Lorentz transformation on mode functions

[Haorong Wu]

TL;DR

How to do a Lorentz transformation on mode functions?

Suppose we can boost from a frame $S$ to another frame $S'$ by using a Lorentz transformation $\Lambda$. Also, $\phi(x^\mu;\omega,\mathbf k)$ is a mode function of a scalar field in frame $S$. Then, how do we express this mode function in frame $S'$? Here is my attempt.

First, the Klein-Gordon equation is Lorentz invariant, so we can use $x^\mu=\Lambda^{~~\mu}_\nu x'^\nu$ to change the coordinates in $\phi(x^\mu;\omega,\mathbf k)$.

Second, the four wave vector is also transformed by $k^\mu=\Lambda^{~~\mu}_\nu k'^\nu$.

In total, I simply transform the coordinates and four wave vector in the mode function $\phi(x^\mu;\omega,\mathbf k)$ to $\phi(x'^\mu;\omega',\mathbf k')$. Is this correct?

Thanks ahead.

 

[Ibix]

I'm not too familiar with this, so please forgive the partial answer.

First, if you have a scalar function $\phi(x^a)$ on spacetime, obviously it's legitimate to Lorntz transform the coordinates and write $\phi(x'^a)$. Physically, we're saying that the scalar field has the same value at some event $E$ whether you give $E$ the coordinates $x^a$ or $x'^a$.

Second, I think mode functions are a specific set of functions. From symmetry, they presumably have the same functional form in any inertial frame. So if the mode functions defined in frame $S$ were the infinite set of functions $\sin(\omega t)$ (a set parameterised by $\omega$) then the mode functions defined in frame $S'$ would be $\sin(\omega t')$ (again, $\omega$ is merely a parameter so I'm not worrying about primes).

Assuming that's correct, the answer to your question depends on the form of the mode functions. I think they are themselves scalar fields, so it's obviously legitimate to just transform their coordinate dependence and you get the scalar field expressed in some other coordinates. But is that field a mode function in the other coordinates? If the mode functions are something like $\sin(\omega t)$ then the answer is no - the transformed version of this field will have a spatial dependence so can't be written $\sin(\omega t')$. If, on the other hand, they're something like $\sin(k_ax^a)$ and any $k^a$ is allowed then yes since $k_ax^a$ is a scalar itself. But if there's some restriction on the $k^a$ you're allowed to use it depends if the allowed vectors in $S$ and $S'$ map onto each other.

 

[PeterDonis]

From symmetry, they presumably have the same functional form in any inertial frame.

No, they can't. For example, suppose we have a function $\sin \omega t$ in one frame. A Lorentz transformation will obviously make this $\sin \omega \gamma \left( t' - v x' \right)$ in a different frame.

 

[Ibix]

No, they can't. For example, suppose we have a function $\sin \omega t$ in one frame. A Lorentz transformation will obviously make this $\sin \omega \gamma \left( t' - v x' \right)$ in a different frame.

Sure, but I get the impression that the mode functions are the basis of a decomposition (possibly just a Fourier decomposition) of the scalar function. So if I pick frame $S$ and get a decomposition in terms of a weighted sum (or integral) of a one-parameter family of functions (e.g. $\sin(\omega t)$) then if I follow the same process in $S'$ I'll get a decomposition in terms of a weighted sum/integral of $\sin(\omega t')$. That's the symmetry I was meaning, and those hypothetical mode functions are not related by a Lorentz transform, as you note.

So the answer to the OP's question depends on whether the Lorentz transform of an $S$ mode function is an $S'$ mode function. That could be the case if mode functions look like (e.g.) $\sin(k_ax^a)$, but need not be. With the limited information I have about how a mode function is defined, I don't know.

 

[PeterDonis]

the mode functions are the basis of a decomposition (possibly just a Fourier decomposition) of the scalar function.

Yes, they are, but they're still functions, and their arguments still have to transform the way I described.

if I pick frame $S$ and get a decomposition in terms of a weighted sum (or integral) of a one-parameter family of functions (e.g. $\sin(\omega t)$) then if I follow the same process in $S'$ I'll get a decomposition in terms of a weighted sum/integral of $\sin(\omega t')$.

Ah, I see. Yes, you can do the same process in both frames, but the functions you end up won't be Lorentz transforms of each other, as you note.

the answer to the OP's question depends on whether the Lorentz transform of an $S$ mode function is an $S'$ mode function. That could be the case if mode functions look like (e.g.) $\sin(k_ax^a)$, but need not be.

I think we need a reference from the OP to resolve this.

 

[PeterDonis]

Is this correct?

Can you give a reference for how the mode functions you are asking about are defined? Without that we can't answer your question since we don't know what the mode functions are functions of.

 

[Haorong Wu]

Hi, @Ibix, and @PeterDonis. I am recently interested in boosting a scalar field, so I found this paper, Relativistic Hall Effect. I am sorry it is not open-access, so I am not sure whether you can access it or not.

The mode functions, Eq. (9), are a little complicated as the authors use the Bessel functions. Around Eq. (10), the authors argue that

Since the Klein-Gordon equation is Lorentz-invariant, one can find the form of the Bessel beam in the moving reference frame by substituting the Lorentz transformation into Eq. (9). Then, the scalar wave function in the moving frame becomes $\psi'(\mathbf r',t')=\psi[\mathbf r(\mathbf r',t'),t(\mathbf r',t')]$.

The authors do not say anything about the wave four-vector.

As you have discussed, the mode functions of a scalar field should give the same value at the same point irrespective of how the coordinates are defined. For simplicity, let us consider the plane wave solutions, $\psi(x^\mu)=C\exp (ik_\mu x^\mu)$, in frame $S$, and a boost $\Lambda^{\mu'}_{~~\nu}$ to another frame $S'$. The coordinates are transformed as $x^{\mu'}=\Lambda^{\mu'}_{~~\nu} x^\nu$, and the mode functions are transformed into $\psi'(x^{\mu'})$. Since $\psi'(x^{\mu'})=\psi(x^\mu)$, we should transform the wave four-vector at the same time---i.e., $k_{\mu'}=\Lambda_{\mu'}^{~~\nu}k_\nu$, such that $\exp (ik_\mu x^\mu) \rightarrow \exp (ik_{\mu'} x^{\mu'})=\exp (ik_{\mu} x^\mu)$. The normalization constant $C$ is proportional to $k^{-1/2}_0$, so it should be altered, as well.

I think the above transformation of the mode functions does not depend on the explicit form of the functions, though after substituting the new coordinates and wave four-vector, the form may be altered.

 

[Ibix]

So, by "mode functions" you mean the modes of a transmission line? If not, please say what you do mean.