Solving the Spaceship Paradox: A New Explanation

[Mentz114]

No, it isn't. It's a rank-4 tensor, just like any other tensor in spacetime. It just happens to be a rank-4 tensor that, in a particular frame (the comoving frame), has only "space-space" components nonzero, so it can be treated, *within that frame*, as a purely spatial rank-3 tensor.

OK, but $|\sigma_{\mu\nu}|=0$.

There's another way to put this: the rank-3 tensor, which is what has the direct physical interpretation, is really the rank-4 tensor projected into the spacelike 3-surface that is orthogonal to the 4-velocity of the spaceship. That projection is done using the projection tensor $h_{\mu \nu} = g_{\mu \nu} + u_{\mu} u_{\nu}$. So what you should be looking at for physical interpretation is not $\sigma_{\mu \nu}$; it's $\sigma_{ab} = h^{\mu}{}_a h^{\nu}{}_b \sigma_{\mu \nu}$
..
...
you have to actually compute $h_{\mu \nu}$ and contract it with $\sigma_{\mu \nu}$ as above. If you do this, you should end up with the same rank-3 tensor $\sigma_{\hat{\mu} \hat{\nu}}$ as in the comoving frame (by a process similar to the one by which we showed that the trace is frame-invariant)..

In response to your post I calculated $\sigma_{ab} = h^{\mu}{}_a h^{\nu}{}_b \sigma_{\mu \nu}$ with the components of $\sigma_{\mu\nu}$ in my last post, and found $\sigma_{ab}=\sigma_{\mu\nu}$. I was surprised too. Maybe I misinterpret what you're saying.

In short, the rank-3 tensor in the comoving frame is the one that has the direct physical interpretation. In any other frame, just looking at the tensor components isn't a good way to think about physical interpretation, because what is physically meaningful, in frame-invariant terms, is the contraction of the shear tensor with the projection tensor. The comoving frame is just the one in which that contraction works out to just give the tensor components in that frame directly.

Firstly, I understand that the physics is in the invariants. That is why in my calculation no invariant has been hurt in any way. This is what I take to be the meaning of 'covariant calculation'.

If two bodies collide and make a big explosion - the physics is in the details of the explosion and is independent of which frame we choose. But each body could claim that the energy for the explosion came from the kinetic energy of the *other* body. So components can tell us how different observers slice and dice (attribute) the components of the physics.

 

[PeterDonis]

OK, but $|\sigma_{\mu\nu}|=0$.

I don't understand; what is $|\sigma_{\mu\nu}|$? And which frame are we talking about?

In response to your post I calculated $\sigma_{ab} = h^{\mu}{}_a h^{\nu}{}_b \sigma_{\mu \nu}$ with the components of $\sigma_{\mu\nu}$ in my last post, and found $\sigma_{ab}=\sigma_{\mu\nu}$. I was surprised too. Maybe I misinterpret what you're saying.

Again, which frame are we talking about? [STRIKE]What you should find is that $\sigma_{ab} = \sigma_{\hat{\mu} \hat{\nu}}$; note carefully the hats on the indexes, indicating that projecting the shear tensor, in any frame, into the spacelike hypersurface orthogonal to the spaceship's 4-velocity, as expressed in that frame, will give you back the shear tensor components in the comoving frame.[/STRIKE] [Edit: actually, not; see follow-up post.] But I'm not sure you're saying that's what you got.

each body could claim that the energy for the explosion came from the kinetic energy of the *other* body.

Kinetic energy by itself isn't really a "component"; but that's a minor point. See below.

So components can tell us how different observers slice and dice (attribute) the components of the physics.

Not in themselves, no. To give physical meaning to any "component", you have to express it as an invariant. For example, if two bodies collide, assuming they have equal rest mass $m$, the kinetic energy each one sees the other to have is $K = E - m = u^a p_a - \sqrt{ p^a p_a } = \left( \gamma - 1 \right) m$, where $u^a$ is the "observing" body's 4-velocity and $p^a$ is the "observed" body's 4-momentum. We can view the total energy $E$ as a "component" of the observed body's 4-momentum in the observing body's frame, but that only has physical meaning because we can express it as the invariant $u^a p_a$.

In other words, what tells us how different observers slice and dice the physics are the invariants we form by contracting a particular vector, tensor, or whatever with the vectors that describe the observer (his 4-velocity and the spatial vectors of his frame). So if you want to try to give a physical interpretation to shear tensor components in some frame other than the comoving frame, you'll need to figure out what observer, other than the comoving observer, you're interested in, and what it means, physically, to contract that observer's frame vectors with the shear tensor. In general I'm not sure this will be very useful for the shear tensor, because the whole point of that tensor is to capture properties measured by the comoving observer.

 

[PeterDonis]

What you should find is that $\sigma_{ab} = \sigma_{\hat{\mu} \hat{\nu}}$; note carefully the hats on the indexes, indicating that projecting the shear tensor, in any frame, into the spacelike hypersurface orthogonal to the spaceship's 4-velocity, as expressed in that frame, will give you back the shear tensor components in the comoving frame.

I just realized that this can't be right as it stands, because the shear tensor components in the original inertial frame, $\sigma_{\mu \nu}$ (no hats on the indexes) are obtained by applying the Lorentz transform to the components in the comoving frame, $\sigma_{\hat{\mu} \hat{\nu}}$ (with hats on the indexes). But applying the (inverse, to go from original frame to comoving frame) Lorentz transform is not the same as contracting with the projection tensor. So I need to think about this some more.

 

[PeterDonis]

So I need to think about this some more.

Ok, I was confusing myself about the projection tensor. Now I think I'm un-confused.

Contracting with the projection tensor $h_{\mu \nu}$ gives the "transverse" part of a tensor, i.e., the part that is orthogonal to the 4-velocity $u^{\mu}$. But the shear tensor is already transverse; i.e., it is already orthogonal to $u^{\mu}$. So contracting with the projection tensor should leave the shear tensor unchanged. That is, in any frame, if you express the shear tensor in that frame, and contract it with the projection tensor expressed in that frame, the shear tensor components should be unchanged. Sorry for my previous confusion about this.

 

[Mentz114]

Peter, thanks again.

Just to clear up the notation - the un-hatted indexes are for the coordinate basis, and the hats are the local frame basis of the accelerating congruence.

My point about the determinant of $\sigma_{\mu\nu}$ being zero, is that as a bilinear transformation it is 3-dimensional ( there are 3 non-zero eigenvalues). I would expect this after some thought because the transformation with the vierbien into the local frame basis gives a rank-3 tensor.

Regarding the possible physical meaning of $\sigma_{\mu\nu}$, I have some ideas but no time to explicate now.

 

[PeterDonis]

Just to clear up the notation - the un-hatted indexes are for the coordinate basis, and the hats are the local frame basis of the accelerating congruence.

Yes.

My point about the determinant of $\sigma_{\mu\nu}$ being zero, is that as a bilinear transformation it is 3-dimensional ( there are 3 non-zero eigenvalues).

Ah, ok. Yes, if you view $\sigma_{\mu \nu}$ as a 4x4 matrix, it has zero determinant because only three eigenvalues are nonzero. That will be generally true of any transverse tensor, since evaluating the eigenvalues basically amounts to transforming into the comoving frame, where only the space-space components can be nonzero.

 

[WannabeNewton]

Regarding the possible physical meaning of $\sigma_{\mu\nu}$...

Wasn't this already answered?

 

[Mentz114]

Wasn't this already answered?

Not in my understanding. In the local frame the meaning is clear, but what does it mean when written in the coordinate basis ? I've been doing a lot of thinking and calculation with a Born congruence ( ie $\dot{u}_{\hat{\mu}}= k\ dx$, $k$ constant ) where the velocity depends on $t$ and $x$. Some interesting things have emerged.

 

[WannabeNewton]

In the local frame the meaning is clear, but what does it mean when written in the coordinate basis ?

It has no meaningful physical interpretation in that case. The problem is that $\sigma_{\mu\nu}$ is defined in terms of $h_{\mu\nu}$ which physically measures spatial distances (between events) relative to the instantaneously comoving inertial frames of the spaceships i.e. it is only defined relative to the spaceship frames. It has no direct relationship to other frames, not even the background global inertial frame in which the spaceships are accelerated simultaneously. As such, while the components of $\sigma_{\mu\nu}$ relative to a given arbitrary frame (field) are measurable quantities in the mathematical sense, they do not necessarily entail a physical interpretation unless the frame field consists of the local Lorentz frames attached to the orbits of the Bell congruence.

This isn't peculiar to the shear tensor. A lot of quantities in the kinematic decomposition only have meaningful physical interpretation relative to the instantaneous (local) inertial frames of observers in the congruence. Take for example the vorticity $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\partial_{\alpha}\xi_{\beta}$. In an instantaneously comoving local inertial frame, we can use the fact that $\epsilon^{ijk} \Leftrightarrow \epsilon^{\mu \nu\alpha\beta}\xi_{\mu}$ and $\xi^{\mu} = \delta^{\mu}_0$ to see that $\omega^{\mu}\xi_{\mu} = 0$ (purely spatial vector in the local inertial frame) and $\vec{\omega} = \vec{\nabla}\times \vec{\xi}$ which means the vorticity, in the local inertial frame, is just a measure of local rotation relative to local gyroscopes if the congruence is Born rigid. But what is the interpretation of $\omega^{\mu}$ in other local Lorentz frames? Well if we boost from the instantaneously comoving local inertial frame of an observer in the congruence to the rest frame of a coincident observer (say along the $x$-axis of the local inertial frame) then the boost will rotate the local $t-x$ plane which means that $\omega^{\mu}$ relative to the new frame will pick up a time-like component and hence will no longer be a purely spatial vector in this frame: a measure of local rotation with a time-like component...yeah we can't interpret that physically.

 

[Mentz114]

OK, that makes sense.

I got to thinking about the Doppler shift between the spaceships. Obviously there will be a kinematic frequency shift between the ships because they are not comoving. What causes the shift in the global inertial frame, where the ships are comoving ?

The only way I can make sense of this is to introduce a gravitational field which brings the spaceships to rest wrt each other in all frames. With $\vec{u}= \sqrt{u^2+1} \ \partial_t + u\ \partial_x$, if we impose the condition that the local acceleration vector is $\dot{u}_{\hat{\mu}}= k\ d\hat{x}$, then the acceleration vector in the coordinate basis is $\dot{u}_\mu = -u\ k \ dt + \sqrt{1+u^2}\ k\ dx$. $u$ can be a function of $t$, $x$ or both.

So the gravitational field (pulling in the -x direction) must create acceleration $\sqrt{1+u^2}k$. The acceleration (force?) ratio between two points $P_1$ $P_2$ is then $\sqrt{1+u^2}|_{P_1}/\sqrt{1+u^2}|_{P_2}$ and from the 4-velocity the ratio of clock rates $d\tau_1/d\tau_2$ is the reciprocal of the acceleration ratio. Given that $P_1$ $P_2$ are the end points of a null geodesic connecting two spaceship worldlines, I suspect that the frequency shift due to the field is identical to the kinematical shift when the field is not present.

If it is true then I don't see immediately how to apply it to the strings breaking question, although a growing weight could be relevant.

 

[PAllen]

The Doppler in the global inertial frame is trivially interpreted kinematically. The rule for all Doppler is based on the velocity of emitter at event of emission compared to velocity of target at event of reception. Since the ships are accelerating, then for rear to forward ship, the emission speed is lower than the reception speed in the same direction, so relative speed is 'away' so redshift. For forward to rear, the relative speed between emission and reception is 'toward' so blueshift. This would all be true even for non-relativistic Doppler.

A gravitational or pseudo-gravitational redshift is typically introduced for a non-inertial frame, where (if you want to use coordinate velocities), you have redshift/blueshift between objects maintaining constant coordinate positions.

 

[Mentz114]

Thanks for clarifying that. It will be interesting to see if the frequency shift can be decomposed into a non-separation (delay) component and a relative velocity component. This is going off topic though so I'll post anything interesting in a new thread.

[Edit]
On reflection - won't the frequency shifts remain constant in the inertial frame ? The delay between sending and receiving will be the same because the separation is the same. But in the local frame the frequency shift is changing. A back of the envelope calculation suggests the the rate of change of the frequency ratio is $\theta + a$, i.e. the sum of the expansion scalar and the acceleration.

 

[PAllen]

Thanks for clarifying that. It will be interesting to see if the frequency shift can be decomposed into a non-separation (delay) component and a relative velocity component. This is going off topic though so I'll post anything interesting in a new thread.

[Edit]
On reflection - won't the frequency shifts remain constant in the inertial frame ? The delay between sending and receiving will be the same because the separation is the same. But in the local frame the frequency shift is changing. A back of the envelope calculation suggests the the rate of change of the frequency ratio is $\theta + a$, i.e. the sum of the expansion scalar and the acceleration.

The frequency shift would be constant between congruence lines (in all frames, or we have contradictions) for the Born rigid congruence. For the congruence of constant proper acceleration at all times in a given inertial frame (starting frame), the shift increases with time in all frames. This is strictly a relativistic effect. The coordinate accelerations are hyperbolic, the shift has be computed with relativistic velocity addition. Due to the hyperbolic coordinate acceleration and constant light speed, the time for light from the rear to catch the front increases with time in the inertial frame. Then, with velocity addition, the shift increases with time. Conversely for light from front to back.

 

[Mentz114]

Yes, the scalars say that the shift is always increasing, so my question was a bit uncalled for.

But the pseudogravitational field provides a satisfactory understanding (to me, anyway) of the spaceship scenario, so I'm happy to leave the subject, finally.

 

[Mentz114]

Mentz, a constant proper acceleration means that the experienced "tension as if in a gravitational field" will be constant, not getting stronger with time. It means that the string can always be made strong enough to not break due to this effect alone.

At the time this was posted I was still incorrectly assuming that the proper acceleration of the spaceships is constant in both frames. This is not so and the acceleration of the ships in S is $a\sqrt{1+a^2 t^2}$. So the 'tension of the graviatational field' is increasing without limit. Furthermore, in the ship frame, the separation is increasing as $a\sqrt{1+a^2 t^2}$ ( with $a$ being the initial separation), so effects caused by the separation in the ship frame can be attributed to the increasing pseudo-gravitational field caused by the increasing acceleration in S.

This looks very plausible to me

 

[WannabeNewton]

I have no idea how you came to that conclusion. The proper accelerations of both spaceships are equal and constant. This is obviously true in all frames because proper acceleration is frame-independent.

 

[PeterDonis]

I was still incorrectly assuming that the proper acceleration of the spaceships is constant in both frames. This is not so

As you state it, yes, it is. Proper acceleration is an invariant. The acceleration you appear to be looking at is coordinate acceleration, and it does change, yes, but I don't think it increases with time in frame S; see below.

the acceleration of the ships in S is $a\sqrt{1+a^2 t^2}$.

This doesn't look right. I think it should be $a / \sqrt{1 + a^2 t^2}$, where $a$ is the (constant) proper acceleration. The coordinate acceleration in S should decrease with time, because the change in speed of the ships, with respect to S, must get smaller as they get closer to the speed of light.

So the 'tension of the graviatational field' is increasing without limit.

Not if my correction given just now is right.

effects caused by the separation in the ship frame can be attributed to the increasing pseudo-gravitational field caused by the increasing acceleration in S.

Again, this won't work if the correction I gave above is right.

 

[stevendaryl]

This doesn't look right. I think it should be $a / \sqrt{1 + a^2 t^2}$, where $a$ is the (constant) proper acceleration. The coordinate acceleration in S should decrease with time, because the change in speed of the ships, with respect to S, must get smaller as they get closer to the speed of light.

I think your expression is not quite right, either. I think it's actually:

\dfrac{d^2 x}{d t^2} = \dfrac{a}{(a^2 t^2 + 1)^{\frac{3}{2}}}

 

[Mentz114]

I have no idea how you came to that conclusion. The proper accelerations of both spaceships are equal and constant. This is obviously true in all frames because proper acceleration is frame-independent.

The norm of the proper acceleration vector is constant, but the components of $\dot{v}_a=v^b\partial_b v_a$ are different in different coordinates.

PeterDonis and stevendaryl - I think my calculation of $\dot{v}_a$ is correct.

However, there is an obvious contradiction in my case for the pseudo-gravitational so it's back to the old envelope.

 

[WannabeNewton]

The norm of the proper acceleration vector is constant...

Proper acceleration is the norm.

 

[PeterDonis]

I think my calculation of $\dot{v}_a$ is correct.

I'm not sure what you are trying to calculate. If you are calculating the $t$ and $x$ components of the 4-acceleration vector $\dot{v}^a$ in frame S, you should be giving two expressions, because there are two components, but you only gave one. The two expressions will be $\dot{v}^t = d v^t / d \tau$ and $\dot{v}^x = d v^x / d \tau$. Since $v^t = \gamma$ and $v^x = \gamma v$, where $v = dx / dt$, we have (note that $dv / d \tau = (dt / d \tau) dv / dt = \gamma d^2 x / dt^2$):

$$
\dot{v}^t = \frac{d \gamma}{d \tau} = \gamma^3 v \frac{dv}{d \tau} = \gamma^4 v \frac{d^2 x}{d t^2}
$$

$$
\dot{v}^x = \frac{d (\gamma v)}{d \tau} = \left( \gamma^3 v^2 + \gamma \right) \frac{dv}{d \tau} = \gamma^3 \frac{dv}{d \tau} = \gamma^4 \frac{d^2 x}{d t^2}
$$

Note that neither of these components is equal to $d^2 x / d t^2$. (It looks like the expression you gave may be equal to $\dot{v}^x$, if we substitute for $\gamma$ using my expression for $dv / dt$ below.)

If you are calculating the coordinate acceleration $d^2 x / d t^2$ in frame S, which is what stevendaryl and I were giving expressions for, here's a quick computation of that:

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

$$
\frac{dx}{dt} = \frac{a t}{\sqrt{1 + a^2 t^2}}
$$

$$
\frac{d^2 x}{d t^2} = \frac{a}{\sqrt{1 + a^2 t^2}} - \frac{a^3 t^2}{\left(1 + a^2 t^2\right)^{3/2}} = \frac{a}{\left(1 + a^2 t^2\right)^{3/2}} \left[ \left( 1 + a^2 t^2 \right) - a^2 t^2 \right] = \frac{a}{\left(1 + a^2 t^2\right)^{3/2}}
$$

So it looks like stevendaryl was correct, and I should have actually done the calculation instead of guessing off the top of my head.

 

[PeterDonis]

(It looks like the expression you gave may be equal to $\dot{v}^x$, if we substitute for $\gamma$ using my expression for $dv / dt$ below.)

Confirmed: from the expression I gave for $dx / dt = v$ in my last post, we have

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

This gives

$$
\dot{v}^t = \gamma^4 v \frac{d^2 x}{d t^2} = \left( 1 + a^2 t^2 \right)^2 \frac{at}{\sqrt{1 + a^2 t^2}} \frac{a}{\left( 1 + a^2 t^2 \right)^{3/2}} = a^2 t
$$

$$
\dot{v}^x = \gamma^4 \frac{d^2 x}{d t^2} = \left( 1 + a^2 t^2 \right)^2 \frac{a}{\left( 1 + a^2 t^2 \right)^{3/2}} = a \sqrt{1 + a^2 t^2} = a^2 x
$$

So the expression Mentz114 gave corresponds to $\dot{v}^x$, but that expression doesn't stand by itself; it goes with the other component of the same vector, $\dot{v}^t$, given above.

Note, by the way, that, following on from the symmetric way I wrote the 4-acceleration above (by showing that $\dot{v}^t = a^2 t$ and $\dot{v}^x = a^2 x$), we can write the 4-velocity in a more symmetric way as well:

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

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

 

[Mentz114]

Peter, we are in partial agreement with the acceleration vector. But the norm should be $a$, I think.

My efforts to explain the 'paradox' have lead nowhere, so I started looking at the problem more closely, looking for hidden assumptions etc. I did find a good candidate.

Let the endpoints of the thread be P₁ and P₂, for the trailing and leading ends respectively. At $t=0$ all clocks are set to zero.

Clocks comoving with P₁ and P₂ cannot be synchronised, and the elapsed time on the leading clock is greater than that on the trailing clock ( this last assertion is crucial and only true for non-inertial rockets).

This means that if P₁ and P₂ are expressed in the coordinates of the inertial frame S, they cannot have the same $t$ value (see ** below). Thus a line joining the spaceship worldlines which has the same $t$ value, is not connecting P₁ with P₂, and cannot be the representation (in S) of the thread. That would be like the line AC on the picture.

The hidden assumption is that the clocks at the ends of the thread are synchronised, as they would be in inertial motion.

The horizontal line is the length measured using an inappropriate clock synchronisation scheme and could well have no physical meaning.

With the correct mapping from the worldlines to S, it is clear that the separation between P1 and P2 (in S coordinates) is increasing.



** Assuming the transformations from the worldline proper $\tau$ to coordinate $t$ and $x$ is
<br /> \begin{align*}<br /> t_k&amp;=\int_0^{\tau_k} \frac{dt}{d\tau}d\tau\\<br /> x_k &amp;= x_{k0}+\int_0^{\tau_k} \frac{dx}{d\tau}d\tau<br /> \end{align*}<br />

 

[WannabeNewton]

There is no "hidden" assumption in the statement of the paradox that the clocks comoving with the spaceships remain synchronized. The fact of the matter is the separation between the spaceships is not increasing in the inertial frame as measured on the simultaneity hyperplanes of the inertial frame, and this is trivially true by construction. This has nothing to do with the fact that the clocks comoving with the spaceships fail to remain synchronized, as the separation between the spaceships, in the inertial frame, is measured using the synchronized clocks at rest in the inertial frame.

At the instant the spaceships are simultaneously accelerated in the inertial frame, the clocks comoving with the spaceships will become desynchronized so that in the rest frame of either spaceship, the other spaceship beings to accelerate before the one we're in the rest frame of and we see that the distance between them is increasing in this instantaneously comoving inertial frame but this is obviously not happening in the background inertial frame, wherein the acceleration was applied simultaneously, and wherein, as already stated, the distance between the spaceships remains constant.

 

[PeterDonis]

the norm should be $a$, I think.

Yes, it is $a$. We have $\sqrt{\eta_{ab} \dot{v}^a \dot{v}^b} = \sqrt{- (\dot{v}^t)^2 + (\dot{v}^x)^2} = \sqrt{ a^4 \left( x^2 - t^2 \right) } = a^2 \left( 1 / a \right) = a$.

Clocks comoving with P₁ and P₂ cannot be synchronised, and the elapsed time on the leading clock is greater than that on the trailing clock ( this last assertion is crucial and only true for non-inertial rockets).

Agreed.

This means that if P₁ and P₂ are expressed in the coordinates of the inertial frame S, they cannot have the same $t$ value

This doesn't make sense as it stands, because a given $t$ value is assigned to a particular event on a worldline, and there are certainly events on both worldlines that will be assigned any $t$ value you choose, so if we pick an event on the worldline of $P_1$ and look at its $t$ value, there will certainly be *some* event on the worldline of $P_2$ that has the same $t$ value.

A better way of saying what I think you are trying to say here is that events with the same $t$ value on the worldlines of $P_1$ and $P_2$ will *not* have the same $\tau$ value (i.e., proper time) according to either $P_1$ or $P_2$ (except for the events with $t = 0$). However, this is not correct as it stands either; see below.

Thus a line joining the spaceship worldlines which has the same $t$ value, is not connecting P₁ with P₂

A better way of saying this, as above, would be that a line of constant $t$ is *not* a line of constant $\tau$ according to either $P_1$ or $P_2$. But there are still complications; see below.

and cannot be the representation (in S) of the thread.

Because a "representation" of the thread (meaning, more precisely, a spacelike curve that represents the thread "at an instant of its proper time") would have to be a line of constant $\tau$ according to either $P_1$ or $P_2$.

However, there is a complication here as well. Suppose I pick a particular event on the worldline of $P_1$, and extend a line of constant $\tau$ through that event until it meets the worldline of $P_2$. Call the value of $\tau$ according to $P_1$ that corresponds to this line $\tau_1$. The value $\tau_2$ of the proper time according to $P_2$ at the event where the line intersects the worldline of $P_2$ will *not* be the same as $\tau_1$ (we will always have $\tau_2 > \tau_1$). This, of course, is just another way of saying that the clocks at the two ends of the thread can't be synchronized. (But even here there are subtleties--see below.)

Furthermore: suppose we look at points on the worldlines of $P_1$ and $P_2$ that have the same $t$ values. Do they have the same $\tau$ values? The answer, for this particular congruence, is *yes*. It would be "no" for the Rindler congruence, but different worldlines in that congruence have different proper accelerations--the proper acceleration decreases as you move "up" the congruence, i.e., to larger $x$ values. The congruence we're talking about here is the Bell congruence, in which each worldline has the *same* proper acceleration; and that means that $\tau$ as a function of $t$ is the same for every worldline. So the horizontal lines in the spacetime diagram of frame S *do* in fact connect points on the worldlines of $P_1$ and $P_2$ that have the *same* $\tau$!

What those horizontal lines do *not* do is connect points on the worldlines of $P_1$ and $P_2$ that are *simultaneous*, according to either $P_1$ or $P_2$. For that, we need the "tilted" lines such as the one you drew; if we pick a point on either worldline with $t > 0$, the line of simultaneity for that worldline passing through that point will be tilted up and to the right on the spacetime diagram of frame S. But note that if, as above, we extend the line of simultaneity from one worldline to the other, say from $P_1$ to $P_2$, if it is a line of simultaneity for $P_1$ at the point where it intersects $P_1$, it will *not* be a line of simultaneity for $P_2$ at the point where it intersects $P_2$! That, of course, is because, in the instantaneous rest frame of either ship (or either end of the thread), the other ship (or the other end of the thread) is moving ($P_2$ is moving forward relative to $P_1$; $P_1$ is moving backward relative to $P_2$).

So one has to be quite careful in describing the kinematics of this scenario.

 

[Mentz114]

Thanks to you both.

Peter, I tried to understand what you are saying ( I appreciate the effort you obviously put in), but it doesn't actually contradict what I'm asserting. If we pick a P₁ then P₂ is defined, and the separation of the spaceships is the separation of P₁ and P₂. The horizontal line does not join P₁ and P₂ and so is not the separation. It is something else which happens to remain the same. This does *not* imply that the separation is constant.

If two objects were approaching on collision course then obviously they cannot be brought to rest wrt each other without a paradox. All the ss paradox shows is that if this is allowed there is a paradox !

The separation between the ships is always increasing. Everything points to this, especially the expansion scalar and shear tensor. I cannot accept that there is any frame where this is not true.

If there is something which looks like the separation of the spaceships in S coords, which is not increasing, it should be recognised as a non-physical coordinate effect which needs no explanation.

 

[WannabeNewton]

What is your exact (mathematical) definition of "separation" between the spaceships?

 

[PeterDonis]

If we pick a P₁ then P₂ is defined

How? What are $P_1$ and $P_2$? Are they events? Worldlines? Spatial points? If they're spatial points, in what frame and at what time in that frame? Your statement does not answer any of those questions; that's why I was trying to clarify it and restate it in an unambiguous way.

and the separation of the spaceships is the separation of P₁ and P₂.

Same questions here: what are $P_1$ and $P_2$, and what does their "separation" mean? That's not a standard term in relativity, so you have to define what you mean by it.

The horizontal line does not join P₁ and P₂ and so is not the separation.

This implies that by "separation" you mean "spatial separation in a frame in which one of the two spaceships is instantaneously at rest". But even here there is an ambiguity: which spaceship? Except at the initial instant, when the spaceships are just beginning to accelerate, there is no common rest frame for the two ships; in any frame in which one is at rest, the other is moving. Once again, your statement does not resolve all these ambiguities. That's why I tried to restate what I think you were trying to say in an unambiguous way.

If two objects were approaching on collision course then obviously they cannot be brought to rest wrt each other without a paradox. All the ss paradox shows is that if this is allowed there is a paradox !

I don't understand how this relates to the spaceship paradox at all.

The separation between the ships is always increasing. Everything points to this, especially the expansion scalar and shear tensor. I cannot accept that there is any frame where this is not true.

For your definition of "separation", you are correct; it is increasing in all frames (because your definition basically corresponds to the expansion scalar being positive, which is an invariant). But "separation" does not have a unique definition in relativity.

If there is something which looks like the separation of the spaceships in S coords, which is not increasing, it should be recognised as a non-physical coordinate effect which needs no explanation.

Huh? Suppose there are a whole family of observers all at rest in frame S, at different spatial coordinates, and all with synchronized clocks. Each one of them records the time on his clock at which each spaceship passes his spatial position. Then we collect all this data. The data will show that, for any pair of observers A and B, if A records that one spaceship passes him at time $t$ by his clock, and B records that the other spaceship passes him at the same time $t$ by his clock, A and B's spatial separation will be the same constant value. This is a concrete physical realization of the "separation" being constant in frame S.

 

[Mentz114]

How? What are P1 and P2? Are they events? Worldlines? Spatial points? If they're spatial points, in what frame and at what time in that frame? Your statement does not answer any of those questions; that's why I was trying to clarify it and restate it in an unambiguous way.

Let me clarify something that I have not put across properly. I hope we can agree on
\begin{align*}
t_k&=\int_0^{\tau_k} \frac{dt}{d\tau}d\tau\\
x_k &= x_{k0}+\int_0^{\tau_k} \frac{dx}{d\tau}d\tau
\end{align*}
with the clocks set to zero at $t=0$.

If we choose a point at time $\tau_1$ on the trailing worldline, then the leading ship has $\tau_2=\tau_1+\gamma\beta(x_{20}-x_{10})$ where $\gamma, \beta$ are from the LT joining the inertial frame and a spaceship frame. I hope this removes any gaps about what P₁ and P₂ are - they are moving events on the worldlines.

So, having picked P₁ we have no choice about where P₂ is. These are the 4D positions of the ships in S coordinates and the separation $x_2-x_1$ is increasing.

The thing that does not increase is the distance between the trailing ship and the point where the leading ship would have been $\gamma\beta(x_{20}-x_{10})$ seconds earlier than $\tau_1$. This 'retarded length' is not a useful measurement and the fact that it does not change is not physically important.

That's the nub of it.

I haven't had time to think long about the rest of your last post but I think the examples you give all use the 'retarded length'.

Thanks, I greatly appreciate your input.

 

[PeterDonis]

I hope we can agree on
\begin{align*}
t_k&=\int_0^{\tau_k} \frac{dt}{d\tau}d\tau\\
x_k &= x_{k0}+\int_0^{\tau_k} \frac{dx}{d\tau}d\tau
\end{align*}
with the clocks set to zero at $t=0$.

To be clear, these are equations for the coordinates $(t_k, x_k)$ of spaceship $k$, in frame S, as a function of $\tau_k$, the proper time of spaceship $k$, correct?

If we choose a point at time $\tau_1$ on the trailing worldline, then the leading ship has $\tau_2=\tau_1+\gamma\beta(x_{20}-x_{10})$ where $\gamma, \beta$ are from the LT joining the inertial frame and a spaceship frame.

Only because you have *defined* things this way. Your equations above give no connection between $(t_1, x_1)$ and $(t_2, x_2)$, nor between $\tau_1$ and $\tau_2$. So your statement just quoted is an additional constraint, over and above the equations for $(t_k, x_k)$ in terms of $\tau_k$.

It looks to me like your intent with this additional constraint is to pick out the point on spaceship 2's worldline that is simultaneous, with respect to spaceship 1, with the point on spaceship 1's worldline having proper time $\tau_1$. Note, once again, that it is a *different* constraint from, for example, picking out the point on spaceship 1's worldline that is simultaneous, with respect to spaceship 2, with the point on spaceship 2's worldline having proper time $\tau_2$. It is also a different constraint from picking out points on the two spaceship worldlines that have the same proper time $\tau$, i.e., picking out two points such that $\tau_1 = \tau_2$.

I hope this removes any gaps about what P₁ and P₂ are - they are moving events on the worldlines.

Yes, that's clear, but note, once again, that your definitions are not the only possible ones. I gave two other possible ones above.

So, having picked P₁ we have no choice about where P₂ is.

Given your definitions, yes.

These are the 4D positions of the ships in S coordinates and the separation $x_2-x_1$ is increasing.

Given your definitions, yes.

The thing that does not increase is the distance between the trailing ship and the point where the leading ship would have been $\gamma\beta(x_{20}-x_{10})$ seconds earlier than $\tau_1$.

In the instantaneous rest frame of spaceship 1, yes.

This 'retarded length' is not a useful measurement and the fact that it does not change is not physically important.

I disagree. I gave a concrete physical realization of this "retarded length" and what the fact that it does not change means, physically, in my last post. Whether or not that measurement is "useful" or "physically important" depends on what you are trying to do with it; and the same is true for the measurement you are defining as the "separation". Your "separation" happens to be the one that is useful for explaining why the string breaks; but that is by no means the only possible physical use for a measurement.