Light building the standard model

[elegysix]

Something has bothered me for a long time about our model of light: that it has no mass.
Just for a moment, forget what you know about light and let's call it the unknown, X, and suppose we want to model it.

Here's what we know for certain:
X has momentum, as seen by radiation pressure
X is affected by gravity, as seen by gravitational lensing
X can act as a particle and a wave

So we take something with momentum that is affected by gravity and we can easily conclude that it has mass. And since it has mass, we can easily say that it must be a particle... And systems of particles can show wave behavior.

So then we conclude that light is a particle with mass.

How exactly did we come to the conclusion that Light has no mass?

 

[Pengwuino]

How did you come to the conclusion that light has mass?

X has momentum, which only means it obeys conservation laws. X is affected by gravity, which only means that it has energy as per general relativity. X acts as a wave/particle, how does one conclude anything about its mass from this? How do you conclude that photons have mass from ANY of those 3 criteria?

You're just taking an elementary knowledge and trying to claim it must generalize to everything else, which is wrong. What it really should be telling you is that momenta, gravity, and wave/particle duality are more complicated than you think.

 

[ecneicS]

Something has bothered me for a long time about our model of light: that it has no mass.
Just for a moment, forget what you know about light and let's call it the unknown, X, and suppose we want to model it.

Here's what we know for certain:
X has momentum, as seen by radiation pressure
X is affected by gravity, as seen by gravitational lensing
X can act as a particle and a wave

So we take something with momentum that is affected by gravity and we can easily conclude that it has mass. And since it has mass, we can easily say that it must be a particle... And systems of particles can show wave behavior.

So then we conclude that light is a particle with mass.

How exactly did we come to the conclusion that Light has no mass?

I guess one way of answering your question is... experimental data. Light is nothing more than a ripple in the electromagnetic field, why should that have mass?

 

[Dale]

How exactly did we come to the conclusion that Light has no mass?

By measuring it very carefully.

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#photon_mass

 

[Pengwuino]

Actually yah, I should point out my post was more of the theory side. Experimentally the upper limit on the photon mass is on the order of 10^-54 kg.

http://pdg.lbl.gov/2009/tables/rpp2009-sum-gauge-higgs-bosons.pdf

 

[I like Serena]

How exactly did we come to the conclusion that Light has no mass?

The mass of any object increases if its speed increases to near the speed of light.
At the speed of light an object would have infinite mass, which is not possible.
Hence a photon can not have rest mass.

However photons do have a relativistic mass given by E=mc².
This is called the mass-energy-equivalence relation.
That is, relativistic mass is seen as equivalent to energy.

 

[elegysix]

How did you come to the conclusion that light has mass?

X has momentum, which only means it obeys conservation laws. X is affected by gravity, which only means that it has energy as per general relativity.

Because momentum has been defined as mV in classical mechanics, and only things with mass are affected by gravity, (F=GMm/R^2...). That gives you two different reasons to think it has mass.

Everything physically tangible obeys classical mechanics... but when we look at light and anything on the atomic level, we throw out the logic of classical mechanics. Applying the logic from classical mechanics, it is obvious to me that it has mass. Yet I know that this is not so. How did we decide that classical mechanics did not apply here?

I would like to get at the history of this model, how we came to this conclusion.


an upper limit on photon mass? Why would they try to measure it if they are sure it doesn't exist?

 

[Pengwuino]

The mass of any object increases if its speed increases to near the speed of light.
At the speed of light an object would have infinite mass, which is not possible.
Hence a photon can not have rest mass.

However photons do have a relativistic mass given by E=mc².
This is called the mass-energy-equivalence relation.
That is, relativistic mass is seen as equivalent to energy.

That's not exactly true. The 'universal speed limit', as one might want to put it, is the ~3x10^8m/s speed limit we know of as the speed of light. What that speed is is actually the speed of a massless particle. The photon happens to be the only massless particle we know of at this point, so people typically say speed of light when technically, you should say "speed of a massless particle". The photon could have a mass and it would be so small that the deviation from the "speed of a massless particle" would be so insignificant that it would be completely undetectable in every experiment we've ever done at this point.

 

[elegysix]

The mass of any object increases if its speed increases to near the speed of light.

This is a major part of what I am questioning. From classical mechanics alone, I see no reason to believe this. We know classical mechanics works. So starting with classical mechanics, How do you figure that these things happen?

 

[Pengwuino]

Because momentum has been defined as mV in classical mechanics, and only things with mass are affected by gravity, (F=GMm/R^2...). That gives you two different reasons to think it has mass.

No, that is not how momentum is defined in classical mechanics. That is how momentum is defined in a very specific case. Something as simple as angular momentum is not defined like that. Get into classical electrodynamics and you'll see even more complicated forms of momenta. So in general, P = mv is NOT the whole story.

Everything physically tangible obeys classical mechanics... but when we look at light and anything on the atomic level, we throw out the logic of classical mechanics. Applying the logic from classical mechanics, it is obvious to me that it has mass. Yet I know that this is not so. How did we decide that classical mechanics did not apply here?

We didn't just decide one day to throw out the logic of classical mechanics. Many many experiments were able to show that classical mechanics just makes no sense at what we now consider the atomic and quantum scale. For example, Rutherford (don't quote me on who actually did this, my memory of physics history is appalling) showed that the atom was in fact, a hard solid object (the proton/nucleus). So classical mechanics makes one assume that maybe electrons and protons are like planets, the electron orbits the proton in the same way the Earth orbits the Sun. Well, one of the problems with that is that accelerating charges radiate away energy. So applying classical mechanics and classical electrodynamics to the atom, the electron should radiate away all its energy and collapse into the proton within a fraction of a second.

Clearly that is not what happens. So classical mechanics is not the way to approach things at the quantum scale. Of course, there are a LOT of other ways to show classical mechanics does not work on the atomic scale, but I wouldn't want to spell them all out.

an upper limit on photon mass? Why would they try to measure it if they are sure it doesn't exist?

Because that's how science works. No one is sure about anything. All we can say is that "X theory holds true up to an experimental limit of Y". All our experiments show that the photon has pretty much no mass and we can build our theories safely by assuming it has no mass. If it turns out it DID have mass, the consequences would be so small that it would have almost no bearing on present day physics.

Another example is, like you mentioned, Newton's law of gravitation. We don't actually know for sure that the law goes like F = {{GMm}\over{r^2}}. One could say, well, what if Newton's gravitational law goes like F = {{GMm}\over{r^{2+\epsilon}}} where \epsilon is incredibly small. You could measure the orbits of the planets and be able to put upper or lower limits on what that \epsilon could be. Obviously since we know Newton's law of gravitation works very well, we know it would have to be very close to 0 so that it wouldn't be detectable over all these centuries.

Of course, general relativity already shows us that that form of gravity is not exactly right either, but that's kind of another way of looking at how we talk about theories vs. what our experiments tell us.

 

[Dale]

only things with mass are affected by gravity, (F=GMm/R^2...).

That is not even true in classical Newtonian mechanics. Only test particles with mass have a force on them, but classically it doesn't take any force to accelerate a massless particle. The acceleration due to gravity is independent of the mass of the test particle.

an upper limit on photon mass? Why would they try to measure it if they are sure it doesn't exist?

That's the scientific method. You have a theory, the theory makes some prediction, so you perform an experiment to test the hypothesis.

So we have measured it over and over with increasing precision, and to the best precision possible it is 0.

 

[DeG]

"X has momentum, which only means it obeys conservation laws. X is affected by gravity, which only means that it has energy as per general relativity." - Pengwuino
"However photons do have a relativistic mass given by E=mc2." - I Like Serena
If it's true we measure the momentum of light via radiation pressure, does this pressure increase with the energy/ frequency of light? Since light (affected by gravity) has energy as per general relativity, a relativistic mass given by E=mc^2, and energy given by E=hf, could you say that light has a mass proportional to it's frequency given by m=hf/c^2?

 

[Drakkith]

Don't think of relativistic mass as actual mass. The momentum of an object increases as it's velocity increases, but its mass (IE it's rest and gravitational mass) do not increase to my knowledge. I believe it is part of how people tend to explain the fact that you can't get to c, because your mass increases. However I think this is inaccurate. If I'm incorrect someone let me know please.

 

[Pengwuino]

"X has momentum, which only means it obeys conservation laws. X is affected by gravity, which only means that it has energy as per general relativity." - Pengwuino
"However photons do have a relativistic mass given by E=mc2." - I Like Serena
If it's true we measure the momentum of light via radiation pressure, does this pressure increase with the energy/ frequency of light? Since light (affected by gravity) has energy as per general relativity, a relativistic mass given by E=mc^2, and energy given by E=hf, could you say that light has a mass proportional to it's frequency given by m=hf/c^2?

No. Photons (assumed to be massles) do not have relativistic masses either, contrary to what I Like Serena stated. The relativistic mass of a particle is given by m = \gamma m_0 where m_0 is the rest mass of the particle and \gamma = {{1}\over{\sqrt{1-{{v^2}\over{c^2}}}}}. Photons do not have a rest mass and their velocity is equal to c, so this equation is clearly ill-defined.

The momentum of light is given by the full relativistic energy equation, E^2 = p^2c^2 + m_0^2c^4. Light has no rest mass so the second term on the right hand side is 0 and solving for the momentum, you find it is given by p = {{E}\over{c}}. The energy of a photon is given by E = \hbar \omega where \omega is the angular frequency. So the momenta is propotional to the frequency, but there is no mass to speak of.

 

[elegysix]

Here's what I'd like to see...

solve for the mass by calculating the momentum from radiation pressure.

Take that mass at the speed of light and solve for the trajectory in the vicinity of a planet (non-relativistically, through classical mechanics).

Then check the trajectory against that of current model.

If it doesn't match up, clearly we'd be wrong about assigning a mass to a photon...

If you've got the time and the curiosity, I'd really like to see this done. I've tried over and over again, but I always get stuck on some of the math... :(

 

[Char. Limit]

Here's what I'd like to see... solve for the mass by calculating the momentum from radiation pressure.
Take that mass at the speed of light and solve for the trajectory in the vicinity of a planet (non-relativistically, through classical mechanics). Then check the trajectory against that of current model. If it doesn't match up, clearly we'd be wrong about assigning a mass to a photon...

If you've got the time and the curiosity, I'd really like to see this done. I've tried over and over again, but I always get stuck on some of the math... :(

Of course we'd be wrong about assigning a mass to a photon. That's why we DON'T. We instead say that a photon is massless. This happens to be consistent with all known experiments to all known precisions.

 

[Pengwuino]

Here's what I'd like to see...

solve for the mass by calculating the momentum from radiation pressure.

Take that mass at the speed of light and solve for the trajectory in the vicinity of a planet (non-relativistically, through classical mechanics).

Then check the trajectory against that of current model.

If it doesn't match up, clearly we'd be wrong about assigning a mass to a photon...

If you've got the time and the curiosity, I'd really like to see this done. I've tried over and over again, but I always get stuck on some of the math... :(

That's probably already been done. The easiest comparison would be to compare against the deflection of light given GR's calculation. This is common enough knowledge. Then do the calculation of a massive particle being deflected by the Sun in some 1/r potential.

There are some fairly simple reasons why I expect this won't even accidentally be right considering GRs calculation is independent of the photon's momentum whereas the deflection in this "classical mechanics" calculation would have to depend on the momentum. Of course, GR has been confirmed experimentally.

 

[I like Serena]

No. Photons (assumed to be massles) do not have relativistic masses either, contrary to what I Like Serena stated. The relativistic mass of a particle is given by m = \gamma m_0 where m_0 is the rest mass of the particle and \gamma = {{1}\over{\sqrt{1-{{v^2}\over{c^2}}}}}. Photons do not have a rest mass and their velocity is equal to c, so this equation is clearly ill-defined.

The momentum of light is given by the full relativistic energy equation, E^2 = p^2c^2 + m_0^2c^4. Light has no rest mass so the second term on the right hand side is 0 and solving for the momentum, you find it is given by p = {{E}\over{c}}. The energy of a photon is given by E = \hbar \omega where \omega is the angular frequency. So the momenta is propotional to the frequency, but there is no mass to speak of.

So how do you calculate the momentum p of an object then?

I thought you calculate p from E^2=p^2 c^2 + m_0^2 c^4 by combining it with E=m c^2 and m = \gamma m_0.
For low speeds this can be approximated with p=mv.

Since the rest mass of a photon is zero (or insignificant ), the formula m = \gamma m_0 breaks down, giving an indeterminate result.
However, for a photon you can calculate the energy from the formula E = \hbar \omega.

Afaik, the mass-energy-equivalence relation does not break down.
So a photon has a relativistic mass of m=p/c (classically speaking).
And yes, a photon has to have equivalent mass (classically speaking) for gravity to act on it.

 

[Pengwuino]

So how do you calculate the momentum p of an object then?

I thought you calculate p from E^2=p^2 c^2 + m_0^2 c^4 by combining it with E=m c^2 and m = \gamma m_0.
For low speeds this can be approximated with p=mv.

Since the rest mass of a photon is zero (or insignificant ), the formula m = \gamma m_0 breaks down, giving an indeterminate result.
However, for a photon you can calculate the energy from the formula E = \hbar \omega.

Afaik, the mass-energy-equivalence relation does not break down.
So a photon has a relativistic mass of m=p/c (classically speaking).
And yes, a photon has to have equivalent mass (classically speaking) for gravity to act on it.

Yes, that's how you calculate the momentum of a particle with some rest mass m_0. It's just instead of the energy given by E = mc^2 = \gamma m_0 c^2, the energy is given by E = \hbar \omega. E = \gamma m_0 c^2 only works for massive particles. The right hand side of the relativistic energy equation is fine, the left side is the one you tweak for photons (or equivalently you can define the momenta of a photon as \hbar \omega /c).

And yes, you can calculate an "equivalent classical mass", but it's meaningless beyond the one scenario of gravitational pull. Hell, I don't even think it says anything about the actual mass of a photon beyond that because the photon's mass might be irrelevant due to Equivalence.

You can do a quick calculation and find any normal photon has a "mass" 20 orders magnitude higher than the current experimental upper limit.

 

[Dale]

Take that mass at the speed of light and solve for the trajectory in the vicinity of a planet (non-relativistically, through classical mechanics).

Then check the trajectory against that of current model.

If it doesn't match up, clearly we'd be wrong about assigning a mass to a photon.

This wouldn't tell you anything about the mass. Remember, gravitational acceleration is independent of the mass. The trajectory depends only on the initial velocity, not the mass. This experiment would allow you to distinguish between different models of gravity, but not different masses of the test particle.

In any case, since we have measured it's mass to be 0, then clearly we would be wrong about assigning a mass to a photon in any case.

 

[Dale]

And yes, a photon has to have equivalent mass (classically speaking) for gravity to act on it.

No it doesn't. This is a common fallacy. The acceleration due to gravity is independent of the mass classically.

 

[I like Serena]

And yes, you can calculate an "equivalent classical mass", but it's meaningless beyond the one scenario of gravitational pull. Hell, I don't even think it says anything about the actual mass of a photon beyond that because the photon's mass might be irrelevant due to Equivalence.

You can do a quick calculation and find any normal photon has a "mass" 20 orders magnitude higher than the current experimental upper limit.

No it doesn't. This is a common fallacy. The acceleration due to gravity is independent of the mass classically.

For a gravitational lens the equivalent mass of a photon is indeed irrelevant.
And I am aware that classical mechanics does not quite predict the right angle, although I seem to recall it is not an order of a magnitude off.

But suppose you make a box containing perfect mirrors containing many photons.
How much would the box weigh?
Doesn't the mass-energy-equivalence relation give you this?

And what kind of quick calculation would find a "mass" 20 orders magnitude higher?
In what context?

 

[elegysix]

The trajectory depends only on the initial velocity, not the mass.

Ummm so what you're saying is that the acceleration is given by a=GM/r^2, which accelerates even massless objects? Wouldn't this imply that F=GMm/r^2 is wrong?

 

[russ_watters]

Ummm so what you're saying is that the acceleration is given by a=GM/r^2, which accelerates even massless objects? Wouldn't this imply that F=GMm/r^2 is wrong?

No, you can derive the second from the first by substituting F=ma.

Bottom line here for your initial question is simple: the assumption that the photon has [measurable] mass just plain doesn't match observations.

Anyway, that was one of the great triumph's of early physics, the idea that acceleration due to gravity was not dependent on the mass of the dropped object.

 

[elegysix]

I think i meant to say that dividing by m you get the relations
a= F/m=GM/r^2
letting m=0 implies a=inf, GM/r^2=inf, and F=0,
but also you get the relation a=GM/r^2, which contradicts a=inf

but how can we accelerate something without a force? that doesn't seem to make sense.

 

[Drakkith]

I think i meant to say that dividing by m you get the relations
a= F/m=GM/r^2
letting m=0 implies a=inf, GM/r^2=inf, and F=0,
but also you get the relation a=GM/r^2, which contradicts a=inf

but how can we accelerate something without a force? that doesn't seem to make sense.

That's because photons are not accelerated. The always move at c. Hence one reason why gravity is thought of as a curvature of space and not a force like the other forces.

 

[Pengwuino]

For a gravitational lens the equivalent mass of a photon is indeed irrelevant.
And I am aware that classical mechanics does not quite predict the right angle, although I seem to recall it is not an order of a magnitude off.

But suppose you make a box containing perfect mirrors containing many photons.
How much would the box weigh?
Doesn't the mass-energy-equivalence relation give you this?

And what kind of quick calculation would find a "mass" 20 orders magnitude higher?
In what context?

After some discussion with someone else, Einstein (amongst many others), did do the calculation and to first order, the deflection is 1/2 of what it should be.

As for the "mass" calculation, the mass should naively be m = p/c. Figure a 1x10^9 Hz photon or something and you can show the mass is way way above the experimental upper bound.

 

[Pengwuino]

I think i meant to say that dividing by m you get the relations
a= F/m=GM/r^2
letting m=0 implies a=inf, GM/r^2=inf, and F=0,
but also you get the relation a=GM/r^2, which contradicts a=inf

but how can we accelerate something without a force? that doesn't seem to make sense.

You do realize you're going to run into a lot of things that make no sense if you refuse to believe there is a purpose to general relativity, right?

That's because photons are not accelerated. The always move at c. Hence one reason why gravity is thought of as a curvature of space and not a force like the other forces.

And to preempt one who might ask "Then how does light curve if it's not accelerated", accelerations are basically deviations from what are called 'geodesics' . Since energy/momentum curves space-time, the path a photon would travel in flat-space looks different than it would if there were a sun/planet nearby curving the space-time.

 

[Dale]

Ummm so what you're saying is that the acceleration is given by a=GM/r^2, which accelerates even massless objects?

Yes.

Wouldn't this imply that F=GMm/r^2 is wrong?

Nope. It only implies that it doesn't take any force to accelerate a massless object. Classically this is clear by f=ma.

 

[Dale]

I think i meant to say that dividing by m you get the relations
a= F/m=GM/r^2
letting m=0 implies a=inf, GM/r^2=inf, and F=0,
but also you get the relation a=GM/r^2, which contradicts a=inf

Your math is a little off. 0/0 is not infinite, it is undefined.

 

[Dale]

But suppose you make a box containing perfect mirrors containing many photons.
How much would the box weigh?
Doesn't the mass-energy-equivalence relation give you this?

A single photon is massless, but a system of two or more photons may have mass.

Here I am referring to the invariant mass which is the norm of the four-momentum. Since the four-momentum is conserved, so is the invariant mass.

 

[elegysix]

I guess I'm just too ingrained with classical mechanics... in that course, everything was straight forward. A few basic rules to follow, and you can model the behavior of just about any system of masses and forces you can imagine, with the only limitation being my knowledge of calculus...

but when it comes to light, If you begin to ask about even the simplest properties from mechanics, such as momentum, you get these totally different, somewhat ambiguous answers...

Light is treated completely differently, and I can't help but ask why.

I naturally assume that something must be wrong with our model, because I feel like we have come up with all of these seemingly round-about answers to these basic questions. Where as in mechanics, momentum could easily be used to determine the velocity or mass, without question, for any system.

Yet with light, this is simply "wrong".

Special sets of rules for a few particular things in our universe? Never!
"there can only be one!" lol

 

[russ_watters]

Light is treated completely differently, and I can't help but ask why.

It's pretty simple, really: light is treated totally different because it is totally different.

I naturally assume that something must be wrong with our model, because I feel like we have come up with all of these seemingly round-about answers to these basic questions.

When your understanding of those models is extremely limited, it is counterproductive and illogical to assume that thousands of experienced physicists over the past 100 years made a serious, simultaneous error. It's best to give them the benefit of the doubt and try to learn what they figured out.

Special sets of rules for a few particular things in our universe? Never!
"there can only be one!" lol

There is only one set of rules - you just haven't learned all of it yet. If you are talking about classical vs relativistic mechanics, though, its worse than you think: the only set that's right is relativistic mechanics. Newtonian physics is wrong most of the time, but it's still used because it's not too wrong for most purposes.

 

[Pengwuino]

Yet with light, this is simply "wrong".

Special sets of rules for a few particular things in our universe? Never!
"there can only be one!" lol

And remember, it's not "wrong", it's just wrong. This isn't opinion, it's simply fact.

What's more surprising is the further you go into physics, the more you find that light is not the only thing that, in your opinion, requires a "special set of rules". You're going to find that the things that can be fully understood using just classical mechanics are so few in number that you may start considering classical mechanics as a "special case".

Then again, at the end of the day, if you take your studies far enough, you'll find that there's nothing special at all about light or gluons or general relativity or what have you. They all simply are more advanced treatments of simpler ideas.

 

[I like Serena]

As for the "mass" calculation, the mass should naively be m = p/c. Figure a 1x10^9 Hz photon or something and you can show the mass is way way above the experimental upper bound.

Isn't that the experimental upper bound for the "rest mass" of a photon (not the relativistic mass)?

A single photon is massless, but a system of two or more photons may have mass.

Here I am referring to the invariant mass which is the norm of the four-momentum. Since the four-momentum is conserved, so is the invariant mass.

Hmm, interesting!

If I understand correctly (http://en.wikipedia.org/wiki/Four-momentum" ) we have -||P||^2=m_0^2 c^2 for a single object, which is zero for a photon.

The box would have: ||P||^2=-(\sum E/c)^2, since the total 3-momentum would be zero.
So the invariant mass would be: m_0=\sum E/c^2.

For photons this becomes: m_0=\sum p/c.

So the invariant mass of the box with photons would be the sum of their "relativistic masses".

Right?

Light is treated completely differently, and I can't help but ask why.

I naturally assume that something must be wrong with our model, because I feel like we have come up with all of these seemingly round-about answers to these basic questions. Where as in mechanics, momentum could easily be used to determine the velocity or mass, without question, for any system.

Yet with light, this is simply "wrong".

Special sets of rules for a few particular things in our universe? Never!
"there can only be one!" lol

The fact that there were a few difficulties with light that just wouldn't fit in the "simple-rule model", such as an invariant light speed, is the very reason relativity theory was thought up!
And when photons were found to bend in a gravitational lens just like relativity theory predicted, which was more accurate than the simple-rule model, this was one of the grand confirmations of relativity theory!

 

[Dale]

A few basic rules to follow, and you can model the behavior of just about any system of masses and forces you can imagine, with the only limitation being my knowledge of calculus.

Same with light. Just a few basic rules to follow and you can model the behavior of just about any system of masses and charges and EM fields you can imagine, with the only limitatin being your knowledge of calculus.

The rest of your post is basically nonsense. There is never any rule to science that nature should be completely described in 4 equations or less. These additional equations aren't added for the hell of it, but in order to match what is experimentally observed. If you don't like how complicated this universe is then you are certainly welcome to try to find a simpler one.

 

[Pengwuino]

Isn't that the experimental upper bound for the "rest mass" of a photon (not the relativistic mass)?

I thought I had written it down, but the upper limit of the photon mass is ~10^-51kg. That calculation should give a mass for that photon of ~ 10^-29kg or so.

 

[I like Serena]

I thought I had written it down, but the upper limit of the photon mass is ~10^-51kg. That calculation should give a mass for that photon of ~ 10^-29kg or so.

I checked back on the quoted articles from DaleSpam:
http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#photon_mass

and from you:
"Actually yah, I should point out my post was more of the theory side. Experimentally the upper limit on the photon mass is on the order of 10^-54 kg."
http://pdg.lbl.gov/2009/tables/rpp20...ggs-bosons.pdf

Interesting reads btw! Thanks for that! However, in DaleSpam's article I found:
"# Luo et al., “New Experimental Limit on the Photon Rest Mass with a Rotating Torsion Balance”, Phys. Rev. Lett, 90, no. 8, 081801 (2003).
A limit of 1.2×10−51 g (6×10−19 eV/c2)."Note the reference to "Photon Rest Mass", not just "mass" or "relativistic mass".
Relativistic mass (or perhaps I should say "energy equivalent mass") is not bounded as I understand it, it's just a matter of how much energy you put into the photon (how high its frequency is).

 

[DeG]

Could anyone give me the mathematics behind the blue/red shift of light as it travels through a gravitational potential difference?

 

[Dale]

Note the reference to "Photon Rest Mass", not just "mass" or "relativistic mass".
Relativistic mass (or perhaps I should say "energy equivalent mass") is not bounded as I understand it, it's just a matter of how much energy you put into the photon (how high its frequency is).

That is correct. When modern physicsts use the unqualified word "mass" they are generally referring to the invariant mass (aka rest mass). And specifically, that is the mass that is implied when we say that a photon is massless.

The term "relativistic mass" is not usually used any more, instead physicists usually simply speak of "total energy" which is the same as relativistic mass times c^2.

 

[Dale]

If I understand correctly (http://en.wikipedia.org/wiki/Four-momentum" ) we have -||P||^2=m_0^2 c^2 for a single object, which is zero for a photon.

Yes, where P is the four-momentum.

The box would have: ||P||^2=-(\sum E/c)^2, since the total 3-momentum would be zero.

Yes. You can see this explicitly for a pair of photons in the CoM frame:
P1=(E/c,p,0,0)
P2=(E/c,-p,0,0)
||P1+P2||=||(2E/c,0,0,0)||=2E/c

So the invariant mass of the box with photons would be the sum of their "relativistic masses".

Plus the mass of the box, and only in the rest frame of the box, yes.

 

[DeG]

You say only in the rest frame of the box, does this imply that it wouldn't take a force to "accelerate," i.e change the motion, the photons in the box if you pushed the box? In other words, to someone moving the box it would seem as though the mass of the system is just the mass of the box, not the box plus the relativistic mass of the photons?

 

[I like Serena]

No.
To someone moving the box it would seem as if the mass of the system is the mass of the box plus the relativistic masses of the photons.

 

[Dale]

You say only in the rest frame of the box, does this imply that it wouldn't take a force to "accelerate," i.e change the motion, the photons in the box if you pushed the box? In other words, to someone moving the box it would seem as though the mass of the system is just the mass of the box, not the box plus the relativistic mass of the photons?

Sorry, occasionally the math-to-english translation gets fuzzy.

In some reference frame (and in units where c=1) if you have a box with a four-momentum
P_B=(m,0,0,0)

Enclosing two photons with four-momenta
P_1=(E,-E,0,0)
P_2=(E,E,0,0)

Then the mass of the box is:
|P_B+P_1+P_2|=|(m+2E,0,0,0)|=m+2E

So the invariant mass is the sum of the mass of the box and the total energy (aka relativistic mass) of the photons in the rest frame of the box. However, in another reference frame:

P'_B=\left(\frac{m}{\sqrt{1-v^2}}, \frac{m v}{\sqrt{1-v^2}}, 0,0\right)
P'_1=\left(E\sqrt{\frac{1+v}{1-v}},-E\sqrt{\frac{1+v}{1-v}},0,0\right)
P'_2=\left(E\sqrt{\frac{1-v}{1+v}},E\sqrt{\frac{1-v}{1+v}},0,0\right)

So

m+2E \ne m + E\sqrt{\frac{1+v}{1-v}} + E\sqrt{\frac{1-v}{1+v}}

Therefore the invariant mass is not the sum of the mass of the box and the total energy of the photons in another frame.

The photons do contribute to the inertia of the box in all reference frames, but not necessarily by an amount equal to the "relativistic mass" of the photons.

 

[DeG]

I see. Thank you.

 

[I like Serena]

Yes, thanks for the clarification!


Apparently it matters whether the box you have is filled with photons, or filled with sand.
In one frame the 2 boxes will have the same invariant mass and the same energy.
But in another frame, the 2 boxes still have the same invariant mass, but different energies.


Since I fiddled around with the formulas I do have a couple of questions/comments.

Shouldn't the multipliers you have in P'₁ and P'₂ be interchanged?
And I think the final formula should have a couple of √(1-v²) in it, since I think we're comparing relativistic masses (energies) both in the other frame here.

 

[Dale]

Apparently it matters whether the box you have is filled with photons, or filled with sand.
In one frame the 2 boxes will have the same invariant mass and the same energy.
But in another frame, the 2 boxes still have the same invariant mass, but different energies.

The above conclusion still holds for sand.

Since I fiddled around with the formulas I do have a couple of questions/comments.

Shouldn't the multipliers you have in P'₁ and P'₂ be interchanged?

Oops, you are correct , my apologies. The correct four-momenta should be:
P'_1=\left(E\sqrt{\frac{1-v}{1+v}},-E\sqrt{\frac{1-v}{1+v}},0,0\right)
P'_2=\left(E\sqrt{\frac{1+v}{1-v}},E\sqrt{\frac{1+v}{1-v}},0,0\right)

And I think the final formula should have a couple of √(1-v²) in it, since I think we're comparing relativistic masses (energies) both in the other frame here.

That is just a matter of algebra.
\frac{1+v}{\sqrt{1-v^2}}=\frac{1+v}{\sqrt{(1-v) (1+v)}}=\sqrt{\frac{1+v}{1-v}}

 

[I like Serena]

And I think the final formula should have a couple of √(1-v²) in it, since I think we're comparing relativistic masses (energies) both in the other frame here.

That is just a matter of algebra.
\frac{1+v}{\sqrt{1-v^2}}=\frac{1+v}{\sqrt{(1-v) (1+v)}}=\sqrt{\frac{1+v}{1-v}}

That's not what I meant.
I meant that instead of:

m+2E \ne m + E\sqrt{\frac{1+v}{1-v}} + E\sqrt{\frac{1-v}{1+v}}

it should be
\frac {m+2E} {\sqrt{1-v^2}} \ne \frac {m} {\sqrt{1-v^2}} + E\sqrt{\frac{1+v}{1-v}} + E\sqrt{\frac{1-v}{1+v}}
That is, in another frame, the mass-energy of a box with sand is not the same as the mass-energy of the box with photons.

 

[Dale]

That's not what I meant.
I meant that instead of:

m+2E \ne m + E\sqrt{\frac{1+v}{1-v}} + E\sqrt{\frac{1-v}{1+v}}

it should be
\frac {m+2E} {\sqrt{1-v^2}} \ne \frac {m} {\sqrt{1-v^2}} + E\sqrt{\frac{1+v}{1-v}} + E\sqrt{\frac{1-v}{1+v}}
That is, in another frame, the mass-energy of a box with sand is not the same as the mass-energy of the box with photons.

No, what I wrote is correct. Remember, this is a response to DeG's question regarding my statement that the mass of the box with photons is the sum of the relativistic masses of the photons plus the mass of the box, but only in the rest frame of the box.

The mass of the box with photons was determined to be m+2E in the rest frame. Since mass is invariant, that is the mass in all frames, so the left hand side is correct. Similarly, the mass of the box is m in all frames, and only the relativistic mass (total energy) of the photons is different in the moving frame. So the right hand side is also correct, and not equal to the left hand side.

 

[I like Serena]

Hmm, the question of DeG was: "to someone moving the box it would seem as though the mass of the system is just the mass of the box, not the box plus the relativistic mass of the photons?"So what happens if we push against the box with some force until the box is accelerated to some speed v?

Then to accelerate only the box, you would need an impulse \frac {mv} {\sqrt{1-v^2}} \approx mv.
But to accelerate the box with photons, wouldn't you need a bigger impulse?
Wouldn't that impulse be:
\frac {mv} {\sqrt{1-v^2}} + E\sqrt{\frac{1+v}{1-v}} - E\sqrt{\frac{1-v}{1+v}}\approx (m + 2E)v
In other words, the "mass" that needs to be accelerated is not just the box, but it is approximately the box plus the relativistic masses of the photons in the rest frame?
EDIT:
And now that I think some more about it, don't we have:
\frac {mv} {\sqrt{1-v^2}} + E\sqrt{\frac{1+v}{1-v}} - E\sqrt{\frac{1-v}{1+v}} = \frac {(m+2E)v} {\sqrt{1-v^2}}
So the force to accelerate the box with photons in the rest frame, is the same as the force to accelerate a box with sand.

Finally, this looks more like the mass-energy equivalence that I was looking for and expecting!