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How can we prove that ##k^\mu=(\frac{\omega}{c}, \vec{k})## is a four-vector?
One way is to consider the invariance of light velocity. If we postulate ##k^\mu## to be a four-vector then the scalar product of ##k^\mu\,x_\mu=\vec{k}\cdot\vec{r}-\frac{\omega}{c}\,t## is invariant, if it constant in one frame it will also be constant in any other inertial frame.
However this is only a sufficient contition. In general it does not have to be the same constant value to be consistent with the same velocity of light.

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Ibix
2020 Award
Why not just do a Lorentz transform and show that the result is consistent with the wave four-vector in another frame?

Dale
Why not just do a Lorentz transform and show that the result is consistent with the wave four-vector in another frame?
Because this is only a sufficient condicion and not a necesary one. My question is : Can this be rigorously proved? or we have no choice other than postulating it

Orodruin
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The wave 4-vector is defined as the gradient of the phase function. Hence it is a 4-vector.

vanhees71 and Dale
The wave 4-vector is defined as the gradient of the phase function. Hence it is a 4-vector.
That's correct and my question is how can we prove that the phase funtcion is a scalar? or we must content ourselves postulating it

Orodruin
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That's correct and my question is how can we prove that the phase funtcion is a scalar? or we must content ourselves postulating it
Are you suggesting that wave crest spacetime position depends on the coordiate system? That cannot be the case. Consider a wave in a scalar field. The field is Lorentz invariant and so the spacetime events where it takes its maximal value are as well. You can do similar arguments for EM waves.

Dale
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Because this is only a sufficient condicion and not a necesary one.
I disagree. It is necessary and sufficient.

Are you suggesting that wave crest spacetime position depends on the coordiate system? That cannot be the case. Consider a wave in a scalar field. The field is Lorentz invariant and so the spacetime events where it takes its maximal value are as well. You can do similar arguments for EM waves.
Probably you're right. Barut's book justifies the necesary part like you. However it is not clear to me how this work for an EM wave. In this case magnetic and electric fields are mixed. I will try to understand it. Thank you

I disagree. It is necessary and sufficient.
Can you prove the "necessary" part with a different argument from @Orodruin 's

Dale
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Can you prove the "necessary" part with a different argument from @Orodruin 's
A four-vector is defined by how it transforms under the Lorentz transform. So if one takes a mathematical object, applies the Lorentz transform, and shows that it transforms correctly (as @Ibix recommended), then that object is by definition a four-vector. Showing that something satisfies the definition is both necessary and sufficient to prove it.

Orodruin
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In this case magnetic and electric fields are mixed
Whether the amplitude is a scalar, a vector, or a tensor has no bearing whatsoever on the argument.

A four-vector is defined by how it transforms under the Lorentz transform. So if one takes a mathematical object, applies the Lorentz transform, and shows that it transforms correctly (as @Ibix recommended), then that object is by definition a four-vector. Showing that something satisfies the definition is both necessary and sufficient to prove it.
Ok, but the original question was: can we prove that the object transforms like a four-vector. The object in this case is ##k^\mu##. One way to prove that the object is a four-vector is to show that the wave phase is a scalar and this is @Orodruin 's argument.

PeterDonis
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the original question was: can we prove that the object transforms like a four-vector.
@Ibix already told you how to prove this in post #2. Have you tried what he suggested there?

Dale
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Ok, but the original question was: can we prove that the object transforms like a four-vector. The object in this case is ##k^\mu##.
And @Ibix immediately provided a correct approach for doing so, which you incorrectly rejected!

vanhees71
Gold Member
The issue is not as simple as suggested by the answers to the OP. Whether four numbers (I use units with ##c=1## for convenience) ##(\omega,\vec{k})## transform like the components of a four-vector is not so clear from just writing these four components down. It's a question of context. The notation suggests that it is frequency and wave vector of some wave. Thus there is some wave equation describing these waves. This implies that there's a dispersion relation dependent on the wave equation. If the wave equation can be written in Lorentz covariant form this dispersion relation must be covariant, and the only invariant you can built of four numbers ##(\omega,\vec{k})## assumed (sic!) to be components of a four-vector without further physically relevant vectors or tensors around (i.e., a wave propagating in vacuum) is ##k_{\mu} k^{\mu}=m^2## with a scalar quantity ##m^2 \geq 0##. Since this is a Lorentz-invariant expression supposed the four numbers ##(\omega,\vec{k})## transform like four-vector components, you have a Lorentz-covariant dispersion relation.

This implies that the phase factor, which is part of the mode decomposition of the field with respect to momentum eigenmodes (where momentum is defined as the Noether charge under spatial translations), ##\exp(-\mathrm{i} x_{\mu} k^{\mu})## is a scalar either, as is immediately clear by the notation since by assumption the ##k^{\mu}## and the ##x^{\mu}## transform as comonents of four-vectors under Lorentz transformations.

Whether or not a theory makes sense within relativistic physics is, whether you can find a transformation law for all involved quantities under the symmetry transformations underlying the (special) relativistic space-time structure, i.e., it must be possible to define transformation rules of the fields under consideration (and the natural way to express relativistic natural laws is in terms of local (quantum) field theory) realizing the proper orthochronous Lorentz group.

The most simple idea are scalar, vector and arbitrary tensor fields. As QT tells us one can also consider spinor representations of the Poincare group.

The most simple example is, of course, just a scalar field, and a nice covariant equation is the Klein-Gordon equation,
$$(\Box + m^2) \phi(x)=0.$$
With the ansatz ##\phi(x)=\phi_0 \exp(-\mathrm{i} k_{\mu} x^{\mu})## you get the above postulated dispersion relation, and everything is manifestly covariant.

For electromagnetism it's somewhat more complicated, but you can realize it in terms of the four-vector potential with components ##A^{\mu}##, assumed to transform under Lorentz transformations as four-vector fields. The physical quantities are of course the electric and magnetic field components, of put in the usual real-valued relativistic formulation the Faraday tensor components
$$F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}.$$
Then the equation of motion reads
$$\partial^{\mu} F_{\mu \nu}=j_{\nu},$$
where ##j^{\nu}## are the components of the electric-charge four-current ##(\rho,\vec{j})##.

Since this doesn't uniquely determine ##A_{\mu}##, because the physical quantities ##F_{\mu \nu}## don't get changed when adding an arbitrary four-gradient field (gauge invariance), i.e., the two fields ##A_{\mu}## and ##A_{\mu}'=A_{\mu} + \partial_{\mu} \chi## are equivalent for any scalar field ##\chi##, you can impose an arbitrary gauge-fixing condition. A particularly convenient and manifestly covariant one is the Lorentz-gauge condition
$$\partial_{\mu} A^{\mu}=0.$$
With that condition the equation of motion becomes
$$\Box A^{\mu}=j^{\mu}.$$
For free fields, i.e., ##j^{\mu}=0##, you can impose even one more constraint, like, e.g., ##A^0=0##, and you see that the plane-wave solutions are transverse vector fields with the dispersion relation given by ##k_{\mu} k^{\mu}=0##.

Since everything is manifestly covariant, assuming that ##A^{\mu}## and ##j^{\mu}## transform as components of four-vector fields, you have a proper relativistic field theory.

Ibix