## Lorentz contraction of box filled with gas

Consider what happens when we accelerate a box filled with gas. We have to expend a certain amount of energy to accelerate the box, In Newtonian mechanics, this energy goes into the kinetic energy of the box: as its speed increases so does its kinetic energy.
This happens in relativity too, of course, but in addition, Do we have to spend some extra energy because the box contracts and its pressure goes up? How does the box know it's moving?

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire

Blog Entries: 6
 Quote by Xeinstein Consider what happens when we accelerate a box filled with gas. We have to expend a certain amount of energy to accelerate the box, In Newtonian mechanics, this energy goes into the kinetic energy of the box: as its speed increases so does its kinetic energy. This happens in relativity too, of course, but in addition, Do we have to spend some extra energy because the box contracts and its pressure goes up? How does the box know it's moving?
The pressure inside the box does not increase. I will try and explain why. Pressure is defined as force divided by area. Force is defined as m*a = m*dv/dt = dp/dt where dp is change in momentum. If we have a box with n particles, each of mass m, conveniently bouncing straight up and down with velocity w and colliding with top of the box every t seconds then the total force of the particles colliding with the top of the box n*m*w/t. When the box is moving from left to right with velocity v with respect to us, the transverse component of the particles velocities is reduced by gamma (v) and the tranverse mass of each particle increases by a factor of of gamma(v). The time interval t also increases by gamma(v) from our point of view so that overall the force acting on the top of the box is n*(my)*(w/y)/(t*y) = (n*m*w/t)/y where y is gamma(v) or 1/sqrt(1-v^2/c^2). So the overall force is reduced by gamma. Since pressure is force divided by area and the surface area of the top of the box is also reduced by gamma (due to length contraction) the pressure is the same from our point of view as it is to to an observer that is stationary with respect to the box. You can do a similar analysis for the sides of the box and arrive at the same conclusion.

 Quote by kev The pressure inside the box does not increase. I will try and explain why. Pressure is defined as force divided by area. Force is defined as m*a = m*dv/dt = dp/dt where dp is change in momentum. If we have a box with n particles, each of mass m, conveniently bouncing straight up and down with velocity w and colliding with top of the box every t seconds then the total force of the particles colliding with the top of the box n*m*w/t. When the box is moving from left to right with velocity w with respect to us the transverse component of the particles velocities is reduced by gamma (v) and the tranverse mass of each particle increases by a factor of of gamma(v). The time interval t also increases by gamma(v) from our point of view so that overall the force acting on the top of the box is n*(my)*(w/y)/(t*y) = (n*m*w/t)/y where y is gamma(v) or 1/sqrt(1-v^2/c^2). So the overall force is reduced by gamma. Since pressure is force divided by area and the surface area of the top of the box is also reduced by gamma (due to length contraction) the pressure is the same from our point of view as it is to to an observer that is stationary with respect to the box. You can do a similar analysis for the sides of the box and arrive at the same conclusion.
Very nice, kev

I am getting the same result with a slightly improved mathematical formalism.

In the frame of the box, the mass of the gass is m_0 and the speed of the molecules is w so, the momentum is

p=m_0*w

The force exerted by molecules is

F=dp/dtau=m_0*dw/dtau where tau is the proper time as measured in the box frame

The crossection of the top of the box is A=a*b

The pressure in the box frame is:

Pr=F/A

In the observer frame , assuming the crossection is:

A'=a'*b'=a/gamma*b=A/gamma

where a is the dimension of the box side moving along the box movement, b is the dimension perpendicular on the movement, gamma=1/sqrt(1-(v/c)^2) and v is the box speed wrt the observer. Lorentz transforms say that the molecules move with speed

w'=w/gamma

p'=gamma*m_0*w'=gamma*m_0*w/gamma=m_0*w=p! (no real surprise here, it is quite intuitive)

But:

F'=dp'/dt=m_0*dp/dt=m_0*dp/dtau*dtau/dt=F*dtau/dt

dt=gamma* dtau (time dilation) so dtau/dt=1/gamma so:

F'=F/gamma

Pr'=F'/A'=F/A=Pr (Q.E.D)

Recognitions:
Staff Emeritus

## Lorentz contraction of box filled with gas

There is really not much need for calculation, *if* you measure the pressure in the reference frame of the box itself. In that case, the pressure will not be affected by the velocity of the box. (This should be obvious from the fact that velocity is relative and not absolute).

The box does not contract in its own frame, and the pressure in its own frame does not increase. In fact, the box cannot tell if it is moving or not.

If this is not obvious, it might be helpful to watch, for example

http://www.onestick.com/relativity/

And note that Al cannot tell if his train is moving, or if he is moving.

Measuring the pressure in some other frame is possible, but would require a detailed discussion of the stress-energy tensor. For introductory pedagogical purposes, I think the simpler treatment is all that is necessary.

 Quote by pervect There is really not much need for calculation, *if* you measure the pressure in the reference frame of the box itself. In that case, the pressure will not be affected by the velocity of the box. (This should be obvious from the fact that velocity is relative and not absolute). The box does not contract in its own frame, and the pressure in its own frame does not increase. In fact, the box cannot tell if it is moving or not. If this is not obvious, it might be helpful to watch, for example http://www.onestick.com/relativity/ And note that Al cannot tell if his train is moving, or if he is moving. Measuring the pressure in some other frame is possible, but would require a detailed discussion of the stress-energy tensor. For introductory pedagogical purposes, I think the simpler treatment is all that is necessary.
You are right , of course.It is still nice to have a formal proof, especially in the context of the question by OP. While the kinetic energy of the gas is frame dependent, its pressure is not (it is a constant in all frames).

Blog Entries: 6
 Quote by 1effect You are right , of course.It is still nice to have a formal proof, especially in the context of the question by OP. While the kinetic energy of the gas is frame dependent, its pressure is not (it is a constant in all frames).
Yes, the proof is interesting and along the way demonstates that tranverse force is reduced by gamma (which is sometimes questioned) and that tranverse mass is increased by gamma. If we analyse the pressure on the sides of the box we find that force parallel to the motion of the box is invarient and that longitudinal inertial mass behaves as if it has increased by gamma^3. The concept of different tranverse and longitudinal inertial masses for the same object is unpleasant and this is usually wrapped up in a momentum term.

On the subject of kinetic energy of the gas (that both the OP and 1effect alluded to) it is interesting to look at the classical gas law PV/T = P'V'/T'. In the frame moving wrt the box, the box has contracted in volume (V) by gamma but the pressure (P) is unchanged. If the classic gas law holds in the relativistic context then the temperature (T) must have cooled by a factor of gamma. Temperature is classically related to average kinetic energy of the gas particles. This implies a loss of energy. However it is not too surprising if we compare it to a flywheel that is moving with relativistic speed wrt to us. The flywheel has to slow down by a factor of gamma (it is after all a simple form of clock) so the flywheel's angular kinetic energy must have reduced from our point of view. On the other hand the kinetic energy of the box or flywheel due to its linear motion relative to us has increased. Presumably if we factor in the energy used to accelerate the box (or flywheel) and the momentum of particles ejected by a rocket used to accelerate the box, then the overall energy and momentum of the two reference frames is conserved.

[EDIT] Perhaps I should add that to an observer in the reference frame of the box, would of course not detect any change in volume, pressure or temperature of the gas.

 In the lab-frame (moving wrt the box), the Lorentz contraction is real and inevitable: the faster the box goes, the shorter it gets. But this shorting does Not come for free.The box is filled with gas, and if we shorten the box we reduce the volume occupied by the gas. This compression is resisted by pressure, and the energy required to compress the gas has to come from somewhere. It can only come from the energy exerted by the applied force. This means the force has to be larger (for the same increase in speed) that it would be in Newtonian mechanics, and this in turn means that the box has a higher inertia, by an amount proportional to the pressure in the box

 Quote by Xeinstein In the lab-frame (moving wrt the box), the Lorentz contraction is real and inevitable: the faster the box goes, the shorter it gets. But this shorting does Not come for free.The box is filled with gas, and if we shorten the box we reduce the volume occupied by the gas.
Correct, both kev and I have shown you this , mathematically.

 This compression is resisted by pressure, and the energy required to compress the gas has to come from somewhere. It can only come from the energy exerted by the applied force. This means the force has to be larger (for the same increase in speed) that it would be in Newtonian mechanics,
Incorrect: both kev and I have shown you that the speed of the gas molecules decreases. See w'=w/gamma.
This results into F'=F/gamma and that results, in turn, into:

Pr'=Pr

Please review the mathematics posted by kev and I, they both show where you are going wrong in your reasoning.

Mentor
 Quote by Xeinstein This compression is resisted by pressure, and the energy required to compress the gas has to come from somewhere.

What about a solid rod of steel? Even fairly small changes in length (strain) of a steel bar result in enormous changes in pressure (stress) within the bar. At relativistic speeds the stress and strain would be far beyond the failure point of the steel.

Since you cannot have something failing in one frame and being unstressed in another frame then you must come to the conclusion that Lorentz-contraction does not cause material stress (pressure) in general.

 Quote by pervect Measuring the pressure in some other frame is possible, but would require a detailed discussion of the stress-energy tensor.

Just to take this discussion on a more interesting tack: how would you use the stress-energy tensor in order to do the calculations for a steel rod. I'd love to see the equations.

 Quote by DaleSpam The other posters have already demonstrated the mistake here quite well, but I would encourage you to think about this further. What about a solid rod of steel? Even fairly small changes in length (strain) of a steel bar result in enormous changes in pressure (stress) within the bar. At relativistic speeds the stress and strain would be far beyond the failure point of the steel. Since you cannot have something failing in one frame and being unstressed in another frame then you must come to the conclusion that Lorentz-contraction does not cause material stress (pressure) in general.
Excellent point. In other words, the distance between atoms does not decrease in the direction of motion. I asked pervect if he could show the equations (similar to the ones I showed for the gas-filled box). This would be very interesting. Can you show them? (I don't know how and I would like to learn).

Mentor
 Quote by 1effect Excellent point. In other words, the distance between atoms does not decrease in the direction of motion. I asked pervect if he could show the equations (similar to the ones I showed for the gas-filled box). This would be very interesting. Can you show them? (I don't know how and I would like to learn).
No, the distance between the atoms does decrease in the direction of motion, but in a way that does not stress the material.

I don't have any equations for you, but here is a hand-waving analysis. Since relativity is based on EM phenomenon you know that the EM field around an isolated atom will length-contract as it attains relativistic velocities. The unstressed length of a piece of metal is determined by the spacing of atoms that yields the lowest energy state, which is in turn determined by the fields generated by the atoms. If the field length-contracts then the lowest energy state spacing will be correspondingly smaller and the unstressed length will also be correspondingly smaller. Thus you have physical length contraction without any material stress.

 Quote by DaleSpam No, the distance between the atoms does decrease in the direction of motion, but in a way that does not stress the material. I don't have any equations for you, but here is a hand-waving analysis. Since relativity is based on EM phenomenon you know that the EM field around an isolated atom will length-contract as it attains relativistic velocities. The unstressed length of a piece of metal is determined by the spacing of atoms that yields the lowest energy state, which is in turn determined by the fields generated by the atoms. If the field length-contracts then the lowest energy state spacing will be correspondingly smaller and the unstressed length will also be correspondingly smaller. Thus you have physical length contraction without any material stress.
Sorry, this is indeed armwaving :-)
I cannot parse without some equations to look at :-)
Let's hope that pervect can come up with the math.
BTW, I doubt that the distance between atoms decreases, there is no direct test for length contraction to date: http://www.edu-observatory.org/physi...th_Contraction

 Mentor Blog Entries: 1 You seriously think that length can be contracted without the distance between atoms decreasing? Or did you mean something else?
 if you measure the pressure in the reference frame of the box itself then this doesn't change. What is calculated from any other frame is purely just a calculation and surely incorrect if it doesn't come to the same answer, it's incorrect because of an observed length contraction, not a real one...

 Quote by Doc Al You seriously think that length can be contracted without the distance between atoms decreasing? Or did you mean something else?
Doc,

I meant exactly what I wrote. There is no experimental evidence that the distance between atoms contracts, nor is there any evidence that the atoms' radius contracts either.
Now, I used length contraction in my detailed post, just as a convenient mathematical tool.

Blog Entries: 6
 Quote by Magic Man if you measure the pressure in the reference frame of the box itself then this doesn't change. What is calculated from any other frame is purely just a calculation and surely incorrect if it doesn't come to the same answer, it's incorrect because of an observed length contraction, not a real one...
The explanation for the famous null result of the Michelson Morley interferometer experiment is due to the length contraction of the arm parallel to the direction the interferometer is moving. I am curious if you think the length contraction of the parallel arm is is real or imaginary. I hope you agree there is something intrinsically unsatifactory about basing physcs on imaginary phenomena.

Time dilation is considered real because we place two clocks that had relative motion alongside each other and see that different times have elapsed on the two clocks. When we place two rulers that had relative motion alongside each other we do not see a difference in length and this leads some people to conclude that length contaction is imaginary.

Here is a thought experiment that might demonstrate length contraction is real. Imagine a wheel with a spike on its perimeter. A narrow tape is fed to the wheel at the same speed as a point on the perimeter of the spinning wheel so that a hole is punched in the tape every time the the wheel completes one rotation. When the wheel stops spinning we can directly measure the distance between the holes on the stationary tape. We would find that the holes are spaced at intervals (2*pi*r*gamma) that are greater than the rest perimeter of the wheel. This is because the holes were spaced at intervals of 2*pi*r from our point of view when the tape and wheel were moving.

[EDIT} Perhaps I should make it clear that the wheel is spinning but not moving linearly with respect to us. Somebody at rest with the tape when the wheel is spinning would see the wheel as rolling (without slipping) along the tape, which is the statinary "road" in his frame.