Is pressure an invariant in Special Relativity?

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What you said regarding the pressure in the stress-energy tensor is incorrect. ... 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.
I think you are missing my point and no doubt the fault is mine due to my informal language. When I said "the pressure used in the stress-energy tensor" I was referring to the directional components of p' rather than the isotropic (proper) pressure represented by the symbol p. By analogy if I said length is not a Lorentz invariant in SR, I am referring to L' rather than the proper length ##L_0## in the length contraction formula ##L' = L_0/\gamma##.

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## ... whereas ##p'_y = p'_z = p##.
Clearly the quantity ##p'_x## you describe here is not a Lorentz invariant and is not isotropic and this is the quantity I meant when I said "the pressure" which was a bit sloppy of me. It is this non isotropic transformed pressure that can be thought of as the total pressure that includes terms related to static pressure and dynamic pressure. The isotropic pressure (p) measured by a comoving observer can be thought of as the static pressure. When v is zero, the dynamic pressure term of the covariant total pressure (p') vanishes and you are left with only the static isotropic pressure (p). While static pressure is defined as the measurement made by a comoving observer, it can be calculated or even physically measured by a non comoving observer. The air speed indicator on an aircraft measures the static pressure of the air (even though the instrument is moving relative to the air) and compares this to the total pressure which includes static and dynamic pressure. The difference is the dynamic pressure which is a measure of the air speed.

... 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.
Here Peter is using the phrase "the pressure" in the same way that I used the same phrase, to mean the covariant transformed pressure, (that includes terms related to dynamic pressure).
... 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 ...
When I said "the pressure" in the s-e tensor is not a Lorentz invariant, I was trying to suggest that the components do not transform in an invariant way under a Lorentz boost as you describe here. Your terminology is confusing to me here. I thought the definition of a "scalar" quantity was a quantity that does not have directional components?

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.
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.
I am not sure which of my two statements you are saying saying is incorrect. (probably both :). My statement that "static pressure as used in the gas laws is isotropic, scalar and Lorentz invariant" is very similar to your statement that the scalar field ##p## is scalar, isotropic and and invariant by definition. I guess the subtle difference is the difference between the meanings of a proper quantity and Lorentz invariant quantity. Perhaps you could clarify.

Perhaps your objection is to my statement "The total pressure used in the stress energy tensor is none of those things", in which case, hopefully I have explained what I meant by that statement in the above replies. I have no formal training in the language used here, so hopefully you will be generous and skilled enough to interpret my intended meaning from the context.

Finally and most importantly in the context of the OP, given:

##P' = f\frac{nRT_0}{(V_0/\gamma)}## and ##P' =g P_0##

what are the functions f and g?
 
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  • #27
WannabeNewton
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Clearly the quantity ##p'_x## you describe here is not a Lorentz invariant and is not isotropic and this is the quantity I meant when I said "the pressure" which was a bit sloppy of me. It is this non isotropic transformed pressure that can be thought of as the total pressure that includes terms related to static pressure and dynamic pressure.
Alright so I see now what quantity you are referring to but let me point to something I said earlier: ...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.

Now above pervect took the special case of a pressure-less perfect fluid (also called dust in relativity) and showed that if one takes the static pressure ##p## measured in the fluid rest frame and Lorentz boosts the associated components of the stress-energy tensor, one gets a quantity which in the Newtonian limit resembles the dynamic pressure. But does the interpretation carry over unscathed in the fully relativistic regime?

Furthermore if the perfect fluid is not pressure-less then we have ##p'_x = \gamma^2 (\frac{v^2}{c^2} c^2\rho + p)##. In the Newtonian limit, ##c^2 \rho >> p## but ##\frac{v^2}{c^2} << 1## so one cannot ignore the ##p## term in ##p'_x## in this limit. How would one interpret ##p'_x## now?

Your terminology is confusing to me here. I thought the definition of a "scalar" quantity was a quantity that does not have directional components?
Yes so ##p## is a scalar field in exactly that sense but it also comprises three components of the stress-energy tensor in the fluid rest frame, namely ##T^{xx} = T^{yy} = T^{zz} = p## which is what I was referring to.

Perhaps your objection is to my statement "The total pressure used in the stress energy tensor is none of those things", in which case, hopefully I have explained what I meant by that statement in the above replies.
If your statement refers to the "pressure" that you were referring to above then I don't object to your statement but rather I have yet to see a source that actually gives physical/operational meaning to the Lorentz boosted quantities like ##p'_x## in a fully relativistic regime. This is why I immediately thought you were referring to ##p## because this is what one normally calls "the pressure" when dealing with the stress-energy tensor of perfect fluids.

Finally and most importantly in the context of the OP, given:

##P' = f\frac{nRT_0}{(V_0/\gamma)}## and ##P' =g P_0##

what are the functions f and g?
Wouldn't this require one to adopt a transformation rule for temperature first? This goes back to the thorny nature of relativistic thermodynamics mentioned earlier. Furthermore is the ideal gas law Lorentz covariant?
 
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Now above pervect took the special case of a pressure-less perfect fluid (also called dust in relativity) and showed that if one takes the static pressure ##p## measured in the fluid rest frame and Lorentz boosts the associated components of the stress-energy tensor, one gets a quantity which in the Newtonian limit resembles the dynamic pressure.
It appears pervect chose the case of a pressure-less fluid to isolate the dynamic pressure term for clarity.

Furthermore if the perfect fluid is not pressure-less then we have ##p'_x = \gamma^2 (\frac{v^2}{c^2} c^2\rho + p)##. In the Newtonian limit, ##c^2 \rho >> p## but ##\frac{v^2}{c^2} << 1## so one cannot ignore the ##p## term in ##p'_x## in this limit. How would one interpret ##p'_x## now?
Wikipedia http://en.wikipedia.org/wiki/Static_pressure#Static_pressure_in_fluid_dynamics defines stagnation pressure as ##(\rho v^2)/2 + p##. This is basically the total pressure that I mentioned earlier with head pressure ignored.

Your equation for ##p'_x## reduces to ##p'_x = \rho v^2 + p## in the non relativistic regime which is pretty close to the expression for stagnation pressure, except for the factor of 2 in the dynamic pressure that pervect mentioned.

Since ##p'_x## is effectively the total pressure, then when v/c=0 then ##p'_x## reduces to ##p## exactly as it does in non relativistic hydronamics according to the Bernoulli equation.


... Wouldn't this require one to adopt a transformation rule for temperature first? This goes back to the thorny nature of relativistic thermodynamics mentioned earlier. Furthermore is the ideal gas law Lorentz covariant?
Pretty much what I am trying to find out in the parallel thread on relativistic temperature. Also see #25.
 
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