Kinematic Decomposition for "Rod and Hole" Relativity Paradox

[PeterDonis]

TL;DR Summary

Spin-off from previous thread to present math for the kinematic decomposition:

https://www.physicsforums.com/threads/the-paradox-of-relativity-length-contraction.1010138/

In a recent thread, I said that if there was interest, I would post in a separate thread the calculations for the kinematic decomposition of the congruence of worldlines describing the rod in the "rod and hole" relativity paradox discussed in that thread. Since there was interest, I am posting that separate thread.

(Note: Many physicists would probably say that the kinematic decomposition is a graduate level, i.e., "A" level in PF terms, topic. However, for this problem at least, the math is fairly simple and I think it can be presented at an undergraduate, i.e., "I" level. So that is the level I have assigned to this thread.)

The basic definition of the kinematic decomposition is given in this Wikipedia article. The basic steps, at least as I will present them here, are as follows:

(1) Determine the vector field $U^a$ that describes the congruence of worldlines of interest. (The Wikipedia article uses $X$, but I am using $U$ since that is a more common notation for a 4-velocity field, which is what this is.)

(2) Compute the tensor $U_{a;b}$, i.e., the covariant derivative of the 1-form $U_a$ corresponding to the vector field $U^a$. Since we will be working in an inertial frame, the covariant derivative is just the partial derivative, so all we have to compute is $U_{a,b}$.

(3) Compute the tensor $K_{ab} = h^m{}_a h^n{}_b U_{m;n}$, i.e., the derivative $U_{a;b}$, projected orthogonally to $U$. (The projection tensor $h$ is defined in the Wikipedia article.)

(4) Separate the tensor $K$ (the notation $K$ is my own, to give the entire tensor a name; the literature only names the pieces of it that I'm about to describe) into three pieces: its trace (the expansion scalar $\theta$), its traceless symmetric part (the shear tensor $\sigma_{ab}$), and its antisymmetric part (the vorticity tensor $\omega_{ab}$). It is a mathematical theorem that any 2nd-rank tensor can be fully described by these three pieces, i.e., there is no information left over once we've extracted them. The three pieces are the kinematic decomposition.

Two other notes: first, if the expansion and shear are zero, the congruence can be said to be "Born rigid"; physically, this means the local distances between neighboring worldlines, in the hypersurfaces locally orthogonal to $U$, are constant. Second, if the vorticity is zero, the congruence can be said to be "hypersurface orthogonal"; physically, this means that we can construct a single set of spacelike hypersurfaces that are everywhere orthogonal to every worldline in the congruence (i.e.,orthogonal not just locally but globally).

With those preliminaries out of the way, here is a brief description of the scenario:

We have a surface along which a rod is moving horizontally in the positive $x$ direction with speed $v$, until the rod encounters a hole in the surface. We will use the Rindler version of the scenario, in which a trapdoor is placed over the hole and only removed once the entire length of the rod, as seen in the hole/surface rest frame, is over the hole. The instant that occurs, the trapdoor is removed, in such a way that it does not impair the rod free-falling into the hole due to gravity, and in such a way that all parts of the trapdoor are removed at the same time in the hole/surface rest frame, which we will call frame A. We will call that time $t = 0$.

In the rest frame of the rod before it starts to fall, which we will call frame B, the surface and the hole are moving in the negative $x'$ direction with speed $v$. Each piece of the rod has a constant $x'$ coordinate in this frame. In this frame, the pieces of the trapdoor are removed at different times, and hence the pieces of the rod start to fall freely at different times. The time $t' = t_0$ at which a piece of the rod at $x' = x_0$ starts falling freely in this frame is given by the condition $t = 0 = \gamma \left( t' + v x' \right)$, which gives $t_0 = - v x_0$. This is telling us that the parts of the rod at larger $x_0$ (i.e., further to the right, so they encounter the hole first) start to fall earlier, which is what we expect.

The 4-velocity of each piece of the rod therefore starts out being at rest in this frame (i.e., only a $t'$ component), but then acquires an increasing negative $z'$ component once that piece starts to fall freely. The coordinate acceleration will thus be zero up to time $t_0$ (which, as above, is a function of the position $x_0$ of the piece), and $- a$ after. (We will leave $a$ as an undetermined constant in this analysis, although of course in the scenario as originally posed we expect it to be determined by the Earth's gravity. But we will want to consider various possible values later on.)

The components of $U^a$ are therefore (giving them in order $t$, $x$, $z$ and dropping the primes on the coordinates):

$$
U^a = \left( \sqrt{ 1 + a^2 \left( t - t_0 \right)^2 }, 0, - a \left( t - t_0 \right) \right)
$$

Substituting $t_0 = - v x_0$, and realizing that $x_0 = x$ since each piece of the rod always has the same $x$ coordinate in this frame, we obtain

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

Since these components are in an inertial frame in flat spacetime, so the metric is $\eta_{ab}$, it is simple to confirm that $\eta_{ab} U^a U^b = -1$, so $U$ is a timelike unit vector field, as desired.

That completes our first step, and our second step is to compute the tensor $U_{a,b}$. To do that, we need the 1-form $U_a$, which since the metric is $\eta_{ab}$ has the same components as $U^a$ except that the first component's sign is changed. So $U_a$ has only $t$ and $z$ components, and the components only depend on $t$ and $x$; hence, we have four nonzero components of the derivative:

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

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

$$
U_{z, t} = \frac{\partial}{\partial t} \left( - a \left( t + v x \right) \right) = - a
$$

$$
U_{z, x} = \frac{\partial}{\partial x} \left( - a \left( t + v x \right) \right) = - a v
$$

That completes the second step. I'll continue with the two remaining steps in a follow-up post.

 

[PeterDonis]

Here is the follow-up completing the kinematic decomposition.

For the third step, we need the projection tensor $h^a{}_b$, which is given by (note that I am raising an index from the formula given in the Wikipedia article):

$$
h^a{}_b = \delta^a{}_b + U^a U_b
$$

The nonzero components of this are:

$$
h^t{}_t = 1 + U^t U_t = - a^2 \left( t + v x \right)^2
$$

$$
h^t{}_z = U^t U_z = - a \left( t + v x \right) \sqrt{ 1 + a^2 \left( t + v x \right)^2 }
$$

$$
h^z{}_t = U^z U_t = a \left( t + v x \right) \sqrt{ 1 + a^2 \left( t + v x \right)^2 }
$$

$$
h^z{}_z = 1 + U^z U_z = 1 + a^2 \left( t + v x \right)^2
$$

$$
h^x{}_x = 1
$$

[Edit: What follows is not correct; see this post and its follow-ups for corrections.]

Combining these with the components of $U_{a, b}$ from the second step, we get the following nonzero components of $K_{ab}$:

$$
K_{tt} = h^t{}_t h^t{}_t U_{t, t} + h^z{}_t h^t{}_t U_{z, t}
$$

$$
K_{tz} = h^t{}_t h^t{}_z U_{t, t} + h^z{}_t h^t{}_z U_{z, t}
$$

$$
K_{zt} = h^t{}_z h^t{}_t U_{t, t} + h^z{}_z h^t{}_t U_{z, t}
$$

$$
K_{zz} = h^t{}_z h^t{}_z U_{t, t} + h^z{}_z h^t{}_z U_{z, t}
$$

$$
K_{tx} = h^t{}_t h^x{}_x U_{t, x} + h^z{}_t h^x{}_x U_{z, x}
$$

$$
K_{zx} = h^t{}_z h^x{}_x U_{t, x} + h^z{}_z h^x{}_x U_{z, x}
$$

Expanding these out gives (assuming I've done the algebra right):

$$
K_{tt} = \frac{a^4 \left( t + v x \right)^3}{\sqrt{ 1 + a^2 \left( t + v x \right)^2 }}
$$

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

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

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

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

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

Looking at these components, we can see that the trace, traceless symmetric, and antisymmetric parts are all nonzero, so we expect the expansion, shear, and vorticity to all be nonzero. I'll save the explicit calculations of the decomposition pieces for a follow-up post.

 

[PeterDonis]

[Edit: The previous post, on which this one is based, is not correct; see this post for corrections.]

This is my next follow-up post to give the final part of the kinematic decomposition. The three pieces are:

$$
\theta = \eta^{ab} K_{ab}
$$

$$
\sigma_{ab} = \frac{1}{2} \left( K_{ab} + K_{ba} \right) - \frac{1}{3} \theta \left( \eta_{ab} + U_a U_b \right)
$$

$$
\omega_{ab} = \frac{1}{2} \left( K_{ab} - K_{ba} \right)
$$

Expanding these out gives:

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

$$
\sigma_{tt} = \frac{2}{3} \frac{a^4 \left( t + v x \right)^3}{\sqrt{ 1 + a^2 \left( t + v x \right)^2 }}
$$

$$
\sigma_{tz} = \sigma_{zt} = \frac{2}{3} a^3 \left( t + v x \right)^2
$$

$$
\sigma_{zz} = \frac{2}{3} a^2 \left( t + v x \right) \sqrt{ 1 + a^2 \left( t + v x \right)^2}
$$

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

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

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

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

So we do indeed have nonzero expansion, shear, and vorticity. To briefly describe the physical meanings of each of these:

The expansion is positive, which means the rod is being stretched.

There is shear in the $x$ - $z$ plane, which means that adjacent pieces of the rod (i.e., pieces separated in the $x$ direction) are moving relative to each other in the $z$ direction.

There is vorticity in the $x$ - $z$ plane, which means that the rod is rotating in this plane.

(Note that the shear and vorticity tensors also have nonzero $t$ components in this frame; this is simply an indication that the spacelike hypersurfaces that are locally orthogonal to the congruence are not surfaces of constant coordinate time in this frame.)

 

[PAllen]

Great, thanks!

With a little more algebra on the expansion, and noting v is approximately 1 and a is huge under conditions discussed in the other thread, shear, expansion, and vorticity are all of order a. All similar in magnitude and huge, but the expansion is biggest (contrary to a very crude calculation I did).

 

[PAllen]

If occurs to me that the congruence (and the problem specification) are invalid after a short time. One can't have magnitude of $U_z$ greater than c. But this is certainly good for analyzing what happens initially along the rod, which is what we are most interested in.

 

[PeterDonis]

If occurs to me that the congruence (and the problem specification) are invalid after a short time.

No, it isn't. Remember these are 4-velocity components. As long as $U^a U_a = -1$, it's a valid 4-velocity vector. There is no limit on the values of the individual components.

One can't have magnitude of $U_z$ greater than c.

Yes, one can, because $U_z$ is not $v_z$, it's $\gamma_z v_z$. The magnitude of that is unlimited. If you want the ordinary $z$ velocity in the given frame, it's $U_z / U_t$, which is easily shown to always be less than $1$ (I'm using units where $c = 1$).

 

[PeterDonis]

No, it isn't.

More precisely, it doesn't become invalid because of any limits on the magnitude of the components of $U$. However, the congruence is constructed on the assumption that the acceleration $a$ is a constant proper acceleration, which means it's really more appropriate for the version of the problem where the downward acceleration is caused by a non-gravitational force. In the "acceleration due to gravity' version, even if we are more careful and formulate the scenario inside a huge rocket accelerating upward in flat spacetime to avoid any tidal gravity issues, the assumption that the frame in which the congruence is formulated is inertial will not hold indefinitely, so in that sense yes, the congruence as I have written it becomes invalid after a sufficient time.

 

[PAllen]

No, it isn't. Remember these are 4-velocity components. As long as $U^a U_a = -1$, it's a valid 4-velocity vector. There is no limit on the values of the individual components.Yes, one can, because $U_z$ is not $v_z$, it's $\gamma_z v_z$. The magnitude of that is unlimited. If you want the ordinary $z$ velocity in the given frame, it's $U_z / U_t$, which is easily shown to always be less than $1$ (I'm using units where $c = 1$).

Ok, but now I have a different issue. I get a different 4 velocity altogether:

$$ \frac {1} {\sqrt {1-a^2(t-t_0)^2}} \left( 1,0,-a(t-t_0) \right) $$

and in this form, there is again a limit on how long before $U_z/U_t$ becomes greater than 1 (and $\gamma$ would be imaginary). The problem was initially specified using a pure coordinate acceleration (see e.g. Rindler's paper referenced in post #2 of the original thread).

 

[PeterDonis]

I get a different 4 velocity altogether

How are you obtaining that?

 

[PAllen]

How are you obtaining that?

Derivative by proper time of the coordinates of a world line with constant coordinate acceleration in the -z direction, no motion in x. Coordinates assumed standard inertial coordinates in SR. World line coordinates simply $(t, x_0,- .5 a(t-t_0)^2)$

 

[PeterDonis]

The problem was initially specified using a pure coordinate acceleration

Yes, I know, and it is true that constant coordinate acceleration in the given frame (or indeed in any inertial frame) cannot be maintained. I was relying on the fact that, for short enough times, constant coordinate acceleration can be approximated by constant proper acceleration. I was groping towards saying that in post #7 but didn't phrase it very well.

An alternative formulation would be the one in the accelerating rocket that I described; in this formulation, the constant proper acceleration would be stopped at different times in the "rod rest frame" (which now would not be inertial, it would be Rindler coordinates with the surface with the hole in it at constant $z$, boosted in the $x$ direction--we had a thread some time ago about a sliding block in an accelerating elevator where some math for that was developed, but it was rather hairy). So in this formulation, the expansion, shear, and vorticity would come from different parts of the rod being released into free fall at different times.

It could be interesting to do this formulation, but I won't have time to do any of the calculations for a while.

 

[PAllen]

Derivative by proper time of the coordinates of a world line with constant coordinate acceleration in the -z direction, no motion in x. Coordinates assumed standard inertial coordinates in SR. World line coordinates simply $(t, x_0,- .5 a(t-t_0)^2)$

It occurs to me that Rindler gave a in hole frame, so just replace a in the above and in my 4 velocity with $a\gamma^2$, wherever it appears. Or just redefine this to be a.

 

[Ibix]

It does seem to me that using constant coordinate acceleration means that $U^a$ at any chosen constant $x,y$ in the rod rest frame describes Bell's spaceships in the z direction (constant proper acceleration would describe Rindler observers). Perhaps some of the stress comes from there?

Relatedly, $U^a$ is uniquely defined on the rod by tracking the motion of atoms or whatever, but is it uniquely defined off the rod? I would think that $U'^a=\exp(y)U^a$ would be equally valid, at least for a thin rod lying at $y=0$. So is there some slack in the partial derivatives? Perhaps more generally, how does one handle the physical bounds of an object in this kind of analysis?

Sorry for the barrage of questions...

 

[PAllen]

Ok, but now I have a different issue. I get a different 4 velocity altogether:

$$ \frac {1} {\sqrt {1-a^2(t-t_0)^2}} \left( 1,0,-a(t-t_0) \right) $$

and in this form, there is again a limit on how long before $U_z/U_t$ becomes greater than 1 (and $\gamma$ would be imaginary). The problem was initially specified using a pure coordinate acceleration (see e.g. Rindler's paper referenced in post #2 of the original thread).

And, for constant proper acceleration, it would be the same form except replace $-a(t-t_0)$ with $v_z(t-t_0,a)$, where $v_z$ is a rather complicated function that asymptotes to 1, and a is now the chosen constant proper acceleration.

 

[PAllen]

It does seem to me that using constant coordinate acceleration means that $U^a$ at any chosen constant $x,y$ in the rod rest frame describes Bell's spaceships in the z direction (constant proper acceleration would describe Rindler observers). Perhaps some of the stress comes from there?

Correct, but that is what is described as given in the formulation of post #2 of the original thread (and, indeed, this would be one source of stress). I claim the only form that would have no stress would be acceleration varying in the z direction per a Rindler family of observers, and constant across any line of constant x along the rod, for a given t in the rod initial rest frame. As long as the acceleration is instead constant along x for given t in the hole rest frame, there must be stress. And this would only change my prior post by having some form of $v_z((t-t_0),z,a)$.

Relatedly, $U^a$ is uniquely defined on the rod by tracking the motion of atoms or whatever, but is it uniquely defined off the rod? I would think that $U'^a=\exp(y)U^a$ would be equally valid, at least for a thin rod lying at $y=0$. So is there some slack in the partial derivatives? Perhaps more generally, how does one handle the physical bounds of an object in this kind of analysis?

Sorry for the barrage of questions...

I don't know what you mean here. The congruence only exists within the rod. We take everything else to be vacuum.

 

[Ibix]

I don't know what you mean here. The congruence only exists within the rod. We take everything else to be vacuum.

I was thinking that stresses are usually different at the surface of a material object to in the body, which is not the case here. On further thought, though, I think this is just that this is a somewhat idealised model of a rod that doesn't have this feature because it doesn't have a full material model.

 

[PeterDonis]

I think this is just that this is a somewhat idealised model of a rod that doesn't have this feature because it doesn't have a full material model.

Yes, that's correct. The model assumes no internal forces between the different parts of the rod. The basis for that assumption is that the external forces involved, for the kinds of numbers implied by the OP, are so much larger than any internal forces in the rod that the latter can be ignored.

 

[PAllen]

Just to get back to the question of the 4-velocity field of a congruence, I claim the following for any congruence given description in standard inertial coordinates:

We define each element of the congruence by some position $(x_0,y_0,z_0)$ at $t_0$. Then, the world line of each congruence element is defined by $(t,x(t,x_0,y_0,z_0),y(t,x_0,y_0,z_0),z(t,x_0,y_0,z_0))$, and the 4 velocity by:
$$\frac {1} {\sqrt {1-v^2}} (1,v_x,v_y,v_z)$$
where the v terms have the obvious meanings (coordinate time derivative for components, pythagorean expression for speed v).

 

[PAllen]

And for the proper acceleration version of this, with a Rindler observer distribution in the z direction, so that the only obstacle to a rigid motion is the non-simultaneity of the start of motion in the x direction, we have (noting u is the speed of the hole with respect to the rod):
$$z=-\sqrt{z_0{}^2+(t+ux_0)^2}$$
with $v_z$ as the time derivative of this, and 4-velocity then simply:
$$\frac{1}{\sqrt{1-v_z{}^2}}(1,0,0,v_z)$$

This should provide the minimum 'stress' consistent with maintaining horizontal position in the hole fame.

 

[PAllen]

And for the proper acceleration version of this, with a Rindler observer distribution in the z direction, so that the only obstacle to a rigid motion is the non-simultaneity of the start of motion in the x direction, we have (noting u is the speed of the hole with respect to the rod):
$$z=-\sqrt{z_0{}^2+(t+ux_0)^2}$$
with $v_z$ as the time derivative of this, and 4-velocity then simply:
$$\frac{1}{\sqrt{1-v_z{}^2}}(1,0,0,v_z)$$

This should provide the minimum 'stress' consistent with maintaining horizontal position in the hole fame.

Note that $z=z_0$ along a line of simultaneity in the hole frame, though we are working in a frame where the rod has constant x position. This is as desired.

 

[pervect]

Peter's congruence

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

is a different congruence than the sliding block congruence I am familiar with. The non-zero expansion and shear of Peter's congruence bothers me, I'm not sure I understand the difference.

I am also not sure how to express the sliding block congruence in Peter's choice of coordinates. The coordinate choice I used are in geometrized Rindler coordinates, with a line element of:

$$-z^2 dt^2 + dx^2 + dy^2 + dz^2$$

The sliding block can be considered to be located at z=1.

The unnormalized 4-velocity given by dt/dt, dx/dt, dy/dt, dz/dt has components
$$[1,\beta,0,0]$$

Normalizing to the 4-velocity $dt/d\tau, dx/d\tau, dy/d\tau, dz/d\tau$ gives the following non-zero components.

*correction*
$$u^t = \frac{1}{\sqrt{z^2 - \beta^2}} \quad u^x = \frac{\beta}{\sqrt{z^2 - \beta^2}}$$

The 4-accleration of this congruence has one non-zero component

$$a^a = \frac{z} { z^2 - \beta^2}$$

Peter's congruence has a more complex 4-acceleration vector, having a nonzero t component and a nonzero z component, but the magnitude of his 4-accleration is constant and has zero components in the x and y directions, as does mine, so they should be closely related. One obvious difference is the z-dependence of the magnitude of the 4-accleration of my congruence, but I'm unclear if this is the only difference. I can certainly see why having the same proper acceleration at the top and bottom of a thick block causes expansion, this is just Bell's spacheship paradox, but I'm not sure if that's all that is going on here.

The sliding block congruence should however be similar to Peter's congruence for thin blocks, as points on the thin block have constant downward acceleration.

The "sliding block" congruence has an expansion and shear tensor of zero. The vorticity tensor has a three dimensonal hodges dual which is a vector, called the vorticity vector, which is calculated by the program I use. It is nonzero, and has components of

$$[0, 0, -\frac{\beta}{{\beta^2 - z^2}},0]$$

I don't think the values matter so much as the fact that it's nonzero, but we could wade into the vorticity more if it looks to be of interest. I'm more interested in understanding the shear and expansion of Peter's congruence, and if it's entirely eliminated simply by making the top of a thick block accelerate downwards slightly less than the bottom.

 

[PeterDonis]

is a different congruence than the sliding block congruence I am familiar with

Yes, there are two key differences from the previous "sliding block" problem:

(1) I have assumed that the "table" frame is inertial, not accelerated (i.e., the coordinates for this frame are standard inertial coordinates, not Rindler coordinates), and that the rod is accelerated downward by some non-gravitational force (so the rod has nonzero proper acceleration, not the table). In the "sliding block" scenario discussed in previous threads, the "table" frame (the frame of the table along which the block is sliding) is accelerated (Rindler coordinates, not inertial coordinates).

(2) My congruence describes the rod, being accelerated downwards relative to the table as well as moving horizontally relative to it. The "sliding block" congruence describes a block moving horizontally relative to the "table" but not moving vertically relative to it.

 

[PAllen]

Yes, there are two key differences from the previous "sliding block" problem:

(1) I have assumed that the "table" frame is inertial, not accelerated (i.e., the coordinates for this frame are standard inertial coordinates, not Rindler coordinates), and that the rod is accelerated downward by some non-gravitational force (so the rod has nonzero proper acceleration, not the table). In the "sliding block" scenario discussed in previous threads, the "table" frame (the frame of the table along which the block is sliding) is accelerated (Rindler coordinates, not inertial coordinates).

(2) My congruence describes the rod, being accelerated downwards relative to the table as well as moving horizontally relative to it. The "sliding block" congruence describes a block moving horizontally relative to the "table" but not moving vertically relative to it.

But even given this, I come up with a different congruences, whether assuming constant coordinate acceleration or constant proper acceleration (both given in earlier posts). Can you describe how you arrive at your congruence.

Thanks.

 

[PeterDonis]

Can you describe how you arrive at your congruence.

Remember that my congruence is expressed in the rod rest frame (the frame in which the rod is at rest before it starts to fall). I am assuming that frame is inertial and that when the rod starts to fall it does so in response to a force that gives it nonzero proper acceleration downward. (I call this force "gravity" in the OP but I am not actually modeling it that way, as I clarified in post #7.) The components of the rod's 4-velocity are then simply $\gamma$ ($t$ component) and $\gamma v$ ($z$ component) as given by the relevant relativistic rocket equations for constant proper acceleration $a$, with a minus sign for the $z$ component because the acceleration is downward. The $\left( t - t_0 \right)$ factor accounts for the fact that the downward proper acceleration starts at different coordinate times in this frame for different points along the rod.

 

[PeterDonis]

I am assuming that frame is inertial and that when the rod starts to fall it does so in response to a force that gives it nonzero proper acceleration downward. (I call this force "gravity" in the OP but I am not actually modeling it that way.)

As I mentioned in post #11, an alternative formulation would be to have the "table" (and the hole) accelerating upward (nonzero proper acceleration) and the rod free-falling once it is unsupported. I still have not had time to try the calculations for that case.

 

[PAllen]

Remember that my congruence is expressed in the rod rest frame (the frame in which the rod is at rest before it starts to fall). I am assuming that frame is inertial and that when the rod starts to fall it does so in response to a force that gives it nonzero proper acceleration downward. (I call this force "gravity" in the OP but I am not actually modeling it that way, as I clarified in post #7.) The components of the rod's 4-velocity are then simply $\gamma$ ($t$ component) and $\gamma v$ ($z$ component) as given by the relevant relativistic rocket equations for constant proper acceleration $a$, with a minus sign for the $z$ component because the acceleration is downward. The $\left( t - t_0 \right)$ factor accounts for the fact that the downward proper acceleration starts at different coordinate times in this frame for different points along the rod.

But that is what I have in my congruence and not what you have. Look at your first post In this thread. Compare to my post #8, which implements what you describe here (for coordinate acceleration). For proper acceleration, see my post #19.

 

[PeterDonis]

that is what I have in my congruence and not what you have.

I don't understand. The relativistic rocket equation has (from this page, in units where $c = 1$):

$$
\gamma = \sqrt{ 1 + a^2 t^2 }
$$

$$
v = \frac{at}{\sqrt{ 1 + a^2 t^2 }}
$$

Multiplying these gives

$$
\gamma v = a t
$$

My $U$ is then just $\gamma$ and $\gamma v$ from the above, with the $\left( t - t_0 \right)$ factor in place of just $t$ for the reason I explained.

 

[PAllen]

I don't understand. The relativistic rocket equation has (from this page):

$$
\gamma = \sqrt{ 1 + a^2 t^2 }
$$

$$
v = \frac{at}{\sqrt{ 1 + a^2 t^2 }}
$$

Multiplying these gives

$$
\gamma v = a t
$$

My $U$ is then just $\gamma$ and $\gamma v$ from the above, with the $\left( t - t_0 \right)$ factor in place of just $t$ for the reason I explained.

Ok, then I see yours is valid, as are both of mine, for three different cases:

1) constant coordinate acceleration, my post #8
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)

 

[PeterDonis]

yours is valid, as are both of mine, for three different cases

Yes.

 

[pervect]

Yes, there are two key differences from the previous "sliding block" problem:

(1) I have assumed that the "table" frame is inertial, not accelerated (i.e., the coordinates for this frame are standard inertial coordinates, not Rindler coordinates), and that the rod is accelerated downward by some non-gravitational force (so the rod has nonzero proper acceleration, not the table). In the "sliding block" scenario discussed in previous threads, the "table" frame (the frame of the table along which the block is sliding) is accelerated (Rindler coordinates, not inertial coordinates).

(2) My congruence describes the rod, being accelerated downwards relative to the table as well as moving horizontally relative to it. The "sliding block" congruence describes a block moving horizontally relative to the "table" but not moving vertically relative to it.

The "table" isn't physical, though. It's just a matter of coordinates, or if you prefer, frames of reference. It should be possible to transform both congruences to the "lab frame" coordinates, then find the appropriate integral curves in each coordinate system. However, I haven't done this. I'm not sure I'll get around to it, unfortunately, though I'll feel unsatisfied with the answers unless I do.

At the moment, neither of the congruences is expressed in the lab frame coordinates. Your coordinates move relative to the lab frame at a constant velocity, the table coordinates move relative to the lab frame at an accelerating velocity.

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.

Additionally, some external stress energy tensor is presumably responsible for the acceleration of the rod in either case, a stress-energy tensor which Rindler called the "hail", but nobody has written that down. I think the simplest stress-energy tensor might be a pure pressure field in the lab frame, which of course would violate the weak energy condition. Presumably there are other possible stress energy tensors that do not violate the weak energy condition that would also suffice.

A "static" non-time varying hail in the lab frame can and probably should result in a time-varying proper acceleration. This is easiest to see in the "hail" is a laser beam driving a light sail, the proper acceleration of the light sail for a static, constant intensity laser in the lab frame decreases with time.

 

[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.