
#1
Dec2907, 06:17 AM

P: 68

Let's speaking about special relativity.
A fourvector is a physical quantity with 4 components that, when passing from one inertial frame to another, transform those components applying the Lorentz matrix, as the event [latex](ct, x, y, z)[/latex] do. Now i define the 4potential as [latex](\phi, \vec{A})[/latex]. For being sure that IS a 4vector i have to change frame and show that the new 4potential is obtained applying the Lorentz matrix, but i dont know how to do it. So i put in Lorentz's gauge and write Maxwell equations that results to be: [latex]\square A^{\mu} = j^{\mu}[/latex] where [latex]j^{\mu}[/latex] is the 4current (that i know is a 4vec). So because the d'alembert operator is invariant, or rather, it doesnt change the algebra of the quantity applied it i obtain the 4vec nature of [latex]A^{\mu}[/latex]. But... 1) Does the 4potential is a 4vector ONLY in Lorentz gauge? Because i've seen in 2nd quantitation that it has be used as a 4vec in Coulomb's gauge. 2) How can i proof that the d'alembert operator is "invariant"? Really thank you for attention, Ll. 



#2
Jan108, 07:12 AM

P: 455

[tex]\partial^\mu[/tex] is a four vector. In order for the continuity equation to be a scalar, this requires [tex]j^\mu[/tex] to be a four vector.
Since [tex]\partial^\mu[/tex] is a four vector, the d'Alembertian [tex]\partial^\mu\partial_\mu[/tex] is a scalar. The wave equation then shows that [tex]A^\mu[/tex] is a four vector in the Lorentz gauge. 



#3
Jan108, 04:25 PM

HW Helper
P: 1,273

Start in the Lorentz gauge, where you already know that the 4vec potential A_\mu is a 4vector. Then change to a different gauge... and how to do this? Well, we know how... Introduce a scalar function 'f(x)' and in the new gauge the 4vector potential is A_\mu + \partial_\mu f(x) Is this still a fourvector? Yes, because \partial_\mu f(x) is a fourvector and the sum of two fourvectors is a fourvector. Cheers. 



#4
Jan108, 05:08 PM

P: 1,983

About the 4 potentialAnyway, the current density [itex]j^{\mu}[/itex] can be assumed to be a four vector for other reasons too. "If we assume that the equation of motion is Lorentz invariant, then it follows that the four potential must transform as a four vector." Alternatively we could draw the conclusion in the other direction: "We postulate that the four potential transforms as a four vector, and therefore the equation of motion is Lorentz invariant." Whatever you like. IMO the transformation properties of different fields are something that should be postulated, and the invariance of the equations are something that should be proven. 



#5
Jan108, 07:05 PM

HW Helper
P: 1,273

Of course, any tensor of all zero entries will have all zero entries in any frame and so then any subset I pick out to make up a "thing" will also always be zero in any frame. I think that perhaps Pam just misspokeshe must have meant to say, "... In order that the LHS of the continuity equation transform as a Lorentz scalar, [itex]j_\mu[/itex] must be a 4vector." 



#6
Jan508, 05:10 AM

Sci Advisor
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P: 11,863





#7
Jan508, 07:14 AM

P: 1,983

[tex] T^{\mu 0}\mapsto \Lambda^{\mu \nu} T_{\nu}{}^0, [/tex] which would be a four vector transformation, but instead like [tex] T^{\mu 0}\mapsto \Lambda^{\mu\nu}\Lambda^{0 \alpha} T_{\nu\alpha} [/tex] It could be that if it is a nonzero scalar, then [itex]j^{\mu}[/itex] would have to be a four vector. 



#8
Jan508, 08:46 AM

Sci Advisor
HW Helper
P: 11,863

I'm afraid you're confusing things. To help out a bit: How many components does [tex] T^{\mu 0} [/tex] have ? And how many will it have after performing a Lorentz transformation on its argument, "x" ?




#9
Jan508, 08:53 AM

P: 1,983

Do you know why energy is not considered to be a scalar? 



#10
Jan508, 08:56 AM

Sci Advisor
HW Helper
P: 11,863

Yes, you're right. My bad.




#11
Jan508, 09:00 AM

P: 1,983





#12
Jan1208, 08:33 PM

Sci Advisor
P: 817

Actually, the vector potential 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 field of hilicity zero. But [itex]F_{\nu \nu}[/itex] (which does transform as a tensor) is composed of two terms of hilicity 1 and 1.
Since [itex]F_{\mu \nu} = \partial_{\mu}A_{\nu}  \partial_{\nu}A_{\mu}[/itex], thus under Lorentz transformations [itex]A_{\mu}[/itex] does not transform as a vector but is supplemented by an additional gauge term: [tex] U( \Lambda ) A_{\mu}(x) U^{1}( \Lambda ) = \Lambda^{\nu}{}_{\mu} A_{\nu}( \Lambda x ) + \partial_{\mu} \lambda ( x , \Lambda ) [/tex] So, there is no ordinary Lorentzvector field for massless particles of hilicity [itex]\pm 1[/itex]. regards sam 



#13
Jan1208, 08:41 PM

Sci Advisor
P: 817

[QUOTE=bigubau;1561050]
[tex]\int d^{3}x T^{\mu 0}[/tex] is a 4vector, if the tensor T is conserved, i.e., if [tex]\partial_{\nu} T^{\mu \nu} = 0[/tex] regards sam 



#14
Jan1208, 08:57 PM

HW Helper
P: 1,273





#15
Jan1208, 10:10 PM

Sci Advisor
P: 817

[QUOTE=olgranpappy;1568937]
[tex]A_{i}(x) = 0[/tex] it follows that [tex]UA_{i}(x)U^{1} = 0[/tex] for any unitary transformation U. The structure of [tex]UAU^{1} = \Lambda A( \Lambda x) + \partial \lambda[/tex] guarantees that the gaugeinvariant Maxwell equations are Lorentz covariant. Thus two Lorentz observers O and O' who construct in their respective frames a quantum electrodynamics in any gauge are assured of a unitary transformation relating the states of O & O'. regards sam 


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