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Weight of a box of PHOTONS 
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#19
Nov1812, 03:30 PM

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I'd interpret this situation as the box being in brownianlike 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 nonequilibrium due to the birds flapping their wings in a sealed box / truck. See http://mythbustersresults.com/episode77 


#20
Nov1812, 05:28 PM

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(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?) 


#21
Nov1812, 05:52 PM

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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.



#22
Nov1812, 06:07 PM

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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. 


#23
Nov1812, 06:38 PM

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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). 


#24
Nov1812, 08:15 PM

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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 P_{0} 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. 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 , noninteractive up and down paths? Thanks 


#25
Nov1812, 08:32 PM

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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?? How about 100 photons??? 


#26
Nov1812, 08:41 PM

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#27
Nov1812, 09:18 PM

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Quote by Austin0
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 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. 


#28
Nov1812, 09:25 PM

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#29
Nov1812, 10:18 PM

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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 upwardsmoving 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. 


#30
Nov1912, 06:06 AM

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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. 


#31
Nov1912, 07:15 PM

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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 


#32
Nov1912, 07:25 PM

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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. But thanks for your help. 


#33
Nov1912, 09:23 PM

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Besides, the process of calculating the result is an important learning experience that all of the experts on the forum have gone through in order to become experts. I don't understand why so many nonexperts seem annoyed when experts recommend using the most effective method for gaining expertise. I am sad that you deprived yourself of the opportunity and seem oblivious to the enormous value in it. 


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