What distributions of electromagnetic oscillating fields are Lorentz invariant? How can oscillating electromagnetic fields look the same regardless of inertial frame?

According to Marshall and Boyer a cubic distribution like

$ \rho(\omega) d\omega = \frac{\hbar\omega^3}{2 \pi^2c^3} d\omega \tag{1}$

is Lorentz invariant. How?

The cubic distribution is somewhat explained on WikiPedia page on zero-point energy.

Boyer just assumes that the spectrum is Lorentz invariant. This of course tells nothing. There might not be any such spectrum at all.

Quotes from Boyer.

From the abstract:

The hypotheses assume (a) the existence, at the absolute zero of temperature, of classical homogeneous fluctuating electromagnetic radiation with a Lorentz-invariant spectrum;

Page 1375

Making no initial assumptions about the form of this energy beyond the Lorentz invariance of the spectrum...

Page 1377

We have postulated that the zero-point spectrum is Lorentz-invariant

Marhsall makes similar assumptions.

From the abstract:

The random pure radiation field postulated in an earlier paper is set up in a relativistically invariant manner.

Page 538

In effect, the requirement that the zero-temperature energy spectrum be left invariant under a Lorentz transformation ...

Page 545

The random field there postulated has here been defined in a relativistically invariant manner.

Where the previous paper is MARSHALL, T. W. Proc. Roy. Soc. Ser. A, 276 (1963), 475-491.

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    $\begingroup$ An excellent explanation of this appears in a Scientific American article from the 1980's called The Classical Vacuum by Boyer. $\endgroup$ Commented Dec 14, 2018 at 22:55
  • $\begingroup$ Without access to the articles you link it is very hard to understand what exactly this question is about, which in turn makes it hard to evaluate the answers. Please elaborate whether or not the $\rho(\omega)\mathrm{d}\omega$ is "the spectrum" here, or something from which the spectrum is derived. In either case, please explicitly state what $\rho(\omega)$ is and what the spectrum is. $\endgroup$
    – ACuriousMind
    Commented Dec 23, 2018 at 13:54
  • $\begingroup$ Please edit such relevant information into your question. Comments are ephemeral and should not contain information essential to making the question accessible. $\endgroup$
    – ACuriousMind
    Commented Dec 23, 2018 at 14:01
  • $\begingroup$ The problem is to show that $du \propto \omega^3 \, d\omega$ is invariant for any inertial observer, where $du$ is the energy density of radiation (in $\mathrm{J}/\mathrm{m}^3$) in a frequency range $d\omega$. Substituting the Lorentz transformation of $\omega$ isn't enough, since $du$ isn't a Lorentz scalar: $du$ is part of the radiation energy-momentum : $du = dT_{00}$, so it transforms covariantly. Whatever the frame, we want $d\tilde{T}_{00} \propto \tilde{\omega}^3 \, d\tilde{\omega}$. $\endgroup$
    – Cham
    Commented Dec 23, 2018 at 15:34
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    $\begingroup$ In order prove that ZPF is Lorentz Invartiant first you have to prove that it is a product of QFT's quantum fields almost completelly cancelling out in free space with a tiny residue left out thus ZPF noise. I don't know is such a proof is possible and also backed up by any empirical data. On the other hand it could be simply that ZPF of the vacuum is dark energy noise in our spacetime. If this is the case since we have not a complete theory about dark energy we cannot even claim that ZPF is Lorentz Invariant. $\endgroup$
    – Markoul11
    Commented Feb 19, 2023 at 17:51

3 Answers 3


This isn't a complete answer, just some remarks about what should be invariant, if there really is something invariant here. I use $c = 1$ to simplify things.

I'm expecting that the EM radiation energy-momentum tensor should be of the following shape (incoherent superposition of EM plane waves): \begin{equation}\tag{1} T_{ab} \propto \int k_a \, k_b \, f(\omega) \, d^3 k, \end{equation} where $k^a$ is a ligth-like 4-vector (so $k_a \, k^a = 0$). The energy-momentum of EM radiation should have a null trace. Expression (1) satisfy this constraint, and cannot gives something like a Lorentz invariant enery-momentum $T_{ab}^{\text{vac}} = u \, \eta_{ab}$ (which implies pressure $p = -\, u$). The function $f(\omega)$ is arbitrary, but is defined in the special frame for which radiation is isotropic. Here: \begin{equation}\tag{2} \omega = u_a^{\text{iso}} \, k^a \end{equation} is the frequency in this special frame and could be defined to be a Lorentz invariant (the frequency as measured by the observer of 4-velocity $\smash{u_{\text{iso}}^a}$, for which radiation is isotropic!). It's easy to verify that pressure of the radiation is $p = \smash{\frac{1}{3}} \, u$, in this frame: \begin{align} u \equiv T_{00} &\propto \int \omega^2 \, f(\omega) \, d^3 k = 4 \pi \int \omega^4 \, f(\omega) \, d\omega. \tag{3} \\[12pt] T_{0 i} &\propto \int \omega^2 \, n_i \, f(\omega) \, d^3 k = 0. \qquad \text{(from isotropy)} \tag{4} \\[12pt] T_{ij} &\propto \int \omega^2 \, n_i \, n_j \, f(\omega) \, d^3 k = \int \omega^4 \, \big( \frac{4 \pi}{3} \, \delta_{ij} \big) \, f(\omega) \, d\omega \equiv \frac{1}{3} \, u \, \delta_{ij}. \tag{5} \end{align} Expression (5): $T_{ij} = \frac{1}{3} \, u \, \delta_{ij}$, implies that pressure is $p = \frac{1}{3} \, u$, as it should for random radiation, and that the full energy-momentum of incoherent EM radiation cannot be Lorentz invariant (i.e $T_{ab}^{\text{rad}} \ne u \, \eta_{ab}$). In any other inertial frame (frame $\tilde{\mathcal{R}}$, for which $u_{\text{iso}}^a \ne \{ 1, 0, 0, 0 \}$), the radiation has a dipolar pattern since (2) gives \begin{equation}\tag{6} f(\omega) = f\big(\gamma \, (1 - v \cos{\vartheta}) \, \tilde{\omega} \, \big), \end{equation} which implies an anisotropy of radiation in that frame. Then $\tilde{T}_{0i} \ne 0$ and there's a directional flux.

However, it doesn't mean that expression (3) cannot have a Lorentz invariant spectrum. We recover \begin{equation}\tag{7} du \propto \rho(\omega) \, d\omega \propto \omega^3 \, d\omega \end{equation} in the special case $f(\omega) \propto \omega^{- 1}$ (which I think is weird, since it implies some infrared divergence for $f(\omega)$, and ultraviolet divergence for the total energy density $u$!). This isn't enough to show that this special spectrum of $u$ is Lorentz invariant. This is why my answer is currently incomplete.

Notice that, according to expression (1), the product $f(\omega) \, d^3 k$ is a Lorentz scalar, while $f(\omega)$ and $d^3 k$ separately are not Lorentz scalars. Also, it can be proved that $d^3 k / \omega$ is a Lorentz invariant measure in the momentum space of the radiation. Then $f(\omega) \, \omega$ should be a Lorentz scalar too. If $f(\omega) \propto \omega^{-1}$, then obviously $f(\omega) \, \omega = \text{cste}$!

I feel personally unsure and puzzled by the previous paragraph.

  • $\begingroup$ Just a comment to try to remove the ultraviolet divergence of (3). If I introduce some temporary ultraviolet cutoff at $\omega = \omega_{\text{max}}$ (at the risk of destroying the Lorentz invariance of the distribution), I could use the function \begin{equation}f(\omega) = \frac{\mathcal{C}}{\pi \, \omega_{\text{max}}^4 \, \omega}\end{equation} to get $u = \mathcal{C}$. Under the limit $\omega_{\text{max}} \rightarrow \infty$ (to recover Lorentz invariance of the distribution), then $f(\omega) \rightarrow 0$ while still getting $u = \mathcal{C} \ne 0$. Hmm, I know this is clumsy! $\endgroup$
    – Cham
    Commented Dec 26, 2018 at 2:34
  • $\begingroup$ A remark about the energy-momentum tensor of pure radiation. Formally, its trace must cancels : $\eta^{ab} \, T_{ab} = u- 3 p = 0$, which the expression $T_{ab}^{\text{vac}} = u \, \eta_{ab}$ doesn't satisfy ($\eta^{ab} \, T_{ab}^{\text{vac}} = 4 \, u$). However, in the case of the Lorentz invariant spectrum, the full energy density and pressure are infinite! Thus the energy-momentum trace cannot be defined: $\infty - \infty$ is ill-defined. We cannot conclude that the Lorentz invariant radiation energy-momentum isn't Lorentz-invariant! This tensor cannot even be found from an action... $\endgroup$
    – Cham
    Commented Dec 28, 2018 at 19:06
  • $\begingroup$ ... so it cannot be included in Einstein's equation from a variation of the full action! This seems to suggest that the zero-point of Lorentz-invariant radiation cannot gravitate, and shouldn't be included into Einstein's equation! There's a kind of exit door here! $\endgroup$
    – Cham
    Commented Dec 28, 2018 at 19:07
  • $\begingroup$ What about adding something about cutoff, and how the speed dependency of the interaction with matter can affect the eventual invariance. It is actually the entire radiation pressure to matter that needs to be invariant, not the spectrum itself, in order to preserve Newtons first law. $\endgroup$ Commented Jan 7, 2019 at 22:47

I will construct a response in the only way I see possible, given the concerns of access to the source as mentioned in the comments. I will provide an in detail outline of how photon spectra transform under LTs, and try to directly address the question concerning the Lorentz invariance of the photon spectrum.

In general a photon spectrum ${dN_\gamma \over dE_\gamma d\Omega_\gamma}$ is not a lorentz invariant (as I will explain further below). However, there is an invariant that is close to this, which may be what the authors are getting at.

In particle physics, if we want to know the photon spectrum in the lab frame ${dN_\gamma \over dE_\gamma d\Omega_\gamma}$ , what we do is use the fact that the two observers must agree on the number of photons emitted $dN_\gamma$ in order to transform the (usually easier to find) distribution in the rest frame of, say, a decaying particle $S'$, back to to the lab frame.

So how doest this work? Well, suppose that the photon spectrum is given by function ${dN_\gamma' \over dE'_\gamma d\Omega'_\gamma}$ in the parent particle's rest frame, and suppose that the spectrum in the lab frame is given by ${dN_\gamma \over dE_\gamma d\Omega_\gamma}$. Using the fact that $dN_\gamma$ is an invariant (i.e. $dN_\gamma = dN'_\gamma$), it follows then that

$$dN_\gamma = dN_\gamma' \implies {dN_\gamma \over dE'_\gamma d\Omega'_\gamma} dE_\gamma' d\Omega_\gamma' = {dN_\gamma \over dE_\gamma d\Omega_\gamma} dE_\gamma d\Omega_\gamma$$

Now this could be precisely what the authors meant by Lorentz invariant spectrum. However, this wouldn't be very useful if we didn't know how to get from one distribution to the other. Luckily--given that $dN_\gamma = dN_\gamma'$--we do, as is beautifully outlined here in Jacobians and Relativistic Kinematics. The summary is that the Lorentz transformation of a photon spectrum is just a coordinate transformation subject to the constraint that $dN_\gamma = dN_\gamma'$. Then, we can relate the two distributions via the jacobian that they pick up. That is, (please allow me to drop the subscripts $\gamma$'s) we have that

$$\boxed{ \frac{dN}{dE d\Omega}= \left\lvert \frac{\partial(E', \Omega')}{\partial(E, \Omega)} \right\rvert \frac{dN}{dE' d\Omega'}}$$

This relation is exactly how the two spectra are related, and is of immense usefulness (as it has been to me when calculating photon spectra for dark matter annihilation spectra).

At any rate: Based on what I have just detailed, it seems to me that by "Lorentz invariant photon" spectrum is somewhat of a misnomer, it is not the distribution ${dN_\gamma \over dE_\gamma d\Omega_\gamma}$ that is Lorentz invariant (as it picks up a jacobian), but rather it is the number of photons that is invariant considered in some interval of phase space.

Edit: Using the Lorentz Invariant Phase Space element (LIPS). Note if instead we choose to parametrize our distribution with momentum instead of energy, then all of the previous arguments are valid if we replace $dE_\gamma d\Omega_\gamma \to d^3 k$ (same for the primes). This induces the substitution

$$ {dN_\gamma \over dE_\gamma d\Omega_\gamma} \to {dN_\gamma \over d^3k} $$

and then we may reuse the arguments above. Hence, our photon distributions now are just

\begin{align*} \begin{cases} & {dN_\gamma \over d^3 k} \qquad \text{(Lab Frame)}\\ & {dN_\gamma \over d^3 k'} \qquad \text{(Rest Frame of Parent Particle)} \end{cases} \end{align*}

It therefore follows that the invariant quantities are now

$$dN_\gamma = dN_\gamma' \implies {dN_\gamma \over d^3 k'} d^3k' = {dN_\gamma \over d^3 k} d^3k$$

where of course the phase space elements $d^3 k \neq d^3 k'$ and hence are not in general Lorentz invariant. Now, this is inconvenient so in field theory what we do is define the Lorentz Invariant Phase Space (LIPS) element by

$$ (d\Pi)_{LIPS} \equiv \prod_{i=1}^n \frac{d^3k_i}{2E_{k_i}}$$

for $n$ final state particles, and $E_{k_i} = \sqrt{\mathbf{k}_i^2 + m_i^2}$. It can be shown, as it is in this answer,that this indeed is a lorentz invariant quantity. Therefore, using this as the phase space measure, we have the relation that

$$dN_\gamma = dN_\gamma' \implies {dN_\gamma \over d^3 k'} \frac{d^3k'}{2E_{k'}} = {dN_\gamma \over d^3 k} \frac{d^3k}{2E_{k}}$$

for a single particle final state (i.e. $n=1$). What this means is that the two observers must agree on the quantities $ \frac{d^3k'}{2E_{k'}}$ and $\frac{d^3k}{2E_{k}}$. However, this does do not imply that need to agree on the quantities ${dN_\gamma \over d^3 k'}$ and ${dN_\gamma \over d^3 k}$. But from particle conservation, they still must agree on the product

$$dN_\gamma =\overbrace{ \underbrace{{dN_\gamma \over d^3 k} }_{\text{Not Lorentz Invariant}}\frac{d^3k}{2E_{k}}}^{\text{Lorentz Invariant}}. $$

This is what is meant by "Lorentz Invariant Distribution". It would seem that if they agree on the phase space element and on the product of the phase space then surely they must agree on ${dN_\gamma \over d^3 k}$, right? Well, no. Two different integrands can yield the the same number.

Edit 2: As @Cham pointed out in the comments, given the above facts, we can then consider your distribution. Namely, that

$$ \hbar \omega \underbrace{\frac{1}{c^3}\omega^2 \ d\omega d\Omega}_{d^3k} = \hbar\omega \ d^3 k $$

So we see that

$$\frac{dN_\gamma}{d^3 k} = \hbar \omega$$

Also, our LIPS element is given by

$$ \frac{d^3k}{2\omega}$$

which means that

$$dN_\gamma =\frac{d^3k}{2\omega} \frac{dN_\gamma}{d\omega d\Omega_\gamma} = \hbar\omega \frac{d^3k}{2\omega} = d^3 k \frac{\hbar}{2}. $$

So using the conservation of particle number we have that

$$ \hbar\omega \frac{d^3k}{2\omega} = \hbar\omega' \frac{d^3k'}{2\omega'} \implies \hbar\ d^3k = \hbar\ d^3k'.$$

So now we can see that when using the Lorentz Invariant Phase Space that the "distribution" $\propto \hbar$ is invariant. I use quotation marks because technically the distribution is w.r.t the phase space element and so really is linear in $\omega$.

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    $\begingroup$ Hmm, I don't think this gets to the heart of the issue. We do want to know why the distribution is invariant. If it is, say, only cubic in one frame, that picks out a special frame and breaks Lorentz invariance. $\endgroup$
    – knzhou
    Commented Dec 25, 2018 at 12:14
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    $\begingroup$ I think that the supposed invariance of the spectral distribution should be formulated after integration on all angles. It should be the spectre of the local intensity of light which is supposed to be invariant. $\endgroup$
    – Cham
    Commented Dec 26, 2018 at 0:18
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    $\begingroup$ @knzhou, what I understand (or not) of this issue, is that the intensity spectral distribution is invariant, not the flux in a given direction (how could it be, from the Doppler and abberation effects?). There should be a dipolar Lorentz effect, if an observer is moving relative to the frame in which flux is isotropic. Or else, if the flux is also invariant, I don't understand how the radiation energy-momentum could be Lorentz invariant, i.e of the form $T_{ab} = u \, \eta_{ab}$. The trace of this expression cannot be $0$, so it's not radiation. $\endgroup$
    – Cham
    Commented Dec 26, 2018 at 0:43
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    $\begingroup$ If the vacuum is Lorentz invariant, then it's of the form $T_{ab} \propto \eta_{ab}$ (pressure is $p = -\, u$, which isn't EM radiation). Then the trace is $T =4 u \ne 0$. The energy-momentum of EM radiation (even on average) should have a trace equal to $0$ (and pressure $p = \frac{1}{3} \, u$), unless there's something I really don't understand here. $\endgroup$
    – Cham
    Commented Dec 26, 2018 at 0:46
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    $\begingroup$ Boyer is still alive. Any chance we can get him to offer his perspectives on this? $\endgroup$ Commented Dec 27, 2018 at 3:22

Assuming that the vacuum energy consists of radiation that is Doppler shifted when moving relative to leads to the spectral distribution not being Lorentz invariant. The longitudinal Doppler shift

$$ \omega' {\displaystyle ={\sqrt {\frac {1-\frac{v}{c} }{1+\frac{v}{c} }}}\,\omega} $$

leads the spectrum to become

$$\rho(\omega') d\omega' = {\left(\frac {1-\frac{v}{c}}{1+\frac{v}{c} }\right)}^{\frac{3}{2}} \frac{\hbar\omega^3}{2 \pi^2c^3} d\omega $$

This is not Lorentz invariant. It differs with a factor ${\left(\frac {1-\frac{v}{c}}{1+\frac{v}{c} }\right)}^{\frac{3}{2}}$. Serious consequences are that vacuum would have drag due to differences in radiation pressure and that there is only one inertial system.

This result is commonly avoided by postulating Lorentz invariance but what kind of fields could vacuum consist of if not radiation fields that can be Doppler shifted?

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    $\begingroup$ 1. Please do not leave comments asking for explanations for a downvote, they are noise. If the downvoter would have wanted to explain themselves, they would have left a comment right away. 2. Precisely because the answer seems to trivial it is hard to believe that it is correct - it suggests that either the sources are trying to claim something other than "$\rho(\omega)\mathrm{d}\omega$ is Lorentz invariant" when they say the spectrum is invariant, or that the transformation behaviour is not as you assume. Why are you using longitudinal Doppler shift here - what are source and observer here? $\endgroup$
    – ACuriousMind
    Commented Dec 23, 2018 at 14:09
  • $\begingroup$ Sorry, but this "answer" is pure gibberish. It doesn't make sense at all. $\endgroup$
    – Cham
    Commented Dec 23, 2018 at 15:23
  • $\begingroup$ What is wrong with this derivation? $\endgroup$ Commented Jun 27, 2023 at 22:39

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