Does Light Have Mass? Astrophysics Discussion

[Young Seeker]

I am new in this forum, and I'm just getting started. I am a very big admirer of astrophysics and I'm seeking answers and at the same time proposing new possibilities for some of the awesome mysteries of the universe. I'm young, so sometimes there is a great possibility tht i might add some stupid ideas, please understand and forgive me on such occasions :)

Ok, so to my first question,

Does light have mass?

I have thought a lot abt it and i can't find a better view to see light as a unique, not yet identified substance, which can at some conditions exhibit mass, and at other conditions, massless

Please share ur views abt this...

 

[Dale]

Hi Young Seeker, welcome to PF!

You may want to read the several relevant FAQ posts that we have on the topic:
https://www.physicsforums.com/threads/do-photons-have-mass.511175/
https://www.physicsforums.com/threads/how-can-light-have-momentum-if-it-has-zero-mass.512541/
https://www.physicsforums.com/threads/why-are-massless-photons-affected-by-gravity.511173/

 

[Drakkith]

Welcome to PF! First and foremost, I think you should understand that PF's primary goal is to help people understand physics as it is currently understood and practiced by mainstream professional scientists. Personal ideas and theories are not usually acceptable here because they typically run counter to the mainstream view and thus counter to PF's primary goal.

So while we can help you understand why light is massless according to current science, we cannot discuss any personal ideas you may have on it.

So to answer your question, light is massless when it exists as a freely propagating electromagnetic wave. I've heard of strange things happening when light travels in certain mediums, but as far as I know these phenomena are not described as normal light, but as an interaction between the EM wave and the medium giving rise to these effects. For the vast majority of cases you can consider light to be massless.

 

[Young Seeker]

ok, thanks for the reply, but can u tell me how can an objest, being massless interact with something that has mass? ( I'm talking about light getting reflected off a surface)

 

[ZapperZ]

ok, thanks for the reply, but can u tell me how can an objest, being massless interact with something that has mass? ( I'm talking about light getting reflected off a surface)

First of all, what rule is there that something must have mass for it to interact with something that has mass? Do you think an electric field has a mass? After all, that can interact with anything with a charge.

It is a good time here to re-examine your assumption.

Zz.

 

[Orodruin]

Why would you think that they could not? Physics is a descriptive science where we introduce theories that describe what we can observe and we observe that light interacts with matter so naturally we create theories where this happens. We do not seek any deeper philosophical reason. Then you can of course ask why they interact in a given theory, in the case of light it is described by Maxwell's equations and the Lorentz force law.

 

[Drakkith]

Sure. Light is an Electromagnetic wave. This means that light is a fluctuation within the EM field. What's fluctuating you may ask? The answer is something we call a "field vector". A field vector is a mathematical way of describing how a charged particle will interact with a field. In the case of the electric field, this field vector tells us which way the field is pointing and what the magnitude of the field is at this location. In other words, a field vector tells us which way a charged particle will move in response to the field and how much force is applied to the particle by the field.

Since an EM wave is desribed as a fluctuating field vector, this means that the wave exerts a force on a charged particle that alternates in direction and varies in strength as the wave passes over it the particle.

In other words, light is an EM wave of alternating force that acts on charged particles. No mass is required for the EM wave to interact with matter.

 

[Dale]

how can an objest, being massless interact with something that has mass

Interactions have very little to do with whether or not something has mass, and more to do with whether or not it has charge. Photons are the carriers of the EM force so they interact with things that have EM charge. Gluons are the carriers of the strong force, so they interact with things that have color charge. And so forth.

 

[Young Seeker]

hmm, thank you very much guys, for your useful feedbacks :)

 

[DrStupid]

Does light have mass?

That depends on the specific conditions. Due to

E^2 = m^2 c^4 + p^2 c^2

light is massless for p=E/c (e.g. in the case of a single photon). Otherwise (e.g. in the case of a stationary radiation field) it has mass.

 

[PeterDonis]

Otherwise (e.g. in the case of a stationary radiation field) it has mass.

Can you explain? This doesn't seem right.

 

[DrStupid]

Can you explain?

Let's take a very simple example: Two photons with equal energy moving in opposite directions. This system has energy but no momentum and according to the equation above it's mass is m=E/c². Light is only massless if all photons are moving in the same direction. In any other case it has mass.

 

[Matterwave]

Let's take a very simple example: Two photons with equal energy moving in opposite directions. This system has energy but no momentum and according to the equation above it's mass is m=E/c². Light is only massless if all photons are moving in the same direction. In any other case it has mass.

In what practical context would you ever consider this 2 body system together rather than simply treating the photons individually? This is not a bounded system, so I don't see how one could, for example, apply a force to the center of mass of this system.

 

[Orodruin]

In what practical context would you ever consider this 2 body system together rather than simply treating the photons individually? This is not a bounded system, so I don't see how one could, for example, apply a force to the center of mass of this system.

I would plot the event rate as a function of this "mass" in the ATLAS and CMS detectors, to find the Higgs. But the invariant mass of the two photons aside, this does not mean that the photons (or light) has mass.

 

[Matterwave]

I would plot the event rate as a function of this "mass" in the ATLAS and CMS detectors, to find the Higgs. But the invariant mass of the two photons aside, this does not mean that the photons (or light) has mass.

You mean collide the two photons and hope they have enough energy to produce a Higgs? Seems unlikely even if the two photons were both gamma rays...and what if the two photons don't have enough CoM energy to produce a Higgs?

In addition, I have no idea what this has to do with the "mass" of the 2 photon system?

 

[A.T.]

In what practical context would you ever consider this 2 body system together rather than simply treating the photons individually? This is not a bounded system, so I don't see how one could, for example, apply a force to the center of mass of this system.

Put the photons into an ideal mirror box.

 

[Matterwave]

Put the photons into an ideal mirror box.

So then you've created a bounded system. I have no problem with that and calling it the mass of the bounded system, box+photons. This is the same concept as a hydrogen particle in a higher energy state should have slightly more mass than a hydrogen particle in a lower energy state. Or that a hot box is slightly heavier than a cold, identical in every other way, box. However, I don't see any practical reason to consider two free photons traveling in different directions as one system (in particular, to consider the invariant mass of such a system), but I can of course be missing something!

 

[Orodruin]

You mean collide the two photons and hope they have enough energy to produce a Higgs? Seems unlikely even if the two photons were both gamma rays...and what if the two photons don't have enough CoM energy to produce a Higgs?

In addition, I have no idea what this has to do with the "mass" of the 2 photon system?

The Higgs was discovered mainly in the decay channel $H\to \gamma\gamma$. By observing the photon pairs and computing their invariant mass, an excess over background was found around 125 GeV. Of course, this is the mass of the Higgs, not the mass of the photons, but it does have physical significance even if the photons are not bound.

 

[PeterDonis]

This system has energy but no momentum and according to the equation above it's mass is m=E/c².

Ah, ok. Was this sort of thing what you meant by a "stationary radiation field"? I ask because an actual radiation field doesn't have a description anything like this; what you've described is a highly idealized system.

 

[Matterwave]

The Higgs was discovered mainly in the decay channel $H\to \gamma\gamma$. By observing the photon pairs and computing their invariant mass, an excess over background was found around 125 GeV. Of course, this is the mass of the Higgs, not the mass of the photons, but it does have physical significance even if the photons are not bound.

You mean by computing the CoM energy and then setting $m\equiv E_{com}/c^2$?

EDIT: I guess I should have been clearer in this question. I meant, did they somehow find the invariant mass of the system of both photons (in one measurement), or did they find the Energy of each individual photon separately, but in the center of mass frame and simply do the above calculation?

In other words, were the 2 photons treated as a system, or treated separately, but in one inertial frame?

 

[Orodruin]

You mean by computing the CoM energy and then setting $m\equiv E_{com}/c^2$?

Yes.

EDIT: I guess I should have been clearer in this question. I meant, did they somehow find the invariant mass of the system of both photons (in one measurement), or did they find the Energy of each individual photon separately, but in the center of mass frame and simply do the above calculation?

In other words, were the 2 photons treated as a system, or treated separately, but in one inertial frame?

No, clearly you have to measure the photons separately. However, from those measurements you can obtain the invariant mass of the two-photon system (in fact, both collaborations have used $m_{\gamma\gamma}$ as the label of their x-axis). This is of course a property of a system of any number of particles (as long as everything is not moving in the same direction, the total 4-momentum will correspond to an invariant mass, the CoM energy) and a photon gas is going to be different from a gas of massive particles (unless the massive gas has a temperature high enough for the mass to be irrelevant) in the sense that its pressure is going to be comparable to its energy density.

 

[DrStupid]

Was this sort of thing what you meant by a "stationary radiation field"?

I mean a radiation field without momentum. A practical example would be the black body radiation inside a hollow body. The full energy of this radiation contributes to the mass of the system.

 

[PeterDonis]

I mean a radiation field without momentum. A practical example would be the black body radiation inside a hollow body.

Ok, but this only works because the radiation is confined by something that isn't radiation, so that the system as a whole (a) is bound, and (b) has a well-defined rest frame in which its net momentum vanishes.

As a physical model, this is fine; but as a response to the OP's question, I think it's a little misleading. The fact that light (or radiation in general) can be one component of a system that, considered as a whole, has nonzero mass, does not mean light, considered as a fundamental thing, has nonzero mass. The electromagnetic field is a fundamentally massless field, and that doesn't change when it's part of a system.

 

[jtbell]

Using the invariant mass defined by the relation $E^2 = (mc^2)^2 + (pc)^2$, the mass of a system generally does not equal the sum of the masses of its components. Radiation confined in a container is one example of this.

 

[DrStupid]

but this only works because the radiation is confined by something that isn't radiation

In case of the cosmic background radiation it works without confinement.

 

[PeterDonis]

In case of the cosmic background radiation it works without confinement.

Yes, in the sense that at any event there is a local "rest frame" in which the CMBR is isotropic. But you can't extend that globally because of the universe's expansion.

 

[DrStupid]

But you can't extend that globally because of the universe's expansion.

You don't need to extend it globally. Just select a finite volume. There is always a frame of reference where the containing background radiation has no momentum but there is no frame of reference where it has no mass. That's not limited to background radiation or black body radiation. It applies to almost every radiation. Light without mass is an exceptional case (e.g. single photons or laser) or just a good approximation (e.g. for sunlight).

 

[PeterDonis]

Just select a finite volume. There is always a frame of reference where the containing background radiation has no momentum

No, there isn't, because if you select a finite volume that's large enough for spacetime curvature to be significant, there is no well-defined "momentum" at all. Defining "momentum" for an extended system (and hence defining an overall 4-momentum vector for the CMBR, so you can take its invariant length and verify that it's nonzero) requires flat spacetime--or at least asymptotically flat spacetime. The universe, on distance scales large enough for curvature to be significant, is neither.

(Also, you haven't responded at all to the main point of my post #23, which is that the EM field is a fundamentally massless field, and that's what the OP was really asking about.)

 

[Matterwave]

Yes.
No, clearly you have to measure the photons separately. However, from those measurements you can obtain the invariant mass of the two-photon system (in fact, both collaborations have used $m_{\gamma\gamma}$ as the label of their x-axis). This is of course a property of a system of any number of particles (as long as everything is not moving in the same direction, the total 4-momentum will correspond to an invariant mass, the CoM energy) and a photon gas is going to be different from a gas of massive particles (unless the massive gas has a temperature high enough for the mass to be irrelevant) in the sense that its pressure is going to be comparable to its energy density.

Then it seems more like a mathematical construct rather than a concrete "thing" (mass) that we are used to. Of course, this is beginning to head over to the philosophy side, so I'll just drop it. ;)

 

[Dale]

In what practical context would you ever consider this 2 body system together rather than simply treating the photons individually?

I have done this in positron emission tomography (PET). The momentum of the electron-positron pair is conserved in the photon pair. This results in a small deviation from the ideal line of response and is one of the fundamental factors limiting the spatial resolution of PET (although not the dominant factor).

 

[DrStupid]

No, there isn't, because if you select a finite volume that's large enough for spacetime curvature to be significant, there is no well-defined "momentum" at all.

There is currently no indication for a global space-time curvature.

the EM field is a fundamentally massless field

That would require that the absolute values of momentum density and energy density differ by the factor c only. Are you really sure that this is always the case?

, and that's what the OP was really asking about.

The OP asks about the mass of light.

 

[PeterDonis]

There is currently no indication for a global space-time curvature.

No, there is currently no indication of a global spatial curvature. Big difference.

That would require that the absolute values of momentum density and energy density differ by the factor c only.

No, it would require that the fundamental field that appears in the Lagrangian has zero mass. You can "build" objects with nonzero mass out of this fundamental field, but that doesn't change the fact that the field itself is massless.

The OP asks about the mass of light.

Then I guess the answer depends on what the OP meant by "light". I (and, it seems, everyone else except you who is participating in this thread) think the OP meant the fundamental EM field. You think he meant something else. I guess we'll have to see how the OP responds, if he's still watching the thread.

 

[DrStupid]

That would require that the absolute values of momentum density and energy density differ by the factor c only. Are you really sure that this is always the case?

No, [...]

E²/c² = m²c² + p² does not apply to EM fields? That's hard to believe.

You think he meant something else.

I always expect that everybody means what he writes.

 

[PeterDonis]

E²/c² = m²c² + p² does not apply to EM fields?

It applies to the fundamental EM field, sure (the one that appears in the QED Lagrangian), if it's viewed in its particle aspect (photons), with $m = 0$. 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?

 

[Dale]

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.

 

[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.