- #1

yuiop

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__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

[tex]P = \frac{NMU^2}{AL\sqrt{1-\frac{U^2}{c^2}}}[/tex]

__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:

[tex] F=\frac{(V+U)}{(1+\frac{UV}{c^2})}[/tex]

and the backward velocity is:

[tex] B=\frac{(V-U)}{(1-\frac{UV}{c^2})}[/tex]

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

[tex] T=\frac{L}{y}\left(\frac{1}{(V-B)}+\frac{1}{(F-V)}\right)[/tex]

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

[tex]T=\frac{2L}{U \sqrt{1-\frac{V^2}{c^2}}}[/tex]

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:

[tex]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)[/tex]

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:

[tex]\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)[/tex]

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

[tex]\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)[/tex]

-->

[tex]\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)[/tex]

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

[tex]\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)[/tex]

-->

[tex]\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)[/tex]

-->

[tex]\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)[/tex]

-->

[tex]\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)[/tex]

-->

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

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

[tex]\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)[/tex]

-->

[tex]\frac{NMU^2}{AL\sqrt{1-U^2}} = \frac{NMU^2}{AL\sqrt{(1-U^2)}}[/tex]

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

Q.E.D.