About the 4 potential

  • Thread starter Llewlyn
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Let's speaking about special relativity.
A four-vector 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 [itex](ct, x, y, z)[/itex] do. Now i define the 4potential as [itex](\phi, \vec{A})[/itex]. 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:

[itex]\square A^{\mu} = j^{\mu}[/itex]

where [itex]j^{\mu}[/itex] 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 [itex]A^{\mu}[/itex]. 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.
 
Last edited:

pam

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[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.
 

olgranpappy

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1) Does the 4potential is a 4vector ONLY in Lorentz gauge?
It's a 4-vector regardless of the gauge. Want to know how to prove it? Okay, I'll tell ya.

Start in the Lorentz gauge, where you already know that the 4-vec potential A_\mu is a 4-vector. 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 4-vector potential is

A_\mu + \partial_\mu f(x)

Is this still a four-vector? Yes, because \partial_\mu f(x) is a four-vector and the sum of two four-vectors is a four-vector. Cheers.
 
2,096
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[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.
In fact it is possible for [itex]\partial_{\mu} j^{\mu} = 0[/itex] to be true in all frames without [itex]j^{\mu}[/itex] being a four vector. For example, there could be a 2nd-rank tensor [itex] T^{\mu\nu}[/itex] that satisfies [itex]\partial_{\mu} T^{\mu\nu}=0[/itex]. If you then define [itex]j^{\mu}:=T^{\mu 0}[/itex], we have an counter example.

Anyway, the current density [itex]j^{\mu}[/itex] can be assumed to be a four vector for other reasons too.

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.
This conclusion was logically valid, but I'll restate it for clarity:

"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.
 
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olgranpappy

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In fact it is possible for [itex]\partial_{\mu} j^{\mu} = 0[/itex] to be true in all frames without [itex]j^{\mu}[/itex] being a four vector.
...
true. good example.

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 misspoke--she 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 4-vector."
 

dextercioby

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In fact it is possible for [itex]\partial_{\mu} j^{\mu} = 0[/itex] to be true in all frames without [itex]j^{\mu}[/itex] being a four vector. For example, there could be a 2nd-rank tensor [itex] T^{\mu\nu}[/itex] that satisfies [itex]\partial_{\mu} T^{\mu\nu}=0[/itex]. If you then define [itex]j^{\mu}:=T^{\mu 0}[/itex], we have an counter example.
But isn't [itex] j^{\mu} =T^{\mu 0} [/itex] also a 4 vector ? You have found no counterexample. If [itex] \partial_{\mu} j^{\mu} [/itex] is a scalar, then [itex] j^{\mu} [/itex] necessarily is a 4-vector.
 
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But isn't [itex] j^{\mu} =T^{\mu 0} [/itex] also a 4 vector ?
No. It doesn't transform like

[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]


If [itex] \partial_{\mu} j^{\mu} [/itex] is a scalar, then [itex] j^{\mu} [/itex] necessarily is a 4-vector.
This conclusion seems to be right on a quick glance, but the previous counter example makes one think more carefully. I suppose that it could be made correct by more rigor assumptions.

It could be that if it is a non-zero scalar, then [itex]j^{\mu}[/itex] would have to be a four vector.
 
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dextercioby

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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" ?
 

dextercioby

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Yes, you're right. My bad.
 

samalkhaiat

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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 Lorentz-vector field for massless particles of hilicity [itex]\pm 1[/itex].


regards

sam
 

samalkhaiat

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But isn't [itex] j^{\mu} =T^{\mu 0} [/itex] also a 4 vector ?
No, but the integral

[tex]\int d^{3}x T^{\mu 0}[/tex]

is a 4-vector, if the tensor T is conserved, i.e., if

[tex]\partial_{\nu} T^{\mu \nu} = 0[/tex]

regards

sam
 

olgranpappy

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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...
hmm. but if I want to get, say, the 4-vector potential in the Lorentz gauge of a uniformly moving point charge, I can start from the rest frame Columb potential [itex]A_\mu= (1/r, \vec 0) [/itex] (charge chosen to be 1) and boost this in the normal way...
 

samalkhaiat

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hmm. but if I want to get, say, the 4-vector potential in the Lorentz gauge of a uniformly moving point charge, I can start from the rest frame Columb potential [itex]A_\mu= (1/r, \vec 0) [/itex] (charge chosen to be 1) and boost this in the normal way
A gauge term is still necessary, since if

[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 gauge-invariant 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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