What is the mechanism behind the weight of a box of photons?

In summary: Mass and gravity are connected in this way.2) How does the interior mass affect the box without a direct connection in order to have the scale register that mass??The weight of the gas affects the scale - similarly, the mass of the photons affects the scale.
  • #1
Austin0
1,160
1
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
 
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  • #2
Austin0 said:
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.
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?
This suggests two questions
1) What exactly is the scale measuring if not total mass??
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.

it does not seem plausible that the velocity difference applied to this small mass could result in a momentum reflecting the total mass.
Personal intuition is not a good guide - do the math.

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

How do the internal photons interact with the box to cause an instantaneous reaction of the scale equivalent to the total number?
Same as the gas - but what makes you think it is instantanious?
The internal pressure increases the stress energy of the box which increases the total weight equivalent to the gas mass.
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.

Since none of the above seem convincing
"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.
 
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  • #3
Austin0 said:
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.
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
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.


DaleSpam said:
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.

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
Austin0 said:
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.

Simon Bridge said:
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?

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

Austin0 said:
This suggests two questions
1) What exactly is the scale measuring if not total mass??

Simon Bridge said:
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.
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.

Austin0 said:
it does not seem plausible that the velocity difference applied to this small mass could result in a momentum reflecting the total mass.

Simon Bridge said:
Personal intuition is not a good guide - do the math.
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??

Austin0 said:
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.

Simon Bridge said:
For a perfectly reflecting surface, the frequency should not change. Momentum is a vector.
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??



Austin0 said:
How do the internal photons interact with the box to cause an instantaneous reaction of the scale equivalent to the total number?

Simon Bridge said:
Same as the gas - but what makes you think it is instantanious?
Because the scale is reflecting the total number of internal photons at an instant

Austin0 said:
The internal pressure increases the stress energy of the box which increases the total weight equivalent to the gas mass.

Simon Bridge said:
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.
Yes I merely mentioned stress energy as a possible explanation for the "instantaneous" effect on the box and scale.
 
  • #6
Austin0 said:
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??

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 let's 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
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
Austin0 said:
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??
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
Just noticing this:
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?
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
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
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.
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
The_Duck said:
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.

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
If the gas is one molecule, is it, therefore, a gas?
How does this apply to OPs question?
 
  • #14
Simon Bridge said:
If the gas is one molecule, is it, therefore, a gas?
How does this apply to OPs question?

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
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
atyy said:
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?
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.
 
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  • #17
atyy said:
Does this mean that if the gas is not in equilibrium then the scale will not read Mg?

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
@DaleSpam and Nugatory, thanks!
 
  • #19
atyy said:
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
 
  • #20
pervect said:
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?)
 
  • #21
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
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
atyy said:
(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).
 
  • #24
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Austin0 said:
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.

The_Duck said:
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.

Simon Bridge said:
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.
`
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.


Austin0 said:
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??
________
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
 
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  • #25
DaleSpam said:
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??

How about 100 photons?
 
  • #26
Austin0 said:
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.
 
  • #27
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?
Nugatory said:
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.

Nugatory said:
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 anyone 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.
 
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  • #28
Why would the momentum imparted by the number hitting the bottom at anyone 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.
 
  • #29
Austin0 said:
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 anyone 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.
 
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  • #30
Austin0 said:
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.
 
  • #31
Austin0 said:
Since none of the above seem convincing I am thinking I must be missing some obvious fundamental factor ?. Any insights welcome

Nugatory said:
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
 
  • #32
DaleSpam said:
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.
But thanks for your help.
 
  • #33
Austin0 said:
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.
I disagree. You knew the correct answer from the OP, but simply did not believe that the magnitude of the effect could correctly account for the necessary answer. That makes it a quantitative question.

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 non-experts 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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1. What is the concept of weight in relation to photons?

The concept of weight in relation to photons is a bit different from our traditional understanding of weight. Weight is typically defined as the force of gravity acting on an object. However, photons do not have mass and therefore are not affected by gravity in the same way as objects with mass. Instead, the weight of a photon is determined by its energy and momentum.

2. How does the energy and momentum of a photon contribute to its weight?

According to Einstein's famous equation E=mc^2, energy and mass are equivalent. This means that even though photons do not have mass, they do have energy. The energy of a photon is directly proportional to its frequency, and the momentum is proportional to its wavelength. Therefore, the energy and momentum of a photon contribute to its weight by giving it a certain amount of "inertia" or resistance to acceleration.

3. What is the mechanism behind the weight of a box of photons?

The weight of a box of photons can be explained by the collective energy and momentum of all the photons within the box. Each individual photon has a small amount of weight, but when combined, their weight becomes significant. This is similar to how the weight of a group of objects is greater than the weight of each individual object.

4. How does the weight of a box of photons compare to the weight of a box of matter?

The weight of a box of photons is significantly less than the weight of a box of matter with the same volume. This is because matter has both mass and energy, while photons only have energy. Additionally, the energy of photons is much higher than the energy of matter, making their weight much smaller in comparison.

5. Can the weight of a box of photons be measured?

Yes, the weight of a box of photons can be measured using a device called a photon scale. This device uses the principles of energy and momentum to measure the weight of photons. However, the weight of a box of photons may be difficult to measure accurately due to the extremely small amount of weight each photon has.

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