Does Light Have Mass? Astrophysics Discussion

[PeterDonis]

That energy contributes to its energy, but does not contribute any momentum to the capacitor

Actually, it can, if you're willing to open the "hidden momentum" can of worms. ;) See, for example, this paper by Griffiths et al:

http://www.ate.uni-duisburg-essen.de/data/postgraduate_lecture/AJP_2009_Griffiths.pdf

The short version: if you want momentum to be conserved, you have to allow for the possibility of static EM fields carrying nonzero momentum.

 

[DrStupid]

Can you give a reference that applies it to something else that's called an "EM field", such as $\vec{E}$ and $\vec{B}$ as they appear in Maxwell's Equations?

What is wrong with the examples I already mentioned above? It seems you agree that a photon is are valid examples for an EM field. Why don't you accept systems of more than one photons? Is the superposition of EM fields not an EM field?

 

[PeterDonis]

It seems you agree that a photon is are valid examples for an EM field.

Only in a certain very limited approximation. If you're using $E^2 - p^2 = m^2$ to describe a system of photons, you're treating the photons as particles, not fields. Each individual photon, as a particle, moves on a null worldline, so its mass, as an individual particle, is zero. If you add the 4-momentum vectors of multiple photons to get a "system" of photons, and use $E^2 - p^2 = m^2$ on the system, you can get $m > 0$, yes, but you're still treating the whole system as a system of particles, not fields.

Fields come into it because, classically speaking, the ultimate justification for treating photons as particles is applying the geometric optics approximation to EM field waves. If you wanted to deal with the fields directly, you could just not do that, and work with the fields without ever talking about photons. But if you do that, you're not using $E^2 - p^2 = m^2$; you're using Maxwell's Equations. At least, that's the way I've always seen it done; perhaps there is someone who has tried to apply $E^2 - p^2 = m^2$ to solutions of Maxwell's Equations directly, but if so, I've never seen it, which is why I asked for a reference.

(Also, since the real world is quantum mechanical, the ultimate justification for using classical EM fields that satisfy Maxwell's Equations is that those equations give the correct classical limit for the quantum EM field as it appears in QED. And, as I said, that fundamental field is massless.)

 

[DrStupid]

If you wanted to deal with the fields directly, you could just not do that, and work with the fields without ever talking about photons. But if you do that, you're not using $E^2 - p^2 = m^2$; you're using Maxwell's Equations.

Why is it impossible to do both? Are Maxwell's Equations not consistent with m² = E² - p²?

 

[PeterDonis]

Are Maxwell's Equations not consistent with m² = E² - p²?

I don't know, because I've never seen them both used in the same model. Have you? That's the question I keep asking.

 

[DrStupid]

I've never seen them both used in the same model. Have you?

No I haven't, but that does not mean that it is not possible. It would be bad news for relativity if Maxwell would violate m² = E² - p².

 

[Matterwave]

I am sort of torn here.

It is clear that you can have classical EM fields which satisfy Maxwell's equations and have mass according to relativity. For example, consider a capacitor. A charged capacitor has more energy than an otherwise identical uncharged capacitor. A charged capacitor can be used to do work to turn it into an uncharged capacitor. That energy contributes to its energy, but does not contribute any momentum to the capacitor, and therefore it increases the capacitor's mass.

However, where I am torn is in whether or not it is reasonable to call such a thing a "stationary radiation field". The term "stationary radiation field" is simply not one that I am familiar with. What I would call EM radiation is never stationary, regardless of whether or not the configuration has mass.

I don't think one would call a static field like that "radiation", or "light", as the terms are usually understood.

But even if one did, that would be analogous to a box of photons trapped by perfect mirrors, a bound state.

 

[jtbell]

It would be bad news for relativity if Maxwell would violate m² = E² - p².

Indeed. Classical electromagnetism doesn't contain photons, but in electromagnetic waves the energy and momentum densities obey the same relation with m = 0 (and inserting a factor of c for SI units). Energy density: $$U = \epsilon_0 E^2$$ Momentum density: $$g = \frac{\epsilon_0 E^2}{c}$$ Therefore U = gc, analogous to E = pc for photons.

 

[samalkhaiat]

Photons are massless regardless of their state of motion. Experimentalists some times use the (rather stupid) notation $m_{ \gamma \gamma }$ to indicate the mass of the DECAYING particle. Take for example the decay $\pi^{ 0 } \rightarrow 2 \gamma$.
The 4-momentum conservation for this decay is
$$P^{ \mu } ( \pi ) = P^{ \mu } ( \gamma_{ 1 } ) + P^{ \mu } ( \gamma_{ 2 } ) .$$
In the frame where the Pion is at rest, the two photons fly apart in the opposite directions. So, in this rest frame, the conservation of energy and 3-momentum become
$$m_{ \pi } c = \frac{ E( \gamma_{ 1 } ) }{ c } + \frac{ E( \gamma_{ 2 } ) }{ c } ,$$
$$\vec{ P } ( \gamma_{ 1 } ) = - \vec{ P } ( \gamma_{ 2 } ) .$$
Now, since
$$m_{ \gamma_{ 1 } } = m_{ \gamma_{ 2 } } = 0, \ \mbox{and} \ | \vec{ P } ( \gamma_{ 1 } ) | = | \vec{ P } ( \gamma_{ 2 } ) | \equiv | \vec{ P } | ,$$
we get
$$E( \gamma_{ 1 } ) + E( \gamma_{ 2 } ) = 2 c | \vec{ P } | ,$$
and
$$| \vec{ P } | = \frac{ m_{ \pi } c }{ 2 } .$$

 

[DrStupid]

I don't think one would call a static field like that "radiation", or "light", as the terms are usually understood.

Has the hollow black body mentioned above a radiation field inside? If yes, is it static? If yes, why not call it a static radiation field?

but in electromagnetic waves the energy and momentum densities obey the same relation with m = 0

Could you please show that for a standing electromagnetic wave?

 

[Matterwave]

Has the hollow black body mentioned above a radiation field inside? If yes, is it static? If yes, why not call it a static radiation field?

I was responding to Dalespam and his situation is a static electric field inside a capacitor. What does a hollow black body have to do with that?

In fact, inside a star, one usually does talk about a radiation field. I have no problem if you want to call that a "static radiation field", but I think most people would just call that the radiation field inside a star. In any case, it would be a radiation field that is bounded inside an object, trapped so that it can not escape, similar to a box of photons trapped with perfect mirrors. I don't think I've ever had a problem with that concept in this thread. In fact, I think I mentioned several times, that in such a case, the photons do in fact contribute to the mass of the box, which is something you can physically measure.

 

[Orodruin]

Photons are massless regardless of their state of motion. Experimentalists some times use the (rather stupid) notation $m_{ \gamma \gamma }$ to indicate the mass of the DECAYING particle.

This is not exactly true. They use it to denote the invariant mass of any pair of two photons (or squared total 4-momentum of the two photons if you prefer) and then plot the number of events they obtain for different values of $m_{\gamma\gamma}$. The spectrum will typically be continuous but will have a peak (for obvious reasons) at the mass of a produced particle that decays into $\gamma\gamma$.

They also do the same for other collision products, for example $m_{4\ell}$ in the case of Higgs to ZZ.

 

[samalkhaiat]

This is not exactly true. They use it to denote the invariant mass of any pair of two photons (or squared total 4-momentum of the two photons if you prefer).

The invariant mass of photon or photons is ZERO. The squared total 4-momentum of the two photons IS equal to the invariant rest mass of the decaying particle. In the $\pi^{ 0 }$ decay (or any other one-particle decay), the squared 4-momentum of the two photons is
$$( P^{ \mu } ( \gamma_{ 1 } ) + P^{ \mu } ( \gamma_{ 2 } ) )^{ 2 } = m^{ 2 }_{ \pi }$$

 

[Orodruin]

The invariant mass of photon or photons is ZERO. The squared total 4-momentum of the two photons IS equal to the invariant rest mass of the decaying particle. In the $\pi^{ 0 }$ decay (or any other one-particle decay), the squared 4-momentum of the two photons is
$$( P^{ \mu } ( \gamma_{ 1 } ) + P^{ \mu } ( \gamma_{ 2 } ) )^{ 2 } = m^{ 2 }_{ \pi }$$

Yes, of each individual photon, nobody is disputing that. You are missing the point that what they call $m_{\gamma\gamma}$ is not the mass of the Higgs/pion but the square of the summed 4-momenta, referred to as the invariant mass or CoM energy of the system. It is not the mass of a decaying particle because a priori we do not know the photons come from a particle decay (that is why we make the experiment!) and in the experiment you will have a lot more photon pairs flying around. Only a few of those will have a CoM energy of the decaying particle mass but enough to make a peak in the $m_{\gamma\gamma}$ spectrum if you produce enough of the particles. Clearly, $m_{\gamma\gamma}$ will be equal to the decaying particle mass if the photons come from a particle decay, but again, all photons typically do not. See for example http://www.atlas.ch/news/images/stories/1-plot.jpg - it is clear from this image that you have photon pairs which belong to a continuous spectrum in $m_{\gamma\gamma}$ and not from Higgs decays. In fact, the photon pairs originating from Higgs decays are in overwhelming minority.

As I have said repeatedly, this does not mean photons have mass.

 

[samalkhaiat]

Yes, of each individual photon, nobody is disputing that. You are missing the point that what they call $m_{\gamma\gamma}$ is not the mass of the Higgs/pion but the square of the summed 4-momenta, referred to as the invariant mass or CoM energy of the system. It is not the mass of a decaying particle because a priori we do not know the photons come from a particle decay (that is why we make the experiment!) and in the experiment you will have a lot more photon pairs flying around. Only a few of those will have a CoM energy of the decaying particle mass but enough to make a peak in the $m_{\gamma\gamma}$ spectrum if you produce enough of the particles. Clearly, $m_{\gamma\gamma}$ will be equal to the decaying particle mass if the photons come from a particle decay, but again, all photons typically do not. See for example http://www.atlas.ch/news/images/stories/1-plot.jpg - it is clear from this image that you have photon pairs which belong to a continuous spectrum in $m_{\gamma\gamma}$ and not from Higgs decays. In fact, the photon pairs originating from Higgs decays are in overwhelming minority.

As I have said repeatedly, this does not mean photons have mass.

I think, I am very well aware of this. This is why I said (theoretically speaking) it is stupid to think that the two-photon system has a non-zero invariant mass. Theoretically, if $A \rightarrow \gamma \gamma$ then, in the rest frame of $A$, you have $( P^{ \mu }_{ 1 } + P^{ \mu }_{ 2 } )^{ 2 } = m^{ 2 }_{ A }$ for THOSE gammas which definitely came from $A$.

 

[Orodruin]

What I am arguing against is your assertion that $m_{\gamma\gamma}$ is a "rather stupid" choice for the squared 4-momentum of the system. It is the mass a particle that decayed into two specific photons that you measured would have had if it existed. The same thing goes for $m_{4\ell}$. Whether you want to call this the invariant mass of the system or CoM energy is just a question of semantics.

Furthermore, you can never know that a pair of photons definitely comes from a particle decay as you always have background (again, look at the ATLAS Higgs plot, the background is overwhelming) and the peak appears on the statistical level.

 

[Dale]

Has the hollow black body mentioned above a radiation field inside? If yes, is it static? If yes, why not call it a static radiation field?

A hollow black body has a radiation field inside, but it is most definitely not static. It fluctuates thermally. I simply cannot think of a situation where what I would call "radiation" is also static. Perhaps it is a technical term which is well defined in some context that I am not familiar with, but to me "static radiation field" seems like an oxymoron.

 

[DrStupid]

but to me "static radiation field" seems like an oxymoron.

Do you have a better term for a radiation field without momentum?

 

[Dale]

How about "radiation field with 0 momentum"?

 

[ChrisVer]

I think, I am very well aware of this. This is why I said (theoretically speaking) it is stupid to think that the two-photon system has a non-zero invariant mass.

Two real photons system will have zero invariant mass... if instead you have virtual photons the invariant mass doesn't have to be zero.

 

[zoki85]

No, light doesn't "have" a mass. It has energy which corresponds to mass of E/c²

 

[bhobba]

No, light doesn't "have" a mass. It has energy which corresponds to mass of E/c²

Hmmmm.

Some people get confused with E=MC^2 thinking it says mass has energy. It says mass is a form of energy - not the other way around. It has energy and no mass - that's it - that's all.

Thanks
Bill

 

[vanhees71]

Here a lot gets confused. The reason is, as quite often, the abuse of the idea of photons. Let's take the case of black-body radiation in a cavity which is among the very few examples, where the quantum theory of free fields is sufficient for an accurat description of an observable phenomenon, namely black-body radiation.

It starts with the free electromagnetic field, which reads in radiation gauge, i.e., the gauge, where $A^0=0$ and $\vec{\nabla} \cdot \vec{A}=0$ on the classical level, and with this gauge constraint you quantize the radiation field. This is most conveniently done in terms of Fock space of the momentum-single-photon space for photons in a cavity, which we choose as a cubic box of length $L$ and impose periodic boundary conditions (you could as well work with rigid boundary conditions, but that's less convenient). The field operator reads
$$\hat{A}^{\mu}(x)=\frac{1}{\sqrt{L^3}} \sum_{\lambda=\pm 1} \sum_{\vec{p}} [\hat{a}(\vec{p},\lambda) \vec{\epsilon}(\vec{p},\lambda) \exp(-[\mathrm{i} \omega t+\mathrm{i} \vec{x} \cdot \vec{k}]+\text{h.c.}].$$
Here $p^0=|\vec{p}|$ (massless (!) photons). The $\lambda$ labels the helicity, i.e., sums over the two circular polarized modes of the em. fields, and $\vec{p} \in \frac{2 \pi}{L}$.
The energy (Hamilton) and momentum operators are calculated from the normal ordered Belinfante energy-momentum tensor (or equivalently from the canonical energy momentum tensor, because total energy and momentum do not depend on this):
$$\hat{P}^{\mu} =\sum_{\vec{p},\lambda} \hat{N}(\vec{p},\lambda) p^{\mu}, \quad \text{with} \quad \hat{N}(\vec{p},\lambda)=\hat{a}^{\dagger}(\vec{p},\lambda) \hat{a}(\vec{p},\lambda).$$
The Hilbert space is spanned by the Fock-space basis vektors
$$|\{N(\vec{p},\lambda) \} \rangle = \prod_{\vec{p},\lambda} \frac{1}{\sqrt{N(\vec{p},\lambda)!}} (\hat{a}^{\dagger}(\vec{p},\lambda))^{N(\vec{p},\lambda)} |\Omega \rangle,$$
where $|\Omega \rangle$ is the vacuum state, which obeys $\hat{a}(\vec{p},\lambda)|\Omega \rangle =0$ for all $(\vec{p},\lambda)$.

Now we can look at the equilibrium state. Since photons carry no conserved charges, we work with the canonical ensemble. The covariant statistical operator for a non-rotating state (0 total angular momentum) reads
$$\hat{R}=\frac{1}{Z} \exp(-u \cdot \hat{P}).$$
Here, $u$ is a time-like vector, appearing as Lagrange parameters for the maximum-entropy principle to keep the average energy and momentum as constraints.

The partition sum is calculated with the above Fock-space (occupation number) basis
$$Z=\mathrm{Tr} \exp(-u \cdot \hat{P})=\prod_{\vec{p},\lambda} \frac{1}{1-\exp(-u \cdot p)}.$$
In the limit of a very big container this partition sum can be expressed in the form
$$\ln Z=-2 \frac{V}{(2 \pi)^3} \int_{\mathbb{R}^3} \ln[1-\exp(-u \cdot p)].$$
The factor $2$ comes from the two helicity states for each mode of the em. field.

The partition sum can be evaluated exactly. Without loss of generality we can assume that $\vec{u}=u_3 \vec{e}_z$ and introducing spherical coordinates gives
$$\ln Z=-2 \frac{V}{(2 \pi)^2} \int_0^{\infty} \mathrm{d} p \int_{-1}^1 \int_{-1}^1\mathrm{d} \eta \ln[1-\exp[-\beta p(u^0-u_3 \eta)]] = \frac{V u_0 \pi^2}{45 (u \cdot u)^2}.$$
On the first glance this looks not manifestly covariant, but the reason is that we evaluated the partition sum in the lab frame, where the cavity with the radiation inside is moving.

We can make the partition sum manifestly covariant, by expressing everything with respect to the restframe of the cavity (which is at the same the of course the frame where the radiation inside the cavity has total 0 momentu either). To set end we set $u=(\beta,0,0,0)$. In the general case $u=\beta/\sqrt{1-v^2}(1,\vec{v})$. Then we have to take into account the fact that the length contraction gives $$V=V_0 \sqrt{1-v^2}=V_0 \frac{\beta}{u^0}.$$
This gives the canonical potential
$$\Omega=-\ln Z=\frac{\pi^2 V_0}{45 (u \cdot u)^{3/2}},$$
where we have written $u \cdot u=\beta^2$. This has the advantage that we can take derivatives with independent variables $u_{\mu}$ to get the average energy and momentum. At the end of the calculation we use $u \cdot u=\beta^2$ again to express everything in the scalar quantities $V_0$ and $\beta=1/T$ ($T$: temperature). This gives finally
$$\langle P^{\mu}_{\text{rad}} \rangle=-\frac{\partial}{\partial u_{\mu}}=\frac{\pi^2 V_0}{45 \beta^4} \frac{1}{\sqrt{1-v^2}} (1,\vec{v}).$$
This shows that the total mass of the radiation is given by its energy in the rest frame, which is
$$m_{\text{rad}}=E_{0,\text{rad}}=\frac{\pi^2 V_0}{45 \beta^4},$$
as is well known from textbooks.

This reflects the true meaning of the famous formula "$E=m$", which of course refers to the rest frame of the moving massive body.

The em. field itself is, of course, a massless field according to the Standard Model of particles, and this assumption is one among the best experimentally tested ones in whole physics.

 

[PeterDonis]

Two real photons system will have zero invariant mass...

Not if the photons are moving in different directions. Invariant mass is not additive: a system composed of multiple particles can have an invariant mass that is different from the sum of the inariant masses of the particles.

 

[Nugatory]

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