# Weight of a box of PHOTONS

by Austin0
Tags: photons, weight
Emeritus
P: 7,204
 Quote by 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?
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
P: 7,399
 Quote by pervect 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
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?)
 Mentor P: 15,555 Yes, the birds in a truck myth busters episode is actually what got me thinking along these lines in the first place. If you look at the weight signal in the episode, the weight is approximately constant after some time, but there is an initial transient as the birds and structure first fall. The center of mass drops, and the signal shows it.
 HW Helper Sci Advisor Thanks P: 8,998 I suppose this could be thought of as a particularly loose example of measuring anything? A solid object is composed of particles jumping around - at any time the particles are all headed in random directions and one could imagine that a sufficiently sensitive scale would read these small bounces. BTW: I saw (I think it was) Magnus Pike do a similar demo with a toy helecopter.
Emeritus
P: 7,204
 Quote by atyy (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?)
Things start to get a lot more complicated in GR. The point where things start to get complicated is where the metric coefficients (usually, g_00) varies enough over the region of the box so as to be "noticeable".

Another, different pitfall here is the relativity of simultaneity. The box isn't an isolated system if you have it on a scale. So if you discuss things from the SR perspective, E^2 - p^2 won't be invariant until you isolate the box (which means taking it off the scale).
P: 1,162
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 Quote by Austin0 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.
 Quote by 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. ##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.
 Quote by 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.
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As you can see although I was talking about momentum rather than pressure I was taking into account the effect of gravity.
And isn't pressure fundamentally impact momentum per area??

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

Here it appears that weight (-Mg) is directly equivalent to the pressure differential ## -P_0A + (P_0 - \rho g h)A = -Mg##

As all other factors are constant doesn't this mean that P0 varies directly with Temperature.' It seems to follow from this that the value of -Mg would then vary linearly with temperature also.

Would you say this was correct??

Although i don't doubt that the input of heat energy would have some infinitesimal increase in mass equivalence but this does not seem to support an increase in weight comparable to an increase in pressure.

 Quote by Austin0 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 number could be affecting the bottom of the box and the scale at any moment correct??
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Nobody has responded to this simple model.
DO you have an explanation how in this model the velocity differential between the top and bottom at any instant could, applied to the small percentage of mass actually acting on the box, be equivalent to the entire mass acting on the box???

Or how this would take place with photons on parallel , non-interactive up and down paths?
Thanks
P: 1,162
 Quote by DaleSpam 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.
SO in the one molecule system the measured weight would fluctuate between two values.
One, the correct total for the system and another actually less than correct total ,yes???

So if the number is increased to say 100 particles the COM would have much less range of deviation assuming random distribution, but the measured total weight would still have some fluctuation but now between two values that are both less than the actual total weight. Or do you see a different relationship pertaining??

P: 2,350
 Quote by Austin0 SO in the one molecule system the measured weight would fluctuate between two values. One, the correct total for the system and another actually less than correct total ,yes???
It would fluctuate between two values, but one would be too high and one would be too low - not one correct and one low. That matters a lot, because..

 So if the number is increased to say 100 particles the COM would have much less range of deviation assuming random distribution, but the measured total weight would still have some fluctuation but now between two values that are both less than the actual total weight.
Less deviation, yes. But because one extreme is too high and the other is too low, less deviation means that it's converging on the actual weight.
P: 1,162
Quote by Austin0
 SO in the one molecule system the measured weight would fluctuate between two values. One, the correct total for the system and another actually less than correct total ,yes???
[QUOTE=Nugatory;4165063]It would fluctuate between two values, but one would be too high and one would be too low - not one correct and one low. That matters a lot, because..

yes you are correct there would be a complex fluctuation determined by the relative impact angles and the propagation time for the momentum to reach the scale and vice versa.
Quote by Austin0
 So if the number is increased to say 100 particles the COM would have much less range of deviation assuming random distribution, but the measured total weight would still have some fluctuation but now between two values that are both less than the actual total weight.
 Quote by Nugatory Less deviation, yes. But because one extreme is too high and the other is too low, less deviation means that it's converging on the actual weight.
Could you explain why the one extreme would be too high??
I.e., Why would the momentum imparted by the number hitting the bottom at any one time or infinitesimal interval be greater than the momentum of the total.

What would you guess the maximum number could be 7 or 8 ?? Even that seems pushing it given random distribution. And of course it is the difference between the top and bottom that is germane.
HW Helper
Thanks
P: 8,998
 Why would the momentum imparted by the number hitting the bottom at any one time or infinitesimal interval be greater than the momentum of the total.
momentum is a vector... on average the momenta will cancel out but for the difference due to gravity. However, there is a non-zero but vanishingly small probability that a lot of particles will happen to strike the bottom at once. This will give a too-high value. However, the more particles you have the lower the deviation.
 That matters a lot, because..., less deviation means that it's converging on the actual weight.
Well done.
P: 2,350
 Quote by Austin0 Could you explain why the one extreme would be too high?? I.e., Why would the momentum imparted by the number hitting the bottom at any one time or infinitesimal interval be greater than the momentum of the total?
We aren't in equilibrium yet, so sometimes there will be an excess of downwards-moving molecules causing the scale to read too high a weight, and other times an excess of upwards-moving molecules causing the scale to read too low a weight. Average over time and these fluctuations will cancel out, and as we approach statistical equilibrium their magnitude will approach zero.

Consider a single molecule, initially moving downwards. It hits the bottom of the box and rebounds upwards - the scale is forced down and registers an increase in weight. But once the molecule starts upwards again, the scale also rebounds, accelerating the box upwards. Thus, at the position where the weight of the box and the upwards force of the scale would be exactly balanced at equilibrium, the box is moving upwards and the scale reads a bit low. And eventually the upwards-moving rebounding molecule hits the top of the box and rebounds again downwards, nudging the box up a bit more, further reducing the reading the scale reading. But now the molecule is heading back down, and gravity is pulling the box back down, so the cycle repeats.

Compute the average over time of the upwards fluctuations and the downwards fluctuations, and it will come out to mg, where m is the mass of the single molecule.

Consider a large number of molecules, all going through this cycle at their own rate with a random statistical distribution of when they're going up and when they're going down, and you'll get an average of Mg where M is the total mass of all the molecules.
Mentor
P: 15,555
 Quote by Austin0 SO in the one molecule system the measured weight would fluctuate between two values. One, the correct total for the system and another actually less than the correct total ,yes??? So if the number is increased to say 100 particles the COM would have much less range of deviation assuming random distribution, but the measured total weight would still have some fluctuation but now between two values that are both less than the actual total weight. Or do you see a different relationship pertaining?? How about 100 photons???
Please work out the math on this as I have suggested previously. Your conclusion is incorrect.

For a single molecule the measured weight fluctuates between three values depending on the acceleration of the COM. One value is higher than the total weight (molecule hitting bottom) and two are lower (molecule hitting top, molecule in free fall).

Since the COM is not moving when averaged over time the measured weight averaged over time is equal to the total weight, including the gas molecule. Similarly when averaging over many molecules.
P: 1,162
 Quote by Austin0 Since none of the above seem convincing I am thinking I must be missing some obvious fundamental factor ??????. Any insights welcome
 Quote by Nugatory Consider a single molecule, initially moving downwards. It hits the bottom of the box and rebounds upwards - the scale is forced down and registers an increase in weight. But once the molecule starts upwards again, the scale also rebounds, accelerating the box upwards. Thus, at the position where the weight of the box and the upwards force of the scale would be exactly balanced at equilibrium, the box is moving upwards and the scale reads a bit low. And eventually the upwards-moving rebounding molecule hits the top of the box and rebounds again downwards, nudging the box up a bit more, further reducing the reading the scale reading. But now the molecule is heading back down, and gravity is pulling the box back down, so the cycle repeats. Compute the average over time of the upwards fluctuations and the downwards fluctuations, and it will come out to mg, where m is the mass of the single molecule. Consider a large number of molecules, all going through this cycle at their own rate with a random statistical distribution of when they're going up and when they're going down, and you'll get an average of Mg where M is the total mass of all the molecules.
As you can see in my OP I had it right.
Somehow I became focused on the instantaneous state of the system and like Zeno's arrow I got stuck and could not see the obvious.
Your qualitative view set me straight where looking at it over time it is glaringly obvious.
It was just such a view of fundamental process I was looking for but escaped me
Thanks
P: 1,162
 Quote by DaleSpam Please work out the math on this as I have suggested previously. Your conclusion is incorrect. For a single molecule the measured weight fluctuates between three values depending on the acceleration of the COM. One value is higher than the total weight (molecule hitting bottom) and two are lower (molecule hitting top, molecule in free fall). Since the COM is not moving when averaged over time the measured weight averaged over time is equal to the total weight, including the gas molecule. Similarly when averaging over many molecules.
Hi thanks for your input . Of course everything you said was perfectly correct except for the shut up and calculate advice.

It was not really a quantitative question. All the final values were given. As far as I can see the gas maths themselves produce statistical net results. Instantaneous state conditions. They do not directly entail the dynamic processes behind those net results.
But that was the source of my question and my problem, looking at the instantaneous state rather than the processes over time.
A conceptual question requiring a conceptual resolution.