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

• LHS1
In summary, the vector potential is not a 4-vector because it does not transform like one under Lorentz transformation. This is illustrated by the fact that the gauge-fixed Maxwell equation is incompatible with the gauge transformation, making it impossible for a_{\mu} to be a 4-vector. Furthermore, the two polarization states of light cannot be described by a genuine 4-vector, making it understandable why the vector potential is not a 4-vector.

#### 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.

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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

pmb_phy said:
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?

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.

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

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.

LHS1 said:
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

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.

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

LHS1 said:
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

LHS1 said:
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.

LHS1 said:
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.

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)

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LHS1 said:
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

$$a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu} + \partial_{\mu}\Omega$$

This means that $a_{\mu}$ is a 4-vector, if and only if;

$$\partial_{\mu}\Omega = 0$$

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

Deriving the transformation law of $a_{\mu}$ from the gauge-fixed Maxwell equation

$$\partial_{\mu}\partial^{\mu} a_{\nu} = J_{\nu}$$

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

$$\partial^{\nu}f_{\mu \nu} = J_{\mu}$$

Lorentz covariance(of this gauge invariant equation) requires

$$a \rightarrow \Lambda a + \partial \Omega$$

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

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

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

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

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

regards

sam

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samalkhaiat said:
LHS1 said:
The vector potential is not a 4-vector!

Under Lorentz transformation, the vector potential transforms according to

$$a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu} + \partial_{\mu}\Omega$$

This means that $a_{\mu}$ is a 4-vector, if and only if;

$$\partial_{\mu}\Omega = 0$$

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

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

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

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samalkhaiat said:
LHS1 said:
Since this is not compatible with the arbitrary nature of the gauge function, $\Omega$, we conclude that $a_{\mu}$ is not a 4-vector.

Deriving the transformation law of $a_{\mu}$ from the gauge-fixed Maxwell equation

$$\partial_{\mu}\partial^{\mu} a_{\nu} = J_{\nu}$$

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 $A^\mu$ does transform like a four vector.Regards, Hans.

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Hans de Vries said:
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 $A^{\mu}$ transforms" under Lorentz group.

See for instance Jackson section 6.3.

Me see Jackson!

In the Lorentz gauge $A^\mu$ does transform like a four vector.

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

regards

sam

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

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

Pete

Phrak said:
I don't understand where $$a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu} + \partial_{\mu}\Omega$$ comes from.

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

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

Gauge transformation tells you that the difference between two potentials is a 4-vector:

$$a_{(2)}^{\mu} - a_{(1)}^{\mu} = \partial^{\mu} \omega$$

Thus, under Lorentz transformation, $(a_{(2)} - a_{(1)})$ transforms like a 4-vector

$$\left( a_{(2)}^{\mu} - a_{(1)}^{\mu} \right)^{'} = \Lambda^{\mu}{}_{\nu} \left( a_{(2)}^{\nu} - a_{(1)}^{\nu}\right)$$

Clearly, this is satisfied by

$$a_{(1,2)}^{\mu^{'}} = \Lambda^{\mu}{}_{\nu} a_{(1,2)}^{\nu} + \partial^{\mu}\Omega$$

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 $\partial . a$ to be relativistic invariant

$$\partial^{'} . \ a^{'} = \partial \ . \ a$$

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

$$\partial^{2}\Omega = 0$$

which is the familiar equation for $\Omega$ in the Lorentz gauge.

The inhomogeneous transformation law

$$a \rightarrow \Lambda a + \partial \Omega$$

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

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

$$R^{(.)}{}_{(.) \mu \nu} = D_{\mu} \Gamma^{(.)}{}_{\nu (.)} - D_{\nu} \Gamma^{(.)}{}_{\mu (.)}$$

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

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

regards

sam

pmb_phy said:
Shhh! No yelling please. Some of us may have a hangover.

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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samalkhaiat said:
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 $\vec{A}$ 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.

samalkhaiat said:
No it does not because Lorentz gauge does not determine the potentials uniquely. However, $\partial_{\mu}A^{\mu}$ is a Lorentz scalar even though $A^{\mu}$ 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 $A^{\mu}$ 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

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samalkhaiat said:
Under Lorentz transformation, the vector potential transforms according to

$$a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu} + \partial_{\mu}\Omega$$

Isn't that the Poincare transformation? Lorentz transformations are purely rotations, the Poincare family are the ones that include translations. Anyway I remember how bad the Coulomb gauge is, and I'm sure if you just restrict yourself to the Lorentz gauge and gauges related to it by a pure rotation, we're fine right?

samalkhaiat said:
LHS1 said:
Under Lorentz transformation, the vector potential transforms according to
$$a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu} + \partial_{\mu}\Omega$$

No, this is not the correct starting point for gauge theory. The potential in a gauge theory is, by definition, a connection 1-form which takes values in the Lie algebra of the gauge group. Under a coordinate transformation it transforms as
$$a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu}$$​
i.e., as a 4-vector. A Lorentz transformation is a particular special case of a coordinate transformation on Minkowski space.

On the other hand, under a local gauge transformation S the potential transforms as
$$a_{\mu} \rightarrow S a_{\mu} S^{-1} + (\partial_{\mu} S) S^{-1}$$​
and in the case of electromagnetism, $$S = exp(i\Omega)$$ and
$$a_{\mu} \rightarrow a_{\mu} + \partial_{\mu} \Omega$$​
(I've lost an i somewhere.) This accounts for the second term in your result. The gauge transformation may be performed independently of any coordinate transformation, and vice versa. Generally there is no requirement to perform them in the specific combination you have mentioned. It is necessary if you wish to maintain a particular (non-covariant) gauge condition under the transformation. QED can be defined in a non-covariant gauge, so it's convenient to define the **operator** a_\mu in this way, which perhaps is what Weinberg is doing. However, prior to quantisation, the potential in any gauge theory is certainly a vector.

Dave

samalkhaiat said:
I should have known this by looking at the "proofs" :rofl:
There is no need to be rude. It is against forum policy in fact. My comment was an obvious joke and not to be taken literally. Large letters are considered yelling and posting them as such goes against acceptable forum posting policy. Please read those rules before posting comments like this again. Especially since your claim is wrong. The 4-potential is most definitely a 4-vector.
Read post #13 when you are sober! Sorry to disturbe your sleep.:zzz:

sam
(sigh!) Yet another rude comment? I could very well point out that you were asleep when you responded to this post because it was exactly post #13 that I was asking about. That is obvious to the even the most casual observer since it was the same post that I commented on the large letters. Do you wish to answer the question or not?

Pete

ps - The snide comments are against forum posting policy. Please refrain from posting such comments. If you must say something like then then send it in a PM to the person.

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Since samalkhaiat didn't answer my question I will have to make an assumption here. It seems clear to me that when samalkhaiat wrote

$$a_{\mu} \Lambda_{\mu}{}^{\nu} a_{\nu} + \partial_{\mu}\Omega$$

he was actually referring to the gauge invariance of the 4-potential but used it incorrectly. It is quite clear that his expression is not meaningful since the right hand side has two terms. The first term, after performing the Lorentz transformation, results in an expression in the new coordinate system while the second term in his expression is still in the original coordinate system. The result is adding terms in different coordinate systems, which, of course, is meaningless.

A gauge transformation adds the derivative of an arbitrary function to the 4-potential without changing Maxwell's equations (the Faraday tensor, aka EM tensor, is defined in terms of the derivatives of the 4-potential). So the actual Lorentz transformation for the expression for a general 4-potential is really the same as any other one for a 4-vector. However it is important to realize that the gauge transformation, must be made first.

In any case all of this holds for a general physically valid coordinate transformation and thus the 4-Potential it is more than just a Lorentz 4-vector. It is a general 4-vector.

Pete

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samalkhaiat said:
Me see Jackson!
That is correct. But the proof of the 4-potential being a 4-vector is in section 11.9. If you did look in Jackson then you'd see where he why the 4-potential is a 4-vector. In the 3rd edition of his EM book Jackson states on page 555 that
The differential operator form in (11.130) is the invariant four-dimensional Laplacian (11.78), while the right hand sides are the components of a 4-vector. Obviously, Lorentz covariance requires that the potentials $$\Phi$$ and A form a 4-vector.
Ohanian's EM text Classical Electrodynamics - Second Edition shows this explicitly in section 8.1 The Four-Vector Potential.

samalkhaiat - It seems clear that you have some serious misconceptions regarding gauge invariance. You'd better review the material in, say, Jackson before you post in the subject again.

Pete

pmb_phy said:
It is quite clear that his expression is not meaningful since the right hand side has two terms. The first term, after performing the Lorentz transformation, results in an expression in the new coordinate system while the second term in his expression is still in the original coordinate system. The result is adding terms in different coordinate systems, which, of course, is meaningless.
Sorry about that but I had it mixed up due to the strange equation used. Although the coordinates are consistent there is no valid reason to assume that expression is meaningful, or at least none has been provided. However what I said above still holds, i.e. a gauge transformation is that transformation which adds the derivative of an arbitrary function to the 4-potential without changing Maxwell's equations (the Faraday tensor, aka EM tensor, is defined in terms of the derivatives of the 4-potential). So the actual Lorentz transformation for the expression for a general 4-potential is really the same as any other one for a 4-vector. However it is important to realize that the gauge transformation, must be made first.

Pete

LHS1 said:
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.
I found a text with just such a derivation. This particular derivation shows that the components of the 4-potential transform in the same way as a 4-vector does and is therefore a 4-vector. Would you like me to scan them in and E-mail the pages to you? If so then please PM an e-mail address to me that you have access to so that I may forward the pages to you.

Pete

Phrak said:
samalkhaiat said:
I don't understand where $$a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu} + \partial_{\mu}\Omega$$ comes from.

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

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

Phrak - You are dead on correct! Bravo sir!

Pete

LHS1 said:
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?
Excellant point. Perhaps you;'re right after all. Let me get back to you on this (in the next few weeks I'll be away so when I get back I'll take another look at this).
Anyway, I found a way to prove that 4-vector potential is indeed a 4-vector. Thank you for your help sincerely.

Pete

Hans de Vries said:
samalkhaiat said:
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 $A^\mu$ does transform like a four vector.

Regards, Hans.
There is no reason that the Coulomb potential can't move instantaneously. All that is required is that changes in the EM field travel at c. This is mentioned in Jackson too (if not the same thing you're talking about). It is also addressed in the American Journal of Physics.

Also, $A^\mu$ transforms as a 4-vector under any gauge. Notice the the partial derivative with respect to mu of a Lorentz scalar is always a 4-vector and can thus always be added to another 4-vector to get an object which transforms as a 4-vector.

Pete

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Pete,

Samalkhaiat was quoting Weinberg vol I. page 251, where Weinberg is addressing
the worries of Feynman's use of the Coulomb gauge (aka. the radiation gauge)
to define the polarization vector of the external photons going in and out a
Feyman diagram.

In the radiation gauge, (the more appropriate name), indeed, $A^\mu$ doesn't transform
like a four-vector. This is impossible to start with since it has only two components...
Most notably, $A^o$ is always set to zero, while $A^\mu$ is always transversal to the momentum.

Internal photons in Feynman diagrams do have a longitudinal component as well as
an electric potential "polarization" component and they play important roles.

Regards, Hans.

Hans de Vries said:
Pete,

Samalkhaiat was quoting Weinberg vol I. page 251, where Weinberg is addressing
the worries of Feynman's use of the Coulomb gauge (aka. the radiation gauge)
to define the polarization vector of the external photons going in and out a
Feyman diagram.
What is the name of this book of Weinberg's of which you refer to?
In the radiation gauge, (the more appropriate name), indeed, $A^\mu$ doesn't transform like a four-vector.
I don't follow you. We're not interested here in whether the radiation gauge transforms as a 4-vector. In fact the gauge itself is merely a scalar function, correct? Hence we're only interested in whether the 4-potential transforms correctly. It appears to me that you're referring to the term that is added to the 4-potential during a gauge transformation to get a new 4-potential, right? Even so then I still don't follow since all one is doing is setting the 3-divergence of the magnetic 3-vector potential to zero. I don't see how that would prevent the gauge transformed 4-potential from transforming as a 4-vector. After all, every single gauge transformation consists only of adding the partial derivative of a 4-scalar, which is always a 4-vector.

Can you elaborate what you believe is wrong with this reasoning? Especially since Jackson makes no mention of what you say.
This is impossible to start with since it has only two components...
Most notably, $A^o$ is always set to zero, while $A^\mu$ is always transversal to the momentum.
Why? Also, what do you mean when you say that ...$A^\mu$ is always transversal to the momentum.

Thanks

Pete

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pmb_phy said:
What is the name of this book of Weinberg's of which you refer to?

"The Quantum Theory of Fields, volume I"
pmb_phy said:
After all, every single gauge transformation consists
only of adding the partial derivative of a 4-scalar, which is always a 4-vector.
Can you elaborate what you believe is wrong with this reasoning?
In the case of the Coulomb gauge one always sets $A^0=0$ , Now Look at expression
11.22 in Jackson which represents the 4-vector transform of $A^\mu$. You see that $A^0$
can not stay zero in any reference frame.

Physicist generally assume that $A^\mu$ does transform like a 4-vector but that they can
make gauge transformations unpunished because they leave the EM fields unchanged,
but that doesn't mean that the Lorentz transform properties aren't spoiled by the
arbitrary gauge transform.

The addition of two 4-vectors in general isn't something which transforms like a 4-vector.
pmb_phy said:
Also, what do you mean when you say that ...$A^\mu$ is always transversal to the momentum.
See for instance figure 7.1 on page 297 of Jackson which shows the two polarization
vectors orthogonal to the momentum. These are electric fields, say E1 and E2 and they
come from d(A1)/dt and d(A2)/dt pointing in the same direction.

If you set $A^0$ to 0 in the case of electromagnetic radiation, then you have
to set the $A^z$ component longitudinal to momentum also to zero to keep the
electric fields unchanged since $\partial_zA^0=-\partial_tA^z$.Regards, Hans

Hans de Vries said:
The addition of two 4-vectors in general isn't something which transforms like a 4-vector.

Can you clarify this statement? (Are there some unstated assumptions?)

robphy said:
Can you clarify this statement? (Are there some unstated assumptions?)

It just depends on the sign and value of $S^2$ which has to be an
invariant under Lorentz transform.

An object transforms as a 4-vector if C is invariant and positive:

$$V^2_0-V^2_1-V^2_2-V^2_3 = S^2 > 0$$

An object transforms as an axial 4-vector if C is invariant and negative:

$$V^2_0-V^2_1-V^2_2-V^2_3= S^2 < 0$$

An object transforms as light like if C is zero:

$$V^2_0-V^2_1-V^2_2-V^2_3= S^2= 0$$Additions can result in $S^2$ not being constant anymore or they can
even change the type of transformation behavior. For instance
(10,1,1,1) + (-10,1,1,1) = (0,2,2,2)
The first two are 4-vectors while the result is an axial vector.Regards, Hans

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