Kinematic Decomposition for "Rod and Hole" Relativity Paradox

[PAllen]

@pervect, I don't have time now for a more detailed response, but starting from a congruence in the rod initial rest frame in which there is never any x velocity, and boosting to a hole frame with a boost of say u, it seems impossible to me that you don't end up with constant rod x component velocity of u in the hole frame. What should change, in the hole frame for such a boost, is the z component velocity. In particular, if a point of the rod has $v_z$ in the rod initial rest frame, and you boost by u in x direction, the speed in the hole frame would be $\sqrt {u^2+v_z^2-u^2v_z^2}$, and since the x direction velocity must by u by definition of the boost, the z component would be derived by pythagorean algebra from this orthogonal velocity addition formula, and would be different from $v_z$ in the rod initial rest frame.

 

[PeterDonis]

2) constant proper acceleration with no attempt to counter Bell spaceship type stress in the vertical dimension, your post #1
3) constant proper acceleration adjusted to have Rindler acceleration dependence in the vertical direction to avoid an inessential contribution to stress, my post #19 ( @Ibix first pointed out this issue in this thread; I mentioned it in the other thread)

I know I said "yes" before, but I've thought about it some more and now I'm not sure.

In the "rod rest frame" (the frame in which the rod is at rest before any part of it starts to accelerate downward), each piece of the rod starts its downward acceleration at the same height. (I am treating the rod as having negligible thickness.) The only difference is the time, in this frame, in which each piece of the rod starts accelerating; the front starts before the back does. But since each piece starts accelerating downward from the same height, they should all have the same proper acceleration when they start. And I don't see any reason why the proper acceleration should change as they fall, so each piece should have the same proper acceleration throughout.

The above is why I did #2. In the light of the above, I'm not sure I understand what #3 is doing. Why would the proper acceleration vary in the vertical direction? (I understand that mathematically it removes a certain component of the stress, but why physically would we expect it to happen?)

 

[PeterDonis]

it seems impossible to me that you don't end up with constant rod x component velocity of u in the hole frame

Remember that a boost boosts 4-vectors. The ordinary velocity components $v_x$ and $v_z$ are not components of a 4-vector, so we cannot expect that $v_x$ will be constant everywhere simply because we are boosting by a constant in the $x$ direction. I have not worked through the detailed math (or found the previous thread where we ran into a similar issue), but off the top of my head I would expect, in general, both $v_x$ and $v_z$ to be affected, since the relevant 4-velocity components are of the form $\gamma v$ and $\gamma$ varies with height even though the boost speed does not. What a constant $x$ boost should lead to is a constant $x$ component of 4-velocity.

 

[PAllen]

In the "rod rest frame" (the frame in which the rod is at rest before any part of it starts to accelerate downward), each piece of the rod starts its downward acceleration at the same height. (I am treating the rod as having negligible thickness.) The only difference is the time, in this frame, in which each piece of the rod starts accelerating; the front starts before the back does. But since each piece starts accelerating downward from the same height, they should all have the same proper acceleration when they start. And I don't see any reason why the proper acceleration should change as they fall, so each piece should have the same proper acceleration throughout.

The above is why I did #2. In the light of the above, I'm not sure I understand what #3 is doing. Why would the proper acceleration vary in the vertical direction? (I understand that mathematically it removes a certain component of the stress, but why physically would we expect it to happen?)

It is precisely to avoid Bell spaceship type expansion in the z direction. We are free to posit any acceleration profile we want, and if the goal is to have one where in the case of the force being applied simultaneously in the rod frame (i.e. the hole and rod do not initially have relative motion and the hole is large) leads to zero expansion, shear and vorticity, one must have the acceleration change in the z direction. There is no physical reason for it, it is simply to find the congruence with minimum stresses, that with u being zero, produces a congruence that is Born rigid. This will demonstrate that no matter what, there must be physical deformation in the case of force applied constantly along a horizontal line of the rod simultaneously in the hole frame. That is, no matter what else you adjust, you cannot remove the fact that the large rod getting through a small hole involves physical deformation of the rod (contrary to some claims in the literature).

 

[PAllen]

Remember that a boost boosts 4-vectors. The ordinary velocity components $v_x$ and $v_z$ are not components of a 4-vector, so we cannot expect that $v_x$ will be constant everywhere simply because we are boosting by a constant in the $x$ direction. I have not worked through the detailed math (or found the previous thread where we ran into a similar issue), but off the top of my head I would expect, in general, both $v_x$ and $v_z$ to be affected, since the relevant 4-velocity components are of the form $\gamma v$ and $\gamma$ varies with height even though the boost speed does not. What a constant $x$ boost should lead to is a constant $x$ component of 4-velocity.

But for $v_x$ of zero, and a boost of u in the x direction, I don't see how this can possibly lead to anything other than $v_x'$ of u in the boosted frame.

 

[PeterDonis]

if the goal is to have one where in the case of the force being applied simultaneously in the rod frame

My understanding was that the force was to be applied to all parts of the rod simultaneously in the hole frame. That's a very different scenario. I would agree that if the force were applied simultaneously to all parts of the rod in the rod frame, then the resulting congruence should have zero expansion, shear, and vorticity. But I did not think that was the case we were discussing. Applying the force to all parts of the rod simultaneously in the rod frame should not even result in the rod going through the hole, since in the rod frame the hole is shorter than the rod and so there is no instant in the rod frame where the entire rod can be pushed through the hole by any force applied to the rod at that instant.

 

[PAllen]

My understanding was that the force was to be applied to all parts of the rod simultaneously in the hole frame. That's a very different scenario. I would agree that if the force were applied simultaneously to all parts of the rod in the rod frame, then the resulting congruence should have zero expansion, shear, and vorticity. But I did not think that was the case we were discussing.

I am saying the congruence I proposed in #19 has the property that if you make u (the relative velocity of hole and rod) be zero, it leads to Born rigid motion. Note, this is not constant proper acceleration everywhere. To get Born rigid motion there must be variation of proper acceleration in z, to avoid Bell spaceship expansion.

 

[PeterDonis]

I am saying the congruence I proposed in #19 has the property that if you make u (the relative velocity of hole and rod) be zero, it leads to Born rigid motion. Note, this is not constant proper acceleration everywhere. To get Born rigid motion there must be variation of proper acceleration in z, to avoid Bell spaceship expansion.

This doesn't clear up the issue I raised in post #41. DId you actually mean to say "simultaneously in the rod frame" in post #39, or was that a typo and it should read "simultaneously in the hole frame"? (Emphasis mine.)

 

[PAllen]

This doesn't clear up the issue I raised in post #41. DId you actually mean to say "simultaneously in the rod frame" in post #39, or was that a typo and it should read "simultaneously in the hole frame"? (Emphasis mine.)

No typo. I have a congruence with u as a parameter. When u=0, simultaneity in the hole frame and the rod initial rest frame are the same - it is the same frame. And I want this case to then be Born rigid motion. And the case of large u to have 'as little deformation as possible'.

Also, I routinely assume that hole is simply as big as needed. That is, the concern is the properties of the congruence, not whether a rod fits through the hole (that can be answered separately). So consider just a rod sliding off a table with different rules about force application producing different congruences.

 

[PeterDonis]

I have a congruence with u as a parameter. When u=0, simultaneity in the hole frame and the rod initial rest frame are the same - it is the same frame. And I want this case to then be Born rigid motion. And the case of large u to have 'as little deformation as possible'.

Ah, ok. I hadn't picked that up from your earlier posts. I'll take another look when I can.

 

[PeterDonis]

I have a congruence with u as a parameter. When u=0, simultaneity in the hole frame and the rod initial rest frame are the same - it is the same frame. And I want this case to then be Born rigid motion. And the case of large u to have 'as little deformation as possible'.

Just to clarify: when $u > 0$, is the force applied simultaneously in the hole frame or the rod frame (i.e., in the case where the two are not the same)?

 

[PAllen]

Just to clarify: when $u > 0$, is the force applied simultaneously in the hole frame or the rod frame (i.e., in the case where the two are not the same)?

In the hole frame. That is why, in the congruence (which is given in the rod initial rest frame) there is a term $+ux_0$.

 

[PeterDonis]

In the hole frame. That is why, in the congruence (which is given in the rod initial rest frame) there is a term $+ux_0$.

Ok, got it.

 

[PAllen]

But for $v_x$ of zero, and a boost of u in the x direction, I don't see how this can possibly lead to anything other than $v_x'$ of u in the boosted frame.

I computed this out explicitly via Lorentz transform, and I get the z velocity changes for a boost in the x direction, but if the x velocity starts out as 0, it simply becomes the boost speed after the boost.

 

[PeterDonis]

$$z=-\sqrt{z_0{}^2+(t+ux_0)^2}$$

What is $z_0$ here? Is it supposed to be the value of $z$ at which the rod starts accelerating downward? Wouldn't that just be $0$?

Also, where does the proper acceleration $a$ appear here?

 

[PeterDonis]

Uh oh...I just realized that the congruence I gave in the OP has nonzero proper acceleration, but the computations I did for the kinematic decomposition are for a geodesic congruence. Correcting for the proper acceleration means subtracting out the dyad $A_a U_b$ from the tensor $K_{ab}$ before doing the rest of the decomposition. I'll follow up with corrections.

 

[PAllen]

What is $z_0$ here? Is it supposed to be the value of $z$ at which the rod starts accelerating downward? Wouldn't that just be $0$?

Also, where does the proper acceleration $a$ appear here?

An element of the congruence is specified by some $(x_0,z_0)$, with the range of $x_0$ values ranging over a difference of L, the rod length, and the $z_0$ values ranging over values with difference of rod thickness. In the given formula, $z_0=0$ would be on the Rindler horizon, so that would be no good. Instead, if one specifies some proper acceleration for $a$ (assumed to apply to the top surface of the rod), then $z_0$ should range between $-1/a$ and $-1/a - thickness$. Sorry I didn't specify this.

 

[PeterDonis]

An element of the congruence is specified by some $(x_0,z_0)$, with the range of $x_0$ values ranging over a difference of L, the rod length, and the $z_0$ values ranging over values with difference of rod thickness.

Ah, ok. My congruence was assuming negligible thickness for the rod (i.e., each piece of the rod starts out at the same $z$). If I were to take thickness into account, I would agree that the proper acceleration should vary with vertical position.

In the given formula, $z_0=0$ would be on the Rindler horizon, so that would be no good.

Yes, given what you are doing, you have to assume some range of nonzero $z_0$ values for the rod.

 

[PeterDonis]

Uh oh...I just realized that the congruence I gave in the OP has nonzero proper acceleration, but the computations I did for the kinematic decomposition are for a geodesic congruence. Correcting for the proper acceleration means subtracting out the dyad $A_a U_b$ from the tensor $K_{ab}$ before doing the rest of the decomposition. I'll follow up with corrections.

Ok, here are the corrections.

The acceleration $A^a$ is easily computed; it's just $dU^a / d\tau = \gamma \partial U^a / \partial t$, where $\gamma$ is just the $t$ component of $U^a$. This gives:

$$
A^a = \left( a^2 \left( t + v x \right), 0, - a \sqrt{ 1 + a^2 \left( t + v x \right)^2 } \right)
$$

It is easily verified that this vector has magnitude $a$, as desired.

We now just lower the index on $A$ (which means flipping the sign of the $t$ component) and form the components of the dyad $A_a U_b$:

$$
A_t U_t = a^2 \left( t + v x \right) \sqrt{ 1 + a^2 \left ( t + v x \right)^2 }
$$

$$
A_t U_z = a^3 \left( t + v x \right)^2
$$

$$
A_z U_t = a \left( 1 + a^2 \left( t + v x \right)^2 \right)
$$

$$
A_z U_z = a^2 \left( t + v x \right) \sqrt{ 1 + a^2 \left ( t + v x \right)^2 }
$$

Now we add these to the corresponding components of $U_{a ; b}$ from the OP, to get a tensor that I should have named in the OP, but didn't; let's call it $Y_{ab}$. Thus, $Y_{ab} = U_{a ; b} + A_a U_b$, and $Y_{ab}$ is now the tensor we will apply the projection $h^{a}{}_{b}$ to to get the tensor $K_{ab}$ that we then decompose into the three pieces of expansion, shear, and vorticity. [Edit: see the strikethrough above. Equations below have been updated to correspond. Further follow-up in post #58 and following below.] So we have [Edit: The first equation has been corrected]:

$$
K_{tt} = U_{t, t} + A_t U_t = a^3 v \left( t + v x \right)^2
$$

$$
K_{tz} = A_t U_z = a^3 \left( t + v x \right)^2
$$

$$
K_{zt} = U_{z, t} + A_z U_t = a^3 \left( t + v x \right)^2
$$

$$
K_{zz} = A_z U_z = a^2 \left( t + v x \right) \sqrt{ 1 + a^2 \left ( t + v x \right)^2 }
$$

$$
K_{tx} = U_{t, x} = - \frac{a^2 v \left( t + v x \right)}{\sqrt{ 1 + a^2 \left ( t + v x \right)^2 }}
$$

$$
K_{zx} = U_{z, x} = - a v
$$

I'll leave the re-computation of $K_{ab}$ for a further follow-up post.

[Edit: See strikethrough above. Further follow-up in post #58 and following below.]

 

[PeterDonis]

At this point I'll leave the decomposition into the three pieces as an exercise for the reader.

Just a comment, though: we still see nonzero expansion, shear, and vorticity for this congruence.

 

[PAllen]

I computed this out explicitly via Lorentz transform, and I get the z velocity changes for a boost in the x direction, but if the x velocity starts out as 0, it simply becomes the boost speed after the boost.

A little more on this point that the congruences @PeterDonis and I have been using do, indeed, have constant horizontal velocity when transformed to the hole frame. In the calculation indicated above, I simply Lorentz transformed the equation of a congruence world line. However, it is readily shown that the same results from transforming a 4-velocity.

Consider a 4 velocity in the rod frame given by: $\gamma (v) (1,0,v)$ with coordinate order (t,x,z). Now boost of u in the x direction gives: $\gamma(u) \gamma(v) (1,u,v/\gamma(u))$. A key thing to note is that from a general formula for composition of $\gamma$ : $\gamma_{res} = \gamma(u) \gamma(v) (1-uv \cos(\theta))$, for the case of u and v orthogonal, we just have $\gamma(u) \gamma(v)$. Thus the transformed 4-velocity in terms of total $\gamma$ of the particle in the boosted frame is just: $\gamma (1,u,v/\gamma(u))$, meaning that coordinate x velocity is simply constant u in this frame, as I have claimed. This is also exactly the same as I got transforming the equation of a congruence world line.

 

[PeterDonis]

Consider a 4 velocity in the rod frame given by: $\gamma (v) (1,0,v)$ with coordinate order (t,x,z). Now boost of u in the x direction gives: $\gamma(u) \gamma(v) (1,u,v/\gamma(u))$. A key thing to note is that from a general formula for composition of $\gamma$ : $\gamma_{res} = \gamma(u) \gamma(v) (1-uv \cos(\theta))$, for the case of u and v orthogonal, we just have $\gamma(u) \gamma(v)$. Thus the transformed 4-velocity in terms of total $\gamma$ of the particle in the boosted frame is just: $\gamma (1,u,v/\gamma(u))$, meaning that coordinate x velocity is simply constant u in this frame, as I have claimed.

Yes, after working through the transformation myself, I agree with this. I think the issue I referred to before that came up in the previous "sliding block" threads was looking at the ordinary velocity in the $x$ direction in the MCIF of the chosen piece of the rod, not in the original rod rest frame.

 

[PeterDonis]

A further follow-up: I realized on reviewing the general equations for kinematic decomposition that the extra step of projecting with the tensor $h$ is not needed. I have edited post #54 accordingly, and have deleted what was the post just after it that contained the projections. The final tensor whose components are given in post #54 is the one that I called $K_{ab}$ before, i.e., the one that gets split into the three pieces of expansion, shear, and vorticity. That makes things look somewhat neater.

 

[PeterDonis]

One more follow-up: I have corrected the $K_{tt}$ equation in post #54, I thought that component would vanish but it doesn't.

Also, we can simplify the appearance of the equations by noting that, since the $t$ and $z$ components of $U^a$ are $\gamma$ and $\gamma v$, we have $\gamma = \sqrt{ 1 + a^2 \left( t + v x \right)^2}$ and $\gamma v = a \left( t + v x \right)$, and the ratio of those two is $v$. That let's us write the components of $K_{ab}$ as:

$$
K_{tt} = a \gamma^2 v^3
$$

$$
K_{tz} = K_{zt} = a \gamma^2 v^2
$$

$$
K_{zz} = a \gamma^2 v
$$

$$
K_{tx} = - a v^2
$$

$$
K_{zx} = - a v
$$

This makes splitting the above into the three pieces easier: we have

$$
\theta = K_{zz} - K_{tt} = a \gamma^2 v \left( 1 - v^2 \right) = a v
$$

$$
\sigma_{tt} = K_{tt} - \frac{1}{3} \theta \left( - 1 + U_t U_t \right) = \frac{2}{3} a \gamma^2 v^3
$$

$$
\sigma_{tz} = \sigma_{zt} = K_{tz} - \frac{1}{3} \theta U_t U_z = \frac{2}{3} a \gamma^2 v^2
$$

$$
\sigma_{zz} = K_{zz} - \frac{1}{3} \theta \left( 1 + U_z U_z \right) = \frac{2}{3} a \gamma^2 v
$$

$$
\sigma_{tx} = \sigma_{xt} = \frac{1}{2} K_{tx} = - \frac{1}{2} a v^2
$$

$$
\sigma_{zx} = \sigma_{xz} = \frac{1}{2} K_{zx} = - \frac{1}{2} a v
$$

$$
\omega_{tx} = - \omega_{xt} = \frac{1}{2} K_{tx} = - \frac{1}{2} a v^2
$$

$$
\omega_{zx} = - \omega_{xz} = \frac{1}{2} K_{zx} = - \frac{1}{2} a v
$$

This is not that different from what was posted back in post #3, but now the structure is clearer.

 

[PAllen]

Just a quick comment on why shear and/or expansion must be expected for this case. Given that everyone agrees that for a rod horizontal in hole frame starting to accelerate downward (no matter whether coordinate, proper, or proper with vertical Rindler distribution) while remaining horizontal in this frame, there must be some vorticity. But then, a generalization of the Ehrenfest paradox is a theorem that says if a congruence has a nonzero Lie derivative of vorticity, it cannot be rigid. This means, as soon as you accept changing vorticity, it must be true that you also have expansion and/or shear.

Here is a reference which whose focus is rigidity in expanding cosmologies, but reviews the whole theory of rigidity in relativity:

https://arxiv.org/abs/2008.11836

P.5, equation (13) gives the result I refer to.

 

[PeterDonis]

If I were to take thickness into account, I would agree that the proper acceleration should vary with vertical position.

Given my latest series of posts updating and correcting my original congruence, the obvious way to make it account for thickness in the manner @PAllen suggested, per my comment quoted above, is to replace $a$ in the equations with $1 / z_0$ (which would be $c^2 / z_0$ in conventional units), since this is the Rindler condition for proper acceleration. I think this will have the effect of introducing a new derivative into the computation since $a$ now will vary with $z$. But since $z_0$ is a constant along any worldline in the congruence, I'm not sure exactly how the $z$ derivative of $a$ will appear.

 

[PAllen]

Given my latest series of posts updating and correcting my original congruence, the obvious way to make it account for thickness in the manner @PAllen suggested, per my comment quoted above, is to replace $a$ in the equations with $1 / z_0$ (which would be $c^2 / z_0$ in conventional units), since this is the Rindler condition for proper acceleration. I think this will have the effect of introducing a new derivative into the computation since $a$ now will vary with $z$. But since $z_0$ is a constant along any worldline in the congruence, I'm not sure exactly how the $z$ derivative of $a$ will appear.

I think after computing the velocity field in terms of congruence parameters and time, you have to invert the functions giving congruence line coordinates in terms of parameters to get parameters in terms coordinates. Then, using these, re-express the vector field in terms of coordinates. Synge mentions another method (he calls Lagrangian) where you can continue to use parameters, but he gives no details on it. Then you are able to apply the standard formulas for the decomposition.

 

[PAllen]

I think after computing the velocity field in terms of congruence parameters and time, you have to invert the functions giving congruence line coordinates in terms of parameters to get parameters in terms coordinates. Then, using these, re-express the vector field in terms of coordinates. Synge mentions another method (he calls Lagrangian) where you can continue to use parameters, but he gives no details on it. Then you are able to apply the standard formulas for the decomposition.

So, going back to my post #19, and carrying this out, I get, for the 4-velocity field:

$$\mathbf {U} = \frac {|z|} {\sqrt {z^2-(t+ux)^2}}(1,0,0,-(t+ux)/|z|)$$

where $x\epsilon [0,L]$, $y\epsilon [0,w]$, and $\sqrt {z^2-(t+ux)^2} \epsilon [1/a,(1/a)+h]$, and also $z \leq -1/a$ by my conventions.

A nice feature of having u (the relative velocity of hole and rod) as an explicit parameter is that I expect that setting u=0 will lead to Born rigid motion.

 

[PeterDonis]

after computing the velocity field in terms of congruence parameters and time, you have to invert the functions giving congruence line coordinates in terms of parameters to get parameters in terms coordinates. Then, using these, re-express the vector field in terms of coordinates.

I agree with the general method, but I'm not sure I'm getting the same result. I'll briefly describe what I get, using my sign conventions for $z$ (which are that $z$ decreases going downward--but I agree that $z = 0$ has to be a distance $1 / a_\text{top}$ above the top of the rod, where $a_\text{top}$ is the proper acceleration of the top of the rod).

The parameter $z_0$ is the $z$ value at which a given rod worldline in the congruence is before it starts falling. Once it starts falling, its $z$ coordinate is $z_0 - d$, where $d$ is the distance traveled in time $t$ at constant proper acceleration $a$. The proper acceleration $a$ is $1 / z_0$ and the distance $d$, from the relativistic rocket equation, is $\left( \sqrt{1 + a^2 t^2} - 1 \right) / a$, where $t$ here is the time the piece has been falling. Since that time for a rod worldline at $x$ is $t + v x$, where $t$ is now coordinate time and $v$ is the relative velocity of rod and hole (what you are calling $u$, but I'll stick here to my notation), we obtain

$$
z = z_0 \left[ 1 - \left( \sqrt{1 + \frac{\left( t + v x \right)^2}{z_0^2}} - 1 \right) \right]
$$

We then have to invert this to get an expression for $z_0$ in terms of the coordinates and $v$. We end up with a quadratic equation for $z_0$:

$$
3 z_0^2 - 4 z z_0 + z^2 - \left( t + v x \right)^2 = 0
$$

This has solutions:

$$
z_0 = \frac{2z \pm \sqrt{z^2 + 3 ( t + v x )^2}}{3}
$$

Since we must have $z_0 > z$, only the positive sign in the above gives a possible solution. So we have

$$
z_0 = \frac{1}{a} = \frac{2z + \sqrt{z^2 + 3 ( t + v x )^2}}{3}
$$

or, in a form that's more useful for plugging into previous formulas,

$$
a = \frac{3}{2z + \sqrt{z^2 + 3 \left( t + v x \right)^2}}
$$

The 4-velocity components are $\gamma = \sqrt{1 + a^2 \left( t + v x \right)^2}$ and $\gamma v = - a \left( t + v x \right)$, which gives (giving just the nonzero $t$ and $z$ components):

$$
U = \left( \sqrt{ 1 + \frac{9 \left( t + v x \right)^2}{\left(2z + \sqrt{z^2 + 3 \left( t + v x \right)^2}\right)^2} } , - \frac{3 \left( t + v x \right)}{2z + \sqrt{z^2 + 3 ( t + v x )^2}} \right)
$$

This simplifies (somewhat) to:

$$
U = \frac{1}{2z + \sqrt{z^2 + 3 ( t + v x )^2}} \left( \sqrt{ 5z^2 + 4 z \sqrt{z^2 + 3 \left( t + v x \right)^2} + 12 \left( t + v x \right)^2} , - 3 \left( t + v x \right) \right)
$$

 

[PAllen]

I will try to post my work tomorrow. Definitely can’t get to it tonite. I started from a different set of equations, specifically what I wrote in post #19, which gives the general Rindler observer world line in Minkowski coordinates. I was very careful in my work, so I am pretty confident in it.

 

[pervect]

A little more on this point that the congruences @PeterDonis and I have been using do, indeed, have constant horizontal velocity when transformed to the hole frame. In the calculation indicated above, I simply Lorentz transformed the equation of a congruence world line. However, it is readily shown that the same results from transforming a 4-velocity.

Consider a 4 velocity in the rod frame given by: $\gamma (v) (1,0,v)$ with coordinate order (t,x,z). Now boost of u in the x direction gives: $\gamma(u) \gamma(v) (1,u,v/\gamma(u))$.

A key thing to note is that from a general formula for composition of $\gamma$ : $\gamma_{res} = \gamma(u) \gamma(v) (1-uv \cos(\theta))$, for the case of u and v orthogonal, we just have $\gamma(u) \gamma(v)$. Thus the transformed 4-velocity in terms of total $\gamma$ of the particle in the boosted frame is just: $\gamma (1,u,v/\gamma(u))$, meaning that coordinate x velocity is simply constant u in this frame, as I have claimed. This is also exactly the same as I got transforming the equation of a congruence world line.

Consider a small piece of the rod in the lab frame, which I believe you refer to as the "hole frame". We'll cut a few corners and say that this small piece of the rod has a mass m, rather than writing down it's stress energy tensor. I think this is much clearer, and I don't think it'll affect the argument.

Let the piece of the rod be moving in the "x" direction in the lab frame with the x component of the velocity being called $v_x$.

If I am understanding you correctly, you are stating that the x-component of the velocity of the rod in the lab frame remains constant with time. Is this correct?

Then the x component of the momentum of the piece of the rod is $\gamma m v_x$. And I think we agree that gamma is increasing as the rod falls. I haven't really thought about your composition law, but assuming your formulation is correct, the increase in gamma follows from your product law.

While this is a logically consistent assumption, is this really the problem you want to analyze? A case where the x component of the momentum of the rod is increasing due to its interaction with the "hail"? I.e. the hail is not just pushing the piece of the rod "down" in the lab frame, it's also pushing it in the "x" direction, where by "pushing" I mean transferring momentum.

 

[PAllen]

I will try to post my work tomorrow. Definitely can’t get to it tonite. I started from a different set of equations, specifically what I wrote in post #19, which gives the general Rindler observer world line in Minkowski coordinates. I was very careful in my work, so I am pretty confident in it.

Well, it looks like I can start posting tonite. I'll do a little at a time through tomorrow. So, first we have:

$$(t,x,y,z) = \left( t,x_0,y_0,-\sqrt {z_0^2+(t+ux_x)^2} \right)$$

In a very straightforward computation I get for $v_z$, the time derivative of the z component above:

$$v_z=-\frac {t+ux_0} {\sqrt {z_0^2+(t+ux_0)^2}}$$

Then, I get for $\gamma (v_z)$ just: $\sqrt {1+(t+ux_0)^2/z_0^2}$

This gives 4-velocity in terms of parameters as follows (note similarity to the relativistic rocket equation):

$$\mathbf {U} = \left( \sqrt {1+(t+ux_0)^2/z_0^2},0,0,-(t+ux_0)/z_0 \right) $$

The core of the remaining work (which I'll post later) is simply to work out that:

$$z_0=\sqrt {z^2-(t+ux)^2}$$

having substituted $x$ for $x_0$. In fact, the rest is pretty easy algebra making the substitution for $z_0$ above. So, unless requested, I don't plan post more intermediate steps. Thus, I remain convinced of my post #63.

 

[PAllen]

Consider a small piece of the rod in the lab frame, which I believe you refer to as the "hole frame". We'll cut a few corners and say that this small piece of the rod has a mass m, rather than writing down it's stress energy tensor. I think this is much clearer, and I don't think it'll affect the argument.

Let the piece of the rod be moving in the "x" direction in the lab frame with the x component of the velocity being called $v_x$.

If I am understanding you correctly, you are stating that the x-component of the velocity of the rod in the lab frame remains constant with time. Is this correct?

Then the x component of the momentum of the piece of the rod is $\gamma m v_x$. And I think we agree that gamma is increasing as the rod falls. I haven't really thought about your composition law, but assuming your formulation is correct, the increase in gamma follows from your product law.

While this is a logically consistent assumption, is this really the problem you want to analyze? A case where the x component of the momentum of the rod is increasing due to its interaction with the "hail"? I.e. the hail is not just pushing the piece of the rod "down" in the lab frame, it's also pushing it in the "x" direction, where by "pushing" I mean transferring momentum.

[edit: several claims below are incorrect. A force that is pure downward in the hole frame will lead to a decrease in horizontal speed of the rod over time. Since this is not what is desired, see my post #71 for a resolution.]

Well, that is the way all problems of this type are posed. It is always assumed that the rod moves at constant horizontal speed with respect to the hole as it 'falls' or 'is forced' through it. In fact, it seems to me, that to assume otherwise, there must be horizontal force acting against the rod, which is not posited. Your reasoning supposes that a strictly downward force slows down the rod's horizontal motion, which I find physically implausible. But the upshot is that all variants of the problem discussed in the other thread, including all of several referenced papers assume the horizontal speed of the rod with respect to the hole does not change (until possibly hitting a wall, in the variants that have a wall). So assuming something else creates a wholly different problem.

 

[PAllen]

Well, that is the way all problems of this type are posed. It is always assumed that the rod moves at constant horizontal speed with respect to the hole as it 'falls' or 'is forced' through it. In fact, it seems to me, that to assume otherwise, there must be horizontal force acting against the rod, which is not posited. Your reasoning supposes that a strictly downward force slows down the rod's horizontal motion, which I find physically implausible. But the upshot is that all variants of the problem discussed in the other thread, including all of several referenced papers assume the horizontal speed of the rod with respect to the hole does not change (until possibly hitting a wall, in the variants that have a wall). So assuming something else creates a wholly different problem.

I think worrying too much about force changes the nature of the problem. It is, indeed well known that a force in SR generally produces an acceleration not in the same direction as the force. To keep the problem simple, and as intended, it is better to think of some unspecified type of force, acting throughout the body (to avoid worrying about instant crushing of the body) such that a specified acceleration profile is produced. Then, it turns out that if acceleration is orthogonal to the direction of relative motion in one frame, it is also in the other, and also the rod simply has constant horizontal velocity in the hole frame equal to the hole velocity in the rod initial rest frame. The z velocity and acceleration differ in magnitude between the two frames, but they remain z directed.

 

[PeterDonis]

I agree with the general method, but I'm not sure I'm getting the same result.

On looking over my result I think I might have made some errors. Will follow up after I've had a chance to check my math via a different method.