The Mysterious, Impossible Box: Questions & Answers

[mrpeach32]

Lets say you had a box who's inside walls were perfectly reflective. In addition to this, because you are just so very awesome, you are somehow able to put a lamp inside that box that doesn't interfere with and of the light rays in the box except to produce more. Now since this is a silly and impossible set up, I'm going to ask some silly and ridiculous questions about it.

1) Would the box get more massive?

2) Would the box eventually have to "pop?"

3) If it was broken open, would there be a flash of light before it dissipated?

I welcome answers, more questions related to this, and snide comments about how pointless it is to wonder about these things.

 

[DaveC426913]

There's really no such thing as perfectly reflective. Eventually the walls will absorb photons and heat up.

How is your lamp powered? If there's no way in for electricity, then the light will eventually stop. If it's battery-powered then you're putting all the "ingredients" for the photons in the box before you close it.

 

[mrpeach32]

The entire idea was based out of it being completely impossible, really. I was just wondering whether it was possible to "fill" something up with light if certain conditions were met. So at the risk of sounding, well, stupid, I guess I don't really know what the lamp is powered by. Maybe this question is a bit too hypothetical for realistic consideration :-D

 

[pallidin]

The following is from: http://en.wikipedia.org/wiki/Total_internal_reflection


"Total internal reflection is an optical phenomenon that occurs when a ray of light strikes a medium boundary at an angle larger than the critical angle with respect to the normal to the surface. If the refractive index is lower on the other side of the boundary no light can pass through, so effectively all of the light is reflected. The critical angle is the angle of incidence above which the total internal reflection occurs."

----------------------- And further in the article:


"Total internal reflections can be demonstrated using a semi-circular glass block. A "ray box" shines a narrow beam of light (a "ray") onto the glass. The semi-circular shape ensures that a ray pointing towards the centre of the flat face will hit the curved surface at a right angle, this will prevent refraction at the air/glass boundary of the curved surface. At the glass/air boundary of the flat surface, what happens will depend on the angle. Where θc is the critical angle (measured normal to the surface):

If θ < θc, as with the red ray in the above figure, the ray will split. Some of the ray will reflect off the boundary, and some will refract as it passes through.
If θ > θc, as with the blue ray, the entire ray reflects from the boundary. None passes through. This is called total internal reflection."

--------------------

Just something to offer your question.

 

[Phrak]

1)yes
2)what does 'pop' mean?
3)yes

A laser cavity is light-in-a-box, when supplied with both mirrors fully reflective.

 

[pallidin]

It is also important to note that your type of question/scenario is one that has been and continues to be of considerable interest.

For example, at 50% efficiency, a 1 second pulse of photons contained as such and then released at one time(in-phase or not) in one direction, will effect a photonic impact pulse 90,000 times that of the initial input.

Of course, the pulse will necessarily be 90,000 times shorter in pulse-length.

Basically, this is photon density amplification, and should not be confused with a laser, though lasers are often used for the initial pulse.

 

[pallidin]

On a related note, have you ever had one of those plastic "neon" hand-held clipboards, which emit a very bright light from its edges even in low-light surroundings? Doesn't work in total darkness of course.

This is due to internal reflectance from ambient light on the 2 flat surfaces, then exiting the edges.

Not "perfect" internal reflectance or efficiency, but quite remarkable.

I have one here as we speak.

 

[Nick89]

1)yes
2)what does 'pop' mean?
3)yes

1) How does it get more massive, because the energy of the photons can be seen as mass in a relativistic view or something?

2) I think by 'pop' he means that the box shatters or breaks. I would have to say yes, because photons do carry momentum. If you have a large number of photons bouncing off the walls of a box the walls will be pushed outwards by a tiny amount. If you just put in enough photons they would eventually break, depending of course on how much they can take. Note that the momentum of photons is very small and you would need a hell of a lot of photons to achieve this...

 

[DaveC426913]

I wouldn't break, it would just start radiating heat as the impacts started heating the walls. If it couldn't radiate fast enough, it would eventually break down whatever the reflective surface is made of.

 

[Phrak]

1) How does it get more massive, because the energy of the photons can be seen as mass in a relativistic view or something?

In this way: m=F/a. Put it on a scale, it weights more. Try to push on the box, it has more inertia. And in this way: m^2=p^2 for a standing wave. (odd, that it wouldn't work for massless fermions)

 

[Gear300]

Lets say you had a box who's inside walls were perfectly reflective. In addition to this, because you are just so very awesome, you are somehow able to put a lamp inside that box that doesn't interfere with and of the light rays in the box except to produce more. Now since this is a silly and impossible set up, I'm going to ask some silly and ridiculous questions about it.

1) Would the box get more massive?

2) Would the box eventually have to "pop?"

3) If it was broken open, would there be a flash of light before it dissipated?

I welcome answers, more questions related to this, and snide comments about how pointless it is to wonder about these things.

I wouldn't have to say its completely pointless...its sort of like shifting the open end of the box so that the Sun is contained in it, then closing the open part...

 

[Nick89]

I wouldn't break, it would just start radiating heat as the impacts started heating the walls. If it couldn't radiate fast enough, it would eventually break down whatever the reflective surface is made of.

Well ok, but I guess you could see that last thing as the box breaking no?

In this way: m=F/a. Put it on a scale, it weights more.

Weight is not the same as mass though... If you put the box on a scale and press down on it that wouldn't make the box more massive, but it would make the scale read a larger value.

Try to push on the box, it has more inertia.

Why?

I'm not trying to say it's not going to be more massive (for all I know it could), I just don't completely understand how.

 

[DaveC426913]

In this way: m=F/a. Put it on a scale, it weights more. Try to push on the box, it has more inertia. And in this way: m^2=p^2 for a standing wave. (odd, that it wouldn't work for massless fermions)

Yeah, I'm not exactly convinced of this.

 

[Phrak]

Yeah, I'm not exactly convinced of this.

Convinced of what?

 

[DaveC426913]

Convinced of what?

You can't apply classical Newtonian physics to photons and get sensical results.

 

[Phrak]

You can't apply classical Newtonian physics to photons and get sensical results.

That's worth thinking about.

By the way, I can't believe I screwed that up. p=0, so that E^2=m^2

 

[HooDude]

Just my 2 cents, but if photons are massless particles, how can an increase in their concentration cause an increase in mass?

I would think the box would see an increase in pressure and temperature but not an increase in mass?

 

[Phrak]

Weight is not the same as mass though... If you put the box on a scale and press down on it that wouldn't make the box more massive, but it would make the scale read a larger value.

What, if not mass, is responsible for the deflection of the scale?

You have a 2 kilo, ideally reflective box, containing 10 kilo of matter and 10 kilo of antimatter. It is suspended by a balloon capable of lifting 7 kilo. Placed on a scale, the scale reads 15 kilo. You enable the contents to annihilate. The scale still reads 15 kilo.

Let's imagine for a moment that the mass of the contents vanishes to zero upon annihilation. The balloon lifts the box and an additional weight to some astonishing altitude. Some means is provided to return the box to it's initial state. It decends, and the process is repeated, where another weight is lifted.

Now we can object, and say that the annihilation process is thermodyamically irreversible and the box will not decend. This places us in the position of claiming that conservation of energy is a result of the second law; we would making the interesting claiming that energy is only statistically conserved.

 

[Nick89]

What, if not mass, is responsible for the deflection of the scale?

The force that the photons exert on the walls of the box.
(But now I think about it, on average that force would be zero so the weight of the box would not increase).

Anyway as I've said, weight and mass are two different things. Your weight can increase while your mass does not. The topic starter (and me) asked about the mass, not the weight.

 

[HooDude]

Mass and weight are two different things, this is correct. Weight is relative. For example, something with a certain mass will have a a different weight on the moon than on the earth. Although the mass does not change, the weight does. In this case, weight is relative to gravity.

 

[Phrak]

The force that the photons exert on the walls of the box.
(But now I think about it, on average that force would be zero so the weight of the box would not increase).

Actually, the force is greater on the bottom of the box than the top.

Anyway as I've said, weight and mass are two different things.

Of course, that's not the issue. How is mass defined? How is mass measured?

 

[Nick89]

Actually, the force is greater on the bottom of the box than the top.

Because of gravity? Or why would it be greater?

Of course, that's not the issue. How is mass defined? How is mass measured?

But it is an issue here (at least for my question): The weight of the box can increase due to the force of the momentum of the photons, but at the same time, it is perfectly fine for the mass to stay the same. So is the mass changing, or is the weight changing (and the mass not changing)?

 

[pallidin]

Maybe look at it this way...

Place a 1 lb. block of steel on a bathroom scale. Let it stabilize.
Now drop from, say, a height of 3 feet a standard baseball onto that steel block.

Notice that the scale will increase it's "reading" during impact, and that the peak reading exceeds a reading of a different scenario... where the the baseball is gently placed on top of the steel block(as opposed to being dropped from 3-feet).

I hope I'm not confusing you here.

 

[Phrak]

Because of gravity? Or why would it be greater?

Because of gravity. The light at the bottom of the box has a higher frequency as verified by the Mossberg Effect. Any photons with verticle trajectories will recoil from the bottom surface of the box with a greater change in momentum than those photons impinging on the top of the box.

Horizontal trajectories will be bent downward.

But it is an issue here (at least for my question): The weight of the box can increase due to the force of the momentum of the photons, but at the same time, it is perfectly fine for the mass to stay the same. So is the mass changing, or is the weight changing (and the mass not changing)?

The more light that is confined to the box, the more it will weigh on the scale. The light in the box plus the box has ponderable mass greater than the empty box alone. It has inertia greater than the empy box alone.

It might come easier to you if you forgot about gravity and just accellerated the box in free space.

Every body continues in its state of rest, or of uniform motion in a right [straight] line, unless it is compelled to change that state by forces impressed upon it. -Newton

We can replace state of rest or uniform motion with the 'freely falling', and 'right line' with the geodesic of general relativity, and it remains true, when gravity is removed as a force.

I can't think of any reason why the light by itself, considered as a system, doesn't have inertial mass, except that it requires the presence of the box for the system of radiation be as massive as it is.

 

[Phrak]

DaveC.

I've found a way to eliminate the Newtonian elements. In fact, it's an old thought experient of '97 or so. Little did I know that moving it to a scale would clutter the problem.

Consider a massless box contains equal parts matter and antimatter in orbit (freely falling). Upon annihilated, will the box acquire a rectilinear trajectory, remain in it's original orbit, or something else?

 

[atyy]

I welcome answers, more questions related to this, and snide comments about how pointless it is to wonder about these things.

It is pointless to wonder about such things. You should spend your time more wisely by reading this paper, so you can explain it to me.

"let us imagine confining the beam to a mirrored box"
Carlip, Kinetic Energy and the Equivalence Principle
http://arxiv.org/abs/gr-qc/9909014

 

[Phrak]

It is pointless to wonder about such things. You should spend your time more wisely by reading this paper, so you can explain it to me.

"let us imagine confining the beam to a mirrored box"
Carlip, Kinetic Energy and the Equivalence Principle
http://arxiv.org/abs/gr-qc/9909014

I'll read it.

Where is the momentum found in the electromagnetic field equations, do you know?

 

[atyy]

Where is the momentum found in the electromagnetic field equations, do you know?

I only understand the flat spacetime case. The EM energy-momentum tensor is T_m_n = -F^a_mF_a_n + \frac{1}{4}g_m_nF^a^bF_a_b. In flat spacetime, with cartesian coordinates, T^0^m are the components of the Poynting 3-vector S=EXB, which I guess is the intuitive momentum.

 

[Phrak]

For light in a box, flat spacetime should be sufficient. I was looking at the Ponyting vector myself. So for a planar wave of directionally polarized light, the Poynting vector propagates in longitudinal waves. For circularly polarized light S is constant in magnitude.
For standing waves...

 

[atyy]

For standing waves...

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.

 

[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.