# Proof of invariance of gas pressure

• yuiop
In summary, the conversation discusses the proof that pressure is invariant on the front and back faces of a moving box. This is shown by using the relativistic expression for pressure in the rest frame and comparing it to the expression for pressure in the moving frame. The conversation also includes calculations for the time interval and pressure in both frames. There is some disagreement on the simplified expression for pressure in the moving frame.
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.

kev said:
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’:

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

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'.

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

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

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

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).

1effect said:
The problem is that according to the way the stress energy tensor transforms, p'_xx is not equal to p_xx.

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:
pervect said:
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. .

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.

pervect said:
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

OK, that is basically the same as my

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

pervect said:
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.

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 $$\frac{1}{\sqrt{1-v^2}}$$

pervect said:
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).

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

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).

pervect said:
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).

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:
1effect said:
The problem is that according to the way the stress energy tensor transforms, p'_xx is not equal to p_xx.

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.

1effect said:
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.

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 similar method to the one I used (taking the interval dt to be the mean interval between collisions) in various textbooks 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.

1effect said:
Actually, I think that you and kev now agree on an incorrect result.
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.

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

So, the pressure is not an invariant.

Actually force tranverse to the motion of the frame is reduced by the Lorentz factor but force parallel to the motion is invarient (sometimes referred 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.

kev said:
Actually force tranverse to the motion of the frame is reduced by the Lorentz factor but force parallel to the motion is invarient (sometimes referred 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.

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.

kev said:
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.

I agree, see the discrepancy that was uncovered.
I have seen pressure analysed by using a similar method to the one I used (taking the interval dt to be the mean interval between collisions) in various textbooks and it seems a reasonable approach to take.

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

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:
1effect said:
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.

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

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 )

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:
kev said:
The full formula for x component of the notion of force youare 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 $$P_x=p_x$$.

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" :-)

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.

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:
1effect said:
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 $$u_x=0$$.

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.

1effect said:
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.

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:
kev said:
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.

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:
1effect said:
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 :-)

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:
kev said:
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.

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.

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

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:
1effect said:
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 of 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?

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:
kev said:
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.

So? this is well known in relativity, I do not see how it is relevant.

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.

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.

But this contradicts the way you derived the forces in first place, remember? You were dividing the variation of momentum by T.
What made change your mind and decide that the forces are "static"? Just because I showed you that the transforms of dynamic forces render your method unusable?
I am also noticing that the transform of static forces is very curious, where did you get it from? The paper that you just cited?

The crux of the matter is:

Is gas pressure a scalar quantity or not?

As I said earlier: I don't know but I would very much like to find out. Maybe there is someone out there who knows.

Last edited:
kev said:
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:

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

I missed the above earlier, it lloks like an arbitrary derivation (what I tend to call rectal extraction).

The area in pressure is $$a=l_xl_y$$ why use the vector area $$a=\sqrt{l_x^2+l_y^2}$$?

From the strange choice for proper area you ellaborate $$A=\sqrt {\left( {l_x^2+l_y^2(1-v^2/c^2)\right)}$$.

From the above you further ellaborate $$F_{xy}=\sqrt {\left( {f_x^2+f_y^2(1-v^2/c^2)\right)}$$ and further $$F_{xyz}=\sqrt {\left( {f_x^2+(f_y^2+f_z^2)(1-v^2/c^2)\right)$$.

I will say that if you start with something arbitrary, you can arrive to any conclusion you wish. :-)

1effect said:
But this contradicts the way you derived the forces in first place, remember? You were dividing the variation of momentum by T.
What made change your mind and decide that the forces are "static"? Just because I showed you that the transforms of dynamic forces render your method unusable?
I am also noticing that the transform of static forces is very curious, where did you get it from? The paper that you just cited?

Nope, I do not have access to that paper. (I wouldn't mind a synopsis from some who does ). I derived my formula from the postulate (that I postulated) that gas pressure is scalar and and invariant under transformation as shown in a earlier post. From that postulate the quantity F/A is invariant so all we have to do is calculate how surface area transforms and by definition gas pressure force must transform in the same way.

1effect said:
I missed the above earlier, it lloks like an arbitrary derivation (what I tend to call rectal extraction).

The area in pressure is $$a=l_xl_y$$ why use the vector area $$a=\sqrt{l_x^2+l_y^2}$$?

From the strange choice for proper area you ellaborate $$A=\sqrt {\left( {l_x^2+l_y^2(1-v^2/c^2)\right)}$$.

From the above you further ellaborate $$F_{xy}=\sqrt {\left( {f_x^2+f_y^2(1-v^2/c^2)\right)}$$ and further $$F_{xyz}=\sqrt {\left( {f_x^2+(f_y^2+f_z^2)(1-v^2/c^2)\right)$$.

I will say that if you start with something arbitrary, you can arrive to any conclusion you wish. :-)

I use the vector area as that is how area is defined in physics when considering gas pressure. I don't think that is arbitary.

1effect said:
So? this is well known in relativity, I do not see how it is relevant.

The angle of the surface is very relevant as force exerted by pressure is defined af the force normal to the surface. To find that normal compnent of the force we need to the angle of the surface concerned and is that angle changes due a transform we have to take account of it.

1effect said:
But this contradicts the way you derived the forces in first place, remember? You were dividing the variation of momentum by T.
What made change your mind and decide that the forces are "static"? Just because I showed you that the transforms of dynamic forces render your method unusable?

I consider this a static situation because the body of the gas as whole is co-moving with the surface is it acting on. They are not moving relative to each other.

An example of a dynamic situation is particle that is being accelerated by a particle accelerator. Because the velocty of the particle is always changing relative to the accelerator the rate that force transfers to the particle is continuously changing. This is one interpretation of relativistic mass. Rather than consider that the mass of the particle is increasing we can consider there is a limit to the rate that force can be applied to the particle making it increasingly difficult to accelerate the particle.

A more apropriate example of a dynamic situation is a bullet being fired from a rifle. The bullet is moving away from the hot gas of the propellant and this is a dynamic force situation. The volume of gas and the bullet are not co-moving and we cannot use the static pressure force equation to analyse this situation.

I consider this a static situation because the body of the gas as whole is co-moving with the surface is it acting on. They are not moving relative to each other.

You simply made this up in a hurry when I showed you that the solution based on dynamic forces has so many restrictions. Besides, what makes you think that "gas as whole is co-moving with the surface is it acting on"? Gas particles have a random motion and have a random distribution of velocities last I checked :-)

kev said:
I use the vector area as that is how area is defined in physics when considering gas pressure. I don't think that is arbitary.

We are going nowhere with this, you just switched the definitions of forces and areas when I showed you that your method is not general.
This discussion has reached its limits in terms of productivity, maybe someone else can solve this problem in clean way. Pervect?

Last edited:
1effect said:
You simply made this up in a hurry when I showed you that the solution based on dynamic forces has so many restrictions. Besides, what makes you think that "gas as whole is co-moving with the surface is it acting on"? Gas particles have a random motion and have a random distribution of velocities last I checked :-)

Even a brick that is not at absolute zero degrees Kelvin has random vibrations of its molecules but we can consider the brick as single body when it is moving wrt us and and we can define a frame that is comoving with the brick even if all the brick's molecules are not completely at rest with reference frame. The random vibrations of the molecules in the brick or the random velocity vectors of the gas molecules in a container cancel out (for a large numebr of molecules) in a way that we can consider the brick or the body of the gas as a whole to have a definable co-moving or rest reference frame.

1effect said:
We are going nowhere with this, you just switched the definitions of forces and areas when I showed you that your method is not general.
This discussion has reached its limits in terms of productivity, maybe someone else can solve this problem in clean way. Pervect?

I gave up trying to demonstrate the pressure in terms of particle motions because you would not accept the method of "deviding by T". Unless you accept that concept there is point in continuing down that path so I switched the conversation to discuss the problem in your terms you considered more acceptable. That is reasonable, is it not?

Foul! Referee!

kev said:
I gave up trying to demonstrate the pressure in terms of particle motions because you would not accept the method of "deviding by T". Unless you accept that concept there is point in continuing down that path so I switched the conversation to discuss the problem in your terms you considered more acceptable. That is reasonable, is it not?

Foul! Referee!

You didn't , as soon as I pointed out to you that the correct relativistic transformation of forces creates serious restrictions for your approach you switched to another transformation of forces, that you made up on the spot. Red card! :-)

kev said:
$$a=\sqrt{l_x^2+l_y^2}$$.

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

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

Should we assume that $$F=F_{xyz}$$?

If that is the case, how do you prove this claim :

$${F \over A}={f_x \over a_x}$$

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

It would be very interesting to see how you do your calculations.

Last edited:
1effect said:
Should we assume that $$F=F_{xyz}$$?

Yes.

1effect said:
If that is the case, how do you prove this claim :

$${F \over A}={f_x \over a_x}$$

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

It would be very interesting to see how you do your calculations.

First, I should thank you helping me get the formalism right (as always ) The equation I gave:

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

was a little confusing I admit and would be better written as:

$${\bf A}=\sqrt {\left( {{\bf a_x^2}+{\bf a_y^2}(1-v^2/c^2)\right)}$$

where $${\bf a_x}$$ is the vector area of the surface that is orthogonal to the x-axis (which is the surface the pressure force parallel to the x-axis is acting on.) and $${\bf a_y}$$ is the vector area of a surface that is orthogonal to the y axis. Assuming a simple square surface, the surface area can be defined in terms of two vector distances $${\bf l_y}$$ and $${\bf l_z}$$ which can be written as (0,L,0) and (0,0,L) respectively in terms of x,y,z coordinates. The vector area $${\bf a_x}$$ is the cross product of the two vector distances so $${\bf l_y \times l_z}$$ = (L^2,0,0) which is a vector of magnitude L^2 pointing along the x axis. Inserting $${\bf a_x} = {\bf l^2}$$ and $${\bf a_y}=0$$ into the area transform equation we get :

$$A=\sqrt {\left( {a_x^2+a_y^2(1-v^2/c^2)\right)}=\sqrt{l^4+0}=l^2=a_x$$

The transformation of static force is given as :

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

so

$$F =\sqrt {\left( {f_x^2+0)\right)}= f_x$$

So $$F=f_x$$ and $$A=a_x$$ and therefore

$${F \over A}={f_x \over a_x}$$

Last edited:
kev said:
Yes.

First, I should thank you helping me get the formalism right (as always ) The equation I gave:

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

was a little confusing I admit and would be better written as:

$${\bf A}=\sqrt {\left( {{\bf a_x^2}+{\bf a_y^2}(1-v^2/c^2)\right)}$$

where $${\bf a_x}$$ is the vector area of the surface that is orthogonal to the x-axis (which is the surface the pressure force parallel to the x-axis is acting on.) and $${\bf a_y}$$ is the vector area of a surface that is orthogonal to the y axis. Assuming a simple square surface, the surface area can be defined in terms of two vector distances $${\bf l_y}$$ and $${\bf l_z}$$ which can be written as (0,L,0) and (0,0,L) respectively in terms of x,y,z coordinates. The vector area $${\bf a_x}$$ is the cross product of the two vector distances so $${\bf l_y \times l_z}$$ = (L^2,0,0) which is a vector of magnitude L^2 pointing along the x axis. Inserting $${bf a_x} = {\bf l^2}$$ and $${\bf a_y}=0$$ into the area transform equation we get :

$$A=\sqrt {\left( {a_x^2+a_y^2(1-v^2/c^2)\right)}=\sqrt{l^4+0}=l^2=a_x$$

The transformation of static force is given as :

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

so

$$F =\sqrt {\left( {f_x^2+0)\right)}= f_x$$

So $$F=f_x$$ and $$A=a_x$$ and therefore

$${F \over A}={f_x \over a_x}$$
In other words you manipulate $$a_y=a_z=0$$ and $$f_y=f_z=0$$ and poof!, you get $$F_x=f_x$$ and $$A=a_x$$.
Magic, eh?

Last edited:
kev said:
Static force transforms as $$F_{xyz}=\sqrt {\left( {f_x^2+(f_y^2+f_z^2)(1-v^2/c^2)\right)$$

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/

How did you make up this "transform" again? Did you find it in a book or in the paper that you cited earlier? Did you get it from here ?

Why do areas add up and transform in such a funny way? How did you derive this:

$${\bf A}=\sqrt {\left( {{\bf a_x^2}+{\bf a_y^2}(1-v^2/c^2)\right)}$$Why are the force and the are "transforming" differently? The force has the term in $$f_z$$ while the area is missing the term in $$a_z$$. This is not symmetric. Where are you getting all this stuff? Did you find this in any book or did you make it up?

Last edited:
1effect said:
How did you make up this "transform" again? Did you find it in a book or in the paper that you cited earlier? Did you get it from here ?

Nope. I do not have a relativity book nor access to the content of that publication site. I am not sure what to make of the formula y' = y' (x,y) is y' = u x -1 y, at x' = 0 quoted on that page without seeing the rest of the article.

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

was based on a analysis we did of how the force of a static spring transforms in another thread.

Using Hooke's Law

$$F= EA\left({\Delta L \over L}\right)$$

Where E is Young's modulus, A is the cross sectional area of the spring and $${\left({\Delta L \over L}\right)$$ is the ratio of the spring extension to its relaxed length.

From that we got:

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

for the longitudinal force and

$$F_y= E \left(A\sqrt{1-v^/c^2}\right){\left({\Delta L \over L\right)=f_y\sqrt{1-v^2/c^2}$$

$$F_z= E \left(A\sqrt{1-v^/c^2}\right){\left({\Delta L \over L\right)=f_z\sqrt{1-v^2/c^2}$$

for the transverse case.

The magnitude of the transformed force for any orientation is obtained from the resultant (or vector sum) of the force vector components (basically Pythagorous):

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

1effect said:
Why do areas add up and transform in such a funny way? How did you derive this:

$${\bf A}=\sqrt {\left( {{\bf a_x^2}+{\bf a_y^2}(1-v^2/c^2)\right)}$$

Why are the force and the area "transforming" differently? The force has the term in $$f_z$$ while the area is missing the term in $$a_z$$. This is not symmetric. Where are you getting all this stuff? Did you find this in any book or did you make it up?

$${\bf A}=\sqrt { {{\bf a_x^2}+{\bf a_y^2}(1-v^2/c^2))}$$ should be used with $${\bf F}=\sqrt { {{\bf f_x^2}+{\bf f_y^2}(1-v^2/c^2)}$$

and $$A=\sqrt{a_x^2+(a_y^2+a_z^2)(1-v^2/c^2)$$ with $$F=\sqrt{f_x^2+(f_y^2+f_z^2)(1-v^2/c^2)$$

I started with the 2D case as that is easier to analyse. Initially I considered only forces in the x,y plane. The z component of the surface the forces are acting on is always at right angles to the x-axis so it does not length contract. Since Lz is constant through the transformation we can give it a nominal value and ignore it.

Have a look at the attached diagram. The vector area Ax is the same as looking at the white face from the x-axis and does not change when the cube is length contracted. The vector area Ay is the same as looking at the white face from above and it appears length contracted. The vector area Az is looking at the white face from along the z axis and all you see is a line so the value of Az is zero (When we only consider rotations of the cube around the z axis.

In the diagram the cube on the right is at rest. the scalar area of the white face is j*k. When we convert the the area into coordinate terms the scalar area is j*k = $$\left(\sqrt{d_x^2+dy^2}\right)d_z$$

The cube on the left represents the length contracted cube moving parallel to the x axis.
The scalar area of the white face of the length contracted cube is J*K = $$\sqrt{\left(d_x\sqrt{1-v^2/c^2}\right)^2+dy^2+d_z^2}$$

If you look at the vector area normal to the scalar surface (shown by the blue vectors) it is easy to see that the vector area x component is invariant and that it is the y (and z) components that are reduced by the Lorentz factor when we are considering the normal to the physical surface. The vector area therefore transforms as $$A=\sqrt{a_x^2+(a_y^2+a_z^2)(1-v^2/c^2)}$$

I do not agree with the statement in the Springer link abstract "We show that static forces, as defined by Hooke's law, do not transform in the same way as the usual dynamical forces, thus giving rise to the paradoxes. This inconsistency justifies the present search for alternative theories such as the modern ether theories." I think Special Relativity IS consistent and I gave reasons in an earlier post as why static and dynamic forces transform differently.

#### Attachments

• box2.GIF
30.3 KB · Views: 389
Last edited:
kev said:
Nope. I do not have a relativity book nor access to the content of that publication site. I am not sure what to make of the formula y' = y' (x,y) is y' = u x -1 y, at x' = 0 quoted on that page without seeing the rest of the article. The formula $$F_{xyz}=\sqrt {\left( {f_x^2+(f_y^2+f_z^2)(1-v^2/c^2)\right)$$

was based on a analysis we did of how the force of a static spring transforms in another thread.

Using Hooke's Law

$$F= EA\left({\Delta L \over L}\right)$$
Where E is Young's modulus, A is the cross sectional area of the spring and $${\left({\Delta L \over L}\right)$$ is the ratio of the spring extension to its relaxed length.

From that we got:

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

for the longitudinal force and

$$F_y= E \left(A\sqrt{1-v^/c^2}\right){\left({\Delta L \over L\right)=f_y\sqrt{1-v^2/c^2}$$

Wait a sec! The $$\Delta L$$ in the $$F_x$$ component of the force is not the same as the $$\Delta L$$ in the transverse components. I do not know which thread you are referring to but, if you picked the proof from that thread, it is wrong.
So, when you add squares of the F components $$\Delta L$$ doesn't factor out.

As an aside, I don't think that a spring models correctly the gas particles, I have told you this before.

1effect said:
Wait a sec! The $$\Delta L$$ in the $$F_x$$ component of the force is not the same as the $$\Delta L$$ in the transverse components. I do not know which thread you are referring to but, if you picked the proof from that thread, it is wrong.
So, when you add squares of the F components $$\Delta L$$ doesn't factor out.

Of course they do not factor out. Why on Earth 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.

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

1effect said:
As an aside, I don't think that a spring models correctly the gas particles, I have told you this before.

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).

Last edited:

• Special and General Relativity
Replies
55
Views
4K
• Special and General Relativity
Replies
13
Views
1K
• Special and General Relativity
Replies
1
Views
2K
• Special and General Relativity
Replies
24
Views
2K
• Special and General Relativity
Replies
27
Views
2K
• Special and General Relativity
Replies
11
Views
598
• Special and General Relativity
Replies
29
Views
1K
• Special and General Relativity
Replies
14
Views
1K
• Special and General Relativity
Replies
28
Views
2K
• Special and General Relativity
Replies
6
Views
985