Does pressure resist acceleration?

[AstroRoyale]

Unlike 1effect, I don't have any intuition as to why the pressure of an ideal gas should be an invariant, in doubt I'd take that stance on the safe side and suppose that it isn't an invariant. I haven't looked at the proof carefully yet, but I can and did offer the proof based on the stress/energy tensor that the pressure is not an invariant for an ideal gas.

\Lambda^{T} T \Lambda shows that p_xx does change during a boost. I think that when it comes to this, it really is best to discuss T^{\mu \nu} instead of 4 vectors. For instance, when talking about lorentz transforming the electric and magnetic fields, we need the field tensor F^{\mu \nu} to properly describe how the fields change, and how a pure electric field can become an electromagnetic field. The stress energy tensor is not just a General relativity object either, it applies in special relativity too of course, we can make a stress/energy tensor for the EM fields out out the field tensor for example. See Jackson for example. And the T^{\mu \nu} I gave is the same in general or special relativity anyway.

 

[1effect]

Unlike 1effect, I don't have any intuition as to why the pressure of an ideal gas should be an invariant, in doubt I'd take that stance on the safe side and suppose that it isn't an invariant. I haven't looked at the proof carefully yet,

Hi,

The proofs are all flawed, as pointed out earlier. We couldn't come to any agreement that any of the proofs are valid. Moreover, as you (and dharis before you), pointed out, the results contradict the way the stress-energy tensor transforms. So, we are hopelessly stuck.

 

[AstroRoyale]

Well, until someone shows me a proof that I can trust, I do believe that the pressure of an ideal gas is not a Lorentz invariant. For continuous media/fields (ie non point particles), the stress-energy tensor seems required if we want to talk about what happens when we boost these things. Is that a correct assumption?

I suppose that we could always use the stress-energy tensor in any case, but it takes the form of T_{00}=m and the other 15 components are zero for a point-like particle in the rest frame of the paticle so it can essentially be ignored.

 

[Xeinstein]

Well, until someone shows me a proof that I can trust, I do believe that the pressure of an ideal gas is not a Lorentz invariant. For continuous media/fields (ie non point particles), the stress-energy tensor seems required if we want to talk about what happens when we boost these things. Is that a correct assumption?

I suppose that we could always use the stress-energy tensor in any case, but it takes the form of T_{00}=m and the other 15 components are zero for a point-like particle in the rest frame of the paticle so it can essentially be ignored.

Suppose we have a box at rest that is filled with a uniform gas. We denote the volume by V and the pressure by p. Suppose next that we apply a small force to the box and accelerate it until it has a speed v. If the pressure of the box increase during acceleration in the observer's frame in which the box is moving, then eventually it could burst the box. But viewed from box own frame in which the box is at rest, the pressure will always be p since it's at rest, so it can't burst the box.
Since in relativity you cannot have something failing in one frame and being normal in another frame then you must come to the conclusion that Lorentz-contraction does not cause pressure to increase in general.

 

[matheinste]

Hello Xeinstein.

This is a scenario of the type used in many imagined paradoxes of SR. I think care must be taken in using it because on the face of it it may imply that nothing can happen to the contents of the box because, if anything did, any change is potentially detectable by the observer at rest in the frame of the box. The very fact that this discussion is happening implies that some effects are (or may be) detectable by an observer in a frame moving relative to the frame of the box. We know some things are detectable by the moving observer. The volume, as seen by the observer in the moving frame, changes because the length of one side contracts (in the standard configuration) and as nothing is detectable in the stationary observer's frame then this is obviously not detectable by a length measurement, or any other measurement, in the stationary observer's frame.

What i am really asking is "what is the nature of the effects that can be measured in the moving obsever's frame but not in the stationary observer's frame".

And as pointed out to me earlier in this thread, in the case of a gas the physics are a little more complicated than appears if you do not know, as i do not, all the mechanisms involved and take account of them.

The absolute fundamentals, even from a logical point of view, of what effects can and cannot be detected or seen by an observer in a frame moving with respect to a physical system may be worth discussion in another thread.

Matheinste.

 

[pervect]

Hi,

The proofs are all flawed, as pointed out earlier

I don't agree - I attempted to work the problem out myself, and got the same answer as Kev did and a different answer from 1effect. However I would say there is a lot of semantic confusion about what is being proved, and would agree that nobody has come up with any good papers or textbook references to address the point in question.

For the most part, people simple do not use the concept of "pressure" except in the rest frame of the fluid, so far as I'm aware. This makes it difficult to find said textbook references.

I also still think there is some confusion on 1effects part where he assumes that the stress-energy tensor is the same thing as pressure. This is again a semantic issue about what the term "pressure" really means. As nearly as I can tell, it probalby means what kev says, and not what 1effect says.

 

[1effect]

I don't agree - I attempted to work the problem out myself, and got the same answer as Kev did and a different answer from 1effect.

I did not get any answer,if we except this one. I just tried to help kev debug his many solutions. To date, they all have unresolved errors. I would be interested in seeing your solution.

However I would say there is a lot of semantic confusion about what is being proved, and would agree that nobody has come up with any good papers or textbook references to address the point in question.

This is the most important part, finding a good reference, there has to be one somewhere.

I also still think there is some confusion on 1effects part where he assumes that the stress-energy tensor is the same thing as pressure. This is again a semantic issue about what the term "pressure" really means. As nearly as I can tell, it probalby means what kev says, and not what 1effect says.

There are quite a few people that believe the same thing (dhris, AstroRoyale). What disturbs me are two things:

-all of the "published" solutions have errors
-we get disagreement with the only referenced solution, the one that uses the stress-energy tensor

You (pervect) cannot simply waive your magic wand and declare the discrepancy a "confusion" in terms of use of concepts. The reason is that there is no valid solution for the non-tensor treatment. Unless you have one :-) This is why, I'd like to see it.

 

[AstroRoyale]

Well, I think we've happened upon a conundrum. I've been looking around, and apparently there is still considerable discussion as to how even the temperature changes in a Lorentz transformation. Apparently, there are lines of reasoning that lead to increases, decreases and invariance of temperature as measured in the boosted frame. The paper below claims that the pressure IS invariant, but I'm still not convinced personally. I'm going to ask around though, this is really bugging me for some reason. Einstein and statistical thermodynamics. I. Relativistic thermodynamics

P T Landsberg 1981 Eur. J. Phys. 2 203-207

Not terribly recent, but I see some more very recent articles (2007) where the same inconsistencies are seen.

 

[AstroRoyale]

And I just found an article of astroph that claims that the pressure increases as

p=\gamma^{2}p

http://arxiv.org/abs/0712.3793

 

[1effect]

And I just found an article of astroph that claims that the pressure increases as

p=\gamma^{2}p

http://arxiv.org/abs/0712.3793

Interesting, thank you!

 

[yuiop]

And I just found an article of astroph that claims that the pressure increases as

p=\gamma^{2}p

http://arxiv.org/abs/0712.3793

In that article the author states that T ' =\gamma T and that p ' =\gamma^{2}p


Ice (and many other substances) have the property that the melting point is reduced by increasing pressure. Say we place a block of ice at 0 deg C (273K) in a box of air at -1 deg C (272K) and one atm pressure. An observer seeing the box moving relative to him at a velocity such that the gamma factor is 10 will see the temperature of the air as 2447 deg C (2720K) and the pressure in the box as 100 atm or 10 MPa. At that pressure (see http://en.wikipedia.org/wiki/Image:Melting_curve_of_water.jpg) the melting point of ice will be less than 0 deg C so the ice should have melted (even if no heat transfers to the ice from the hot air). The observer at rest with the box sees the the ice as a solid block while the moving observer sees a puddle of water in the box.

 

[1effect]

In that article the author states that T ' =\gamma T and that p ' =\gamma^{2}p


Ice (and many other substances) have the property that the melting point is reduced by increasing pressure. Say we place a block of ice at 0 deg C (273K) in a box of air at -1 deg C (272K) and one atm pressure. An observer seeing the box moving relative to him at a velocity such that the gamma factor is 10 will see the temperature of the air as 2447 deg C (2720K) and the pressure in the box as 100 atm or 10 MPa. At that pressure (see http://en.wikipedia.org/wiki/Image:Melting_curve_of_water.jpg) the melting point of ice will be less than 0 deg C so the ice should have melted (even if no heat transfers to the ice from the hot air). The observer at rest with the box sees the the ice as a solid block while the moving observer sees a puddle of water in the box.

Very interesting counterargument, kev, I like it.
The only problem I can see with your line of argument is that, due to the noblinearity of the curve cited above (see http://en.wikipedia.org/wiki/Image:Melting_curve_of_water.jpg) any transformation for p and T except from the identity:

p'=p and T'=T

will generate the same exact discrepancy between what observers see from different frames.

 

[AcidBathSDMF]

I'm not sure the ideal gas law takes relativistic effects into account and doesn't apply except in the rest frame. I'm not an expert by any means, but it seems to me that pressure, as a macroscopic model, should be dissected into what's going on in the smaller level. Are we assuming that these gas molecules are billiard balls bouncing around? Assuming this is legitimate, the frame at rest with the box has <v> = 0, where v is velocity of molecules. In a frame that's moving in x-direction, <v_x> will not be zero, and should equal u. v'=x'/t', and v= (v'+u)/(1+uv'/c^2). Supposing for a second that it were possible to take many molecule velocity measurements, one could find <v> of the molecules in the moving frame. This value could be u, and the deviations from it are the random molecular movements that give a pressure. Subtracting this u from the above expression for v to find the molecule speeds in the box, it gives v=v'(1-u^2/c^2)/(1+uv'/c^2). In other words, the measured velocity of the molecular movement is less than that in the rest frame, even after subtracting out the relative velocity of the observer. My intuition tells me that the pressure effects of length contraction of the box are compensated for by the slower moving molecules, and a similar argument for temperature.

 

[yuiop]

I'm not sure the ideal gas law takes relativistic effects into account and doesn't apply except in the rest frame. I'm not an expert by any means, but it seems to me that pressure, as a macroscopic model, should be dissected into what's going on in the smaller level. Are we assuming that these gas molecules are billiard balls bouncing around? Assuming this is legitimate, the frame at rest with the box has <v> = 0, where v is velocity of molecules. In a frame that's moving in x-direction, <v_x> will not be zero, and should equal u. v'=x'/t', and v= (v'+u)/(1+uv'/c^2). Supposing for a second that it were possible to take many molecule velocity measurements, one could find <v> of the molecules in the moving frame. This value could be u, and the deviations from it are the random molecular movements that give a pressure. Subtracting this u from the above expression for v to find the molecule speeds in the box, it gives v=v'(1-u^2/c^2)/(1+uv'/c^2). In other words, the measured velocity of the molecular movement is less than that in the rest frame, even after subtracting out the relative velocity of the observer. My intuition tells me that the pressure effects of length contraction of the box are compensated for by the slower moving molecules, and a similar argument for temperature.

Your statement "In other words, the measured velocity of the molecular movement is less than that in the rest frame" is one way of defining temperature which is more formally something like temperature is proportional to the root mean square kinetic energies of the molecules. So that is another way of saying the gas appears to cool down in the moving box. That is a possibility that can not be excluded as the topic of relativistic thermodynamics appears to be hotly debated amongst the experts with peer reviewed published papers contradicting each other.

Does the ideal gas law PV= nRT hold up as a universal law of physics in any inertial reference frame?

The authors of the paper mentioned in the posts above seem to think it is.

The relativistic version of the gas law (according to them) would appear to be:

(P\gamma^2)\left({V \over \gamma}\right) = nR(T\gamma) which reduces to PV = nRT

If we assume that pressure is invariant and also assume the ideal gas law is universal then:

(P)\left({V \over \gamma}\right) = nR({T \over \gamma}) which also reduces to PV = nRT

However at this point we are not absolutely sure the ideal gas law is universal in the relativistic context without modification. The discussion between 1effect and myself that any change of P or T with relative motion would require all melting points and boiling points of all known materials to be adjusted up and down in complicated ways which can not be ruled out, but is certainly far from elegent. However, nature makes the rules and if nature is not elegant we just have to live with that :P

If we assume the ideal gas law is open to modification to make it a universal law, then it could end up as something like this:

P\left({V \over \gamma}\right) = nRT

I suspect the final law will be something altogether different just as the relativistic version of kinetic energy looks nothing like the Newton formula for kinetic energy.

In this very recent paper http://arxiv.org/PS_cache/gr-qc/pdf/9505/9505045v1.pdf by Matsas he states “In particular, the question of how temperature transforms under Lorentz transformations led some distinguished physicists to reach exactly the opposite conclusion of other equally distinguished ones.”

In that paper it not clear (to me) if he is talking about a thermometer moving relative to a (stationary?) background radiation or a thermometer co-moving with the background radiation. He does however make the interesting point that there will be differences between thermometers that just measure the infra red part of the spectrum and thermometers that measure the entire spectrum. It also seems that he talking about temperature of a moving object that is measured by a thermometer that is measuring the blackbody radiation of the moving body which is another interpretation of temperature.

Any final definition of temperature would have to be consistent with:

A relativistic ideal gas law.
A relativistic black body radiation law.
A relativistic definition of entropy and enthalpy.

Defining temperature in terms of Entropy can be complicated by issues of combining (or keeping separate?) thermal entropy and “information” entropy. For an example if information entropy and temperature we can look at Hawking’s definition of the temperature of a black hole in terms of its surface area.

For now, it would instructive to investigate further why we have a discrepancy between the pressure computed by the stress-energy tensor and gas pressure evaluated by more simple means. A start would be for someone to give a clear prediction from the stress-energy tensor of how “pressure” varies with velocity.

Does the stress-energy tensor predict P ‘ = P/\gamma or P ‘ = P\gamma or P ‘ = P\gamma^2 or something else?

 

[Xeinstein]

You seem to have a misconception about the nature of the Lorentz contraction. You seem to be thinking of it as an active process, as opposed to a transformation of measurements between frames. Example: Imagine a can of beer sitting on the table. A rocket ship goes by at high speed and--amazingly enough--observers on the rocket are able to perform measurements of the beer can as they fly by. Is the can and its contents Lorentz contracted? Of course! Does that somehow increase the pressure in the can? Do you expect anything unusual to happen to the can of beer? Do you think that pressure can somehow increase so as to burst the can as viewed in the frame of the rocket? I'm being a bit facetious, but I hope you see the point.

Of course, subtle things can happen when you accelerate an object. (See all the discussions about the Bell spaceships.) But your first sentence said "We all agree that "in its own frame", the box has not shrunk." That implies that you have accelerated the box in such a way as to preserve its proper length (as opposed to stretching or squashing it), thus you have introduced no stresses on the box or its contents whatsoever.

No, I don't think I have a misconception about the nature of the Lorentz contraction. The Lorentz contraction is Not an active process but it's physically real, The big difference is that the Lorentz-contraction does not increase the pressure in the can as you stated, so pressure would Not burst the can. But it Does shorten the length of the can and therefore change its volume which has physical consequence.

In his book: Gravity from the Ground Up, Schutz indicate that the pressure of box is a constant in the observer's frame in which the box is moving. But making a box smaller when it contains a gas/fluid with pressure p requires one to do work on it, in other words to put some energy into the gas. This extra energy represents the extra inertia of the gas. The difference between inertial mass and rest mass is this extra inertia of pressure. This is purely a consequence of special relativity.

Here is the link to Schutz's book http://books.google.com/books?id=P_...ts=eYBnh8oGoa&sig=ZkzEIBINItUiFyMW5-uvjt1kMus

 

[Xeinstein]

You seem to have a misconception about the nature of the Lorentz contraction. You seem to be thinking of it as an active process, as opposed to a transformation of measurements between frames. Example: Imagine a can of beer sitting on the table. A rocket ship goes by at high speed and--amazingly enough--observers on the rocket are able to perform measurements of the beer can as they fly by. Is the can and its contents Lorentz contracted? Of course! Does that somehow increase the pressure in the can?

Yes, Lorentz-contraction does increase the pressure

Do you expect anything unusual to happen to the can of beer? Do you think that pressure can somehow increase so as to burst the can as viewed in the frame of the rocket?

No, pressure can't increase so as to burst the can. You forget time dilation.
Time dilation slows down random motion of gas particles. Both effects, Lorentz-contraction and time dilation, cancel each other, so the pressure is a constant in the observer's frame in which the can is moving

see Schutz book's http://books.google.com/books?id=P_...ts=eYBnh8oGoa&sig=ZkzEIBINItUiFyMW5-uvjt1kMus

 

[Doc Al]

No, I don't think I have a misconception about the nature of the Lorentz contraction. The Lorentz contraction is Not an active process but it's physically real, The big difference is that the Lorentz-contraction does not increase the pressure in the can as you stated, so pressure would Not burst the can. But it Does shorten the length of the can and therefore change its volume which has physical consequence.

As I stated earlier, there's a difference between doing a Lorentz transformation from one frame to another and worrying about the details of how one accelerates something. If you're just transforming to another frame, which is hard enough because all the forces and accelerations must be transformed, no stresses whatsoever are induced in the object (of course).

For example, imagine a stick at rest in one frame. All the molecules are in some equilibrium position as the stick is under no particular stress. If I transform to a frame in which the stick is moving, it will be Lorentz contracted. I would think that even in that new frame, the molecules must be in equilibrium. Of course, working out the details of how the intermolecular forces transform might not be trivial, but the answer must be.

Of course when you accelerate the stick, you have to be careful about how you do it. If you just push too hard on one end, you will create stresses in the stick. Not caused by the "Lorentz transformation" but by the forces applied when producing the acceleration.

In his book: Gravity from the Ground Up, Schutz indicate that the pressure of box is a constant in the observer's frame in which the box is moving. But making a box smaller when it contains a gas/fluid with pressure p requires one to do work on it, in other words to put some energy into the gas. This extra energy represents the extra inertia of the gas. The difference between inertial mass and rest mass is this extra inertia of pressure. This is purely a consequence of special relativity.

Here is the link to Schutz's book http://books.google.com/books?id=P_...ts=eYBnh8oGoa&sig=ZkzEIBINItUiFyMW5-uvjt1kMus

As I've mentioned before, I don't quite follow Schutz's reasoning. (If I ever get hold of his book, I'll read his full argument.)

He's saying (if I understand him) that if you have two boxes with invariant mass m, the one with a pocket of gas inside will be harder to accelerate because you have to increase the internal energy of the gas as well as increase its kinetic energy. A very interesting conclusion! (And, no, the extra inertia due to pressure is not simply the difference between inertial and rest mass.)

If you understand his argument, why don't you spell it out in detail? I suspect that you're not sure yourself, else you wouldn't keep bringing it up. (Over and over again. And what's the deal with you posting, deleting, then reposting... over and over again?)

 

[Prologue]

He's saying (if I understand him) that if you have two boxes with invariant mass m, the one with a pocket of gas inside will be harder to accelerate because you have to increase the internal energy of the gas as well as increase its kinetic energy. A very interesting conclusion! (And, no, the extra inertia due to pressure is not simply the difference between inertial and rest mass.)

This is interesting but I am still thinking there must be something missing. If a box with gas in undergoing acceleration then the gas molecules will tend to pile up on one side of the box, the end opposite of the acceleration. So if you accelerate the box then try to measure the pressure, the density will not be uniform throughout. You would have to measure the total pressure in the box which should sum to the initial pressure.

This does bring up something that others may be more familiar with than I, but how is the temperature affected in the total volume of the gas in the box? Would the temperature at the front of the box (low density) be lower than the back (high density). If so, will the total change in temperature be zero for the gas (still under acceleration)?

 

[pervect]

This thread gets a bit confused for several reasons, one of which is because Schutz is talking about a contrafactual case. Specifically, Schutz says:

"For simplicity, we will only ask about the gas, not about the container." But you can't really talk about the gas and not the container.

So as I remarked much earlier, if you have an actual container of gas in flat space-time, you do not see this effect. The pressure inside the gas is exactly balanced by the tension in the container, and there is no effect of pressure on the mass of the whole system.

You will find that in flat space-time, the Komar mass formula gives no different answer than the SR mass formula. The SR formula does not incorporate pressure, but in any closed system in flat space-time, the integral of the pressure terms over the entire system tuns out to be zero.

If you have a star, you have no container as Schutz also remarks. However, you are also not in flat space-time, so the previous remarks do not apply. The SR concept of mass does not apply in flat space-time, however the Komar mass does, as long as the star is static, that is.

See http://en.wikipedia.org/w/index.php?title=Komar_mass&oldid=170982833 for more information on the Komar mass - note that while it is an intergral of (rho+3P), so that both density and pressure contribute to it, this integral is a *weighted* integral, it's weighted by the redshift factor K, which can be formally defined as \xi^a \xi_a, where \xi is a Killing vector of the static space-time.

 

[Doc Al]

This thread gets a bit confused for several reasons, one of which is because Schutz is talking about a contrafactual case. Specifically, Schutz says:

"For simplicity, we will only ask about the gas, not about the container." But you can't really talk about the gas and not the container.

So as I remarked much earlier, if you have an actual container of gas in flat space-time, you do not see this effect. The pressure inside the gas is exactly balanced by the tension in the container, and there is no effect of pressure on the mass of the whole system.

Thanks for clarifying this, pervect. It was driving me nuts! Schutz's statement (or my misinterpretation of it) seemed to contradict what you had shown earlier. (Schutz is no fool, so I knew I must have missed something. )

 

[ardenmann]

As I've mentioned before, I don't quite follow Schutz's reasoning. (If I ever get hold of his book, I'll read his full argument.)

He's saying (if I understand him) that if you have two boxes with invariant mass m, the one with a pocket of gas inside will be harder to accelerate because you have to increase the internal energy of the gas as well as increase its kinetic energy. A very interesting conclusion! (And, no, the extra inertia due to pressure is not simply the difference between inertial and rest mass.)

If you understand his argument, why don't you spell it out in detail? I suspect that you're not sure yourself, else you wouldn't keep bringing it up. (Over and over again. And what's the deal with you posting, deleting, then reposting... over and over again?)

I was puzzled by Schutz's argument at first, but gradually I realize he was so right.
Now, suppose we have a box not filled with uniform gas but has just only one gas molecule in it. Let's accelerate it to speed v and ask the same question: will the box contract? If yes, will the box compress the gas in it, in other words, will the gas resist the contraction of the box? Since the gas molecule is a point particle, we can say nothing happen to it. It's clear the gas as a whole will not Lorentz-contract (since there is only one gas molecule in the box), so it will resist the contraction of the box. Now we can add another gas molecule and do the same analysis. I hope you get the ideal

 

[yuiop]

I was puzzled by Schutz's argument at first, but gradually I realize he was so right.
Now, suppose we have a box not filled with uniform gas but has just only one gas molecule in it. Let's accelerate it to speed v and ask the same question: will the box contract? If yes, will the box compress the gas in it, in other words, will the gas resist the contraction of the box? Since the gas molecule is a point particle, we can say nothing happen to it. It's clear the gas as a whole will not Lorentz-contract (since there is only one gas molecule in the box), so it will resist the contraction of the box. Now we can add another gas molecule and do the same analysis. I hope you get the ideal

Say we had a solid cube made of 27 molecules, and declare the molecule at the the centre as "the gas". When the box is accelerated to a new constant velocity, there is no reason why the moleculae at the centre of the box should behave any differently from the other 26 moleculses that make up the walls of the box. The centre gas molecule will length contract along with the wall molecules.

Point particles are just convenient mathematical entities to simplify calculations. There is no such thing as a real point particle. A point particle has no height, width or depth and so is not able to resist anything as it occupies no volume. In the example given, all the molecules of the cube could be declared point particles, and then by extension of your argument, no solid object would length contract because it made up of point particles (molecules).

However, that is not to say Schutz is wrong either. The interactions of temperature, pressure, vloume, entropy, enthalpy and internal enrgy in the relativistic context is a complex subject with the experts disagreeing with each other, so Schutz can not be dismissed easily.

In one sense pressure does resist acceleration. If we take a sealed box containing a gas and heat it up it will weigh more if you had it on scales and it will take more energy to accelerate it to a given velocity compared to accelerating the cold low pressure box. Whether it actually resists length contraction is another matter. Now say we had a cold box low pressure box next to a hot high pressure box of the same length in a lab and accelerated the lab. If the cold box length contracted more than the hot high pressure box then we (might) have a problem, because we could probably calculate absolute motion from that.

 

[Prologue]

If we take a sealed box containing a gas and heat it up it will weigh more if you had it on scales and it will take more energy to accelerate it to a given velocity compared to accelerating the cold low pressure box.

Is there an explanation that you are willing to give (or can point out where to get it) behind this, it is very curious to me.

 

[yuiop]

Is there an explanation that you are willing to give (or can point out where to get it) behind this, it is very curious to me.

This was based on the following observations:

A hot brick has more energy than a cold brick, and will weigh more than a cold brick and require more energy to accelerate to a given velocity than the cold brick.

In another thread we compared a box with particles bouncing around inside with a cold solid mass (brick) that weighed the same as the box of particles when both are stationary with respect to the observer. When measured by an observer with motion relative to the box and brick the box of particles has greater total energy, but less rest mass (energy) than the brick. The calculation of rest mass is based on the M^2 = E^2-P^2 relationship which is normally an invariant, but not in this case when we only consider the motion of the particles in the box. This is a little paradoxical and Pervect may have the right solution, in that the tension in the walls of the box under pressure is a form of negative energy (my interpretation) that cancels out the apparent excess energy when we consider only the particles and ignore the tension of the box walls. I am pretty sure Pervect's solution is the correct interpretation of GR.

So from the above I may have been wrong to assume a hot high pressure box behaves the same as a hot brick. To really understand the full situation, you have to really understand the stress energy tensor.

In an earlier post of yours, you stated that the particles in the accelerated box tend to pile up at the rear. That might be true for a split second, but gas particles are perfectly elastic and when the rear wall collides with the particles they rebound with greater velocity and momentum. When the box reaches a stable final velocity everything should even out. If the box remained stationary and the observer accelerated to a constant velocity relative to the box, then the situation should be symmetrical. There is no reason the accelerated observer should see the particles in the static box bunched up at one end.

 

[pmb_phy]

Seems like I probably missed something but this conversation is far too long to read the whole thing. So perhaps the following was already addressed.

Using the SR defintion of mass, pressure does NOT contribute to mass, ever.

That is incorrect. Pressure itself has inertia. The pressure (as well as stress) must be accounted for when calculating the total mass of a system. If one ignores the pressure then errors will pop up. This topic appeared in the American Journal of Physics about 7 months ago. Inertia of stress, Rodrigo Medina, Am. J. Phys. 74,1031 (2006). This article is online at - http://arxiv.org/abs/physics/0609144

Is the SR mass of a box containing a pressurized gas different than the GR Komar mass? The answer to this question is no. The reason that the answer is no is that while the pressure in the interior of the box does contribute to the Komar mass, making it higher, the tension in the walls of the box also contributes to the Komar mass, making it lower. The net effect is that there is no change in the mass of the box due to the pressure terms.

That doesn't mean that one ignores the pressure. In fact ignoring things like this has led to confusion in the past, e.g. electromagnetic mass. When the electromagnetic mass was first calculated in two ways (by energy and then by momentum) the results were different. The reason was that Poincare stresses were ignored. While it is true that the final result is only a function of the inertial mass of body and box that is an end result. It is arrived at by using the stress-energy-momentum tensor. Ignoring the stress in that case leads to errors in the calculation which shows that the result is independant of the pressure/stress.

While it is very tempting to avoid all the above and focus on the end result only it would still not answer the question as asked, i.e. Is it harder to accelerate the gas because ... . I.e. the original question asked about the inertial mass of the gas only. Not the gas/box system. This particular question is answered beginning o page 192 in Gravity from the Ground Up, Bernard F. Schutz. The section is entitled The extra inertia of pressure.

Pete

 

[aachenmann]

Seems like I probably missed something but this conversation is far too long to read the whole thing. So perhaps the following was already addressed.

That is incorrect. Pressure itself has inertia. The pressure (as well as stress) must be accounted for when calculating the total mass of a system. If one ignores the pressure then errors will pop up. This topic appeared in the American Journal of Physics about 7 months ago. Inertia of stress, Rodrigo Medina, Am. J. Phys. 74,1031 (2006). This article is online at -

That doesn't mean that one ignores the pressure. In fact ignoring things like this has led to confusion in the past, e.g. electromagnetic mass. When the electromagnetic mass was first calculated in two ways (by energy and then by momentum) the results were different. The reason was that Poincare stresses were ignored. While it is true that the final result is only a function of the inertial mass of body and box that is an end result. It is arrived at by using the stress-energy-momentum tensor. Ignoring the stress in that case leads to errors in the calculation which shows that the result is independant of the pressure/stress.

While it is very tempting to avoid all the above and focus on the end result only it would still not answer the question as asked, i.e. Is it harder to accelerate the gas because ... . I.e. the original question asked about the inertial mass of the gas only. Not the gas/box system. This particular question is answered beginning o page 192 in Gravity from the Ground Up, Bernard F. Schutz. The section is entitled The extra inertia of pressure.

Pete

Do you agree with Schutz? No one here, even the so-called "guru", agree with him.
If you understand his argument, why don't you spell it out in detail?

 

[1effect]

Do you agree with Schutz? No one here, even the so-called "guru", agree with him.
If you understand his argument, why don't you spell it out in detail?

This has already been answered : https://www.physicsforums.com/showpost.php?p=1605929&postcount=15

 

[Prologue]

A hot brick has more energy than a cold brick, and will weigh more than a cold brick and require more energy to accelerate to a given velocity than the cold brick.

We are talking about inertial mass here right? If you are considering energy to be equivalent exactly to inertial mass (behaves the same way) then I could see how this statement about the bricks could be interpreted as true. Now it comes into the grey area that I am confused with in your statement. This implies that the mass equivalent of the kinetic energy can be summed with the rest mass of the particles inside to produce a new increased inertial mass. To be honest this idea is a new one to me and is very interesting but feels fishy because it depends on the reference frame (because the KE with it's v^2 factor depends on a RF). If your reference frame(1) (10 m/s) is moving twice as fast as another reference frame(2) (5 m/s), then the energy that you measure for your object (15 m/s), measured against that frame (1) (15-10)^2=25 would be a 1/4 of what it is in (2) (15-5)^2=100. So depending on the reference frame you measure with respect to, your relativistic mass changes wildly and would imply that what you are considering weight (KE+Rest Mass) changes just as wildly with the RF. This could be cleared up by an ultimately at rest reference frame, but isn't that a point of SR that you can't determine one. If I am missing something please point it out, I like things to make sense, not a have a confusion about them.

Secondly, as far as I can tell in order to weigh the gas it must be in a container. Once it is in the container, I believe the momentum of the gas now matters because the gas is colliding elastically against the walls, no energy or momentum is transferred because there is no motion of the box. So the gas particles crashing into the left will cancel with the right, and the top will cancel with the bottom*. *Cancel meaning any added impulse (F*t) that could be measured as weight would be the difference between the bottom to top would equal the force of gravity on the gas when the gas is at rest.

This is of course with no actual acceleration of the box. If we accelerate the box then we will have the same (apparent) situation as gravity, a net impulse on one side of the box.

In both of these situations I don't see how you could say that something would weigh more just because it has a higher relative velocity, because of the reference frame dependence.

In an earlier post of yours, you stated that the particles in the accelerated box tend to pile up at the rear. That might be true for a split second, but gas particles are perfectly elastic and when the rear wall collides with the particles they rebound with greater velocity and momentum. When the box reaches a stable final velocity everything should even out. If the box remained stationary and the observer accelerated to a constant velocity relative to the box, then the situation should be symmetrical. There is no reason the accelerated observer should see the particles in the static box bunched up at one end.

I didn't refer to this situation that I know of. The situation the OP is referring to is about the box being in a noninertial frame not the observer.

 

[pmb_phy]

Do you agree with Schutz? No one here, even the so-called "guru", agree with him.

Whether there are "guru"s here who agree with him or not does not make Schutz wrong whatsoever. Schutz is just one source. I also gave you another one, i.e. http://arxiv.org/abs/physics/0609144. There are other sources as well such as Introduction to Special Relativity, Wolfgang Rindler, Oxford University Press (1982). Page 151, Eq. (4.62) and Relativity, Thermodynamics and Cosmology, by Richard C. Tolman, Dover Pub., page 60, Eq. (36.4) and On the Inertia of Energy Required by the Relativity Principle, Albert Einstein, 1907. This is also addressed in Schutz's GR text A first course in general relativity page 110 section 4.7. That section discusses the previous section 4.6 called Aside on the meaning of pressure. On page 110 Schutz states In relativity, (\rho + p) plays the role of 'inertial mass density, in that, from Eq (4.54) the larger (\rho + p), the harder it is to accelerate the object.

If you understand his argument, why don't you spell it out in detail?

I already sent the OP the derivation. Since its a bit complicated I put this on a web page under my web site. Basically what one does is to consider a box which is at rest in the inertial frame S. Therefore no work is done on the box. Forces of equal magnitude and opposite direction are applied to the box at the same time in S. But in a frame moving parallel to the applied force these forces are not applied at the same time. One force is applied before another and as such a total amount of work is done on this box. The energy associated with this work goes into increasing the inertial mass of the box. The result for the mass density is

\rho = \gamma(\rho_0 + pv^2/c^4)

Since posting web pages from personal web sites is not allowed here I could send it to you in PM. However it appears that you've been banned from this forum since I can't PM it to you. I'd be happy to PM it to those who'd like to read it or the material in any of the other sources I've mentioned above. In the mean time I think that page should be modified to make the derivation clearer. I will work on it this comming week.

I'd like to take this moment to mention another importance of understanding how pressure plays its role in inertial mass. Note that when one is deriving an expression for the inertial mass density one cannot avoid inserting the pressure term. This is often overlooked by many people who study relativity. Since nobody I've ever met on the internet has ever mentioned this it took me a great deal of time to find it. Now that I understand it and its derivation its not that difficult to understand. The result also shows why E does not always equal mc². So therefore caution is adviced when calculating the inertial mass density since the pressure term must appear in the most general expression for the inertial mass density.

Pete

 

[pmb_phy]

There is no reason the accelerated observer should see the particles in the static box bunched up at one end.

I beg to differ. There is every reason for the particles to be bunched up. Consider Earth's atmosphere and how the pressure nearer the Earth is greater than the Energy up higher in the Earth's atmosphere. This is due to the fact that the weight of the air above us causes the pressure to be greater due to the mass of the gas above us pressing down on us. The increased pressure results in an increased particle density.

Pete

 

[pmb_phy]

Basically what one does is to consider a box which is at rest in the inertial frame S. Therefore no work is done on the box. Forces of equal magnitude and opposite direction are applied to the box at the same time in S. But in a frame moving parallel to the applied force these forces are not applied at the same time. One force is applied before another and as such a total amount of work is done on this box. The energy associated with this work goes into increasing the inertial mass of the box. The result for the mass density is

\rho = \gamma(\rho_0 + pv^2/c^4)

I neglected to state that the box is a solid object in this derivation.

The above expression can be derived in a much much easier way by simmply transforming the momentum density from S to S' and taking the ratio of momentum tensity to velocity to obtain the mass density.

Pete

 

[yuiop]

I beg to differ. There is every reason for the particles to be bunched up. Consider Earth's atmosphere and how the pressure nearer the Earth is greater than the Energy up higher in the Earth's atmosphere. This is due to the fact that the weight of the air above us causes the pressure to be greater due to the mass of the gas above us pressing down on us. The increased pressure results in an increased particle density.

Pete

I think there has been a misunderstanding. I was considering the case of the box after it stopped accelerating and stabilised to a constant velocity relative to the observer. I agree that while the box is accelerating there would be some bunching. I thought that was clear because I was trying to figure what an observer with constant velocity relative to a stationary box of gas would measure. Would he see a box under increased pressure and walls under increased tension? What would he consider the internal energy of the box to be when he ignores the momentum of the box as a whole because of its linear velocity relative to him?

 

[pmb_phy]

I think there has been a misunderstanding. I was considering the case of the box after it stopped accelerating and stabilised to a constant velocity relative to the observer. I agree that while the box is accelerating there would be some bunching.

Yes. I understood that. I was merely explaining how such a bunching could occur during the accelerating phase of ocket. I was a bit confused as to whether such an explanation was useful or not because it was unclear if the person who mentioned it was speaking of the accelerating phrase. In any case - no harm done.

Would he see a box under increased pressure and walls under increased tension?

Yes. Any box that contains any kind of gas, that has a non-zero pressure, must neccesarily have tension in the walls. Tension has the magnitude of force/area and has a negative value. These two values, the presssure of the gas and the tension in the walls, cancel each other out leaving the system of gas+box having its inerial energy E related to its inertial mass m as E = m².

What would he consider the internal energy of the box to be when he ignores the momentum of the box as a whole because of its linear velocity relative to him?

I'm confused here. In this question you asked about the box and not the closed system of gas+box. You do understand that it is the closed system taken as a whole fpr which E = m², right? If you're asking about the walls alone or the gas alone then its a different story altogether. In such cases E = m² does not hold in general.

Note: In determing the inertial mass of an object one cannot possibly ignore the momentum since the inertial mass is found by first determining the momentum and then dividing by the speed (in general one takes into account the orientation of the box with respect to the different frames).

Pete

 

[yuiop]

...
I'm confused here. In this question you asked about the box and not the closed system of gas+box. You do understand that it is the closed system taken as a whole fpr which E = m², right? If you're asking about the walls alone or the gas alone then its a different story altogether. In such cases E = m² does not hold in general.

Note: In determing the inertial mass of an object one cannot possibly ignore the momentum since the inertial mass is found by first determining the momentum and then dividing by the speed (in general one takes into account the orientation of the box with respect to the different frames).

Pete

My bad, I was talking about the total box + contained gas system.

My interest is not so much in the total energy of the system but the internal energy which has important implications in measurements of pressure, temperature and entropy and would provide some clues to relativistic thermodymanics which is a very unclear subject.

 

[pmb_phy]

My bad, I was talking about the total box + contained gas system.

My interest is not so much in the total energy of the system but the internal energy which has important implications in measurements of pressure, temperature and entropy and would provide some clues to relativistic thermodymanics which is a very unclear subject.

For the total gas+box system the inertial energy is proportional to the mass. As far as the rest is concerned then I'm not sure what concerns you have, espcially as to what is or isn't clear to you. Note that I've never taken the time to study relativistic thermodynamics so I'll be of limited use in that department. Who knows? Maybe it will motivate me to learn this part of relativity!

Pete

 

[yuiop]

For the total gas+box system the inertial energy is proportional to the mass. As far as the rest is concerned then I'm not sure what concerns you have, espcially as to what is or isn't clear to you. Note that I've never taken the time to study relativistic thermodynamics so I'll be of limited use in that department. Who knows? Maybe it will motivate me to learn this part of relativity!

Pete

I have started a new thread on relativistic thermodynamics here https://www.physicsforums.com/showthread.php?t=228758 that gives links to 5 papers on the subject, if you wish to take up that challenge ;)

 

[kahoomann]

As I stated earlier, there's a difference between doing a Lorentz transformation from one frame to another and worrying about the details of how one accelerates something. If you're just transforming to another frame, which is hard enough because all the forces and accelerations must be transformed, no stresses whatsoever are induced in the object (of course).

For example, imagine a stick at rest in one frame. All the molecules are in some equilibrium position as the stick is under no particular stress. If I transform to a frame in which the stick is moving, it will be Lorentz contracted. I would think that even in that new frame, the molecules must be in equilibrium. Of course, working out the details of how the intermolecular forces transform might not be trivial, but the answer must be.

Of course when you accelerate the stick, you have to be careful about how you do it. If you just push too hard on one end, you will create stresses in the stick. Not caused by the "Lorentz transformation" but by the forces applied when producing the acceleration.

As I've mentioned before, I don't quite follow Schutz's reasoning. (If I ever get hold of his book, I'll read his full argument.)

He's saying (if I understand him) that if you have two boxes with invariant mass m, the one with a pocket of gas inside will be harder to accelerate because you have to increase the internal energy of the gas as well as increase its kinetic energy. A very interesting conclusion! (And, no, the extra inertia due to pressure is not simply the difference between inertial and rest mass.)

If you understand his argument, why don't you spell it out in detail? I suspect that you're not sure yourself, else you wouldn't keep bringing it up. (Over and over again. And what's the deal with you posting, deleting, then reposting... over and over again?)

If you agree that, in Bell's spaceship paradox, the distance between two spaceships does Not change in the observer frame. Then the distances between gas particles should Not change either. Since the box contracts and the gas in it does not, so the box must compress the gas in it. In other words, it has to do work to compress the gas in it. So the inertial mass of the gas in the box must increase due to Lorentz-contraction. This is a pure Special Relativity effect.

 

[Doc Al]

If you agree that, in Bell's spaceship paradox, the distance between two spaceships does Not change in the observer frame. Then the distances between gas particles should Not change either. Since the box contracts and the gas in it does not, so the box must compress the gas in it. In other words, it has to do work to compress the gas in it. So the inertial mass of the gas in the box must increase due to Lorentz-contraction. This is a pure Special Relativity effect.

I don't get your reasoning. In Bell's spaceship paradox, the distance between the ships doesn't change in the "observer" frame because the acceleration of each ship was identical in that frame. Are you suggesting that the gas molecules in the accelerating box are somehow given identical accelerations with respect to the observer frame? Why would you think that?

 

[kahoomann]

Seems like I probably missed something but this conversation is far too long to read the whole thing. So perhaps the following was already addressed.

That is incorrect. Pressure itself has inertia. The pressure (as well as stress) must be accounted for when calculating the total mass of a system. If one ignores the pressure then errors will pop up. This topic appeared in the American Journal of Physics about 7 months ago. Inertia of stress, Rodrigo Medina, Am. J. Phys. 74,1031 (2006). This article is online at - http://arxiv.org/abs/physics/0609144

That doesn't mean that one ignores the pressure. In fact ignoring things like this has led to confusion in the past, e.g. electromagnetic mass. When the electromagnetic mass was first calculated in two ways (by energy and then by momentum) the results were different. The reason was that Poincare stresses were ignored. While it is true that the final result is only a function of the inertial mass of body and box that is an end result. It is arrived at by using the stress-energy-momentum tensor. Ignoring the stress in that case leads to errors in the calculation which shows that the result is independant of the pressure/stress.

While it is very tempting to avoid all the above and focus on the end result only it would still not answer the question as asked, i.e. Is it harder to accelerate the gas because ... . I.e. the original question asked about the inertial mass of the gas only. Not the gas/box system. This particular question is answered beginning o page 192 in Gravity from the Ground Up, Bernard F. Schutz. The section is entitled The extra inertia of pressure.

Pete

The Lorentz contraction has shortened the length of the box and therefore changed its volume. My question is this: Does the gas in the box contract also? If not, then the box need to compress the gas in it. Making a box smaller when it contains a gas with pressure requires one to do work on it, in other words to put some energy into the gas. This extra energy represents the extra inertia of the gas.
The key question is: Is it harder to accelerate the gas because it takes work not only to accelerate the existing energy but also to compress the gas as the Lorentz contraction demands? In other words, the moving box will contract but the gas in it will resist the Lorentz-contraction of the box

 

[pmb_phy]

Does the gas in the box contract also?

Yes.

If not, then the box need to compress the gas in it.

That would be true if a process occurred in that frame in which the gas underwent a change in speed and thus undergoing a contraction process. Changing from one frame to another is not such a process.

The key question is: Is it harder to accelerate the gas because it takes work not only to accelerate the existing energy but also to compress the gas as the Lorentz contraction demands? In other words, the moving box will contract but the gas in it will resist the Lorentz-contraction of the box

I'm not sure if you can call it a "cause" (regarding the term "because" used above). But the gas indeed does have more inertia and hence more momentum.

Pete