The Mysterious, Impossible Box: Questions & Answers

[pallidin]

It's not that simple.

When one focuses on "utilizing for some experimental gain" the intervention of destructive wave interference, one must not forget that the constructive waves still and must exist.

 

[Phrak]

He he, I never thought of that. I can't resist putting down my guess: it propagates to the nodes and cancels out there? Let us know when you figure out the right answer.

I hadn't either. I was able to draw it out.

(In case I didn't say so, the perfectly reflextive box is a resonant cavity so contains standing waves, but no propagating waves.)

For two directionally polarized planar waves, having antiparallel velocities, with equal amplitudes: the electric and magnetic fields are antiparallel. So the Poynting vector is identically zero. So if the Poynting vector can be associated with momentum as you say, the momentum is everywhere zero.

For the case of the standing wave of circularly polarized light the Poynting vector is also everywhere zero.

 

[Phrak]

atyy,

have you ever seen this

m^2 = E^2 - p^2

replaced by this

\mu = (E, p_x, p_y, p_z)^T

of something like it? mu is a column vector.

Just curious.

 

[atyy]

For two directionally polarized planar waves, having antiparallel velocities, with equal amplitudes: the electric and magnetic fields are antiparallel. So the Poynting vector is identically zero. So if the Poynting vector can be associated with momentum as you say, the momentum is everywhere zero.

For the case of the standing wave of circularly polarized light the Poynting vector is also everywhere zero.

Wow the textbooks are actually right! (Some people add a term to the Poynting vector which changes its interpretation without changing conservation laws.)

have you ever seen this

m^2 = E^2 - p^2

replaced by this

\mu = (E, p_x, p_y, p_z)^T

of something like it?

Maybe the four-momentum? http://en.wikipedia.org/wiki/Four-momentum

 

[Phrak]

Wow the textbooks are actually right! (Some people add a term to the Poynting vector which changes its interpretation without changing conservation laws.)

I didn't know that. Come to think about it, though, it seems it should be regaugable.

Maybe the four-momentum? http://en.wikipedia.org/wiki/Four-momentum[/QUOTE]

Notice that that article links to one on (Lorentz) invariant mass in order to define the 4-momentum.

http://en.wikipedia.org/wiki/Four-momentum
So the four momentum is defined in terms of the invariant mass,

m = \frac{1}{c^2} \sqrt{E^2-(pc)^2}

The circular defintions we could attribute to different authorship, (If that's not enough, the link to conservation of energy defines mass as one kind of conserved energy) but the appearance of the positive square root is the most irritating part. It tells me something is hidden.

So, for lack of a better name i call this the vector mass, not to be confused with mass:

\mu = \left( \frac{E}{c^2}, \frac{p_x}{c}, \frac{p_y}{c}, \frac{p_z}{c} \right)^T ...

 

[atyy]

So, for lack of a better name i call this the vector mass, not to be confused with mass:

\mu = \left( \frac{E}{c^2}, \frac{p_x}{c}, \frac{p_y}{c}, \frac{p_z}{c} \right)^T ...

Your vector mass is related to the tensor mass http://www.math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html.

 

[Schroecat]

Is one of the assumptions of the thought experiment that the light source/lamp has some magical production properties?

Otherwise I would assume that in producing the light, via chemical reaction, electrical or other, any mass gains from the light would be offset by loss of mass in the means of production.

 

[Nick89]

Is one of the assumptions of the thought experiment that the light source/lamp has some magical production properties?

Otherwise I would assume that in producing the light, via chemical reaction, electrical or other, any mass gains from the light would be offset by loss of mass in the means of production.

That's an interesting question, to which I'm going to ask another question:

If an atom emits a photon when an electron falls back to a lower energy state, does the atom lose mass (energy) because the electron has less energy?

 

[Phrak]

Your vector mass is related to the tensor mass http://www.math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html.

You're precognizant. That's the next question I was going to ask you.

 

[Phrak]

That's an interesting question, to which I'm going to ask another question:

If an atom emits a photon when an electron falls back to a lower energy state, does the atom lose mass (energy) because the electron has less energy?

Nick. I think that's answered to the afirmative in the zrxiv article posted by atyy. http://arxiv.org/abs/gr-qc/9909014

 

[Phrak]

Your vector mass is related to the tensor mass http://www.math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html.

1)On general principles it seems \mu should be related to the stress energy tensor, but I don't see that in the Baez article, somehow.

2)Equating \ \mu^{\rho} \mu_{\rho} = -m^2, is the intent of the equation, but is problematical. (I haven't proved it in any context.)

On the other hand it makes some sense in Minkowski space to adding null masses and getting something other than zero. I should explain this...

3)Is there a mass continuity equation somewhere with general enough applicability?

 

[atyy]

1)On general principles it seems \mu should be related to the stress energy tensor, but I don't see that in the Baez article, somehow.

2)Equating \ \mu^{\rho} \mu_{\rho} = -m^2, is the intent of the equation, but is problematical. (I haven't proved it in any context.)

On the other hand it makes some sense in Minkowski space to adding null masses and getting something other than zero. I should explain this...

3)Is there a mass continuity equation somewhere with general enough applicability?

Edit: It's the total energy that's the gravitational mass, including rest mass (p^up_u) and momentum (p^a, a=1,2,3). For a photon, the energy or gravitational mass is entirely due to its momentum.

I believe it's roughly like this, leaving constants and minus sign out. In 3D cartesian space, the surface integrated Poynting 3-vector (EXB) equals the volume integrated electromagentic energy density (E²+B²). In 4D Minkowski space with cartesian coordinates, the energy-momentum 4-vector p is the top row of the energy-momentum tensor T, and surface integrated p equals volume integrated T (Equation 1 in the Baez article). The Minkowski space integral equation can be formulated as a differential equation (Equation 3 in the Baez article). The differential version, but not the integral version, carries over to general relativity, where T is the gravitational mass.

So I would say that for the photon:
1) Classical rest mass m_o = 0, but the photon is never at rest. This is cute but not formally useful for calculations in classical special relativity, since its only purpose is to assert that E²=p²c²+m_o²c⁴ holds for massive particles and massless waves. However, that is only superficial since the definition of p is not the same for massive particles and massless waves anyway.

2) Quantum rest mass m_o = 0, but the photon is never at rest. This is useful, because massive particles are now waves, so p has the same form for both, and E²=p²c²+m_o²c⁴ holds more than superficially for massive and massless waves.

3) Classical relativistic mass E=mc². holds for massive particles, provides a good heuristic for showing that massive particles can never travel at the speed of light, and makes the relativistic dynamical equations become approximately Newtonian at low speeds. Heuristically, it is an "inertial mass" which by Newton's gravity should have "gravitational mass". We can also formally assign it to the photon, which indicates that a photon should have gravitational mass. However, special relativity doesn't work with gravity, and it is possible to make consistent "general" relativistic gravity theories in which the photon does not have gravitational mass. Nonetheless, nature happens to have chosen general relativity, which is a "general" relativistic gravity theory in which the photon does have gravitational mass via the energy-momentum tensor T.

4) Quantum gravitational mass ... they're working on it ...

 

[Phrak]

Sorry I can't think of an intelligent response to all that atyy. Youi've obviously putting a great deal of thought into it. What I'm missing in all this are the contacts with experiment. What are the experimentally obtained values, and from where obtained, and which are derived.

 

[Dale]

I like the light in a box idea. I once spent some time trying to prove that the light gave the box extra inertia. It is actually not too difficult to do, but I can't find my work right now.

The key principle is the Doppler effect. If you are pushing on a box of light to accelerate it then a photon leaving the front of the box will be blueshifted by the time it reaches the back of the box, and a photon leaving the back of the box will be redshifted by the time it reaches the front. Since the momentum of the photon is proportional to the frequency there is more pressure on the back than on the front. Therefore the light increases the inertia of the box.

By the equivalence principle you get the same effect in a gravitational field, so the box also weighs more.

Also, if the box is accelerating then the wave is not a standing wave in any inertial frame.

 

[Phrak]

Also, if the box is accelerating then the wave is not a standing wave in any inertial frame.

Are you sure? That's a point that's been troubling me. Yet we still have magnetrons and maser cavities, that to good approximation, contain standing waves.

For vertically upward or downward trajectories against the top and bottom of the box, the number of wavelengths are equal.

 

[Dale]

Are you sure? That's a point that's been troubling me. Yet we still have magnetrons and maser cavities, that to good approximation, contain standing waves.

Yes, I am 100% sure of that. Since it is undergoing proper acceleration it must be moving in any inertial frame so by definition it is not a standing wave in any inertial frame.

I don't think your logic of post 32 applies to any non-inertial frame where it is a standing wave.

 

[Phrak]

Yes, I am 100% sure of that. Since it is undergoing proper acceleration it must be moving in any inertial frame so by definition it is not a standing wave in any inertial frame.

I don't think your logic of post 32 applies to any non-inertial frame where it is a standing wave.

Any frame not comoving with the box will have traveling wave components. The same applies between two inerital frames. My posts refer to the comoving frame.

What concerns me are the wave components transverse to the gravitational field.

 

[Dale]

Any frame not comoving with the box will have traveling wave components. The same applies between two inerital frames. My posts refer to the comoving frame.

Do you mean the momentarily co-moving inertial frame or the (always co-moving) non-inertial rest frame?

 

[Phrak]

Always comoving, for the box on the scale.

 

[Phrak]

atyy, it looks like you may have spawned a post on Poynting vectors.

With some mild considerations, the mass units version of the 4-vector momentum, seems both useful and describles mass of a confined electromagnetic wave. It makes contanct with experiment, where the frequency and wave number are measurable.

Adding a 'mass vector' obtains the experimentally measured mass and the particle mass. For two waves of equal amplitude and oppositely directed.

\mu_{(L)} = E, p
\mu_{(R)} = E, -p
------------
\mu_{(\Sigma)} = 2E

It was bothering me that it didn't have an immediate mass continuity equation. But's it's describes a global attribute-- an integrated value over some volume, not a field at a point.

 

[atyy]

The key principle is the Doppler effect. If you are pushing on a box of light to accelerate it then a photon leaving the front of the box will be blueshifted by the time it reaches the back of the box, and a photon leaving the back of the box will be redshifted by the time it reaches the front. Since the momentum of the photon is proportional to the frequency there is more pressure on the back than on the front. Therefore the light increases the inertia of the box.

Very nice! And much easier than reading Carlip's article!

Adding a 'mass vector' obtains the experimentally measured mass and the particle mass. For two waves of equal amplitude and oppositely directed.

\mu_{(L)} = E, p
\mu_{(R)} = E, -p
------------
\mu_{(\Sigma)} = 2E

Very nice too! So rest mass or invariant mass m_o=p^ip_i=0 (4-vectors), and gravitational mass \mu = E = T^{00} as in Carlip's article.

 

[M Grandin]

Several times I have thought about this and aguainted questions. For instance if you have a thermal element in an insulated bottle - what maximum temperature is achieved? I guess
the limit is the temperature of the radiating element. And if light source, the light intensity
never exceeds that of the hypothetical radiating light bulb. I have never found answers to such questions in textbooks, but believe the limit of intensity in the "bottle" is the intensity of the radiating source. So at that saturated stage, the "light bulb" would not deliver further light energy into the insulated volume.

Corresponding should yeld emitted rays from antennas: If for instance a radio wave transmitter emits a certain EM-field of certain magnitude - nothing is emitted if there already is an identical EM-field around the antenna, for instance due to reflected waves. I think
this is the fundamental behind resonant circuits - coupled capacitors/coils - no emitted energy if there already is the EM-field present for other reasons.

But this is just my own (possibly wrong) conclusion - I have never seen this issue discussed by scientist, although they surely have.

 

[Phrak]

Very nice too!

Thank you.

So rest mass or invariant mass m_o=p^ip_i=0 (4-vectors), and gravitational mass \mu = E = T^{00} as in Carlip's article.

(Of course, I know you meant, \ (m_o)^2=p^{\iota}p_{\iota}=0 .)

Carlip's \ \mu is a Lorentz variant scalar; not the same at all.

So I can safely claim that mass is conserved. Neither energy nor momentum are conserved. The non zero mass of a photon is given by the contraction of \ \mu with itself.

Also that the esoteric \ m_o appearing in

m_0^2=E^2 + p^2

is to be associated with the inertial mass.

 

[Dale]

Always comoving, for the box on the scale.

Well, that reference frame is non-inertial. I don't know what form Maxwell's equations take in that frame, in particular how the Poynting vector would be expressed. But in terms of measurable quantities, like the pressure measured on the box wall, or the proper force and proper acceleration, it must agree with the results obtained in the momentarily co-moving inertial frame.

 

[Lok]

1) The box maintains it's mass and clearly it's weight.
The light source loses energy and thus mass. ( a very tiny amount ). The photons have mass, all photons have mass. Their restmass is 0, which means that if you would slow one down (hypoteticaly) it would simply dissapear, because all of it's energy would be absorbed by the slowdown mechanism. It's like imagining a hot object with lots of photons bouncing inside if it( just as heat is transfered), after it cools it looses mass because the system looses energy. The box does not so it keeps it's original mass.

2) If enough photons accumulate, the box is very likely to burst through radiation pressure. If the box is of equal strength distribution then it's likely to burst in one corner.

3) Given the small size (box presumably no bigger than 1 m) in accordance with the speed of light, it's clear that all the photons will exit the box in a very short ( shorter ) amount of time resulting in a big flash. Only if the box breaks evenly will there be a flash in all directions.

 

[Phrak]

Well, that reference frame is non-inertial. I don't know what form Maxwell's equations take in that frame, in particular how the Poynting vector would be expressed. But in terms of measurable quantities, like the pressure measured on the box wall, or the proper force and proper acceleration, it must agree with the results obtained in the momentarily co-moving inertial frame.

Never the less, it's the accelerating lab frame in question, and there is nothing wrong in analyzing the physics in the accelerating frame.

Is this just to give me a hard time over the exact solution of the wave in the box, or do you have some deductive means to support a solution that is other than a standing wave?

This might help. Emperically, the radiation is blue shifted toward the bottom of the box (in the lab frame). An incident wave impinging against the bottom of the box imparts more downward momentum than one incident on the top of the box. (There's more radiation pressure on the bottom of the box than the top; but we already knew this.)

Qualitatively the blue shift is proportional to the height of the box times its acceleration. Without any other variables entering, whatever variation obtained is small compared to v squared for a human scaled box and gravity.

The absolute change in the momenum of the radiation on the bottom of the box is greater than the top. This must be accounted for by emf carrying momentum upward. The

 

[atyy]

The key principle is the Doppler effect. If you are pushing on a box of light to accelerate it then a photon leaving the front of the box will be blueshifted by the time it reaches the back of the box, and a photon leaving the back of the box will be redshifted by the time it reaches the front. Since the momentum of the photon is proportional to the frequency there is more pressure on the back than on the front. Therefore the light increases the inertia of the box.

One thing that was bothering me was I thought that Nordstrom's theory would predict no weight for light in a box because instead of using the whole energy-momentum tensor as the gravitational source, it only uses the trace - and the electromagnetic field is traceless. Nordstrom's theory is consistent with the equivalence principle and predicts redshift, so if that were the case, then the redshift argument must be insufficient to establish weight for light in a box. On the other hand, the redshift argument can be independently argued from energy conservation and de Broglie theory. Contrary to my expectation, Nordstrom's theory apparently predicts weight for light in a box "because electromagnetic waves trapped in a material box with mirrored walls will induce additional stresses in the box’s walls due to radiation pressure. This will increase the weight of the box corresponding to an additional mass m = E_rad/c², where E_rad is the energy of the radiation field. In this sense bound electromagnetic fields do carry weight." That also explains a bit why Carlip's article isn't about why light in a box has weight, but more about why it doesn't weight twice as much!

Giulini, What is (not) wrong with scalar gravity?, http://arxiv.org/abs/gr-qc/0611100

 

[Dale]

Never the less, it's the accelerating lab frame in question, and there is nothing wrong in analyzing the physics in the accelerating frame.

Is this just to give me a hard time over the exact solution of the wave in the box, or do you have some deductive means to support a solution that is other than a standing wave?

Oh, no. I am not trying to give you a hard time. I agree 100% that it is a standing wave in the non-inertial frame and that the analysis can be done there. My comments were only intended to say that I don't know how to work the problem in that frame, so I don't know how to show that the radiation pressure is greater on the bottom without using an inertial frame. So I just did it in an inertial frame with the knowledge that it would agree in terms of any measurable results.

 

[Yor_on]

"If you are pushing on a box of light to accelerate it then a photon leaving the front of the box will be blueshifted by the time it reaches the back of the box, and a photon leaving the back of the box will be redshifted by the time it reaches the front."

Shouldn't it be the opposite?
That the photon leaving the front of the box traveling against its box 'direction' towards the back wall, as seen from an observer, should be redshifted as its following a declining energy gradient, and that the photon emitted from the back towards the front wall then should be blueshifted as its walking an upgrade slope as seen from the observer.

Or am I getting this totally wrong?
Are you placing the observer inside this frame so that the thought observers would be traveling with it and in the photons reference frame?

One observing from the back wall, sort of, and one observing from the front wall.
But then there shouldn't be any 'slope' (in any direction -+) should it?
They are at rest inside their frame of reference, or?
Where would that 'excess energy' of that 'blue shifted' photon hitting the back wall f. ex come from if they share the same reference frame?

 

[Phrak]

Oh, no. I am not trying to give you a hard time. I agree 100% that it is a standing wave in the non-inertial frame and that the analysis can be done there. My comments were only intended to say that I don't know how to work the problem in that frame, so I don't know how to show that the radiation pressure is greater on the bottom without using an inertial frame. So I just did it in an inertial frame with the knowledge that it would agree in terms of any measurable results.

My mistake. Pax.

I've been trying out the argument in the accelerating frame, as I half-gave above, but I have a qualitative, not a quantitative solution as yet. And I'm assumming I can derive the blue shift, delta lambdda over lamba, for an accelerating frame, from first principles, rather than looking it up somewhere.

Did I mention that I agree; in the accelerating frame a standing wave is only a good approximation where the acceleration times box-height product is small?