Light building the standard model

[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!

 

[Dale]

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?"

Which is the question (about mass) that I answered.

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?

This is a related, but slightly different question (about force and acceleration). However, the answer is somewhat complicated in relativity.

It turns out that using ordinary non-relativistic vectors leads to a fairly complicated relationship between force and acceleration that cannot be expressed using a simple single number like "relativistic mass". Specifically, as you say "the mass that needs to be accelerated" is different in the direction parallel to and perpendicular to the velocity. See the section on "Transverse and Longitudinal Mass" here:
http://en.wikipedia.org/wiki/Mass_in_special_relativity#The_relativistic_mass_concept

In general, I would not recommend using ordinary vectors to work with forces and accelerations in special relativity. You are much better off using four-vectors where the equivalent of f=ma holds at all speeds.
http://en.wikipedia.org/wiki/Four-force
http://en.wikipedia.org/wiki/Four-acceleration
http://en.wikipedia.org/wiki/Invariant_mass

 

[I like Serena]

This is a related, but slightly different question (about force and acceleration). However, the answer is somewhat complicated in relativity.

It turns out that using ordinary non-relativistic vectors leads to a fairly complicated relationship between force and acceleration that cannot be expressed using a simple single number like "relativistic mass". Specifically, as you say "the mass that needs to be accelerated" is different in the direction parallel to and perpendicular to the velocity. See the section on "Transverse and Longitudinal Mass" here:
http://en.wikipedia.org/wiki/Mass_in_special_relativity#The_relativistic_mass_concept

In general, I would not recommend using ordinary vectors to work with forces and accelerations in special relativity. You are much better off using four-vectors where the equivalent of f=ma holds at all speeds.
http://en.wikipedia.org/wiki/Four-force
http://en.wikipedia.org/wiki/Four-acceleration
http://en.wikipedia.org/wiki/Invariant_mass

Thanks for the links which provide interesting reading material.
Btw, as you can see I avoided calculation of force or acceleration, restricting myself to impulse, which is the change in momentum (since I was not entirely sure how force or acceleration would turn out ).
Aren't my results correct?

 

[Dale]

Oh, I didn't read carefully enough. Let me look again.

For impulse a more useful link would have been:
http://en.wikipedia.org/wiki/Four-momentum

The conservation of four-momentum contains the concepts of conservation of energy, momentum, and mass, all in one nice neat mathematical package.

 

[I like Serena]

The conservation of four-momentum contains the concepts of conservation of energy, momentum, and mass, all in one nice neat mathematical package.

Yep!
That's why I picked this one to work things out.
It keeps things nicely neat and simple, and shows the relation with classical mechanics, whereas force and acceleration tend to become messy.
And btw, doesn't the relation f = dp/dt hold?
In my example the average force would be the impulse divided by the time to accelerate the box.

 

[Dale]

OK, so looking back at your post I have made a more relevant reply. For a massive system the four-momentum is a four-vector with a length given by the rest mass and a direction given by the four-velocity.

So, assuming that you want the four-velocity to be the same for both the empty and the photon-carrying boxes after the impulse then you can just use a basic "similar triangles" argument to show that the norm of the four-impulse is larger for the photon-carrying box by a ratio equal to the ratio of the (invariant) masses:
\frac{m+2E}{m}

If you boost that into any frame then the similar triangles argument still holds, as you would expect from the fact that the norm is frame invariant. So in all frames the ratio of the norms of the impulses depends on the ratio of the masses of the systems.

 

[I like Serena]

I'm still puzzling on relativistic masses.
I can see from wiki entries that the concept is avoided, and that there are separate parallel and transversal forms.
As it is I can distinguish 6 types of masses.

Let's say we have 4-momentum P = (E, p), where p is the 3-momentum, v is the velocity to a (distant) inertial observer.
I'm setting c=1 for ease of notation.
Let furthermore f = dp/dt be the 3-force.The 6 types of masses I see, are:

1. The rest mass m₀, which is also the invariant mass for a single object.

2. The invariant mass, which is the norm of the 4-momentum P. This one is the same for all observers.

3. The energy-mass E, which is the first component of P.

4. The momentum-speed-ratio \frac {|\boldsymbol p|} {|\boldsymbol v|} = \gamma m_0.

5. The parallel force-acceleration-ratio \frac {f_\parallel} {\dot v_\parallel}.

6. The transversal force-acceleration-ratio \frac {f_\perp} {\dot v_\perp}.Numbers 5 and 6 are messy and probably best avoided.
They can be derived from the time-derivative of the momentum.

Number 3 is ambiguous, since we saw with the box of photons, that it matters to an observer whether we are talking about a box of photons or a box of sand.

The interesting thing is that number 4, the momentum-speed-ratio, appears to match perfectly with the old relativistic mass concept, which appears to be consistent (but not the same) in all frames of reference, and which obeys the rules of classical mechanics as long as it is only being used in combination with momentum and impulse.Please correct me if I'm wrong, because I would really like to know.

 

[TheTommy1]

Correct me if I'm wrong, but my understanding is that light bends around the sun, not because light has mass but because the mass of the sun is so great it "warps" space/time and that light is following this "warped" space/time.

 

[I like Serena]

Correct me if I'm wrong, but my understanding is that light bends around the sun, not because light has mass but because the mass of the sun is so great it "warps" space/time and that light is following this "warped" space/time.

Yes, this is true and it is what is stated in general relativity theory.

If we try to explain it with classical mechanics, the mass of the sun exerts a force of gravity on the photons (whose mass is still irrelevant).
Apparently, as pengwuino stated, the resulting deviations of the photons are off by a factor of about 2.

I'm afraid the discussion in this thread has spun a bit away from the OP.

 

[cragar]

Correct me if I'm wrong, but my understanding is that light bends around the sun, not because light has mass but because the mass of the sun is so great it "warps" space/time and that light is following this "warped" space/time.

Light also has its owns Gravitational field.

 

[Dale]

Light also has its owns Gravitational field.

True, but it is not significant in this case.