# Proof of invariance of gas pressure

1. Feb 4, 2008

### yuiop

Proof that pressure is invariant on the front and back faces of the moving box.

(We are already agreed in another thread that the pressure is invariant on the other faces)

Pressure in the rest frame of the box: (Frame S)

When a gas particle of mass M and velocity U collides with a face (area A) of the box it rebounds at the same velocity (assuming perfect elasticity) so the change in momentum d(mv) of the particle is 2*M*U. The time interval dt between collisions with the same face by the same particle is 2*L/U where L is the length of the box normal to the face. Pressure = force/area = (d(mv)/dt)/A. Using the relativistic expression for momentum (d(mv)y) where y is the Lorentz factor, the pressure in the rest frame for N particles can be expressed as

$$P = \frac{NMU^2}{AL\sqrt{1-\frac{U^2}{c^2}}}$$

Pressure in Frame S’:

When the box is moving relative to the observer the velocity of the particle is given by the relativistic velocity addition formula.

The forward velocity of the particle is:

$$F=\frac{(V+U)}{(1+\frac{UV}{c^2})}$$

and the backward velocity is:

$$B=\frac{(V-U)}{(1-\frac{UV}{c^2})}$$

The time T for the particle to traverse the length of the box and back is given by:

$$T=\frac{L}{y}\left(\frac{1}{(V-B)}+\frac{1}{(F-V)}\right)$$

If we substitute the expressions for F, B and y and simplify we get:

$$T=\frac{2L}{U \sqrt{1-\frac{V^2}{c^2}}}$$

which is the time interval in the rest frame multiplied by the Lorentz factor.

The pressure P’ on the front face of the moving box is then given by:

$$P\prime = \frac{NM}{AT}\left(\frac{F}{\sqrt{1-\frac{F^2}{c^2}}}-\frac{B}{\sqrt{1-\frac{B^2}{c^2}}}\right)$$

To prove that P = P’ it has to be shown that the expression for pressure in the rest frame is equal to the expression for pressure in the moving frame:

$$\frac{NMU^2}{AL\sqrt{1-\frac{U^2}{c^2}}}=\frac{NM}{AT}\left(\frac{F}{\sqrt{1-\frac{F^2}{c^2}}}-\frac{B}{\sqrt{1-\frac{B^2}{c^2}}}\right)$$

Setting the value of c to 1 to simplify things this becomes:

$$\frac{NMU^2}{AL\sqrt{1-U^2}}=\frac{NM}{AT}\left(\frac{F}{\sqrt{1-F^2}}-\frac{B}{\sqrt{1-B^2}}\right)$$

-->

$$\frac{NMU^2}{AL\sqrt{1-U^2}}= \frac{NM}{AT}\left(\frac{1}{\sqrt{\frac{1}{F^2}-1}}-\frac{1}{\sqrt{\frac{1}{B^2}-1}}\right)$$

Substituting the full expressions for F and B (but with c=1)

$$\frac{NMU^2}{AL\sqrt{1-U^2}} = \frac{NM}{AT} \left(\frac{1}{\sqrt{\frac{(1+UV)^2}{(V+U)^2}-1}} - \frac{1}{\sqrt{\frac{(1-UV)^2}{(V-U)^2}-1}}\right)$$

-->

$$\frac{NMU^2}{AL\sqrt{1-U^2}} = \frac{NM}{AT} \left(\frac{1}{\sqrt{\frac{(1+UV)^2-(V+U)^2}{(V+U)^2}}} - \frac{1}{\sqrt{\frac{(1-UV)^2-(V-U)^2}{(V-U)^2}}}\right)$$

-->

$$\frac{NMU^2}{AL\sqrt{1-U^2}} = \frac{NM}{AT} \left(\frac{1}{\sqrt{\frac{(1-U^2)(1-V^2)}{(V+U)^2}}} - \frac{1}{\sqrt{\frac{(1-U)^2(1-V^2)}{(V-U)^2}}}\right)$$

-->

$$\frac{NMU^2}{AL\sqrt{1-U^2}} = \frac{NM}{AT} \left(\frac{(V+U)}{\sqrt{(1-U^2)(1-V^2)}} - \frac{(V-U)}{\sqrt{(1-U^2)(1-V^2)}}\right)$$

-->

$$\frac{NMU^2}{AL\sqrt{1-U^2}} = \frac{NM}{AT} \left(\frac{2U}{\sqrt{(1-U^2)(1-V^2)}} \right)$$

Substitute the expression T = $$\frac{2L}{U \sqrt{1-\frac{V^2}{c^2}}}$$ obtained earlier (setting c=1):

$$\frac{NMU^2}{AL\sqrt{1-U^2}} = \frac{NMU\sqrt{1-V^2}}{2AL} \left(\frac{2U}{\sqrt{(1-U^2)(1-V^2)}} \right)$$

-->

$$\frac{NMU^2}{AL\sqrt{1-U^2}} = \frac{NMU^2}{AL\sqrt{(1-U^2)}}$$

So P = P’ within the valid domain (-c<V<c) and (-c<U<c)

Q.E.D.

2. Feb 4, 2008

### 1effect

The problem is that according to the way the stress energy tensor transforms, p'_xx is not equal to p_xx.

This one is correct because you are averaging the speeds. You can get the above easier by noticing that the average time (2L/U) is time dilated by $$\frac{1}{\sqrt{1-\frac{V^2}{c^2}}}$$
in frame S'.

You can't simply divide by T. You need to calculate the derivative wrt T.
I pointed this out early on in the other thread.

Last edited: Feb 4, 2008
3. Feb 4, 2008

### pervect

Staff Emeritus
I get the following (setting c=1 for convenience)

$P_{rest}$ = pressure in rest frame =

$$P_{rest} = \left( \frac{2 N M u}{\sqrt{1-u^2}} \right) / \left(A \frac{2 L}{u} \right)$$

N = number of particles, M = mass of particle, u = velocity of particle, A = area of front of box, L = length of box

Now in the moving frame:

T = total time to traverse box, in moving frame

$$T = \frac{L}{F-v} + \frac{L}{v-B} = \frac{2L}{u \left(1 - v^2 \right)}$$

this is the same final answer as kev, but kev's setup seems to have typos.

The pressure P is

$$P = \frac{1}{A} \frac{N M}{T} \left( \frac{F}{\sqrt{1-F^2}} -\frac{B}{\sqrt{1-B^2}} \right)$$

However, when I simplify this I get
$$\frac{P}{P_{rest}} = \sqrt{1-v^2}$$

Note that it is unfortunately possible that I've made a mistake somewhere (even though I've used Maple to help with the algebra).

4. Feb 4, 2008

### pervect

Staff Emeritus
As far as I know the definition of pressure that kev used is correct. (It at least seems reasonable) even though it is different from the diagonal element of the stress-energy tensor. It's basically not productive to calculate something different from what kev calculated and then complain that your answer is different than his. Engineering use of the term pressure is different from the GR use of the term pressure due to the convective terms as I mentioned earlier. We need to agree on what we are calculating, first.

I do not, unfortunately, have a textbook reference on this particular problem to help settle the matter more completely. However, I also find that pressure, using kev's definition, is not invariant (per my previous post).

Last edited: Feb 4, 2008
5. Feb 4, 2008

### 1effect

I explained that it is not correct, F is equal to dp/dt, there is no derivative of p with respect to t in kev's derivation (nor is there one in yours). Dividing by t is not the same as calculating the derivative, try it and you will convince yourself. There is no justification in dividing the momentum by the total round trip T either.

6. Feb 4, 2008

### yuiop

OK, that is basically the same as my

$$P = \frac{NMU^2}{AL\sqrt{1-U^2}}$$

When the box is moving with respect to us the length of the box is measured as length contracted. Using L/y rather than just L where y = $$\left(\frac{1}{\sqrt{1-v^2}}\right)$$ for the length of the box in your equation we get:

$$T = \left(\frac{L}{y(F-v)}\right) + \left(\frac{L}{y(v-B)}\right) = \frac{2L}{uy(1 - v^2)} = \frac{2L\sqrt{1-v^2}}{u(1 - v^2)} = \frac{2L}{u \sqrt{1 - v^2 }}$$

As 1effect pointed out $$\frac{T}{T_{rest}}$$ should equal $$\frac{1}{\sqrt{1-v^2}}$$

If you used T = $$\frac{2L}{u \left(1 - v^2 \right)}$$ rather than T = $$\frac{2L}{u \sqrt{1 - v^2 }}$$ in your final answer

that would explain why you get $$\frac{P}{P_{rest}} = \sqrt{1-v^2}$$

rather than $$\frac{P}{P_{rest}} = 1$$

7. Feb 4, 2008

### pervect

Staff Emeritus
OK, with the correction for the change in L, which is in fact needed, I now get the same answer as kev. So I basically agree with his point. I'd still feel better if we had a textbook reference on this point, however. Most textbooks that I have seen do not talk about "pressure" in moving frames at all. Most textbooks make a point of defining and using it only in the rest frame of some particular fluid under consideration.

In this sense "pressure" is a bit of a dead end, because the concept of the stress-energy tensor replaces it.

I also agree that the stress-energy tensor transforms in a different manner than what we have just calculated, however I do not believe that the stress-energy tensor can be regarded as the same as the pressure when one considers moving frames. (The diagonal elements of the stress-energy tenser are the same as the pressure in the rest frame of an isotropic fluid of course).

This issue arises when one tries to explain, for instance, how a mechanical clock ticks at the same rate as a light clock. Thinking that the stress energy tensor gives the pressure leads to incorrect results. (There is some thread where I wandered down this wrong path in the past, I don't recall where it was anymore).

8. Feb 4, 2008

### 1effect

I think that it is easy to derive the correct result in a rigorous way.
Let $$P_x$$ and $$p_x$$ be the pressures in frames K and k respectively.
Let $$F_x$$ and $$f_x$$ be the forces in frames K and k respectively.

$$P_x=\frac{F_x}{A}=\frac{f_x}{A}=p_x$$

So, the "longitudinal" pressure is also invariant. We have already shown that the "transverse" pressure is also invariant. This is interesting, it supports the point that there is discrepancy between the interpretation of the stress-energy tensor and the conventional interpretation of pressure (pervect's original point)

Last edited: Feb 4, 2008
9. Feb 4, 2008

### yuiop

Which is why you have to ask yourself exactly what is the "pressure-like" element that comes out the stress-energy tensor transform. I am convinced it is not a simple pressure as can be measured in a gas with a pressure gauge. I admit I am not familiar with the stress-energy tensor but I usually see it being used in the context of gravity where the energy of a system including stresses contribute to the gravitational curvature of space.

The total change in momentum of the collisions with faces occur in very brief "impulses" folowed by intervals of constant momentum before the next collision. Using calculus is useful for finding what is happening "instantaneously" at any point point of some curve due to some some continuously and smoothly changing function (which is not the case here). I have seen pressure analysed by using a similiar method to the one I used (taking the interval dt to be the mean interval between collisions) in various text books and it seems a reasonable approach to take.

If you are prepared to accept that argument you can easily analyse the invariance of the formulas for P and P' by copying and pasting these formulas into a spreadsheet and labeling some cells with the variable names used in the formuals.:

Paste =(u+v)/(1+u*v/c^2) into a cell labeled f
Paste =(v-u)/(1-u*v/c^2) into a cell labeled b
Paste =1/sqrt(1-v^2/c^2) into a cell labeled y
Paste =L*(1/(f-v)+1/(v-b))/y into a cell labeled t (the time interval in the moving frame)

Define names of empty cells in the name box of the spreadsheet to input variable values:

u (Enter the velocity of the particles in the rest frame here)
v (Enter the velocity of the box relative to the observer here)
c (The speed of light)
L (Enter the length of the box in the rest frame here)
A (Enter the cross sectional area of a tranverse face here)
m (Enter the mass of indivdual particles here)
n (Enter the number of particles here)

Output area:

Paste =n*m*u^2/L/A/sqrt(1-u^2/c^2) for the pressure in the rest frame.
Paste =n*m*(f/SQRT(1-f^2/c^2)-b/SQRT(1-b^2/c^2))/(A*t) for the pressure in the moving frame.

10. Feb 4, 2008

### yuiop

Actually force tranverse to the motion of the frame is reduced by the Lorentz factor but force parallel to the motion is invarient (sometimes refered to as longitudinal force in the literature).

Longitudinal force and tranverse area are both invarient and so therefore is the pressure on the tranverse faces.

11. Feb 4, 2008

### 1effect

You are correct, I was looking at the wrong transformation (4-vector instead of the 3-vector). I edited my post. I still doin't think that the derivation that uses division by T is clean. Besides, it is very long, the one based on the fact that $$F_x$$ is invariant is much cleaner.

12. Feb 4, 2008

### 1effect

I agree, see the discrepancy that was uncovered.

I don't think that the approach is clean. Either way, we have a much better approach now.

There is unfortunately a very big problem with both approaches: there is no reason to consider the force to be aligned with any axis, so, in the general case, you have $$f_x,f_y,f_z$$ transforming into a very ugly way into
$$F_x,F_y,F_z$$. While the approach I outlined can still deal with this situation it becomes clear that you won't get pressure invariance in the general case.

Last edited: Feb 4, 2008
13. Feb 5, 2008

### 1effect

Unfortunately , this is not correct.

$$F_x=f_x$$ only if $$f_y=0$$ and $$f_z=0$$

In the most general case, the transverse pressure is not invariant either because the force is reduced by more than the Lorentz factor alone . This is due to the fact that the general transformation for the transverse force is:

$$F_y=f_y\frac{sqrt(1-(V/c)^2)}{1+\frac{u_xV}{c^2}}$$

where $$V$$ is the constant speed of the box wrt frame K and $$u=(u_x,u_y,u_z)$$ is the variable speed of the particles wrt frame k (the box).
It is obvious that the transverse pressure $$P_y$$ is not invariant in the most general case.

$$P_y=\frac{F_y}{A_y}=\frac{p_y}{1+\frac{u_xV}{c^2}}$$

Likewise:

$$P_z=\frac{F_z}{A_z}=\frac{p_z}{1+\frac{u_xV}{c^2}}$$

In the particular case calculated by us, $$u_x=0$$, so the transverse pressures are indeed invariant.

Last edited: Feb 5, 2008
14. Feb 6, 2008

### yuiop

The full formula for x component of the notion of force you are using is:

$$F_x=f_x-\frac{v\left(f_yu_y+f_zu_z\right)}{c^2\left(1+\frac{u_xV}{c^2}\right)}$$

In the case that $$u_y=0$$ and $$u_z=0$$ then $$P_x=p_x$$.

However you are correct that despite being able to show that $$P_x=p_x$$, $$P_y=p_y$$. and $$P_z=p_z$$ when the faces are exactly parallel or tranverse to the motion of the box those formulas do not hold when we consider faces that are diagonal to the motion. This is surprising and the correction requires closer inspection.

The reason the anomally appears when we consider diagonal faces is that unlike parallel and tranverse faces, diagonal faces change there orientation when the frame is Lorentz transformed. The equations for force that you are using are derived from the effects of an electromagnetic force acting on a charged particle. Those equations give the total force acting on the particles and not just the component of the force that is normal to the face we are considering. When considering static gas pressure force on a surface we consider only the compenents of the pressure force that are normal to the surface. (We are not considering Bernoulli type fluid flow dynamic pressures such as when air flows over an aircraft wing or through a nozzle).

A simple proof that the pressure of a gas in an enclosed vessel is invariant in all directions under Lorentz transformation is to note that pressure is scalar. Knowing gas pressure is scalar then once we have proved pressure is invariant along any axis we have by definition proved gas pressure is invariant under the Lorentz transform in any direction. In fact knowing gas pressure is scalar, it does not make sense to even talk of longitudinal or transverse pressure because a scalar quantity does not have direction, only magnitude.

So if we accept the premise that gas pressure is scalar and invariant under a Lorentz transform (i.e Gas pressure F/A is a Lorentz scalar) then:

$$P={F \over A}={f_x \over a_x}={f_y \over a_y}={f_z \over a_z }={f \over a}=p$$

(Kev's Law :tongue: )

We can easily derive the correct forces due to gas pressure simply considering how the areas of the surfaces the pressure acts on transforms which is much simpler.

Now if the surface area of a face is $$l^2$$ in the rest frame we can break it down into components parallel $$(L_x)$$ and transverse $$(L_y)$$ to the direction to the the x axis. When calculating pressure forces, area is considered to be a vector normal to the surface (strange but true) so when we transform the area vector it is the y component that length contracts and not the x component (parallel to the direction of motion) In the rest frame the area vector is:
$$a=\sqrt{l_x^2+l_y^2}$$.

The transformation of the area vector when the box is moving relative to us in the x direction with velocity v is then:

$$A=\sqrt {\left( {l_x^2+l_y^2(1-v^2/c^2)\right)}$$.

Since F/A is here defined as a Lorentz scalar, pressure force must transform in the same way as Area. The transformation of force due to pressure is then:

$$F_{xy}=\sqrt {\left( {f_x^2+f_y^2(1-v^2/c^2)\right)}$$

or if we include the z plane to get the more general case:

$$F_{xyz}=\sqrt {\left( {f_x^2+(f_y^2+f_z^2)(1-v^2/c^2)\right)$$.

Note: This is exactly how the force of a spring transforms. When we analyse the force of gas pressure acting on a piston held in place by a spring there there is no imbalance of forces on the piston when we transform frames. Any proposal that gas pressure is not invariant is not in agreement with Special Relativity unless it can be shown that there is compensating physical phenomena that transforms so as to compensate for the change in pressure. It is unscientific to say the pressure changes in the moving frame but cannot be detected in the rest frame just because it does. Every other instance of physical change under transformation in relativity is compensated for by another physical change. A simple example is that length contraction is not detected in the rest frame because time dilates in such a way that the length change cannot be measured in the rest frame.

Last edited: Feb 6, 2008
15. Feb 6, 2008

### 1effect

I am aware. This is why I told you that $$F_x=f_x$$ if $$f_y=0$$ and $$f_z=0$$. You replaced my condition with an equally unrealistic one : $$u_y=0$$ and $$u_z=0$$. Both conditions restrict the applicability of the solution, wouldn't you agree?

Maybe we can say that $$F_x=f_x$$ if and only if $$f_yu_y+f_zu_z=0$$. Still, the applicability of the solution is severely constrained since the above cannot be true in the more general case.
Also, you have not addressed my stronger objection, that :

$$F_y=f_y$$ if and only if $$u_x=0$$.

If you want to derive a law, you need to make it work under all conditions, this is why is called a "law" :-)

Good, now we have a clean formalism we can work with.No more "divide by T" :-)
The reason has nothing to do with the orientation of the box facets, they are parallel with the planes of the reference system. The reason is that both force $$f=(f_x,f_y,f_z)$$ and particle velocity $$u=(u_x,u_y,u_z)$$ have all non-zero components. So, the "kev Law" works only under very restrictive conditions, unlike a real law.

Last edited: Feb 6, 2008
16. Feb 6, 2008

### yuiop

I agree that if we use the force formula you are using then $$F_y=f_y$$ if and only if $$u_x=0$$ but as I pointed out earlier that is the total force acting on the particle and not the component that is normal to the face the particle is bouncing off.

The 'law' rests on pressure being a scalar quantity.

Do you not agree gas pressure is a scalar quantity?

Or do you not agree with my statement "Any proposal that gas pressure is not invariant is not in agreement with Special Relativity unless it can be shown that there is compensating physical phenomena that transforms so as to compensate for the change in pressure. It is unscientific to say the pressure changes in the moving frame but cannot be detected in the rest frame just because it does."?

P.S.

There may be risidual components of the particles velocity and force that are not normal (or orthogonal to the face..call it what you will). These residual components act parallel to the face and may not cancel out at the face. However in a closed container there always facers that are opposite and or normal to each other and the residuals cancel out so there is no overall rotation of the box. However there may be stresses acting parallel to the faces and these appear to work in the direction of compression along the x axis which may not be apparent in the rest frame because it is in line with length contraction.

Last edited: Feb 6, 2008
17. Feb 6, 2008

### 1effect

It doesn't matter, all forces transform the same way from frame k to frame K, would you agree? So, my objections to your approach stand, you can't have "laws" that have a lot of restrictions attached. Your method has just too many conditions placed on the force and velocity components. In a gas, the particles have random motion, no? :-)

Last edited: Feb 6, 2008
18. Feb 6, 2008

### yuiop

I disagree. The formulas you are using are not the most general. For example what is $$u_x$$or $$u_y$$ in the context of a spring in a static situation? The force formuals I gave are more applicable to static forces.

A paper was published claiming static and dynamic forces transform differently. I will try and find it if you are interested.

[EDIT] I think this it. http://www.springerlink.com/content/n6573281634614q2/ I do not have access to the content of the article as it "pay per view".

Last edited: Feb 6, 2008
19. Feb 6, 2008

### 1effect

This is not the point (you are wrong, by the way). The point is that , in order to work, your method imposes unrealistic constraints on both force and velocity. This has been pointed out to you repeatedly by dharis.

I looked at the title , it deals with transformation of static forces. I thought that we agreed that the gas pressure is created by molecules changing momentum when they bounced off the walls, so I thought that we agreed that we were working with dynamic forces. (the kind that you got by dividing momentum by T :-)) . So, why are you bringing in a paper that deals with static forces? This further confuses the discussion.

OK, so what would change in your approach? How would your equations change?

Last edited: Feb 6, 2008
20. Feb 6, 2008

### yuiop

I would not change anything. Static force transforms as $$F_{xyz}=\sqrt {\left( {f_x^2+(f_y^2+f_z^2)(1-v^2/c^2)\right)$$

The total forces acting on a particle accelerated by an electromagnetic field transform as described by the equations you are quoting.

The equation I am using considers only the forces normal to the the plane of the surface the pressure force is acting on and take account of the altered orientation of the surface under consideration when the surface is diagonal to the motion. For example a surface that is at 45 degrees to the x axis in the rest frame is not at 45 degrees to the x' axis when the x' axis of the moving reference frame is parallel to the x axis and the motion of the gas vessel is parallel to the x axis.

I consider the pressure force as static as I am considering the force acting on the surface and the surface is not going anywhere in its rest frame.

The crux of the matter is:

Is gas pressure a scalar quantity or not?

Last edited: Feb 6, 2008