I Kinematic Decomposition for "Rod and Hole" Relativity Paradox

[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.

 

[PAllen]

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.

A further thought on this is that we want to apply the force with different simultaneity in different cases (of the problem) but always want it to be downward in the rest frame of a rod element. This means it won’t be vertical in the hole frame, but it will be such as to maintain the rod having constant horizontal speed in the hole frame, with all coordinate acceleration downward.

[edit: consider, if one insists on a 'dust rain' supplying force, that this rain is required be vertical in the local frame of any rod element; this means it won't be vertical in the hole frame. Overall, the dust rain model is not useful for the discussion of this thread since it necessarily applies force only to the top of the rod, and at the relevant magnitudes, the rod is destroyed faster (vertically) than it can respond to the force. For the purposes of examining to what degree a rod can get through the hole non-destructively - the purpose of this thread, we must consider a force 'somehow' able to apply throughout the rod, like gravity.]

 

[PAllen]

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.

Noting that we can say proper acceleration at a point of the congruence is given by:

$$ a(z,x,t)= \frac {1} {\sqrt {z^2-(t+ux)^2} }$$

then the whole congruence can be written:

$$\mathbf {U}=a(z,x,t)(|z|,0,0,-(t+ux))$$

 

[PAllen]

Noting that we can say proper acceleration at a point of the congruence is given by:

$$ a(z,x,t)= \frac {1} {\sqrt {z^2-(t+ux)^2} }$$

then the whole congruence can be written:

$$\mathbf {U}=a(z,x,t)(|z|,0,0,-(t+ux))$$

And for completeness, for the main problem variant under discussion in this thread, we should say the above is true for $t \ge -ux$, else it is:
$$(1,0,0,0)$$

 

[pervect]

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.

Running this congruence through grtensor, and replacing |z| with -z, as z <0 by your conventions, I'm finding the expansion scalar is zero for all values of u, not just u=0. I do prefer using $\beta$ to u, not that that it matters, unless I make a typo and start talking about $\beta$ because my conventions are different. The shear and vorticity tensors are only zero for u equal to zero, however.

Some versions of your post have a typo in the 4-velocity, which slowed me down a bit, the version I quoted is the version I used and is correctly normalized, while the typo version is not correcctly normalized.

Wiki is the best source I've found for calculationg the expansion and shear for non-geodesic congruences. Eric's Poisson's excellent book, "A Relativistis Toolkit", unfortunately only handles geodesic congruences :(.

My general attitude is that I wouldn't care to wade through the necessary procedure without automated tools, but that these results provide a lot of physical insight.

Since it's possible people might be interested, I'll take the risk of making a typo and write the tensors

The shear tensor is (here I HAVE replaced u with $\beta$)

$$\sigma_{ab} = \frac{1}{2d} \begin{bmatrix} 0 & z \beta(t + \beta x) & 0 & 0 \\
z \beta(t + \beta x) & 0 & 0 & -\beta z^2 \\
0 & 0 & 0 & 0 \\
0 & -\beta z^2 & 0 & 0
\end{bmatrix}$$

The vorticity tensor is
$$\omega_{ab} = \frac{1}{2d} \begin{bmatrix}
0 & z \beta(t + \beta x) & 0 & 0 \\
-z \beta(t + \beta x) & 0 & 0 & \beta z^2 \\
0 & 0 & 0 & 0 \\
0 & -\beta z^2 & 0 & 0
\end{bmatrix}$$\

where d, the common denominator is
$$d = \left( z^2 - (t + \beta x)^2 \right)^\frac{3}{2}$$

I'll apologize in advance for any typos

[add] While I have the worksheet up, the 4-acceleration of the congruence is
$$a^a = \frac{1}{z^2 - (t+\beta x)^2} \begin{bmatrix} t+\beta x & 0 & 0 & z \end{bmatrix}$$

which has the magnitude that PAllen has already written in the post currently numbered #72.

And I might as well document the output for $u_{a ;b}$
$$ u_{a ;b} := \frac{1}{d} \begin{bmatrix}
z(t+\beta x) & \beta z (t + \beta x) & 0 & -(t + \beta x)^2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
z^2 & -\beta z^2 & 0 & z(t+\beta x)
\end{bmatrix}$$

The other thing we'd need to calculate the kinematic decomposition following the wiki would be $a_a u_b$, and those can be found from $a^a$ and $u^b$ fairly easily. Looking at the components of $a^a$ and $u^b$ , only the t and z components are nonzero, so this term basically contributes to the 4 "corner" components (1,1) (1,4),(4,1),(4,4). And it appears that these 4 components are zero in the composite sum.

 

[PAllen]

@pervect , Could you let me know which posts have the typo, so I can add a bracketed correction?

Thanks.

 

[PAllen]

It’s not readily available, but Synge’s 1960 GR text covers the kinematic decomposition for general congruences in the context of full GR, not just SR.

 

[pervect]

@pervect , Could you let me know which posts have the typo, so I can add a bracketed correction?

Thanks.

The last line of post #72 is where I noticed it.

 

[pervect]

In an effort to interpret the physical significance of the shear and vorticity tensor results from my pervious calculations, I'd like to draw a comparison to the following congruence in geometrized Minkowskii spacetime with coordinates [t,x,y,z]

$$u^a = \frac{1}{1- \beta^2 x^2} \begin{bmatrix} 1 & 0 & 0 & -\beta x \end{bmatrix}$$

This only makes physical sense when $\beta x < 1$, and it's basically a congruence whose z-velocity depends on x.

This congruence is not accelerated (which is different and simpler), but it does has zero expansion (which is similar).

The results for the nonzero components of the vorticity and shear are similar, though notably the spatial components of the shear and vorticity are constant.

Specifically
$$shear =\frac{1}{2\left(1-\beta^2x^2 \right)^\frac{3}{2}} \begin{bmatrix}
0 & -\beta^2 x & 0 & 0 \\
-\beta^2 x & 0 & 0 & -\beta \\
0 & 0 & 0 & 0 \\
0 & -\beta & 0 & 0 \
\end{bmatrix}$$

$$vorticity =\frac{1}{2\left(1-\beta^2x^2 \right)^\frac{3}{2}}
\begin{bmatrix}
0 & -\beta^2 x & 0 & 0 \\
\beta^2 x & 0 & 0 & \beta \\
0 & 0 & 0 & 0 \\
0 & -\beta & 0 & 0
\end{bmatrix}$$

Visually, the time evolution of a square block of this congruence would look like this as it shears in the x-z plane, the left side of the block would stay fixed, while the right side would move downwards as it shears.

 

[PAllen]

The last line of post #72 is where I noticed it.

That’s exactly the same as the one you used except for trivial algebra.