Proof of invariance of gas pressure

[1effect]

Of course they do not factor out. Why on erath should they? They do not factor out in the rest frame so there is no reason to think they should in the moving frame. Perhaps I should flesh out the equations more fully to make it clear.

In the rest frame:

Say we have 3 different springs all made of different materials with different cross sections and under different strains orientated at right angle to each and connected to common point.

$$f_x= E_xA_x\left({\Delta L_x / L_x}\right)$$

$$f_y= E_yA_y\left({\Delta L_y / L_y}\right)$$

$$f_z= E_zA_z\left({\Delta L_z / L_z}\right)$$

The resultant force in the rest frame is:

$$\sqrt{\left(\left(E_xA_x\left({\Delta L_x / L_x}\right)\right)^2 + \left(E_yA_y\left({\Delta L_y / L_y}\right)\right)^2 + \left(E_zA_z\left({\Delta L_z / L_z}\right)\right)^2\right)}$$

or more simply $$f = \sqrt{f_x^2+f_y^2+f_z^2}$$

There is no implication in that last formula that the Delta L's cancel out. All the above is simple Newtonian physics that can be comfirmed in a school classroom with basic equipment.

In the moving frame:

$$F_x= E_xA_x{\left(\Delta L_x\sqrt{1-v^2/c^2}\right) / \left(L_x\sqrt {1-v^2/c^2}\right)} = E_xA_x\left({\Delta L_x / L_x}\right) = f_x$$

for the longitudinal force and

$$F_y= E_y \left(A_y\sqrt{1-v^/c^2}\right){\left({\Delta L_y / L_y\right) = E_yA_y\left({\Delta L_y / L_y}\right)\sqrt{1-v^2/c^2} = f_y\sqrt{1-v^2/c^2}$$

$$F_z= E_z \left(A_z\sqrt{1-v^/c^2}\right){\left({\Delta L_z / L_z\right) = E_zA_z\left({\Delta L_z / L_z}\right)\sqrt{1-v^2/c^2} = f_z\sqrt{1-v^2/c^2}$$

for the transverse case.

The total magnitude of the transformed forces is done by summing the vectors to get the resultant the same way it was done in the rest frame.

$$F = \sqrt{F_x^2+F_y^2+F_z^2} = \sqrt{f_x^2+\left(f_y\sqrt{1-v^2/c^2}\right)^2+\left(f_z\sqrt{1-v^2/c^2}\right)^2 }=\sqrt{f_x^2+(f_y^2+f_z^2)(1-v^2/c^2)$$

That's better, it looks correct. You don't need the resultant, $$F$$, you already have the transforms for the components $$F_x,F_y,F_z$$. Then, you can use the fact that $$A_x$$ transforms like $$F_x$$, $$A_y$$ transforms like $$F_y$$, so you don't need the funky vector area $$A$$ that you defined in order to prove that
$$P_x=p_x$$, $$P_y=p_y$$, etc. This proof is much cleaner.

Think about this. Imagine a cylinder of gas under pressure is contained at one end by a spring loaded piston. When the system is in equilibrium in the rest frame the pressure of the gas in the cylinder is equal to the (spring force)/(piston area). When we move relative to the cylinder (whatever its orientation) the piston does not move but we measure that the piston area has changed and so has the spring force. The pressure exerted by the spring loaded piston on the gas is (spring force)'/(piston area)' where the prime indicates transformed quantities. If the transformed pressure P' is not equal to (spring force)'/(piston area)' then the piston should move in the rest frame and Special Relativity is broken as a theory.

Therefore (if Special Relativity is a valid theory) we can calculate the transformed pressure in the cylinder simply by calculating the transform of (spring force)/(piston area).

That's not the point,of course SR is valid, no one challenges that. What I challenge is your attempt to model gas molecules as elastic rods (?!). You switched to this when I showed you that regular (dynamic) forces don't work.

 

[yuiop]

That's not the point,of course SR is valid, no one challenges that. What I challenge is your attempt to model gas molecules as elastic rods (?!). You switched to this when I showed you that regular (dynamic) forces don't work.

As far as modelling pressure gas molecules as springs (or rods) that is not exactly what I was doing. I was simply stating when 2 forces are in equilibrium the forces in suitable units (eg Newtons) are equal. If I place a weight on top of a piston enclosing a volume of gas I can show that the pressure within the volume is equal to the gravitational force of the weight per unit area of the piston. That in no way implies modelling the gas particles as gravitational bodies. It is just a statement of the simple law of physics that forces are opposite and equal in a system at equilibrium. The beauty of special relativity is that the laws of physics are the same in all inertial reference frames even when the inertial reference frame is moving relative to the observer making the measurements .

You seem to object to elastic rods being used to analyse Hooke's law. To quote wikipedia "We may view a rod of any elastic material as a linear spring." The elastic rod (or linear spring) is easier to analyse (and visualise) in terms of cross section and extension, but Hooke's law applies equally well to helical springs as it does to elastic rods.

If you at look at post#90 of the "Lorentz contraction of box filled with gas thread https://www.physicsforums.com/showthread.php?t=210634&page=6 you will see I brought up the subject of analysing the pressure in terms of a piston held in place by a spring before I even started this thread. That analysis is fairly simple and the logic is simple and clear. That was why I knew that Pxx was invariant before I even set out to show that by the alternative method of analysing particle collisions (despite the fact dhris was adamant to the point of rudeness that pressure along the x-axis was not invariant). The particle analysis along the diagonals is pretty complex and it is much simpler to analyse in terms of springs and pistons. However. I agree it would more convincing if I could show the pressure is invariant along the diagonal directions in terms of particle collisions using dmdv/dt and I will try and do that as time allows. It might help give us all a clearer understanding of why there appears to be different forms of force transformations. Do static and dynamic forces transform differently because they are fundementally different quantities or is static force just a vector component of dymanic force? I don't think it will be easy, and it would probably be simpler to analyse in terms of 4-vectors by someone who knows how to handle them (like Pervect).

 

[1effect]

You seem to object to elastic rods being used to analyse Hooke's law.

No, you got this wrong, I clearly objected to your using elastic rods for modelling the behavior of gas molecules. My post is pretty clear on this. I think that you will agree with me that a stell rod has very little in common with a gas, no?
While we are on this subject, you can see that you conveniently chose an expression for the force that gurantees pressure invariance. This can be easily seen from:

$$\frac{F}{A}=E \frac{\Delta L}{L}$$ !

Should you choose [/b]real[/b] springs, the Hooke law gives $$F=-kx$$ and this model will not produce invariant pressure.

I object to your modelling choices. I think that you got the answer in your head (invariant pressure, which might as well be correct) and you are trying to fit models that substantiate the answer. This is not right. I am also concerned by the fact that invariant pressure is at odds with the transformation of the stress-energy tensor, an issue that we never solved. So, we have made very little progress in solvong the problem.

 

[yuiop]

No, you got this wrong, I clearly objected to your using elastic rods for modelling the behavior of gas molecules. My post is pretty clear on this. I think that you will agree with me that a stell rod has very little in common with a gas, no?

Obviously you did not read my answer --> " As far as modelling pressure gas molecules as springs (or rods) that is not exactly what I was doing. I was simply stating when 2 forces are in equilibrium the forces in suitable units (eg Newtons) are equal. If I place a weight on top of a piston enclosing a volume of gas I can show that the pressure within the volume is equal to the gravitational force of the weight per unit area of the piston. That in no way implies modelling the gas particles as gravitational bodies. It is just a statement of the simple law of physics that forces are opposite and equal in a system at equilibrium. The beauty of special relativity is that the laws of physics are the same in all inertial reference frames even when the inertial reference frame is moving relative to the observer making the measurements."

While we are on this subject, you can see that you conveniently chose an expression for the force that gurantees pressure invariance. This can be easily seen from:

$$\frac{F}{A}=E \frac{\Delta L}{L}$$ !

Should you choose [/b]real[/b] springs, the Hooke law gives $$F=-kx$$ and this model will not produce invariant pressure.

This is plain silly. They are the same thing.

$${k}=\frac{EA}{L}$$

$${x}={\Delta L}$$

$${kx}=EA\frac{\Delta L}{L}$$

The sign just indicates the force vector is in the opposite direction to the displacement of the extended (or compressed) spring.

I object to your modelling choices. I think that you got the answer in your head (invariant pressure, which might as well be correct) and you are trying to fit models that substantiate the answer. This is not right. I am also concerned by the fact that invariant pressure is at odds with the transformation of the stress-energy tensor, an issue that we never solved. So, we have made very little progress in solvong the problem.

You are contradicting yourself.

You provided the force transformation formula $$F_x=f_x-\frac{v\left(f_yu_y+f_zu_z\right)}{c^2\left(1+\fra c{u_xV}{c^2}\right)}$$

that you say exposes a flaw in my arguments, yet YOUR force transformation disagrees with the stress-energy tensor for Pxx where Fy and Fz =0. So your formula exposes a flaw in the stress-energy tensor (unlikely) or as Pervect has already indicated the pressure element that comes out the stress energy tensor should not be viewed as exactly the same thing as the pressure in a container of gas particles used in the context of the gas laws.

 

[1effect]

Obviously you did not read my answer --> " As far as modelling pressure gas molecules as springs (or rods) that is not exactly what I was doing.

Of course you were, you were using the formula for elastic rods from the wiki page. The model you chose (rods or linear springs) gives your model automatic pressure invariance by virtue of the fact that $$\frac{F}{A}=E \frac{\Delta L}{L}$$. So, there is nothing more to prove.
So, your choice of model seems to be taylored to "fit" the answer you are looking for.
Thinking about this problem a little more, I think that there is a very simple and straightforward answer to the pressure invariance.

This is plain silly. They are the same thing.

$${k}=\frac{EA}{L}$$

$${x}={\Delta L}$$

$${kx}=EA\frac{\Delta L}{L}$$

The sign just indicates the force vector is in the opposite direction to the displacement of the extended (or compressed) spring.

I am not an expert in material science, I saw that the wiki page had two different entries for rods and springs.
The important point is that you chose a formula that guarantees invariance due to the term $$\frac{\Delta L}{L}$$ . This has been worrying me since the very moment you switched to this model.

You are contradicting yourself.

You provided the force transformation formula $$F_x=f_x-\frac{v\left(f_yu_y+f_zu_z\right)}{c^2\left(1+\fra c{u_xV}{c^2}\right)}$$

that you say exposes a flaw in my arguments, yet YOUR force transformation disagrees with the stress-energy tensor for Pxx where Fy and Fz =0.

This is not an issue of "me" vs. "you". I am worried about the fact that all the different methods produce contradictory results and all the results produced so far contradicted the way the stress-energy tensor transforms. Perhaps we should leave the stress-energy tensor aside, there is a need of reconciling the results we've seen so far. That's all.

 

[yuiop]

...

This is not an issue of "me" vs. "you". I am worried about the fact that all the different methods produce contradictory results and all the results produced so far contradicted the way the stress-energy tensor transforms. Perhaps we should leave the stress-energy tensor aside, there is a need of reconciling the results we've seen so far. That's all.

I have just realized we have been interpreting your force transformation equation incorrectly.

The force transformation equations as derived here http://www.sciencebits.com/node/176 and here http://www.geocities.com/physics_world/sr/force_trans.htm

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

$$F_y' =f_y\frac{\sqrt{1-v^2/c^2}} { \left(1-vu_x/c^2\right)}$$

$$F_z ' =f_z\frac{\sqrt{1-v^2/c^2}} { \left(1-vu_z/c^2\right)}$$

where $$f_x, f_y, f_z$$ are the components of the force acting on an object with velocity components $$u_x, u_y, u_z$$ in inertial reference frame S

and $$F_x ', F_y ', F_z '$$ are the components of the force measured by an observer with velocity v relative to frame S.

When the box of gas is moving with velocity v in the x direction relative to the observer, the wall of the box is moving with velocity v relative to the observer (obviously). The object the force (exerted by the gas pressure) is acting on, is the wall of the box. The wall of the box has velocity components $$u_x=v,u_y=0,u_z=0$$ (no matter what orientation the wall under consideration has with respect to the x axis). The transformed force components acting on the wall can now be calculated as:

$$F_x ' =f_x-\frac{v\left(f_y*0+f_z*0\right)}{c^2\left(1+v^2/c^2\right)} = f_x$$

$$F_y' =f_y\frac{\sqrt{1-v^2/c^2}} { \left(1-v^2/c^2\right)} = f_y\sqrt{1-v^2/c^2}$$

$$F_z ' =f_z\frac{\sqrt{1-v^2/c^2}} { \left(1-v^2/c^2\right)} = f_z\sqrt{1-v^2/c^2}$$

which brings your force transformation equations into agreement with the ones I have been using. Don't know why I did not spot that earlier. It is obvious when we take the length contraction of the areas normal to those force components, that pressure of a gas is a Lorentz scalar (Scalar and invariant).

 

[1effect]

where $$f_x, f_y, f_z$$ are the components of the force acting on an object with velocity components $$u_x, u_y, u_z$$ in inertial reference frame S

Correct.

The wall of the box has velocity components $$u_x=v,u_y=0,u_z=0$$ (no matter what orientation the wall under consideration has with respect to the x axis).

You are contradicting the prior statement. $$u=(u_x,u_y,u_z)$$ is the velocity of the gas particle, has nothing to do with the wall. So, it is an arbitrary and statistically random value.

 

[yuiop]

$$u=(u_x,u_y,u_z)$$ is the velocity of the gas particle, has nothing to do with the wall. So, it is an arbitrary and statistically random value.

The general force transformation equations you are using are not specific to a gas particle.

They simply state how any force exerted on any particle transforms.

In this case I am considering how force exerted (by pressure) on a particle (in the wall of the box) transforms.

The original general equations derive from how an electromagnetic force acting on a particle transforms. In this case the force acting on the particle is not calculated by the collision of other particles with it. It only considers how the (electromagnetic) force acting on the particle with velocity $$u=(u_x,u_y,u_z)$$ relative to the observer and becomes a general equation by stating that any force acting on any particle with velocity $$u=(u_x,u_y,u_z)$$ will transform in the same way.

I am choosing to consider the (pressure) force acting on a (wall) particle with velocity $$u=(u_x,u_y,u_z)$$ transforms. The symmetry of the situation is clear.

 

[1effect]

The general force transformation equations you are using are not specific to a gas particle.

They simply state how any force exerted on any particle transforms.

In this case I am considering how force exerted (by pressure) on a particle (in the wall of the box) transforms.

We were talking about the momentum of gas particles.Look at your own definitions in the OP.

"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."

When did all that change to the wall particles?

 

[yuiop]

We were talking about the momentum of gas particles.Look at your own definitions in the OP.

"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."

When did all that change to the wall particles?

Any given problem in maths and science can often be solved in more than one way. Usually any problem solver will pick the easiest way and occasionally also solve it by an alternative but harder method as a comfirmation that no mistakes have made. All that matters is that the methods consistently arrive at the same solution. That is what I have done.

For example if I wanted to know the length of the cross frames for the roof of a house I could calculate using hypotenuse = sqrt(height^2+width^2) or I could calculate using hypotenuse = (width)/sine((apex angle)/2) if I wanted to do it the hard way. I should get the same answer and one method confirms the other.

I have shown that pressure is invariant for the x,y and z components using:

Change in momentum of gas particles during collision method.
Spring force and piston area method.
My force transformation equations.
Your force transformation equations.

and all the methods are consistent with each other (arrive at the same conclusion)

In fact, we could in principle calculate (to first order) how gravity transforms by analysing the force exerted by gas pressure on a piston that is in equilibrium with the force of gravity of a mass placed on the piston. I did a similar analysis some time back, to arrive at the similar conclusions to Matsas in the solution to the relativistic submarine paradox, except Matsas used GR to arrive at his solution. This last paradox is quite interesting and subtle as you have to analyse how a test particle and a gravitational body behave when:

The test particle is moving relative to the observer and the gravitational body is not.
The test particle and the the gravitational body are both moving relative to the observer (co-moving)
The test particle and the the gravitational body are both moving relative to the observer (but not co-moving with each other)

You also have to analyse other factors like how the density of water transforms and how the displaced volume transforms but that is another story for another time ;)