Kinematic Decomposition for "Rod and Hole" Relativity Paradox

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• PeterDonis
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In summary, the kinematic decomposition of the congruence of worldlines describing the rod in the "rod and hole" relativity paradox discussed in that thread is as follows:(1) Determine the vector field ##U^a## that describes the congruence of worldlines of interest.(2) Compute the tensor ##U_{a;b}##, i.e., the covariant derivative of the 1-form ##U_a## corresponding to the vector field ##U^a##.(3) Compute the tensor ##K_{ab} = h^m{}_a h^n{}_b
PAllen said:
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.]

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PAllen said:
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:

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

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PAllen said:
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:
$$(1,0,0,0)$$

PAllen said:
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.

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

Thanks.

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.

PAllen said:
@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.

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.

pervect said:
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.

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