# Is pressure an invariant in Special Relativity?

1. Jan 3, 2014

### vinven7

HI : Consider a cylinder of length L and volume V that contains one mole of an ideal gas. The familiar ideal gas law states that:
PV = RT

Now, if the cylinder were to move with velocity v parallel to the length direction, special relativity requires the length to contract given by:

L' = L √(1-(v^2/c^2))

As the length contracts, the volume must also go down by the same relation:

V' = V√(1-(v^2/c^2))

However, the ideal gas law must continue to hold ( I presume) - which can only be true if the pressure P increases by the same factor. Thus, simply because of it's motion, the pressure experienced the cylinder seems to have gone up by

P' = P /√(1-(v^2/c^2))

if this argument is correct, does this then mean that pressure is not a relativistic invariant but must depend on the frame of reference? A vessel that does not experience any pressure in one frame of reference might be under pressure in another one?
Could anyone confirm or refute this argument? Thanks!!

2. Jan 3, 2014

### WannabeNewton

If we have a fluid with some 4-velocity field $u^{\mu}$ then the pressure $p$ of the fluid is defined as that measured by an observer comoving with a given fluid element.* By definition, an observer comoving with a given fluid element will attribute to that fluid element a 4-velocity $u^{\mu} = (1,0,0,0)$.

In other words $p$ is defined in the rest frame of the fluid, just like the mass density $\rho$ of the fluid is defined in the rest frame of the fluid. So just like $\rho$ is the "invariant mass density" of the fluid, you can think of $p$ as the "invariant pressure" of the fluid because both quantities are by definition measured in the rest frame of the fluid and all observers must agree on measurements made in the fluid rest frame regardless of whether or not they get the same measurement in their own frame.

Note that both $\rho$ and $p$ are components of a single tensor: the stress-energy tensor $T^{\mu\nu}$. So it is clear that neither $p$ nor $\rho$ will remain fixed in value under a Lorentz boost from the rest frame of the fluid to some other frame.** Don't confuse the failure of $\rho$ and $p$ to remain fixed in value under a Lorentz boost with the notion of "invariance" that I talked about above. If I understood your post correctly then what you are referring to is what I have described in this paragraph and not what I described in the preceding paragraph but keep in mind the terminology.

*For simplicity I have dealt with a perfect fluid so that we simply have an isotropic (scalar) pressure $p$ as measured by a comoving observer.

*To make things more concrete, the stress-energy tensor of a perfect fluid is given by $T^{\mu\nu} = (\rho + p)u^{\mu}u^{\nu} + p\eta^{\mu\nu}$ where $\eta^{\mu\nu}$ is the Minkowski metric. In the rest frame of the fluid we therefore have $T^{\mu\nu} = \text{diag}(\rho,p,p,p)$ where "diag" means diagonal matrix. Now let's boost to a frame that's moving with some velocity $v$ along the $x$ axis of the fluid rest frame. Denote by $\Lambda^{\mu}{}{}_{\nu}$ the Lorentz transformation matrix associated with the boost.

Then $p'_x = T^{x'x'} = \Lambda^{x'}{}{}_{\mu}\Lambda^{x'}{}{}_{\nu}T^{\mu\nu} = \rho\Lambda^{x'}{}{}_{0}\Lambda^{x'}{}{}_{0}+ p\Lambda^{x'}{}{}_{x}\Lambda^{x'}{}{}_{x} = \gamma^2 \beta^2 \rho + \gamma^2 p$

and similarly $\rho' = \gamma^2\rho + \gamma^2 \beta^2 p$

whereas $p'_y = p'_z = p$.

3. Jan 3, 2014

### Staff: Mentor

I'm not sure it's exactly correct as you state it, because you haven't taken into account possible effects on the gas's temperature due to relative motion. However, your general point is correct:

Yes. The more general way to see this is to look at the stress-energy tensor [edit: I see WannabeNewton beat me to it ]; pressure appears as the diagonal space-space components of this tensor in a local inertial frame, but if you transform to a different local inertial frame, the tensor components transform too, so the pressure won't be the same in the new frame.

However, you do have to be careful:

No: the vessel "experiencing pressure" appears as stress in the vessel, which is directly observable; you can't change direct observables by changing frames, so if the vessel has zero stress in one frame, it must have zero stress in all frames.

In terms of the transforms given above, you can change the pressure by changing frames and transforming the stress-energy tensor, but the stress experienced by the vessel is given by an invariant combination of SET components which stays the same when you change frames.

4. Jan 3, 2014

### vinven7

Hi - Thanks to the both of you for your reply. I think I kind of understand what you are trying to say but I do not have enough knowledge of tensors to get the mathematical part of it. I was led to his question by another thought-puzzle. I hope that you guys can give an explanation for this:

You probably know what a "phase" is in the material science sense. For example, Iron has a couple of different phases - each distinguished by the structure at the lattice scale. There is BCC iron where all the atoms are arranged at the corners of a cube+ 1 in the center of the cube. There is FCC iron where all atoms are at the corners of the cube + 6 atoms each at the center of every face. The properties of every material is therefore an intrinsic function of it's structure or in other words different phases of the same material have drastically different properties.

Let us say that there is a block of Iron where all the atoms are at the centers of a cube (a x a x a). Let's call this phase 1. This phase has a set of distinct properties say a definite color, electrical conductivity, young's modulus etc.

This block is then put on a rocket and is made to go very fast in one direction at a constant velocity. Now, the side of the cube that is parallel to the direction of velocity will undergo a contraction given by Lorentz formula. So that the basic unit cell of the this piece of Iron is no longer a cube of dimensions a x a x a, but a cuboid of dimension a x a x b. where b< a.

Now however, this material can no longer be called phase 1 iron, it's a different phase altogether. It may have a different color and/or other properties. Let's call it phase 2.

Therefore, for someone inside the rocket, the block of iron is a phase 1 material, whereas for someone on the ground, it's a phase 2 material.

Thermodynamically, phases change only under the action of temperature or pressure or both - that is by a change in it's free energy.

So I find it baffling that something as fundamental as the phase - or the very essence of what a material is, can be a function of the frame of reference.

For example, lets say that the buttons of a coat is made of beta tin. Now when the same thing moves in a different frame of reference, I may observe it as being made of alpha tin, and that's a brittle material that powders easily. I thus have a good coat in one frame and a buttonless coat in another frame.

Again, I deeply appreciate your help

5. Jan 3, 2014

### Staff: Mentor

Your bafflement is justified, because phase isn't a function of the frame of reference.

One way to tell is this: suppose I'm sitting in my lab with my hunk of phase I iron. You are flying by the lab at a relativistic velocity. You will see the iron length contracted along one dimension; but you can't see it as a different phase, because the same hunk of iron can't be in two phases at once. So it must be phase I iron to you as well, even though its spatial arrangement appears to you to be different.

What this means is that phase can't work the way you are assuming it works; it can't be a simple function of things like the distances between atoms that are frame dependent. The phase has to be a function of things that are invariant, so any role that things like the distances between atoms play in determining the phase has to be via something like the stress-energy tensor, or more precisely combinations of components of that tensor that remain the same when you transform the tensor to a different frame (just like the stress in the vessel in my previous post).

6. Jan 3, 2014

### WannabeNewton

Let me point out, vinven, that you have basically opened up a can of worms here since relativistic (covariant) thermodynamics is far, far from trivial in terms of its physical foundations

For example above I noted that the (isotropic) pressure of a perfect fluid is defined in the rest frame of the fluid. Now as noted the scalar pressure comprises three components of the stress-energy tensor in the fluid rest frame and as such transforms under a Lorentz boost in a non-trivial way however one must be very careful in interpreting the transformed quantity that one gets after applying the Lorentz boost to the pressure terms of the stress-energy tensor in the fluid rest frame. I have yet to see a textbook that gives physical/operational meaning to this transformed quantity as the pressure that an observer in the boosted frame would measure of the fluid. In other words, all the textbooks I know of give physical meaning only to the pressure in the fluid rest frame.

There is (for better or for worse) a sea of literature on relativistic (covariant) thermodynamics.

Let me link you to some useful discussions from the past (most of which were started by my good friend yuiop ):

7. Jan 3, 2014

### vinven7

Thanks for the great replies. WannabeNewton, i will go through these links. :)

PeterDonis: I understand your point of view. I have been taught that the very distinction between two phases is the underlying crystal structure. And that the symmetry in the properties of a solid are reflections of the underlying geometric symmetries of the crystal itself.

So, if structure is not the fundamental characterization of a phase, then what should be? (I'm a materials scientist so I am very keen to know)

Coming to the previous example, phase 1 Iron being a cube should have the same physical constants (electrical resistivity, thermal conductivity, stiffness etc in all three directions, while phase 2 iron should be anisotropic (in the direction of velocity). This is purely a reference frame effect. I'm unable to wrap my head around it. !!

8. Jan 3, 2014

### WannabeNewton

Given that phase transition is intimately tied to state variables such as pressure and temperature, this question can be recast in terms of the transformation properties of temperature as a thermodynamic variable under Lorentz boosts (I'm only doing this to open up the profuse of literature that has been written on this in the past century). This is an extremely interesting but thorny subject area as I have stated; in other words your question is a very good one.

See here: http://arxiv.org/pdf/physics/0506214v2.pdf

9. Jan 3, 2014

### vinven7

Thanks so much!

As the last question, what is a good book to learn Special Relativity - that's both interesting and mathematical at the same time? I can learn the math as I go by..

10. Jan 3, 2014

### WannabeNewton

I don't want to derail your thread by going into SR book recommendations, especially since the subject matter of your thread is extremely interesting, so instead let me link you to the textbook subforum where you can ask for SR book recommendations in proper fashion: https://www.physicsforums.com/forumdisplay.php?f=21

11. Jan 3, 2014

### PAllen

I found this question very interesting as well, but I did not feel I had anything authoritative to contribute. In doing some reading to get up to speed on this topic, I found the following references which are related, but with a different slant. This thread is concerned with transformation properties for a system that is non-relativistic in its rest frame; and really boils down to what is the form of equations of state, temperature, pressure etc. for a system moving rapidly relative to some observer (but said system has no relativistic features in its rest frame). I think it is natural that equations of state would be different:

Ultimately, all inter-atomic forces are EM in nature. We know that for a collection of moving charge distribution that is spherical in its rest frame, Maxwell's equations would produce, for a moving set of charges, an equilibrium state that is flattened. Thus I would expect that if you derive the equation of state that says phase X has symmetry Y in its rest frame, from first principles in a frame where the body is moving, you would find that phase X is associated with different symmetry and geometry due precisely to the Lorentz invariance of the fundamental laws.

The references I have started reading address a somewhat different question: what happens to Kinetic theory and thermodynamics for gases with relativistically moving particles. I am not yet sure whether this theory says anything directly useful to the OP questions, but here is what I found:

http://arxiv.org/abs/1303.2899
http://arxiv.org/abs/1108.3744

12. Jan 3, 2014

### Staff: Mentor

Yes, but note that there is an invariant way of stating that. For a perfect fluid, you just write $T_{ab} = \rho u_a u_b + p \left( g_{ab} + u_a u_b \right)$, where $\rho$ is the energy density and $p$ is the isotropic pressure. In the rest frame of the fluid, of course, this just means $T_{00} = \rho$ and $T_{11} = T_{22} = T_{33} = p$; but the equation works in any frame, as long as you transform the metric $g_{ab}$ and the fluid's 4-velocity $u_a$ correctly into that frame. Depending on the frame, that may mean that $T_{ab}$ has off-diagonal components, and that some components may be "mixtures" of $\rho$ and $p$ with various coefficients. As you note, it's difficult to come up with a physical interpretation of individual SET components in an arbitrary frame; but the math still works fine in any frame.

13. Jan 4, 2014

### yuiop

In this very old thread, (I like to think) I demonstrated that pressure (as it is commonly understood) is an invariant in Special Relativity. However, the analysis of the transformations assumed a good old fashioned definition of pressure being force per unit area as a result of molecules bouncing about. I think we agreed in the end that the pressure in the stress-energy tensor is not an invariant, so the S-E tensor definition of pressure must be defined differently to the Newtonian definition of pressure.

Hopefully we can all agree that the pressure indicated by a pressure gauge physically attached to a pressurised vessel represents a 'proper' measurement that all observers can agree on.

Last edited: Jan 4, 2014
14. Jan 4, 2014

### PAllen

Of course I think we can agree on this - anything else would be absurd. The more probelmatic case, is what it means to measure pressure of a moving air mass (for example) with a stationary pressure gauge. Do we want to effectively measure wind force, or factor that out to get some definition of 'moving pressure' ?

15. Jan 4, 2014

### vanhees71

It's easy to see, considering PeterDonis's example of the ideal-fluid energy-momentum tensor. There obviously $\rho$ (I'd rather write $\varepsilon$, because it's the energy density not the mass density) and $p$ are scalar quantities. They are the energy density and pressure of the fluid as measured by an observer comoving with the fluid cell he is looking at, i.e., these are quantities defined in the local restframe of the fluid. In this way you can always define invariant local quantities related to a many-body system: You just go to the "local restframe" of the fluid cell and define your quantities there and then write the quantity in question, here the energy-momentum-stress tensor, in covariant form, where it can be transformed back to any reference frame using a Lorentz transformation.

16. Jan 4, 2014

### PAllen

That's fine for what to put in the stress energy tensor. It is not clear to me how the quantities in the tensor for frame in which fluid is not at rest would relate to measurements by different types of instruments relative to which the fluid is moving.

17. Jan 4, 2014

### yuiop

Here is a simple thought experiment. Consider two boxes with equal volume and air pressure when at rest with respect to each other and at rest wrt reference frame S. There are also some ports to allow the pressure to equalise as in the diagram below.

Now both boxes are accelerated to equal speeds wrt S but in opposite directions. Initially the equalisation ports close, but another set of ports align to allow air to move in either direction. Since in this reference frame, both boxes have length contracted to the same extent, the pressures should still be equal and there is no reason to expect any net flow of air from one box to the other.

Seen from the rest frame of the black box, the red box is length contracted and if we assume a resulting increase in pressure in the red box there should be a net flow of air from red to black. Black should measure an increase in pressure and red should measure a loss of pressure. Black gains a detectable amount of atoms.

Seen from the rest frame of the red box, the black box is length contracted and there should be a net flow of air from black to red. Red should measure an increase in pressure and black should measure a loss of pressure. Red gains a detectable amount of atoms. This a physical contradiction to what is measured in red's rest frame.

Therefore it seems reasonable to conclude that pressure (as understood in the context of the gas laws) is a Lorentzian invariant, or in relativity gas does not flow from high pressure regions to low pressure regions.

P.S. This analysis only demonstrates that transverse pressure is invariant and agrees with the result $p'_y = p'_z = p$ given by WannabeNewton in post #2, but does not address the parallel case $p'_x$. That case was analysed at the molecular level by pervect and myself in the context of gas pressure in the old thread linked above in my earlier post.

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18. Jan 4, 2014

### yuiop

Relativistic thermodynamics is a thorny subject and has been subject to revision over the years.

Around 1905 Planck and Einstein concluded that a moving body would appear colder than its proper temperature. In 1963 Ott asserted that that a moving body would appear hotter than its proper temperature. A few years later Landsberg concluded temperature was a Lorentz invariant.

What they all seemed to agree on, is that the pressure is a Lorentz invariant. See equations (4), (7) and (10) of this historical review.

Last edited: Jan 4, 2014
19. Jan 4, 2014

### WannabeNewton

The isotropic pressure term in the perfect fluid stress-energy tensor is still force per unit area, which follows from the definition of the stress-energy tensor itself.

Given a stress-energy tensor $T_{\mu\nu}$ and a background observer with 4-velocity $u^{\mu}$, the stress tensor associated with the matter field is given by $S_{\mu\nu} = h^{\alpha}{}{}_{\mu}h^{\beta}{}{}_{\nu}T_{\alpha\beta}$ where $h_{\mu\nu} = \eta_{\mu\nu} + u_{\mu}u_{\nu}$ is the orthogonal projection of $u^{\mu}$ (so it's nothing more than the local simultaneity slice of the background observer if we use the $\epsilon = \frac{1}{2}$ simultaneity convention).

If the observer has a local Lorentz frame $\{e_{\alpha}\}$ then the force $dF$ exerted on an infinitesimal space-like area element $dA$, as measured by the background observer, is then given by $dF = (S_{ij}n^{j} dA) e^{i}$ where $n^{j}$ is the unit normal to $dA$.

Specialize this now to a perfect fluid with 4-velocity field $u^{\mu}$ with stress-energy tensor $T_{\mu\nu} = \rho u_{\mu}u_{\nu} + p(\eta_{\mu\nu} + u_{\mu}u_{\nu})$ wherein the background observer is comoving with a given fluid element. Then clearly $S_{\mu\nu} = p(\eta_{\mu\nu} + u_{\mu}u_{\nu})$ hence $dF = (p dA)n_{i}e^{i}$ thus $\frac{\mathrm{d} \left \| F \right \|}{\mathrm{d} A} = p$.

What I have never seen is physical/operational meaning being given to the transformed quantities one gets after applying a Lorentz boost to the individual (isotropic) pressure components of the stress-energy tensor in the perfect fluid rest frame.

20. Jan 4, 2014

### PAllen

There are several problems relating this argument (which I agree with) to what I was actually asking. What this really demonstrates is the fact that you don't expect any flow. To go from here to the conclusion that the reason this is true for the moving box of air is due to its pressure being considered the same as at rest you must assume other things e.g. that pressure is isotropic for a moving box of ideal fluid. While this is true by definition in its rest frame, I see no reason to expect it for a moving mass of perfect fluid. If one allows for non-isotropic pressure for the moving box, this argument says nothing about longitudinal pressure.

Pressure is made a scalar invariant effectively by defining it as a scalar measured by a comoving gauge. What I was asking, is what happens to different types of pressure gauges subject to a moving air mass (they definitely measure more pressure from wind), and whether there is a way to related this to transformed quantities in the stress energy tensor.

21. Jan 4, 2014

### WannabeNewton

This seems to be the exact same question that I have as well. Unfortunately I'm not having any luck finding literature on this.

22. Jan 4, 2014

### pervect

Staff Emeritus
Well, in fluid dynamics, there is such a thing as "dynamic pressure"

http://en.wikipedia.org/wiki/Dynamic_pressure

defined as $q = \frac{1}{2} \rho v^2$

However, if we transform the stress-energy tensor of a pressureless lump of matter, with $T^{00} = \rho$ and all other components zero, the pressure term in the direction of motion becomes.

$T^{11} = \gamma^2 v^2 \rho$

In the Newtonian limit this is $\approx \rho v^2$

So it seems that there is a very close relation between dynamic pressure and the appropriate pressure term in the stress energy tensor (here $T^{11}$), but there appears to be a factor of 2 difference.

23. Jan 4, 2014

### yuiop

After reading up on Pitot tubes, it would seem that the pressure used in the stress-energy tensor is related to the total pressure rather than the static pressure that is usually used in the context of the gas laws. The total pressure is the static pressure plus the dynamic pressure due to the kinetic energy of the air due to its velocity relative to the instrument. This is the additional pressure you feel on one side of your hand when you stick it out of a moving car window due to the velocity of the air past your hand.

http://en.wikipedia.org/wiki/Pitot_tube
http://en.wikipedia.org/wiki/Dynamic_pressure
http://en.wikipedia.org/wiki/Static_pressure

It would seem that static pressure as used in the gas laws is isotropic, scalar and Lorentz invariant. The total pressure used in the stress energy tensor is none of those things.

The pressure we calculated in those old threads was the static pressure. In the context of the OP, the pressure as far as the equation for pressure, temperature and volume is concerned is the static pressure and is Lorentz invariant.

P.S. It seems I cross posted with pervect.

Last edited: Jan 4, 2014
24. Jan 4, 2014

### WannabeNewton

What you said regarding the pressure in the stress-energy tensor is incorrect. As has already been mentioned multiple times in this thread, when we have a perfect fluid (meaning there is no heat exchange between the fluid elements and no viscosity i.e. no shear forces between fluid elements-they perfectly slip along one another) the pressure $p$ that you see in the stress-energy tensor $T_{\mu\nu} = \rho u_{\mu}u_{\nu} + p(\eta_{\mu\nu} + u_{\mu}u_{\nu})$ is defined as the pressure of the fluid measured by an observer $\Leftrightarrow$ pressure gauge comoving with a given fluid element. This means that the pressure is measured in a local Lorentz frame in which the fluid 3-velocity field $\vec{v}$ vanishes: $\vec{v} =0$. In other words, by definition of $p$, the pressure gauge that measures $p$ is at rest with respect to the fluid element.

This is also incorrect I'm afraid, at least when applied to a perfect fluid. See my post above where I show that the scalar field $p$ that appears in $T_{\mu\nu}$ for a perfect fluid is related to the magnitude of the force $dF$ on a space-like infinitesimal area element $dA$ as measured by a comoving observer by $p = \frac{d\left \| F \right \|}{dA}$. Hence the scalar field $p$ is an isotropic pressure field. It is invariant by definition since it's always measured in the rest frame of the fluid.

25. Jan 5, 2014

### yuiop

In arriving at that conclusion, you are of course assuming that temperature T in the ideal gas law is a Lorentz invariant. In this parallel thread on temperature in relativity, so far no one in this forum is prepared to stick their neck out and state temperature is a Lorentz invariant.

You are also assuming that the classical ideal gas law (PV = RT) is valid in relativity, where in fact it might be something like (PV =γRT) where γ is the relativistic gamma factor.

Last edited: Jan 5, 2014