I Kinematic Decomposition for "Rod and Hole" Relativity Paradox

[PAllen]

@pervect , the congruences described here are of the following character, in reference to a link to a Rindler paper I post #2 of that thread:

1) A rod is initially moving inertially, with no forces acting on it. The context is pure SR.

2) A force is suddenly applied to all parts of the rod simultaneously, per a frame in which the the rod is moving horizontally to the right. For the purposes of congruence, we are really talking about accelerations rather than forces.

3) Anything about the hole or table are irrelevant for this congruence.

4) For simplicity, the congruence is specified in the rod initial rest frame, noting that the only consequence of (2) is that the timing of beginning of acceleration is not simultaneous in this frame. @PeterDonis assumes constant proper acceleration over the whole rod, with timing determined by the simualtaneity difference between frame where rod is moving to the right compared to frame where rod is stationary until acceleration begins.

 

[PAllen]

There are some clear differences in the coordinate and frame independent proper acceleration though. Your congruence has a constant proper acceleration for all points in the congruence, this should inevitably yields expansion in the z direction.

In post #19, I gave the congruence which avoids this issue.

 

[PeterDonis]

The "table" isn't physical, though.

It is if it is providing support for the rod until the rod reaches the hole. Which is necessary in at least some versions of the scenario.

It should be possible to transform both congruences to the "lab frame" coordinates

It will be possible, of course, but it might not be very useful since the math might be more complicated in that frame.

 

[pervect]

In post #19, I gave the congruence which avoids this issue.

I re-invented much of the wheel before I figured your post out. But it appears to me that the notion of constant horizontal velocity is fatally flawed, see the details below.

Continuing on, I also have to note that I fixedx a typo (a missing square root) in my earlier post of the congruence.

Having fixed the typo, I can now convert the congruence to the lab frame.

The coordinates I am using for the lab frame are t,x,y,z, with z=0, representing the Rindler horizon which is always above the rod. z is always positive, representing how far below the rod is below z=0. In the lab frame, z^2 - t^2 = constant for rigid motion. You posted much the same relationship, I believe. If z=1 when t=0, and t he rod starts to fall, this gives a unit acceleration.

The lab frame has geometrized Minkowskii coordinates with a metric
$$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$$

Using the chain rule, we can convert the original cogruence to the Minkowskii lab frame. I'll skip over the details and present the result.

$$u^t = \frac{z}{\sqrt{ z^2 - t^2 - \beta^2 } } \quad u^x = \frac{\beta} { \sqrt{ z^2 - t^2 - \beta^2 }} \quad u^z = \frac{t}{\sqrt{ z^2 - t^2 - \beta^2 }} $$

This congruence has zero expansion and shear and nonzero vorticity, changing the coordinates didn't change the congruence.

Using the fact you mention that the 4-velocity is
$\gamma(1, v_x, v_y, v_z)$ we can see that v_x is not constant for this congruence, but has a value of $\beta / z$.

This actually isn't suprising. Let's assume a constant horizontal velocity $v_x = 0.9c$. Then , keeping the horiziontal velocity unchanged at $0.9c$, we accelerate it so that the vertical veloicity $v_z$ is also $0.9c$. The the total velocity would be $\sqrt{v_x^2 + v_z^2}$ > c. So it's impossible to keep the horizontal velocity unchanged, contrary to Peter's assumption. We probably want to keep the horizontal momentum constant, instead. Which is what the above congruence basically does.

 

[PeterDonis]

it's impossible to keep the horizontal velocity unchanged, contrary to Peter's assumption

I actually wasn't assuming anything about horizontal velocity, since I only worked in the rod rest frame (more precisely the rest frame of the rod before it starts being accelerated downward), and I didn't write down any congruence for the table or the hole, only the rod. In the given frame the rod's only motion is the downward proper acceleration.

As far as the "lab frame" is concerned, yes, I believe in that frame the horizontal (i.e., $x$) velocity of the rod will not be constant as it accelerates downward (i.e., in the $z$ direction). What will be constant is the horizontal component of the 4-velocity (which is equivalent to conservation of the horizontal component of the 4-momentum since we are assuming constant rest mass). I believe we ran into this in one of the previous "sliding block" threads.

I haven't checked your congruence itself, so I'll comment on that in a separate post after I've had a chance to work through the math.

 

[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.