Prove that 4 vector potential is really a 4 vector?

[Phrak]

It appears from samalkhaiat's latest comment that he doesn't know the definition of a 4-potential. ...

Be not too quick to judge, my friend.

 

[pmb_phy]

Be not too quick to judge, my friend.

I'm not judging him. How could I judge someone I've never met? I'm only noting that he's not participating in this thead but merely posting material and then refusing to back upp his claims. Participation means to address objections made to statements made.

I rarely judge people, if ever. But he was given proof that the 4-potential is a 4-vector (proof of which is given in many relativity texts) and he has ignored those proofs. He also seems to proud to read/arrogant to pick up and read the proof in Jackson's (or any other) text. All he has done in response to the questions put to him is to claim its garbage and to then failed to/refused to address said proofs. Fine with me. But his attitude needs adjustment.

Also his comments regarding the 4-potential do not appear to be referring to the same object that is the subject of this thread. Also he appears to be speaking about quantum field theory rather than classical relativity, and this is a classical forum. Therein may resides the problem.

Since he's ignored the direct questions put to him regarding clarification of what he means and why all of a sudden gauge theory is part of this thread won't be that forth comming in my opinion.

Pete

 

[pmb_phy]

For some reason this thread has taken a quantum mechanical twist. I wanted to make it clear that this is a classical forum and hence the questions here are answered in a classical sense. Something which may be well-defined and meaningful in classical relativity may be undefined/ill-defined, have a different meaning or even no meaning in quantum theory. E.g. a worldline has a very specific meaning in classical mechanics but has no meaning (or is ill-defined/different) in quantum theory.

Pete

 

[pmb_phy]

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

I was fortunate enough to have someone who could e-mail me the chapter by Feynman (Thank you to that person!). I'm looking at those comments right now. Nothing in that entire section even mentions the 4-potential. The quantity that Weinberg is referring to, i.e. a^{\mu} is not the 4-potential. As you can see at the bottom of page 246 a^{\mu} represents the annihilation operator. I also see no mention of the Coulomb gauge in that section. So again I ask, of what relavence is that section and its contents? Thanks.

Pete

 

[Hans de Vries]

Hans

I was fortunate enough to have someone who could e-mail me the chapter by Feynman (Thank you to that person!). I'm looking at those comments right now. Nothing in that entire section even mentions the 4-potential. The quantity that Weinberg is referring to, i.e. a^{\mu} is not the 4-potential. As you can see at the bottom of page 246 a^{\mu} represents the annihilation operator. I also see no mention of the Coulomb gauge in that section. So again I ask, of what relavence is that section and its contents? Thanks.

Pete

Hi, Pete.

It's good to see that you are delving into the subject. Please be assured that Feynman
and Weinberg are talking about the same thing and that they are all talking about the
four-potential A in the Coulomb gauge.

Samalkhaiat is talking the four potential A under general gauge transformation. He
would not have a phd in theoretical physics, and teach theoretical physics, if he would
not know about A... :^)

Feynman and Weinberg are talking about the polarization vector e of the electro-
magnetic radiation in the Coulomb gauge. In this gauge the polarization 4-vector is the
same as the 3d polarization vector based on the electric field E as discussed in
Jackson section 7.2 where Jackson also talks about the Helicity of the radiation. The e⁰ component of the polarization 4-vector, corresponding to V is zero in
the Coulomb gauge, since the polarization vector is associated with radiation.
Not withstanding the QFT context of these books, this is all just classical field theory.The only real conflict here is about the interpretation of the gauge invariance.
Under gauge invariance the 4-potential is substantially undetermined so you can not
expect it to transform like a Lorentz 4-vector without extra terms involving the gauge
transform itself.

There is a problem in determining in what is really "physical" about the 4-potential
because of the gauge invariance. Opinions about this will vary and thus the opinions
about how the 4-potential "physically" transforms will vary also. As long as everybody
recognizes this then there is no real conflict.What I did was pointing out some of the bizarre properties of the Coulomb gauge
which many physicist (probably most) do not consider to be "physical". These
bizarre properties are the instantaneous propagation of the electric potential V
throughout all of space combined with something even more bizarre, The
instantaneous propagation of a current density throughout all of space.

The effects of both instantaneous effects do cancel each other. They must
cancel because of the gauge invariance. (see Jackson section 6.3) So, the discussion is
(and can only be!) about the interpretation because of the problem of distinguishing
between the different gauges.Regards, Hans.

PS. of course one can find many quotes in QFT books which do describe A as something
which transforms as a Lorentz 4-vector. For instance Peskin and Schroeder on page 37
under figure 3.1, where they actually discuss the Lorentz transform of 4-vectors:

The most familiar case is that of a vector field such as the 4-current j(x) or the vector potential A(x)

 

[pmb_phy]

Hi, Pete.

It's good to see that you are delving into the subject. Please be assured that Feynman
and Weinberg are talking about the same thing and that they are all talking about the
four-potential A in the Coulomb gauge.

Why do you say that Weinberg is talkiung about the 4-potential in section 8.9? There is nothing in that section that even hints to it from what I can see. Is there some relationship between the 4-potential and this polarization vector that you mentioned? If so then of what use is a quantum topic in a classical subject?

Samalkhaiat is talking the four potential A under general gauge transformation. He
would not have a phd in theoretical physics, and teach theoretical physics, if he would
not know about A... :^)

A PhD is not a ticket to being error free.

In Classical Charged Particles, by Fritz Rohrlich the author states on page 72 that

We conclude this section with some remarks concerning the covariant notation and gauges. The gauge transformations (4-22) are covariant

A'_{\mu} = A'_{\mu} + \partial_{\mu}\Lambda

Gauges lose their covariance when non-covariant conditions are imposed on \Lambda and A_{\mu}. Thus, the Lorentz condition is invariant

\partial_{\mu}A^{\mu} = 0

while the Coulomb gauge condition (4-29) is not;...

That seems quite correct to me and not quite what sam as trying to convince us of. I fullly agree with this and in fact this is what I was referring to before. I.e. there is no such thing as a covariant expression for a Coulomb gauge. Is that how you read Rohrlich here?

re sam's PhD - No comment ... yet!

Best regards

Pete

 

[pmb_phy]

Sorry but the edit feature timed out while I was trying to get the Latex right.

In Classical Charged Particles, by Fritz Rohrlich the author states on page 72 that

We conclude this section with some remarks concerning the covariant notation and gauges. The gauge transformations (4-22) are covariant

A'_{\mu} = A_{\mu} + \partial_{\mu}\Lambda

Gauges lose their covariance when non-covariant conditions are imposed on \Lambda or on A_{\mu}. Thus, the Lorentz condition is invariant

\partial_{\mu}[/itex]A^{\mu} = 0<br /> <br /> while the Coulomb gauge condition (4-29) is not;...<br />

<br /> <br /> Best regards<br /> <br /> Pete

 

[Hans de Vries]

Why do you say that Weinberg is talkiung about the 4-potential in section 8.9?

Those who are familiar with (this form of) quantization immediately recognizes this to
be the case.

Is there some relationship between the 4-potential and this polarization vector that you mentioned? If so then of what use is a quantum topic in a classical subject?

See my response in the previous post, and have a look at Jackson 7.2

Feynman and Weinberg are talking about the polarization vector e of the electro-
magnetic radiation in the Coulomb gauge. In this gauge the polarization 4-vector is the
same as the 3d polarization vector based on the electric field E as discussed in
Jackson section 7.2 where Jackson also talks about the Helicity of the radiation.

The e0 component of the polarization 4-vector, corresponding to V is zero in
the Coulomb gauge, since the polarization vector is associated with radiation.
Not withstanding the QFT context of these books, this is all just classical field theory.

Since you agree with me the 4-potential is a 4-vector and sam claims it isn't then where do you think sam is making a mistake? If you don't think he is then its a paradox to me! Please clarify. Can an object be a 4-vector and not a 4-vector simultaneously?.

Again, see my response in the previous post

There is a problem in determining in what is really "physical" about the 4-potential
because of the gauge invariance. Opinions about this will vary and thus the opinions
about how the 4-potential "physically" transforms will vary also. As long as everybody
recognizes this then there is no real conflict.

The Aharonov-Bohm effect comes to mind.

Yes, correctly, There is more physical reality as just the E and B fields. Unfortunately,
the problem is that the Aharonov-Bohm effect is still not enough to uniquely determine
the 4-potential. So, interpretation issues about what is the physical reality remain.Regards, Hans

 

[pmb_phy]

Those who are familiar with (this form of) quantization immediately recognizes this to be the case.

Interesting. Please clarify/elaborate.

Again, see my response in the previous post

I didn't gleen that from your previous post. Hence my question.

Pete

 

[pmb_phy]

Hans,

I'm retracting my previous questions since I'm satisfied with my current knowledge of classical relativity. Thanks.

Regards

Pete