# Weight of a box of PHOTONS

1. Nov 15, 2012

### Austin0

Supposing two cubic containers with perfectly reflective interior surfaces.
In one there is 1 mole of some gas and in the other is an equivalent number NA
of photons of some frequency.
We put the gas on a scale and the scale registers the total weight (mass) of the enclosed gas.
But microscopically the interior gas is moving and interacting in complete vacuum. There is no physical, causal connection to either the box or scale.
This suggests two questions
1) What exactly is the scale measuring if not total mass??

2) How does the interior mass affect the box without a direct connection in order to have the scale register that mass??

The immediate answer to 1) would seem to be that the scale measures the momentum differential between the instantaneous sum of all molecules impacting the interior surface with an upward (positive) vector and the sum of all downward impacts.

I would assume this differential would be some factor of local g and the height of the box.
As this acceleration is reciprocal there should not be an overall increase in the velocity distribution for the temperature so as the percentage of the total mass impacting the surface at any moment is obviously exceedingly small, it does not seem plausible that the velocity difference applied to this small mass could result in a momentum reflecting the total mass.

Looking at the box o' photons velocity is not a factor and the difference in momentum is just a difference in frequency but the same question pertains.

How do the internal photons interact with the box to cause an instantaneous reaction of the scale equivalent to the total number???

The possible answers I have come up with are:

The internal pressure increases the stress energy of the box which increases the total weight equivalent to the gas mass.

The volume mass/energy content acts non-locally as a single geometric entity.

The constant molecular emission of light speed gravitons or GW's result in , not an instantaneous, but a continuous interaction with the box and scale which reflects the total number of emitting particles.

Since none of the above seem convincing I am thinking I must be missing some obvious fundamental factor ??????. Any insights welcome

2. Nov 15, 2012

### Simon Bridge

I'd take issue with that. The gas interacts with the box by bouncing off it's sides. The particles are individually affected by gravity - making the more likely to bounce off the bottom of the box than the top. This is causal if indirect. You want to add the gravitational interaction to the usual kinetic model for gasses?
You know that - the scale measures a balancing force (either by comparing with another pan as in a balance, or by compression as in a spring balance or an electronic scale) from which the weight can be determined - if we know the local force of gravity, then the scale can be designed to read mass units instead of force units.

Personal intuition is not a good guide - do the math.

For a perfectly reflecting surface, the frequency should not change. Momentum is a vector.

Same as the gas - but what makes you think it is instantanious?
Oh well - without resorting to kinetic theory of gasses, you understand that the gas has mass and with just Newtonian gravity the masses are attracting. The pressure of the gas keeps the box away from the gas center of mass. No need for stress-energy.

Of course, we understand gravity (loosely) in terms of the curvature of space according to energy density. The matter in the gas provides a rest-mass energy distribution which curves the local space-time ... the box also does this, and falls into the gravity well of the gas as well - with the pressure of the gas holding it up.

"convincing" and "true" are not always shared by the same proposition. What one finds convincing is neither here nor there. Your problem is that you are trying for an intuitive, qualitative, picture when you need to do the math.

Last edited: Nov 15, 2012
3. Nov 15, 2012

### Staff: Mentor

This is the explanation for both the gas and the photons. As gas/photons go upwards they lose momentum. As gas/photons go downwards they gain momentum. So the momentum of gas/photons hitting the top is less than that hitting the bottom, hence there is more pressure on the bottom and so a scale placed under the filled box detects more weight than with an empty box.

4. Nov 15, 2012

### Austin0

Quote by Austin0

Yes of course I agree as that is essentially exactly what I said but that does not address the question.
Unless you are suggesting that the scale only registers some more weight than an empty box but not the total weight of all the gas then the question remains how does the interior mass affect the scale.
Do you think that there is possibly a great enough increase in velocity through g acceleration that the small number of molecules actually impacting the bottom at any given instant could have a momentum equivalent to the entire mass of gas??
Say all the molecules were on parallel paths simply reflecting up and down with each one in a random position in the cycle.
Only a miniscule portion of the total could be affecting the bottom of the box and the scale at any moment correct?? We must assume that the Maxwell velocity distribution is still in effect so how could this small number have sufficient velocity/momentum to move the scale to reflect the total weight/momentum/mass???

5. Nov 15, 2012

### Austin0

Of course at some point any given molecule will interact with an interior surface.
The point was at any specific time the interior has no physical causal connection with the box

Of course but in actuality I doubt that any kind of scale can distinguish between mass and force. Or more correctly it can only measure force or momentum and cannot measure mass directly.

So you think that the acceleration resulting from the height of the box when applied to the small number of molecular masses directly impacting the bottom of the box at any instant would produce a net momentum (weight) equal to the total mass times the average velocity??

Doesn't it depend on how you view the interior interactions??

If you take the view that the box is accelerating upward then the frequency difference between up and down photons is Doppler shift resulting from reflection,,, yes??

Because the scale is reflecting the total number of internal photons at an instant

Yes I merely mentioned stress energy as a possible explanation for the "instantaneous" effect on the box and scale.

6. Nov 15, 2012

### The_Duck

Suppose we are dealing with a rectangular box. Its height is h and the area of the top and bottom faces is A. It is filled with gas of a density $\rho$. The pressure P of a gas in equilibrium obeys

$\frac{dP}{dz} = -\rho g$

where z is vertical position and g is the acceleration due to gravity. If you're not familiar with this formula you should figure out why it is true.

For simplicity lets assume the density of the gas doesn't change with height. This is more or less true if the box is not gigantic. Call pressure at the bottom of the box $P_0$. Then by a simple application of the above differential equation, the pressure at the top of the box is $P_0 - \rho g h$.

Now, the total upward force exerted on the box by the pressure of the gas on its top and bottom faces is

$F = -P_0A + (P_0 - \rho g h)A = -Ah\rho g = -Mg$

where M is the total mass of the gas. Thus, the box feels a force from gas pressure exactly equal to the total gravitational force on the entire volume of gas.

7. Nov 15, 2012

### Simon Bridge

I was going to answer all that but The_Duck made my main point for me: you have to do the math. It is no good just waving your arms around like that - fact is you don't need all the particles to hit the bottom for the entire weight to register.

Note: I did make a mistake about the photons in the box though - they change momentum due to gravity as well, so there is a frequency shift ... I still wouldn't expect a frequency shift, in the frame of reference of the scales, as a result of reflection. iirc: a photon in an elevator is one of the famous Bohr/Einstein thought experiments.

8. Nov 16, 2012

### Staff: Mentor

Yes. I encourage you to work out the math. It is not that difficult. Do a search for "kinetic theory of gasses" to get started.

9. Nov 17, 2012

### Simon Bridge

Just noticing this:
Hmmm? I don't believe I made any such statement. I suspect it would be higher since not all the particles hitting the bottom of the box have fallen from the top. I also expect that particles will hit the bottom harder than they hit the top... it is the difference between these that counts.

The_Duck has shown you what I meant. BUt you should makes sure you understand where that calculation came from. The difference in pressure between particles imparting down and up momentum to the box will always balance to the total mass of the particles in the box time the acceleration of gravity.

I suppose you could get pedantic and insist that there will be a time in there where there is a one or two-impact mismatch ... if you like. In which case you are faced with working out the statistics: how big an effect is this against the thermal vibrations of the box itself? How accurate would the equipment have to be to detect it as different from Mg?

Do the math and prove me wrong.

10. Nov 17, 2012

### pervect

Staff Emeritus
It sounds to me like the OP is still struggling with understanding the Newtonian view. I'd suggest re-opening the problem, formulated in terms of classical gasses or particles rather than "photons", in the classical physics forum.

I think at this point trying to explain the SR view further would be counterproductive, as would trying to get into the subtilities of GR's notion(s) of mass.

11. Nov 17, 2012

### Simon Bridge

Those models were introduced by OP. Perhaps it is better to just point out that these are different models and OP should focus on one model at a time first, before worrying about the relationship between them?

However, I think the core misunderstanding has been unearthed.

12. Nov 17, 2012

### atyy

Does this mean that if the gas is not in equilibrium then the scale will not read Mg? Like if the gas consisted of just one molecule?

13. Nov 17, 2012

### Simon Bridge

If the gas is one molecule, is it, therefore, a gas?
How does this apply to OPs question?

14. Nov 18, 2012

### atyy

I was wondering how important the equilibrium assumption was, and thought that simplifying to one molecule would illustrate the physics. I guess another way to ask about the assumption is whether small deviations from the average due to a finite number of molecules would make a difference (in principle)?

15. Nov 18, 2012

### Simon Bridge

Sounds like a different question - though it may be what OP was trying to get at by referring to different models.

We know that there are real-world deviations from the classical particles-in-a-box models.
It's partly why we have quantum mechanics.

16. Nov 18, 2012

### Staff: Mentor

If you had just one molecule of a gas then the center of mass would be going up and down and accelerating, and the weight measured on the scale would be consistent with the motion of the center of mass of the system. If the center of mass of the system is not moving then the scale will simply measure the weight of the system, including the gas.

Last edited: Nov 18, 2012
17. Nov 18, 2012

### Staff: Mentor

While the gas is reaching equilibrium the scale may not read Mg (depending on the specific disequilibrum).

This isn't as extraordinarily curious as it sounds though. It's pretty much the same situation as if I were to jump onto a spring-operated scale - first the scale would read heavy as I smashed down on on it, then light as I bounced up, then heavy again as I came down again. But eventually the energy of my jump will be dissipated, the bouncing will stop, equilibrium will be reached, and the scales will read my weight

A sealed box of gas reaches equilibrium pretty quickly.

18. Nov 18, 2012

### atyy

@DaleSpam and Nugatory, thanks!

19. Nov 18, 2012

### pervect

Staff Emeritus
Speaking classical Newtonain physics:

I'd interpret this situation as the box being in brownian-like motion. It's not quite the same, in true brownian motion the random pressure variations push on the outside of the box rather than the inside, but it should be similar.

Because the center of gravity of the box (still speaking classical Newtonian physics) must not move, the statistics will probably be different in detail than in true Brownian motion.

If you average the weight of the box over a long enough time scale, it should remain constant. This is expected by the conservation of momentum.

The Mythbusters episode "Birds in a Truck" might also be interesting and related. Here we have non-equilibrium due to the birds flapping their wings in a sealed box / truck. See http://mythbustersresults.com/episode77

20. Nov 18, 2012

### atyy

Let me see if I understand correctly. The single molecule case is DaleSpam's first case where the CM moves, and the result holds only for long time-averaged values. "Birds in a Truck" is DaleSpam's second case, where the situation is non-equilibrium, but the CM does not move so the result is true, even without averaging?

(Yes, all Newtonian, I assume that's ok from the GR point of view since we expect to be in a regime covered by naive EP?)