Prove that 4 vector potential is really a 4 vector?

[pmb_phy]

"The Quantum Theory of Fields, volume I"

This is a classical forum so why refer to a non-classical text?

In the case of the Coulomb gauge one always sets $A^0=0$ , ..

I see no reason to make such an assumption. The definition of the Coulomb gauge requires only that Div A = 0. Thus to create such a gauge one finds a scalar such that when the gradient of that scalar is added to the original 3-potential the divergence of the new 3-potential becomes zero, in the specific frame chosen. In any case this does not require $A^0=0$. In fact if you are saying that the 4-potential is not a 4-vector then please define this object so that we can see what $A^0=0$ is in other Lorentz coordinate systems. Thanks.

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.

The Coulomb gauge only requires that Div A = 0 in one particular Lorentz coordinate system. It does not required that Div A = 0 holds in all Lorentz coordinate systems.

The addition of two 4-vectors in general isn't something which transforms like a 4-vector.

I highly disagree. In all possible instances the sum of two 4-vectors is itself a 4-vector. This is rather simple to show too.

See for instance figure 7.1 on page 297 of Jackson which shows the two polarization
vectors orthogonal to the momentum.

That diagram addresses only 3-vectors. We're discussing 4vectors, which have an entirely different meaning for "orthogonal" which is that the scalar product of two 4-vectors vanish.

Best wishes

Pete

 

[pmb_phy]

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:

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

Each of those expressions requires only that the object is a 4-vector. You're merely referring to different classifications for 4-vectors.

Additions can result in $S^2$ not being constant anymore or they can
even change the type of transformation behavior.

The addition of two 4-vectors will also satisfy the invariance condition. It is unclear what you mean by "$S^2$ not being constant"? What is S? Is it the magnitude of one of the 4-vectors being added? In anycase one does not require that the magnitude of a 4-vector be constant. All that is required is that it be invariant.

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.

I think you're confusing the classification of 4-vectors with the definition of 4-vectors.

Best wishes

Pete

 

[reilly]

Hans and pmb pretty much have it right, amd pam comes close. In my opinion, a relatively straightforward way to deal with invariance vs. covariance is to look at explicit solutions for the potentials in both the Coulomb and Lorentz gauges; the differences will be very clear.
Regards,
Reilly Atkinson

 

[Hans de Vries]

This is a classical forum so why refer to a non-classical text?

Weinberg's quote (discussing simple electromagnetics) was the subject of your argument...

"The fact that $A^0$ for this image vanishes in all Lorentz frames shows vividly that $A^\mu$ cannot be a four-vector. ...

If Weinberg makes such a statement I would think more careful about what he means with this...

I see no reason to make such an assumption. The definition of the Coulomb gauge requires only that Div A = 0. Thus to create such a gauge one finds a scalar such that when the gradient of that scalar is added to the original 3-potential the divergence of the new 3-potential becomes zero, in the specific frame chosen. In any case this does not require $A^0=0$.

Weinberge is discussing the polarization vector of a photon in the Coulomb gauge.
According to Jackson:

The Coulomb or traverse gauge is often used when no sources are present. Then $\mbox{\Large \Phi=0}$

Thus $\Phi=0$ (!) per definition for radiation in the Coulomb gauge.

In fact if you are saying that the 4-potential is not a 4-vector then please define this object so that we can see what $A^0=0$ is in other Lorentz coordinate systems. Thanks.

I've said many times in this thread that the 4-potential IS a 4-vector.
Use the Liènard Wiechert potentials in the Lorentz gauge if you want it
to have a physical relevance beyond the electromagnetic Field tensor.

The Coulomb gauge only requires that Div A = 0 in one particular Lorentz coordinate system. It does not required that Div A = 0 holds in all Lorentz coordinate systems.

You are turning things around, look at Weinberg's quote:

"The fact that $A^0$ for this image vanishes in all Lorentz frames shows vividly that $A^\mu$ cannot be a four-vector. ...

That diagram addresses only 3-vectors. We're discussing 4vectors, which have an entirely different meaning for "orthogonal" which is that the scalar product of two 4-vectors vanish.

We ARE discussing 4-vectors and we are discussing why radiation does not have
a longitudinal component in the Coulomb gauge, which IS the result of $\Phi=0$ Regards, Hans

 

[robphy]

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?)

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

I think what you are saying is that (for example) the sum of two timelike vectors is not necessarily timelike, the sum of two spacelike vectors is not necessarily spacelike, etc...

The resultant of two vectors is still a vector [i.e. an element of the vector space in which all of these vectors live].
The nature of the vector (with respect to the metric) may change... granted. So, all you're really saying is that "the timelike-vectors don't form a vector-space", etc...

 

[pmb_phy]

Weinberg's quote (discussing simple electromagnetics) was the subject of your argument...

Not really. It was the subject of Samalkhaiat's quote. I merely wanted to know where he got it from. Now that I see its from a QM text I'm curious as to how nonclassical physics pertains to this classical topic. Since I haven't learned QFT yet I'll have to wait on this for a bit. But I was never interested in quantum theory in this thread.

If Weinberg makes such a statement I would think more careful about what he means with this...

I'll have to get a copy of it and read it for myself. Its not clear to me whether Samalkhaiat's misquoted Weingberg. Does Jackson say anything like Samalkhaiat was saying with that quote?

Thus $\Phi=0$ (!) per definition for radiation in the Coulomb gauge.

The Coulomb gauge does not require $\Phi=0$. All it requires is that $\Phi$ be time-independant so that the time derivative vanishes. $\Phi=0$ is merely an example of a Coulomb gauge ... unless I got this all wrong.

I've said many times in this thread that the 4-potential IS a 4-vector.

Sorry. I was confusing what you said with someone else. Did Weinberg say that the 4-potential is not a 4-vector?

You are turning things around, look at Weinberg's quote:

I'll wait until I can get a copy of the relavent pages so I can read it for myself in full context. Thanks Hans.

Best wishes

Pete

 

[LHS1]

Reply to pmb_phy, I am sorry for not responding because I did not visit the site for a while. The answer for this is, in fact, simple. I simply asked the wrong question! I should ask 'if there exist a potential vector as a 4 vector?'. There is a important point I did not realized that 'there is more than one solution satisfies (1) divergence of A =0 and (2) d'Alembertian of A proportional to current density vector. In fact (1) and (2) does ensure 4 vector solution exists. For example, Liénard-Wiechert potential as pointed out by others. As there are more than one solution as a potential function, there is always possible to construct a solution of (1) and (2) that is not a 4 vector just by simple mapping ( because of my poor editing, I do not like to expalin this point in detail, but I believe you can understand it). Therefore either (1) or (2) or both do not ensure A must be 4 vector. Therefore your proof by quotient theorem must be wrong. As pointed out by Samalkhaiat, there exist a solution of A that is not a 4 vector and it is neccesary for consistence with QED. Therefore everyone responded to this question has at least a grain of truth in their answer.

 

[LHS1]

If anyone found my explanation above is not satisfactory, please let me know. Thank you for everyone who made their attempt to help me.

 

[Hans de Vries]

Not really. It was the subject of Samalkhaiat's quote. I merely wanted to know where he got it from. Now that I see its from a QM text I'm curious as to how nonclassical physics pertains to this classical topic. Since I haven't learned QFT yet I'll have to wait on this for a bit. But I was never interested in quantum theory in this thread.

I'll have to get a copy of it and read it for myself. Its not clear to me whether Samalkhaiat's misquoted Weingberg. Does Jackson say anything like Samalkhaiat was saying with that quote?

Hi Pete,

I think Samalkhaiat took Weinberg to literally. In my opinion Weinberg was just
pointing out that Feynman's use of the Coulomb gauge, to simplify the math,
needs some justification to show that it doesn't spoil the Lorentz invariance
of the theory.

Feynman himself says in his own justification of using the Coulomb gauge:

... and let the scalar potential $\Phi=0$. But this is not a unique condition; that is,
it is not relativistically invariant and will be true in only one coordinate
system...

... The "paradox" however is resolved by the fact that gauge transformations
leave the field $F_{\mu\nu}$ unaltered.

Weinberg, in my opinion, just means to illustrate the issue by saying that:
If you do require $\Phi=0$ in all reference frames, then the transformation
behavior of $A^\mu$ comes out wrong….

"The fact that $A^0$ for this image vanishes in all Lorentz frames shows vividly that $A^\mu$cannot be a four-vector. ...

The Coulomb gauge does not require $\Phi=0$. All it requires is that $\Phi$ be time-independent so that the time derivative vanishes. $\Phi=0$ is merely an example of a Coulomb gauge ... unless I got this all wrong.

The fact that $\Phi=0$, for radiation, is not a requirement but a consequence.
Because of the bizarre instantaneous propagation of $\Phi$ in the Coulomb gauge.
If you shake an electron (in the strange Coulomb gauge) then it's entire $\Phi$
field shakes with it simultaneously. If you stop then $\Phi$ doesn't change
anymore, however the vector field $\vec{A}$ keeps on propagating with time
alternating components, but, without an alternating $\Phi$ component.But, to be clear: $A^\mu$ IS supposed to be a genuine four-vector.Regards, Hans

 

[pmb_phy]

Hi Pete,

I think Samalkhaiat took Weinberg to literally. In my opinion Weinberg was just
pointing out that Feynman's use of the Coulomb gauge, to simplify the math,
needs some justification to show that it doesn't spoil the Lorentz invariance
of the theory.

Hi Hans

Thanks. In fact that is the answer to the question I was just about to ask. It appeared to me that assuming that $A^0$ = 0 in all coordinate systems implies that $A^{\mu}$ is not a 4-vector. But since we know that it is a 4-vector then what we have shown is that $A^0$ = 0 in all coordinate systems is a false assumption, i.e. it is a condition that can only be met when there is no field present whatsoever.

Weinberg, in my opinion, just means to illustrate the issue by saying that:
If you do require $\Phi=0$ in all reference frames, then the transformation
behavior of $A^\mu$ comes out wrong….

Yes! Exactly! Thanks for confirming that for me Hans. Much appreciated.

The fact that $\Phi=0$, for radiation, is not a requirement but a consequence.
Because of the bizarre instantaneous propagation of $\Phi$ in the Coulomb gauge.
If you shake an electron (in the strange Coulomb gauge) then it's entire $\Phi$
field shakes with it simultaneously. If you stop then $\Phi$ doesn't change
anymore, however the vector field $\vec{A}$ keeps on propagating with time
alternating components, but, without an alternating $\Phi$ component.

Is this the same thing as saying that since E = -$\Grad \Phi - \partial A/\partial t$ then all that is required is that E propagate at the speed of light? It seems okay for the Coulomb potential to be instantaneoius so long as E isn't. I think that's how Jackson (and the AJP article he cites) explains it.

But, to be clear: $A^\mu$ IS supposed to be a genuine four-vector.

Great! Then we agree on all of this! Whew!

Best wishes

Pete

 

[pmb_phy]

Reply to pmb_phy, I am sorry for not responding because I did not visit the site for a while. The answer for this is, in fact, simple. I simply asked the wrong question! I should ask 'if there exist a potential vector as a 4 vector?'. There is a important point I did not realized that 'there is more than one solution satisfies (1) divergence of A =0 and (2) d'Alembertian of A proportional to current density vector. In fact (1) and (2) does ensure 4 vector solution exists. For example, Liénard-Wiechert potential as pointed out by others. As there are more than one solution as a potential function, there is always possible to construct a solution of (1) and (2) that is not a 4 vector just by simple mapping ( because of my poor editing, I do not like to expalin this point in detail, but I believe you can understand it). Therefore either (1) or (2) or both do not ensure A must be 4 vector. Therefore your proof by quotient theorem must be wrong.

I think I actually agree with you on this point. Seems I made a terrible mistake when I said that. That's the problem when you do something in your head!

As pointed out by Samalkhaiat, there exist a solution of A that is not a 4 vector and it is neccesary for consistence with QED.

It seems clear to me that Samalkhaiat didn't understand what Weinberg was saying. I believe that Weinberg was saying that $A^0 = 0$ cannot hold in all frames except in the trivial case of no EM field. See my comments to Hans on why.

Therefore everyone responded to this question has at least a grain of truth in their answer.

This is very tough stuff. Nobody ever said that relativistic mechanics and tensors were easy, right!

Best wishes

Pete

 

[Phrak]

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

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

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

In gauge, itself, transforms as a vector, of course. A gauge that does not respect linear transforms is coordinate system dependent; like a wind for instance, that always blows over your left shoulder no matter the direction you face.

 

[samalkhaiat]

I will reply to yours, maybe you will understand what I am going to say.

No, this is not the correct starting point for gauge theory.

Starting point?

The potential in a gauge theory is, by definition, a connection 1-form which takes values in the Lie algebra of the gauge group.

Correct. It is a connection not vector. see post#18.

Under a coordinate transformation it transforms as

$$a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu}$$​

i.e., as a 4-vector.

Very wrong;

1) Regarding the tensorial nature of potentials, gauge theories can only say that potentials CARRY a spacetime index. This does not mean that $A_{\mu}$ itself is a vector. In general relativity, Reimannian connection (potential) carries three spacetime indices but it is not a rank-3 tensor. Like $\Gamma^{\rho}{}_{\mu \nu}$, the potential $A_{\nu}$ is a connection. Therefore, only the difference between two such connections, $(A_{2}^{\mu}-A_{1}^{\mu} \ , \Gamma_{2}-\Gamma_{1})$, has a definite tensorial character.see post#18.

2) To understand the whole issue, please pay attention to the following;

Unlike massive vector fields, non-trivial massless vector field is not allowed by Lorentz group. The fundamental theorem on the massless representation of Lorentz group states:
"The only admissible vector field of mass zero is a gradient of a scalar field."

So, if $V_{\mu}$ is a vector field

$$V_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} V_{\nu}$$​

the theorem says that $V_{\mu}$ cannot be massless unless

$$V_{\mu} = \partial_{\mu} \phi$$​

where $\phi$ is a scalar field. But this massless vector is not good for us; it implies an everywhere vanishing field tensor;

$$F_{\mu \nu} = ( \partial_{\mu}\partial_{\nu} - \partial_{\nu}\partial_{\mu}) \phi = 0$$​

Since we know that the electromagnetic field is massless, we therefore conclude that electromagnetism (or any other massless theory) cannot be described by a vector field. That is, according to the above mentioned theorem, the potential of any gauge theory cannot be a 4-vector field.

To those who still think that $A_{\mu}$ is a 4-vector, I say this:

Do yourself a favour and study the representation theory of Lorentz group.
See J. Lopuszanski, "An Introduction to Symmetry and Supersymmtry", World Scientific,1991.
From Page 186;

"the vector potential $A_{\mu}$ cannot be genuinely a vector with respect to the Lorentz transformations as a massless vector must necessarily be given by the gradient of a scalar of helicity zero. Remember that $F_{\mu \nu}$ is composed of two terms of helicity 1 and -1. Thus under Lorentz transformation $A_{\mu}$ does not transform as a vector, although $F_{\mu \nu}$ does transform as a tensor."

The rule is sweet and simple:

"If $A_{\mu}$ is massless, then it cannot be a Lorentz-vector"

3) Now that we know it is not a vector, we ask: How does the potential transform under Lorentz transformation? There are at least five different methods for deriving the transformation law! I will describe the shortest one.
In this method, the vital piece of information comes from the gauge principle which states that ANY two potentials are related by a gradient of a scalar

$$A_{2}^{\mu} = A_{1}^{\mu} + \partial^{\mu}\Omega$$​

From this we conclude that the difference between ANY two potentials (i.e., the object $(V^{\mu}= A_{2}^{\mu} - A_{1}^{\mu}$) is a trivial 4-vector (gradient of a scalar).
Thus, under Lorentz transformation, $V_{\mu}$ transforms according to

$$\left( A_{2}^{\mu} - A_{1}^{\mu} \right) \rightarrow \Lambda^{\mu}{}_{\nu} \left( A_{2}^{\nu} - A_{1}^{\nu} \right) \ \ \ (1)$$​

Please note that this IS NOT a "gauge transformation followed by Lorentz transformation". I used the gauge principle only to INFER the tensorial (vector) nature of the object $( A_{2}^{\mu} - A_{1}^{\mu})$.
Since Lorentz group does not admit massless vectors, we therefore conclude that Eq(1) is satisfied if and only if

$$A_{1}^{\mu} \rightarrow \Lambda^{\mu}{}_{\nu} A_{1}^{\nu} + \partial^{\mu}\lambda$$

$$A_{2}^{\mu} \rightarrow \Lambda^{\mu}{}_{\nu} A_{2}^{\nu} + \partial^{\mu}\lambda$$​

Notice that the scalar function $\lambda$ is completely arbitrary. This means that the above transformation law holds for ANY potential in ANY gauge (COVARIANT or NON-COVARIANT).

In QFT, the transformation of the OPERATOR $A_{\mu}$ becomes

$$A_{\mu} \rightarrow U(\Lambda) A_{\mu} U^{-1}(\Lambda) = \Lambda_{\mu}^{\nu} A_{\nu} + \partial_{\mu}\lambda$$

see Eq(14.25) in Bjorken & Drell, "Relativistic Quantum Fields".
and Eq(5.9.31) in Weinberg, "The Quantum Theory of Fields", VOL I.

For completely CLASSICAL description, see
K. Moriyasu, "An Elementary Primer for Gauge Theory", World Scientific.
On page 42, Moriyasu says

"Thus, the vector potential observed in the two frames are related by

$$A_{\mu}^{'} = L_{\mu}{}^{\nu}A_{\nu} - \partial_{\mu}\lambda \ \ \ III-22$$​

This shows that the vector potential does not transform like an ordinary vector under Lorentz transformation. ...
This interesting fact is well known in quantum field theory but it is rarely mentioned in ordinary electromagnetism."

I have been in the trade for some time and I have identified the sourses of confusion:
1) most students starts their post grad. course with little or no knowledge about the repretentation theory.
2) the fact (potential is not vector) is rarely addressed on the pedestrian level.
3) the misleading name "VECTOR" potential adds to the confusion.
4) in QED, $A_{\mu}$ couples to a CONSERVED current. So, peopel speak of Lorentz ang gauge invariant interaction Lagrangian;

$$\mathcal{L} = A_{\mu}J^{\mu}$$

Here, most students jump to the conclusion;

"$A_{\mu}$ must be a 4-vector"​

Of course in such coupling, it is not a big sin to treat $\mathcal{L}$ as FORMALLY Lorentz-invariant and $A_{\mu}$ as a FORMAL vector.

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.

Quantization means going from a C-number field on spacetime

$$A_{\mu}(x) = \int_{p} \ a(p)f_{\mu}(x;p) + a^{*}(p)f^{*}_{\mu}(x;p)$$

to a Q-number field (operator-valued distribution) on spacetime

$$\hat{A}_{\mu}(x) = \int_{p} \ \hat{a}(p)f_{\mu}(x;p) + \hat{a}^{\dagger}(p)f^{*}_{\mu}(x;p)$$

Notice that the spacetime index $\mu$ is carried by the function f not by the operator a. So, my friend, quantization does not change the tensorial nature of the fields.

regards

sam

 

[pmb_phy]

<Inappropriate commen removed> I will reply to yours, maybe you will understand what I am going to say.

Actually its is your post which seems to contain the most inaccurate comments.
Starting point?

Very wrong;

Nope. Its very right. Since you have simply ignored all the comments which explain your errors I see no reason to assume that you'll understand the reason why you are wrong. Your so-called proof is merely a collection of unproven assertions.

The reason why Jackson explains why the 4-potential is a 4-vector seems to evade you. Tell you what; Go to the proof in Jackson in which he proves that the 4-potential is a 4-vector and post your proof that Jackson is wrong. Otherwise I see no point reading or responding to your continuing erroneous and irritating comments.

Pete

 

[pmb_phy]

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

In gauge, itself, transforms as a vector, of course. A gauge that does not respect linear transforms is coordinate system dependent; like a wind for instance, that always blows over your left shoulder no matter the direction you face.

It appears from samalkhaiat's latest comment that he doesn't know the definition of a 4-potential. Let's start from scratch. The 4-potential A^u is defined as

A^u = ($\Phi$, A)

The spatial and temporal components of this object satisfy certain equations (i.e. Eq. 11.130 in Jackson - too much work to post). The differential operator on the left hand side of said equations is the invariant 4-dimensional Laplacian while the right hand side is a 4-vector. This requires the potentials $\Phi$ and A form a 4-vector ... period! That's all she wrote!

Pete

 

[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

[tex]\partial_{\mu}[/itex][itex]A^{\mu} = 0[/tex]

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

Best regards

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