
#1
May208, 06:22 AM

P: 24

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.




#2
May208, 07:48 AM

P: 2,955

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



#3
May208, 10:12 AM

P: 24





#4
May208, 01:05 PM

P: 455

Prove that 4 vector potential is really a 4 vector?
If you already believe that [tex]j^\mu[/tex] is a 4vector (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 4vector.




#5
May208, 01:09 PM

P: 455

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




#6
May308, 12:51 AM

P: 24

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 Alevel logic tells us that a proposition is true does not mean its converse is true.




#7
May408, 11:44 AM

P: 2,955

Pete 



#8
May408, 01:55 PM

P: 24

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 4vector potential is indeed a 4vector. Thank you for your help sincerely.




#9
May508, 05:14 AM

P: 24

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




#10
May608, 08:44 AM

P: 2,955

Pete 



#11
May608, 01:52 PM

P: 455





#12
May708, 03:22 AM

Sci Advisor
P: 1,136

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) 



#13
May908, 05:26 PM

Sci Advisor
P: 815

[QUOTE=LHS1;1714217]
The vector potential is not a 4vector! 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 4vector, 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 4vector. Deriving the transformation law of [itex]a_{\mu}[/itex] from the gaugefixed Maxwell equation [tex]\partial_{\mu}\partial^{\mu} a_{\nu} = J_{\nu}[/tex] is a wrong practice. The gaugeinvariant 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 4vector transforms. Is it? Genuine 4vectors cannot describe the two polarization states of light. So, it is not a bad thing that the vector potential is not a 4vector. 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 fourvector. ........ .......... .......... Although there is no ordinary fourvector 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 fourvector, ......" 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 fourvector 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 4vector" regards sam 



#14
May1008, 10:22 PM

P: 4,513

[QUOTE=samalkhaiat;1723974]
If I assume [tex]\acute{a}_{\mu}[/tex] is a regauged (dual) 4vector, [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]. 



#15
May1608, 06:01 PM

Sci Advisor
P: 1,136

[QUOTE=samalkhaiat;1723974]
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 Ličnard 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. 



#16
May1708, 07:12 PM

Sci Advisor
P: 815

[QUOTE=Hans de Vries;1733005]
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. regards sam 



#17
May1708, 07:22 PM

P: 2,955

Pete 



#18
May1708, 08:19 PM

Sci Advisor
P: 815

[QUOTE=Phrak;1725452]
[tex]a_{(2)}^{\mu}  a_{(1)}^{\mu} = \partial^{\mu} \omega[/tex] Thus, under Lorentz transformation, [itex](a_{(2)}  a_{(1)})[/itex] transforms like a 4vector [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 gaugeinvariant 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 4vector. 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 1form on principal bundle [itex]\left(R^{4} \times U(1)\right)[/itex]. Connections 1forms 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 2form (= 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 


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