Does pressure resist acceleration?

[matheinste]

Hello again.

Just to clear up some confusion on my part. In answer to a previous query in this thread it was confirmed by Doc Al that an observer in the "stationary" frame would see the separation distances of the gas molecules in the "moving" frame contracted. It was not explicitly stated that the size of the gas molecules is also contracted. I believe this to be so but would like it to be clear, if it is a fect, so we can all accept it as i feel it has relevance to a rerstricted, non accelerated version of the original post.

Matheinste.

 

[Doc Al]

I'd say that any linear dimension along the direction of motion will be measured as contracted, whether its the distance between molecules or the "size" of the molecules themselves.

 

[matheinste]

Thankyou Doc Al.

This means to me that any, and every effect depending only on the linear dimensions and clock rates of a system will not be changed by any inertial relative motion. I am also assuming that every physical process ultimately depends upon these fundamental dimensions of length and time. In other words every term in the mathematiacl formulation of a physical law dependent on these dimensions will either be unaffected or proportionately affected so as to leave the numerical result unaltered. Obviously the observed measurement of length and time being, fundamental and not derived, will be affected.

I have no idea how things behave in accelerated frames so this scenario is probably not relevant to the original post.

Not very well put but, nor put with confidence, but i think you can get the idea.

Mateinste .

 

[yuiop]

Hello again.

Just to clear up some confusion on my part. In answer to a previous query in this thread it was confirmed by Doc Al that an observer in the "stationary" frame would see the separation distances of the gas molecules in the "moving" frame contracted. It was not explicitly stated that the size of the gas molecules is also contracted. I believe this to be so but would like it to be clear, if it is a fect, so we can all accept it as i feel it has relevance to a rerstricted, non accelerated version of the original post.

Matheinste.

Say the box is moving along the x-axis with velocity v relative to the "stationary" frame.

In the rest frame of the box some of the molecules are bouncing up and down the y-axis with velocity u. So even in the rest frame of the box the molecules are length contracted in the y direction because the gas molecules are not at rest except at zero degrees Kelvin. To the observer in the stationary frame the y component of the molecules bouncing up and down is reduced by time dilation so they are less length contracted than they are in the rest frame of the box. In the rest frame of the box the molecules bouncing left to right parallel to the x-axis also have average velocity u. From the stationary frame the forward velocity of the molecules is greater than u but the molecules going from the front wall to rear wall on their return journey at moving at a velocity of less than u from the point of view of the observer in the stationary frame. So to the observer in the stationary frame some of the molecules are actually longer than they are in the rest frame of the box.

You should be able to see that the change in shape of the molecules is complex and not restricted to just the x-axis parallel to the motion of the box. You should also be aware that (as I said in an earlier post) the ideal gas law is independent of the shape, size or mass of the molecules for an ideal gas.

The ideal gas law is PV=nRT

where

P = Pressure of the gas
V = Volume of the gas
n = Number of gas molecules
R = The gas constant
T = Absolute temperature of the gas.

Like I said... no mention of size, shape or molecule mass in the ideal gas law.

I guess you just don't agree that time dilation applies to gas molecules too.

 

[Doc Al]

I'd say that any linear dimension along the direction of motion will be measured as contracted, whether its the distance between molecules or the "size" of the molecules themselves.

Kev makes an important point in his post #39 that I failed to emphasize here: The observed length contraction of a molecule depends on the relative velocity of the molecule, not just the speed of the box.

 

[yuiop]

Thankyou Doc Al.

This means to me that any, and every effect depending only on the linear dimensions and clock rates of a system will not be changed by any inertial relative motion. I am also assuming that every physical process ultimately depends upon these fundamental dimensions of length and time. In other words every term in the mathematiacl formulation of a physical law dependent on these dimensions will either be unaffected or proportionately affected so as to leave the numerical result unaltered. Obviously the observed measurement of length and time being, fundamental and not derived, will be affected.

I have no idea how things behave in accelerated frames so this scenario is probably not relevant to the original post.

Not very well put but, nor put with confidence, but i think you can get the idea.

Mateinste .

First of all, it seems we can answer every question simply by saying nothing changes according to an observer at rest with system under observation. However, this viewpoint has difficulty explaining things like why two twins age differently when brought back together. There may be other physical phenomena that that occur due to relatistic motion in the sense that they can be measured after the event when the system is brought to rest. One example is Thomas precession. A gyroscope sent in a orbit around the Earth will not be pointing in exactly the same direction after each complete orbit. This is a real measurable effect and is in fact being measured the in Gravity Probe B. They are still clearing the noise from the data of that experiment but the precession seems to be there under the noise. Sagnac invented a gyroscope without any moving parts that is in regular comercial use. It is doubtful that the Sagnac effect would have been discovered by someone who had the viewpoint that nothing really changes in a moving system.

 

[matheinste]

Hello Doc Al and Kev.

Your comments have been taken on board of course accepted.

I will follow the this post with a more enlightened interest

Thankyou Matheinste.

 

[pervect]

First of all, it seems we can answer every question simply by saying nothing changes according to an observer at rest with system under observation.

One should perhaps be more formal and say that we can analyze the system in any frame we chose, and that it's usually best to chose the simplest one. This is a simple, but important point, that can hardly be stressed enough. But that leads us to point #2.

However, this viewpoint has difficulty explaining things like why two twins age differently when brought back together.

Actually, it doesn't have any difficulty at all in explaining this. Only one twin is in an inertial frame. Other twins are in a non-inertial frame for at least part of the journey. There are many correct ways of analyzing the later case, and all of them get the same answer.

The difficulty in attempting to explain the Sagnac effect by this line of reasoning also involves the fact that there is no single inertial frame that describes the observer in question.

 

[yuiop]

..

Actually, it doesn't have any difficulty at all in explaining this. Only one twin is in an inertial frame. Other twins are in a non-inertial frame for at least part of the journey. There are many correct ways of analyzing the later case, and all of them get the same answer.

The difficulty in attempting to explain the Sagnac effect by this line of reasoning also involves the fact that there is no single inertial frame that describes the observer in question.

I think it fair to say that newcomers to relativity that are first taught the totally symmetrical and strictly formal version of Special Relativity that is constrained to inertial frames only, are genuinely perplexed by so called paradoxes like the the Twin's paradox. Every year, there are hundreds of questions about the twin's paradox and endless arguments that perhaps could be avoided by introducing earlier in the education process (or stressing more) the concept that acceleration breaks the symmetry. Just my point of view.

Beyond that, I have no quibble with your statement that one of the twins is not in an inertial frame and that the Sagnac effect is not an inertial frame, but in my experience newcomers find these concepts genuinely puzzling when trained to think in terms of symmetrical inertial frames only.

 

[AstroRoyale]

I'm not sure what is being argued here as things have gotten off track, but here's a go of it. The Pressure of a gas is not a lorentz invariant, it does change WHEN MEASURED IN DIFFERENT FRAMES.

[tex]
T^{'}= \Lambda^{T} T \Lambda
with
T=diag[\rho, p, p ,p]
and \Lambda is the boost matrix which I'm too lazy to typeset.
[\tex]

If the pressuse is 1 atm in the rest frame of the gas, then it will not be 1 atm WHEN MEASURED BY AN OBSERVER IN THE MOVING FRAME. However, if you were a physics student living in the box that is moving at some relativistic speed, say 0.5c, you will measure 1 atm every time. This is the essence of relativity, that in the box there is NO experiment to determine if the box is in uniform motion.

 

[yuiop]

I'm not sure what is being argued here as things have gotten off track, but here's a go of it. The Pressure of a gas is not a lorentz invariant, it does change WHEN MEASURED IN DIFFERENT FRAMES.

[tex]
T^{'}= \Lambda^{T} T \Lambda
with
T=diag[\rho, p, p ,p]
and \Lambda is the boost matrix which I'm too lazy to typeset.
[\tex]

If the pressuse is 1 atm in the rest frame of the gas, then it will not be 1 atm WHEN MEASURED BY AN OBSERVER IN THE MOVING FRAME. However, if you were a physics student living in the box that is moving at some relativistic speed, say 0.5c, you will measure 1 atm every time. This is the essence of relativity, that in the box there is NO experiment to determine if the box is in uniform motion.

I assume you are talking about the stress tensor here. We had a long discussion about this in the "Proof of the invariance of gas pressure" thread where we concluded that the pressure calculated in the stress energy tensor is not the same thing as regular gas pressure. Regular gas pressure is a Lorentz invariant. See https://www.physicsforums.com/showthread.php?t=213016

 

[AstroRoyale]

Yes, T is the stress energy tensor, and as far as I know $$T=diag(\rho, p,p,p)$$
is THE stress/energy tensor of an ideal gas. What do you suppose that p in the stress energy tensor is, if it is not the pressure as you claim?

 

[yuiop]

Yes, T is the stress energy tensor, and as far as I know $$T=diag(\rho, p,p,p)$$
is THE stress/energy tensor of an ideal gas. What do you suppose that p in the stress energy tensor is, if it is not the pressure as you claim?

I have no idea what the p in the stess energy tensor is as I am no expert on tensors. Pervect knows more about these things and he tends to agree in the tread I linked to that that stress energy tensor pressure and ideal gas pressure may not be the same thing. Pervect points out that engineers and physicists have different definitions of pressure and that may be the root of the disagreement. As far as I know, Pervect is the main author of thie article on the stress-energy tensor in Wikipedia. http://en.wikipedia.org/wiki/Stress-energy_tensor

Perhaps you can point out some flaws in the arguments I presented in the tread I linked to in my previous post?

 

[1effect]

I have no idea what the p in the stess energy tensor is as I am no expert on tensors. Pervect knows more about these things and he tends to agree in the tread I linked to that that stress energy tensor pressure and ideal gas pressure may not be the same thing. Pervect points out that engineers and physicists have different definitions of pressure and that may be the root of the disagreement. As far as I know, Pervect is the main author of thie article on the stress-energy tensor in Wikipedia. http://en.wikipedia.org/wiki/Stress-energy_tensor

Perhaps you can point out some flaws in the arguments I presented in the tread I linked to in my previous post?

kev,

You already know the flaws, they range from:
-incorrect modelling of the pressure (via springs or rods)

to

-incorrect physics (attributting the speed of the box wall to the molecules of gas)

to

-use of incomplete or conveniently modified formulas for relativistic force transformation in the calculation of momentum

All thse problems are evident in the thread, I have highlighted them as we moved from approach to approach. You are a very creative person but all the solutions have flaws. This is why we haven't managed to reach a conclusion in the calculation of the relativistic transformation for pressure. We have the intuition[/] that pressure might be invariant (intuition contradicted by the way the stress-energy tensor transforms) but we have no valid proof. Does someone have a correct proof? One way or another, I don't care about the outcome (invariant or non-invariant). Isn't there a published paper on this subject?

 

[yuiop]

kev,

You already know the flaws, they range from:
-incorrect modelling of the pressure (via springs or rods)

to

-incorrect physics (attributting the speed of the box wall to the molecules of gas)

to

-use of incomplete or conveniently modified formulas for relativistic force transformation in the calculation of momentum

All thse problems are evident in the thread, I have highlighted them as we moved from approach to approach. You are a very creative person but all the solutions have flaws. This is why we haven't managed to reach a conclusion in the calculation of the relativistic transformation for pressure. We have the intuition[/] that pressure might be invariant (intuition contradicted by the way the stress-energy tensor transforms) but we have no valid proof. Does someone have a correct proof? One way or another, I don't care about the outcome (invariant or non-invariant). Isn't there a published paper on this subject?

Well, you I know I disagree with your objections, so yes, some independent opinions would be welcome ;)

Also, IF the pressure of an enclosed ideal gas in not a Lorentz invariant THEN I can show you a whole host of unresolvable paradoxes that would be created.

 

[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. )