[tex]a \rightarrow \Lambda a + \partial \Omega[/tex]

Well, this is not how a 4-vector transforms. Is it?

Genuine 4-vectors cannot describe the two polarization states of light. So, it is not a bad thing that the vector potential is not a 4-vector.

If this does not convince you, see Weinberg's book "Quantum Field Theory" Vol I, on page 251, he says:

"The fact that [itex]a_{0}[/itex] vanishes in all Lorentz frames shows vividly that [itex]a_{\mu}[/itex] cannot be a four-vector. ........
..........
..........
Although there is no ordinary four-vector field for massless particles of hilicity [itex]\pm 1[/itex], there is no problem in constructing an antisymmetric tensor ........
......
......
[tex]f_{\mu \nu} = \partial_{\mu}a_{\nu} - \partial_{\nu}a_{\mu}[/tex]

Note that this is a tensor even though [itex]a_{\mu}[/itex] is not a four-vector, ......"


See also Bjorken & Drell "Relativistic Quantum Fields", on page 73, they say this:

"Actually, under Lorentz transformation [itex]A_{\mu}[/itex] does not transform as a four-vector but is supplemented by an additional gauge term."

So, my friend, you should have asked the following instead;

"prove that the vector potential is not a 4-vector" :smile:

regards

sam
 
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The vector potential is not a 4-vector!

Under Lorentz transformation, the vector potential transforms according to

[tex]a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu} + \partial_{\mu}\Omega[/tex]

This means that [itex]a_{\mu}[/itex] is a 4-vector, if and only if;

[tex]\partial_{\mu}\Omega = 0[/tex]

Since this is not compatible with the arbitrary nature of the gauge function, [itex]\Omega[/itex], we conclude that [itex]a_{\mu}[/itex] is not a 4-vector.
I don't understand where [tex]a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu} + \partial_{\mu}\Omega[/tex] comes from.

If I assume [tex]\acute{a}_{\mu}[/tex] is a regauged (dual) 4-vector, [tex]\acute{a}=a + d\Omega[/tex],

the Lorentz transformation, or any linear transform for that matter, acts on the entire tensor, [tex]\acute{a}_{\mu}\rightarrow \Lambda_{\mu}\\ ^\nu [a_{\nu} + (\partial_{\nu}\Omega)][/tex].
 
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Hans de Vries

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Since this is not compatible with the arbitrary nature of the gauge function, [itex]\Omega[/itex], we conclude that [itex]a_{\mu}[/itex] is not a 4-vector.

Deriving the transformation law of [itex]a_{\mu}[/itex] from the gauge-fixed Maxwell equation

[tex]\partial_{\mu}\partial^{\mu} a_{\nu} = J_{\nu}[/tex]

is a wrong practice.

OK, but worse things happen if you leave the Lorentz gauge presumed here.
For instance: In the Coulomb gauge the Coulomb potential becomes instantaneous.
(It would propagate with infinite speed....) See for instance Jackson section 6.3.

So I think it would be OK to use the principle of local gauge invariance to find
the interaction terms, but I would take the Lorentz gauge (The one of the
Liènard Wiechert potentials) as the physically most relevant one, and

In the Lorentz gauge [itex]A^\mu[/itex] does transform like a four vector.


Regards, Hans.
 
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samalkhaiat

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OK, but worse things happen if you leave the Lorentz gauge presumed here.
For instance: In the Coulomb gauge the Coulomb potential becomes instantaneous.
(It would propagate with infinite speed....)
The text by Bjorken & Drell uses the Coulomb gauge only! You are free to choose any gauge you want as long as your final results are gauge and Lorentz invariant.

The choice of gauge is a calculation device and it has nothing to do with "how the object [itex]A^{\mu}[/itex] transforms" under Lorentz group.

See for instance Jackson section 6.3.
Me see Jackson! :smile:


In the Lorentz gauge [itex]A^\mu[/itex] does transform like a four vector.
No it does not because Lorentz gauge does not determine the potentials uniquely. However, [itex]\partial_{\mu}A^{\mu}[/itex] is a Lorentz scalar even though [itex]A^{\mu}[/itex] is not a 4-vector.


regards

sam
 
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The vector potential is not a 4-vector!
Shhh! No yelling please. Some of us may have a hangover. :biggrin:
Under Lorentz transformation, the vector potential transforms according to

[tex]a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu} + \partial_{\mu}\Omega[/tex]
Please show me the source of this expression. I don't recall ever seeing it before.

Pete
 

samalkhaiat

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I don't understand where [tex]a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu} + \partial_{\mu}\Omega[/tex] comes from.

If I assume [tex]\acute{a}_{\mu}[/tex] is a regauged (dual) 4-vector, [tex]\acute{a}=a + d\Omega[/tex],

the Lorentz transformation, or any linear transform for that matter, acts on the entire tensor, [tex]\acute{a}_{\mu}\rightarrow \Lambda_{\mu}\\ ^\nu [a_{\nu} + (\partial_{\nu}\Omega)][/tex]
Gauge transformation tells you that the difference between two potentials is a 4-vector:

[tex]a_{(2)}^{\mu} - a_{(1)}^{\mu} = \partial^{\mu} \omega[/tex]

Thus, under Lorentz transformation, [itex](a_{(2)} - a_{(1)})[/itex] transforms like a 4-vector

[tex]\left( a_{(2)}^{\mu} - a_{(1)}^{\mu} \right)^{'} = \Lambda^{\mu}{}_{\nu} \left( a_{(2)}^{\nu} - a_{(1)}^{\nu}\right)[/tex]

Clearly, this is satisfied by

[tex]a_{(1,2)}^{\mu^{'}} = \Lambda^{\mu}{}_{\nu} a_{(1,2)}^{\nu} + \partial^{\mu}\Omega[/tex]

i.e., Lorentz transformation is always associated a gauge transformation so that the same choice of gauge is available for all Lorentz observers. Indeed the above transformation law guarantees that our gauge-invariant theory is Lorentz invariant as well.
You can now see exactly what is involved for a particular choice of gauge. In the Lorentz gauge, for example, we require [itex]\partial . a[/itex] to be relativistic invariant

[tex]\partial^{'} . \ a^{'} = \partial \ . \ a[/tex]

even though a is not a 4-vector. From the above transformation law, you see that this is possible only if

[tex]\partial^{2}\Omega = 0[/tex]

which is the familiar equation for [itex]\Omega[/itex] in the Lorentz gauge.

The inhomogeneous transformation law

[tex]a \rightarrow \Lambda a + \partial \Omega[/tex]

results from the fact that the gauge potential is a connection 1-form on principal bundle [itex]\left(R^{4} \times U(1)\right)[/itex].
Connections 1-forms have the following properties:

1) they transform inhomogeneously under coordinate transformations.
2) if [itex]\Gamma[/itex] is an arbitrary connection, and T is a tensor field, then [itex](T + \Gamma )[/itex] is another connection. Conversely, if [itex]\Gamma_{1}[/itex] and [itex]\Gamma_{2}[/itex] are connections, then [itex]\Gamma_{2} - \Gamma_{1}[/itex] is a tensor field.
3) they enable us to define a covariant derivatives [itex]D = \partial + \Gamma[/itex], and a curvature 2-form (= field tensor):

[tex]R^{(.)}{}_{(.) \mu \nu} = D_{\mu} \Gamma^{(.)}{}_{\nu (.)} - D_{\nu} \Gamma^{(.)}{}_{\mu (.)}[/tex]

You can now see that (not just the gauge potential [itex]a_{\mu}[/itex]) the gravitatinal potential [itex]\Gamma^{\lambda}{}_{\mu \nu}[/itex] ( Reimannian connection) also satisfies the above three conditions.

If I have asked you (and the other participants in this thread) to prove that [itex]\Gamma^{\lambda}{}_{\mu \nu}[/itex] is a (1,2)-type world tensor, would not you (and the others) have laughed at me and said: "but [itex]\Gamma[/itex] is not a tensor!"

regards

sam
 

samalkhaiat

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Shhh! No yelling please. Some of us may have a hangover. :biggrin:
I should have known this by looking at the "proofs" :rofl:

Please show me the source of this expression. I don't recall ever seeing it before.
Read post #13 when you are sober! Sorry to disturbe your sleep.:zzz:

sam
 
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Hans de Vries

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The text by Bjorken & Drell uses the Coulomb gauge only! You are free to choose any gauge you want as long as your final results are gauge and Lorentz invariant.
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 [itex]\vec{A}[/itex] 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.

No it does not because Lorentz gauge does not determine the potentials uniquely. However, [itex]\partial_{\mu}A^{\mu}[/itex] is a Lorentz scalar even though [itex]A^{\mu}[/itex] is not a 4-vector.

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 [itex]A^{\mu}[/itex] 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:

[tex]
{\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
[/tex]

Which depends explicitly on the absolute value of the potentials.


Regards, Hans
 
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