Solving the Spaceship Paradox: A New Explanation

[Mentz114]

I'm sticking my neck out because I just worked this out and may regret this post later.

The spaceship paradox arises because the expansion scalar of the 'Bell' congruence is always positive, implying that the thread will break, even when the ships are not separating.

For this vector field $V=\gamma \partial_t + \gamma\beta \partial_x$ where $\beta$ is a function of $t$ I find

$\Theta=\frac{d\gamma}{dt}=\gamma^3\ B\,\left( \frac{dB}{d\,t}\right)$.

I suggest that there are three cases here, corresponding to

$\Theta<0,\ \Theta=0,\ \Theta>0$.

The 'Bell' congruence is $V$ with $\Theta > 0$, and the case where there is no separation is obviously $\Theta=0$. This ties in with a number of other calculations.

I rest my case.

 

[dauto]

I don't think you rested your case. In fact I don't think you explained your case very well. What in the world are you talking about?

 

[Mentz114]

I don't think you rested your case. In fact I don't think you explained your case very well. What in the world are you talking about?

From the Wiki page

In Bell's version of the thought experiment, three spaceships A, B and C are initially at rest in a common inertial reference frame, B and C being equidistant to A. Then a signal is sent from A to reach B and C simultaneously, causing B and C starting to accelerate (having been pre-programmed with identical acceleration profiles), while A stays at rest in its original reference frame. According to Bell, this implies that B and C (as seen in A's rest frame) "will have at every moment the same velocity, and so remain displaced one from the other by a fixed distance. Now, if a fragile thread is tied between B and C, it's not long enough anymore due to length contractions, thus it will break.

I'm saying that in the case described here, the thread does not break because the B and C are not moving apart.

The only proper justification I have been given for the thread breaking is that the expansion scalar $\partial_\mu V^\mu$ is always positive. I'm saying it can be zero so we cannot use the expansion scalar as a reason why the thread breaks.

Ergo, no paradox.

 

[PeterDonis]

the expansion scalar of the 'Bell' congruence is always positive, implying that the thread will break, even when the ships are not separating.

Huh? The positive expansion scalar is the invariant *definition* of the ships "separating". Any other sense of the word "separating" is frame-dependent.

I suggest that there are three cases here, corresponding to

$\Theta<0,\ \Theta=0,\ \Theta>0$.

The three cases correspond to three *different* congruences. The $\Theta > 0$ case is the Bell congruence. The $\Theta = 0$ case is just an ordinary inertial congruence (i.e., a family of inertial observers all at rest with respect to each other). I haven't seen any discussion in textbooks or articles of the $\Theta < 0$ case, but I think we can leave it out of discussion for now; see below.

(Note: I'm not sure that your $\Theta$ is the same as the expansion scalar; I'll have to check the math when I get a chance. However, I think the *sign* of $\Theta$ will be the same as the sign of the expansion scalar, which is sufficient for this discussion.)

I'm saying that in the case described here, the thread does not break because the B and C are not moving apart.

No, that's not correct; the Wiki page is describing the $\Theta > 0$ case. Both B and C are accelerating relative to a fixed inertial frame (the one in which A remains at rest), so $\gamma$ increases with coordinate time in that frame; i.e., $\Theta = d \gamma / dt > 0$. That means the expansion scalar is positive and the thread breaks.

To see how this is consistent with Bell's description, note the bolded phrase:

According to Bell, this implies that B and C (as seen in A's rest frame) "will have at every moment the same velocity, and so remain displaced one from the other by a fixed distance".

Distance relative to a fixed frame is, obviously, frame-dependent, i.e., it's not an invariant, so it's not a good way to describe the actual physics. The actual physics depends on the distance in the instantaneous rest frame of either B or C, and how that distance changes with proper time along either B or C's worldline. That is what the (invariant) expansion scalar captures.

I have a proposed FAQ on this that's been in the works for some time, which addresses the above; I'll work on getting it published and visible in this forum.

 

[Mentz114]

Thanks, Peter. I'll mull that over. In my original post I suggested it was three congruences and then deleted it.

I've done another calculation that has a nice result. To get rocket C's motion relative to B, I boosted a rest frame with $\beta_1$ ( for B) and boosted a rest frame by $\beta_2$ for C. To get C in B's frame I then boost both with $-\beta_1$, which gives for C

$V=\left( 1-\beta_1\,\beta_2\right) \,\gamma_1\,\gamma_2\partial_t+\left( \beta_2-\beta_1\right) \,\gamma_1\,\gamma_2\partial_x$.

Note that in the rest frame the velocity is $\left( \beta_2-\beta_1\right) /\left( 1-\beta_1\,\beta_2\right)$. The expansion scalar for the composite 4-velocity is ( writing $a_nt$ for $\beta_n$)

$\Theta = \nabla_\mu V^\mu = \frac{ t(a_2-a_1)^2 ( 1-a_1a_2t^2 )} {(1-a_1^2t^2)^{3/2}(1-a_2^2t^2)^{3/2} }$

I think this models the three spaceships and shows that the expansion scalar of the case $a_1=a_2$ is zero. However, $\Theta$ is never negative.

My point is that the string will not break when $a_1=a_2$.

I see that you agree with my view that frame-dependent effects cannot do work - like straining a material.

I'm satisfied that there is no paradox and no need to ascribe the cause of the breaking to length contraction..

 

[WannabeNewton]

I'm satisfied that there is no paradox and no need to ascribe the cause of the breaking to length contraction..

If so, describe how an observer in the inertial frame in which the rockets are accelerated simultaneously with the same proper acceleration would explain the non-vanishing expansion scalar.

You're still seriously misunderstanding the difference between frame-dependent explanations of the non-vanishing of an invariant and the frame-independent consequences of the non-vanishing of an invariant.

 

[PeterDonis]

writing $a_n t$ for $\beta_n$

No, this isn't right, at least not for a constant proper acceleration, which I assume is what you're trying to express here. See below.

My point is that the string will not break when $a_1=a_2$.

If $a_n$ is the proper acceleration of spaceship $n$, then this is not correct; $a_1 = a_2$ means equal proper acceleration, which means the expansion scalar is positive and the string will break.

I've looked at the long-winded derivation of the expansion scalar now, and I agree with your formula $\Theta = \nabla_a u^a$ (i.e., it matches what I get as the general formula for the expansion scalar--short derivation below). Note that, since we're working in an inertial frame, $\nabla_a = \partial_a$, so we just have $\Theta = \partial_a u^a$.

But writing this out, we get

$$
\Theta = \partial_a u^a = \partial_t u^t + \partial_x u^x = \partial_t \gamma + \partial_x \left( \gamma v \right)
$$

For the simplest case of constant proper acceleration (i.e., $a_1 = a_2$ in your notation), everything is a function of $t$ only, so we have

$$
\Theta = \partial_t \gamma = \partial_t \left( 1 - v^2 \right)^{-1/2} = - \frac{1}{2} \left( 1 - v^2 \right)^{-3/2} \left( - 2 v \right) \partial_t v = \gamma^3 v \partial_t v
$$

For constant proper acceleration, we have $\partial v / \partial \tau$ constant (where $\tau$ is the proper time along a given worldline in the congruence); but $\partial v / \partial \tau = \left( \partial t / \partial \tau \right) \partial v / \partial t = \gamma \partial v / \partial t$, so we have $\partial_t v = \gamma^{-1} a$ if $a$ is the constant proper acceleration. (Note, btw, that we are assuming here that both $a$ and $v$ are positive--we are starting from rest and accelerating in the positive $x$ direction. A more sophisticated analysis would account for the other possibilities for the relative signs, to show that $\Theta$ always comes out positive, but I won't go into that detail here.) So we have

$$
\Theta = \gamma^3 v \gamma^{-1} a = \gamma^2 v a
$$

I haven't tried to solve the more general case of letting the proper acceleration vary from worldline to worldline; this would take some time for me to model because in the inertial frame, the proper acceleration itself must be a function of both $t$ and $x$, with the constraint that it must be constant along a given worldline, i.e., $u^a \nabla_a a = 0$ if $a$ is the proper acceleration. But the above is sufficient to show that, for the case of constant proper acceleration (your $a_1 = a_2$), the expansion is positive.

Short derivation of the expansion scalar: for a timelike congruence defined by a vector field $u^a$, the expansion scalar is the trace of the expansion tensor $\theta_{ab}$, which is given by

$$
\theta_{ab} = h^m{}_a h^n{}_b \frac{1}{2} \left( \nabla_n u_m + \nabla_m u_n \right)
$$

where $h_{ab} = g_{ab} + u_a u_b$ is the projection tensor orthogonal to $u_a$. The expansion scalar is just the trace of this tensor, which is

$$
\theta = \theta^a{}_a = h^{ma} h^n{}_a \frac{1}{2} \left( \nabla_n u_m + \nabla_m u_n \right) = \frac{1}{2}\left( g^{ma} + u^m u^a \right) \left( g^n{}_a + u^n u_a \right) \left( \nabla_n u_m + \nabla_m u_n \right)
$$

$$
\ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2} \left( g^{mn} + u^m u^n \right) \left( \nabla_n u_m + \nabla_m u_n \right) = \nabla_m u^m + u^m u^n \nabla_n u_m = \nabla_m u^m
$$

where in the last equality we have used the fact that the 4-acceleration $u^a \nabla_a u_b$ is orthogonal to the 4-velocity.

I see that you agree with my view that frame-dependent effects cannot do work - like straining a material.

Not just in virtue of being frame-dependent effects, no. Any real work done must always correspond to some invariant that is not frame-dependent (such as the expansion scalar).

However, it seems like a lot of people use frame-dependent effects to formulate a physical "interpretation" of what is going on. Bell's discussion of the spaceship paradox, where he says that "length contraction" is what causes the string to break, is an example. The FAQ entry I mentioned in my previous post addresses that point (in fact a previous thread in which the point came up is what prompted me to write the FAQ entry).

 

[Mentz114]

If so, describe how an observer in the inertial frame in which the rockets are accelerated simultaneously with the same proper acceleration would explain the non-vanishing expansion scalar.

The expansion scalar vanishes with equal acceleration as I show in my previous post.

You're still seriously misunderstanding the difference between frame-dependent explanations of the non-vanishing of an invariant and the frame-independent consequences of the non-vanishing of an invariant.

The invariant vanishes.
My calculation is covariant and the final result is a scalar. The way I set up the frames ensures that B and C have the same proper acceleration if $a_1=a_2$.

 

[WannabeNewton]

The expansion scalar vanishes with equal acceleration as I show in my previous post.

It doesn't vanish. If the proper accelerations of the spaceships are equal and simultaneous in the inertial frame then the string will break meaning the expansion scalar must be positive. The observer in the inertial frame attributes this to the string resisting length contraction, simple as that.

 

[Mentz114]

No, this isn't right, at least not for a constant proper acceleration, which I assume is what you're trying to express here. See below.



If $a_n$ is the proper acceleration of spaceship $n$, then this is not correct; $a_1 = a_2$ means equal proper acceleration, which means the expansion scalar is positive and the string will break.

I've looked at the long-winded derivation of the expansion scalar now, and I agree with your formula $\Theta = \nabla_a u^a$
..
..
$$
\Theta = \partial_t \gamma = \partial_t \left( 1 - v^2 \right)^{-1/2} = - \frac{1}{2} \left( 1 - v^2 \right)^{-3/2} \left( - 2 v \right) \partial_t v = \gamma^3 v \partial_t v
$$
..
..
$$
\Theta = \gamma^3 v \gamma^{-1} a = \gamma^2 v a
$$
..
..
However, it seems like a lot of people use frame-dependent effects to formulate a physical "interpretation" of what is going on. Bell's discussion of the spaceship paradox, where he says that "length contraction" is what causes the string to break, is an example. The FAQ entry I mentioned in my previous post addresses that point (in fact a previous thread in which the point came up is what prompted me to write the FAQ entry).

Thanks for that. We seem to be in agreement with the numbers I left in the quote. The way I'm thinking now is that the paradox arises is because the comoving ship frames see the ships B,C separating while A sees them comoving. I have a problem with this.

This $\bar{\gamma}=-U^\mu V_\mu$ is an invariant in SR and has the value 1 if U and V are comoving. Under the initial conditions specified in the Wiki quote of Bell's position, $\bar{\gamma}$ will be 1, since B,C have the same worldline. So this will still be 1 if we transform to either B or C's frame. I could be missing something here but I don't see *how* the ship observer can think the other ship is separating and A thinks they are comoving.

The expansion scalar vanishes with equal acceleration as I show in my previous post.

It doesn't vanish. If the proper accelerations of the spaceships are equal and simultaneous in the inertial frame then the string will break meaning the expansion scalar must be positive. The observer in the inertial frame attributes this to the string resisting length contraction, simple as that.

I don't find it simple. At least the expansion scalar is not frame dependent, but I still think the thread breaks only if the ships separate in all frames.

Anyhow, I'm still mulling all this. Thanks for the input.

 

[WannabeNewton]

I have a problem with this.

Motion and (spatial) distance are relative so I'm not seeing why exactly it's bothering you. Take a rod and Born rigidly accelerate it along its length. Then in the rest frame of any point of the rod the distance to neighboring points is always constant but in the inertial frame through which the rod accelerates the distance between any two points is always decreasing due to "length contraction" (I put it in quotes because it differs from the usual gamma factor length contraction). Is this also troubling to you?

 

[PeterDonis]

The way I'm thinking now is that the paradox arises is because the comoving ship frames see the ships B,C separating while A sees them comoving.

With this interpretation of "comoving", "comoving" is frame-dependent. You have to find a sense of "comoving" which is invariant in order to use it in a physical argument, since we've already agreed that only invariants can be so used. The usual invariant sense of "comoving" is that the congruence is rigid, which means both the expansion and the shear are zero. But the expansion is not zero for the Bell congruence, as I've already shown.

$\bar{\gamma}=-U^\mu V_\mu$

That's not what I was using in my calculation; in my calculation, $\gamma (t) = 1 / \sqrt{ 1 - v(t)^2 }$, where $v(t)$ is the velocity of either spaceship, B or C, in A's rest frame, as a function of coordinate time in that frame. By hypothesis, $v$ is a function of $t$ only *if* we do the calculation in A's frame; but in any other frame, $v$ will be a function of $x'$ as well as $t'$ (I'm using primes for the coordinates in the other frame). So the analysis becomes more complex in any frame other than A's rest frame.

Also, this definition of $\bar{\gamma}$ is not actually invariant, because it is taking the inner product of vectors at different events. See below.

Under the initial conditions specified in the Wiki quote of Bell's position, $\bar{\gamma}$ will be 1, since B,C have the same worldline.

No, they don't have "the same worldline". They have worldlines which have the same path curvature (i.e., the same proper acceleration), but the worldlines are spatially separated, so they're not "the same". This is a critical fact that you have left out of your analysis of what is "invariant".

Your definition of $\bar{\gamma}$ above implicitly evaluates $U^a$ at some time $t$ on B's worldline, and $V^a$ at the same time $t$ on C's worldline. But "at the same time" is frame-dependent. In A's rest frame, at any time $t$, we will indeed have $U^a = V^a$. But if you evaluate $U^a$ and $V^a$ at the same time $t'$ in any other frame, they will *not* be equal; doing that is equivalent to evaluating $U^a$ at some time $t_B$ in A's rest frame and $V^a$ at some different time $t_C$ in A's rest frame, because B and C are spatially separated so relativity of simultaneity comes into play. So in any frame other than A's rest frame, $\bar{\gamma} \neq 1$, and therefore $\bar{\gamma}$ is not an invariant.

 

[Mentz114]

PD and WNB, thanks to both of you for the posts. I have to admit that there may be no invariant definition of relative velocity and that relative velocity ( like the spatial components of a 4-velocity) transforms like a tensor component. In which case the conclusions I thought I drew are not true. But in spite of that I'm still agnostic about '... attributes this to the string resisting length contraction'. I'm also unsure about the causality.

It's necessary to work out how relative velocity transforms. I feel a calculation coming on and I'll be back.

 

[WannabeNewton]

PBut in spite of that I'm still agnostic about '... attributes this to the string resisting length contraction'.

Why? You can't just reject the explanation because it doesn't appeal to your intuitions. It's not that hard to see intuitively how the Lorentz contraction of the string in the inertial frame contributes to the macroscopic and microscopic dynamics in this frame that causes the string to break.

 

[Mentz114]

Why? You can't just reject the explanation because it doesn't appeal to your intuitions. It's not that hard to see intuitively how the Lorentz contraction of the string in the inertial frame contributes to the macroscopic and microscopic dynamics in this frame that causes the string to break.

My current line is to use the equivalence principle. If we dangle a thread in a sufficiently strong gravitational field it will break under its own weight. I believe what the inertial observer sees is the same, with the acceleration gradient playing the part of the gravity. This means the thread breaks because of its inertia and the differential acceleration. Nothing to do ( directly) with time dilation. I'm trying to put some equations together and reasonably optimistic.

 

[pervect]

PD and WNB, thanks to both of you for the posts. I have to admit that there may be no invariant definition of relative velocity and that relative velocity ( like the spatial components of a 4-velocity) transforms like a tensor component. In which case the conclusions I thought I drew are not true. But in spite of that I'm still agnostic about '... attributes this to the string resisting length contraction'. I'm also unsure about the causality.

It's necessary to work out how relative velocity transforms. I feel a calculation coming on and I'll be back.

Parallel transport of a velocity vector is coordinate independent, and in a flat space-time gives the right result for relative velocity (when you compare the parallel-transported vectors).

Unfortunately, it's path dependent in curved space-time, so you aren't guaranteed a unique answer unless you specify a unique path.

For the purposes of Bell's spaceship, I've found that an adequate substitute for "relatively at rest" is having a constant two-way propagation delay for light signals. A static metric is sufficient to cause the two-way light propagation to be independent of time.

So in coordinate dependent terms, if objects have constant spatial coordinates, and none of the metric coefficients is a function of time, we can say they are at rest.

I'm not sure how to express this in coordinate independent language. Something along the lines of all objects whose 4-velocity is orthogonal to a time-like Killing vector field are at rest, I think.

 

[Mentz114]

Parallel transport of a velocity vector is coordinate independent, and in a flat space-time gives the right result for relative velocity (when you compare the parallel-transported vectors).
Unfortunately, it's path dependent in curved space-time, so you aren't guaranteed a unique answer unless you specify a unique path.

This is why I expected $U^\mu V_\mu$ to be a scalar in flat spacetime. I know in curved space it is only a 'local' tensor in a vicinity of a point of coincidence of U and V. But I think PeterDonis revoked my expectation.

Your remarks are apposite, thanks.

 

[PeterDonis]

This is why I expected $U^\mu V_\mu$ to be a scalar in flat spacetime.

Parallel transport between two fixed events in flat spacetime is path-independent, yes. But changing frames changes which two fixed events you are parallel transporting between to compare the vectors $U$ and $V$ (because it changes the surfaces of simultaneity, which define at what points on the worldlines of B and C you evaluate $U$ and $V$). So changing frames changes $U^\mu V_\mu$ because it changes the endpoints, not because it changes the path taken between the same endpoints (the latter would indeed not change $U^\mu V_\mu$ in flat spacetime).

 

[PeterDonis]

Something along the lines of all objects whose 4-velocity is orthogonal to a time-like Killing vector field are at rest, I think.

This defines "at rest" in an invariant way, yes, but it has to be the *same* timelike KVF for all the objects. In Minkowski spacetime, there are two infinite families of timelike KVFs--one for each possible inertial frame, and one for each possible Rindler coordinate chart. So, for example, two inertial observers in relative motion are both following orbits of a timelike KVF, but it's a different KVF for each of them, so they're not both at rest relative to the same definition of "at rest".

Similarly, the two spaceships in the Bell Spaceship Paradox are both following orbits of a timelike KVF, but it's a different timelike KVF for each of them (this time because they are following orbits of different "Rindler" KVFs, i.e., worldlines of constant proper acceleration that asymptote to different Rindler horizons).

 

[WannabeNewton]

If we dangle a thread in a sufficiently strong gravitational field it will break under its own weight.

This is because the tension in the string due to your grip on the near end of the string isn't enough to balance the gravitational force acting on each infinitesimal element of the string.

In other words the string is brought beyond its equilibrium (natural) length to the point of overwhelming elastic stresses.

In the same spirit, in the inertial frame of the Bell spaceship paradox setup there is an interplay between length contraction and equilibrium length when the string is fastened between the spaceships maintaining constant distance between them in said inertial frame.

I'm trying to put some equations together and reasonably optimistic.

It's definitely instructive to try and mess around with this stuff on your own so go for it but I guarantee you that there won't be any way to avoid length contraction when explaining why the string breaks in the inertial frame. If you want a more detailed physical explanation then feel free to ask.

 

[georgir]

On a somewhat possibly related note, what are the correct acceleration profiles that the ships must follow in order for the string to not break and the distance between them to appear constant in their own rest frames? Is that even possible for both ships at once?
On a hunch I'd say if it is possible, then the most likely candidate is a setup where the front ship (according the direction of their acceleration) starts when a signal that the back ship sent when it started reaches it. But I'm at work now and would attract too much weird looks if I started to try to calculate this :(

 

[Mentz114]

This is because the tension in the string due to your grip on the near end of the string isn't enough to balance the gravitational force acting on each infinitesimal element of the string.

In other words the string is brought beyond its equilibrium (natural) length to the point of overwhelming elastic stresses.

Yep, the string breaks.

In the same spirit, in the inertial frame of the Bell spaceship paradox setup there is an interplay between length contraction and equilibrium length when the string is fastened between the spaceships maintaining constant distance between them in said inertial frame.

The first quote above means that it makes no difference if the string is tethered to the trailing ship or not, it will still break. The string will experience tension as if in a gravitational field which gets stronger ( linearly in my first estimation) with time, and so must break.

Now the inertial observer who sees the ships maintaining constant separation can ascribe the string breaking to this inertial lag force. The ship observer can ascribe a string breaking to the increasing separation or the inertal lag or both. This probably depends on the composition of the string.

I find this plausible, since it is expressed in Newtonian terms and fairly intuitive.

I have equations but I've run out of time for some hours and they are not done yet.

 

[georgir]

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.

 

[WannabeNewton]

On a somewhat possibly related note, what are the correct acceleration profiles that the ships must follow in order for the string to not break and the distance between them to appear constant in their own rest frames?

Hi georgir. We call this Born rigidity. If we have a line of spaceships and accelerate them simultaneously longitudinally and want the distances between them to remain constant in all of their rest frames then it turns out we have to accelerate them so that in the rest frame of any of the ships, the 4-velocity field of the line of ships is given by $u = \frac{1}{x}(1,0,0,0)$ where $x$ is the location of each ship in this rest frame. The 4-acceleration is then $a = \frac{1}{x}(0,1,0,0)$and so each ship in fact has a different acceleration which in this rest frame varies with the constant spatial position of each ship. As an side, there's a precise mathematical definition of Born rigidity that I will omit, unless you want to see it.

In the inertial frame in which all the ships were simultaneously accelerated each ship will have a different velocity varying in accordance with the different accelerations of each ship and the line of ships as a whole will length contract continuously in time relative to this frame. The fact that the line is length contracting in the inertial frame is, in an intuitive sense, equivalent to demanding Born rigidity because the resistance to length contraction is what led to for example the string breaking in the inertial frame of the Ball spaceship paradox.

The relationship between $x$ for each spaceship and the coordinates $(X,T)$ of the inertial frame is as it turns out $x^2 = X^2 - T^2$. Note then that $\frac{dX}{dT} = \frac{T}{\sqrt{T^2 + x^2}}$ meaning that the 3-velocity of each spaceship in the inertial frame varies with its $x$ value and similarly so does $\frac{d^2 X}{dT^2} = \frac{1}{ \sqrt{T^2 + x^2}}- \frac{T^2}{(T^2 + x^2)^{-3/2}}$. The length contraction formula between the frontmost spaceship and rearmost spaceship is also easy to calculate but rather messy to write down-just subtract the positions of said spaceships in the inertial frame.

EDIT: see here for more details: http://www.mathpages.com/home/kmath422/kmath422.htm

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.

Thanks! You saved me some typing there you did :)

 

[PeterDonis]

The first quote above means that it makes no difference if the string is tethered to the trailing ship or not, it will still break. The string will experience tension as if in a gravitational field which gets stronger ( linearly in my first estimation) with time

No, it won't; it will be constant with time, as georgir said. This is because only one end of the string has a motion that is constrained by being attached to something else; the other end can move freely (as can all the pieces of the string in between). That changes the congruence of worldlines that the individual pieces of the string follow: they follow a congruence with zero expansion in this case. It is still *possible* for the string to break in this case, but only if the proper acceleration of the ship that is pulling on the string is large enough that the string's weight exceeds its tensile strength.

(Technically, there is another condition as well: the string must be short enough that its trailing end is less than a distance $1 / a$ from the ship, in the ship's instantaneous rest frame, where $a$ is the ship's proper acceleration. If the trailing end is further from the ship than that, the string will break because its trailing end would have to move faster than light to keep up. I've been assuming in this entire thread that we are ruling out this kind of thing.)

 

[Mentz114]

Addressing the time dependence issue - how do you model the constant acceleration ? If a rest frame is boosted by $\beta$ then the proper acceleration in the boosted frame is $\dot{\beta}\gamma^3$ which is time dependent. The expansion coefficient is $\partial_t(\gamma)$. This obviously misleads me about the time dependence.

Peter, the length issue could be a cruncher, so it could be back to the pencil and note book for me.

Aside : This chart of the Minkowski metric $ds^2= -x^2dt^2+t^2dx^2+dy^2+dz^2$ gives an acceleration of $1/x$ and expansion scalar $1/xt$. Does this mean that two tethered comoving observers will see the tether get tighter ? Probably doesn't help my case.

[edit] corrected a typo in expansion scalar.

 

[PeterDonis]

Addressing the time dependence issue - how do you model the constant acceleration ?

Constant path curvature of the worldline; i.e., the 4-acceleration $a = u^a \partial_a u^b$ is constant. In the case of the spaceship paradox, $a$ is furthermore the same for *all* the worldlines (both spaceships, and all of the pieces of the string in between).

If a rest frame is boosted by $\beta$ then the proper acceleration in the boosted frame is $\dot{\beta}\gamma^3$ which is time dependent. The expansion coefficient is $\partial_t(\gamma\beta)$. This obviously misleads me about the time dependence.

That's because "time" is frame-dependent. Which means that what looks like "time dependence" in one frame becomes "time and space dependence" in another frame, even when the invariant path curvature is constant! This is a good illustration of how trying to think of things in terms of frames instead of invariants can cause confusion.

In the original rest frame (A's rest frame), $\beta$ is a function of time (i.e., coordinate time $t$), but not space (i.e., $\beta$ is not a function of $x$). However, in any other inertial frame, $\beta$ will be function of both $t'$ and $x'$. The reason A's original rest frame is different in this respect is that in that frame, both spaceships start accelerating (i.e., firing their rockets) simultaneously (at time $t = 0$). In any other frame, because of relativity of simultaneity, the ships start accelerating at different times.

 

[Mentz114]

... $a$ is furthermore the same for *all* the worldlines (both spaceships, and all of the pieces of the string in between).

Are you sure ? The spaceships have synchronised starts but the strings are being towed so they have a lag.

That's because "time" is frame-dependent. Which means that what looks like "time dependence" in one frame becomes "time and space dependence" in another frame, even when the invariant path curvature is constant!

OK.

This is a good illustration of how trying to think of things in terms of frames instead of invariants can cause confusion.

Preachy. I've been trying to stick to invariants and covariant calculations throughout.

In the original rest frame (A's rest frame), $\beta$ is a function of time (i.e., coordinate time $t$), but not space (i.e., $\beta$ is not a function of $x$). However, in any other inertial frame, $\beta$ will be function of both $t'$ and $x'$. The reason A's original rest frame is different in this respect is that in that frame, both spaceships start accelerating (i.e., firing their rockets) simultaneously (at time $t = 0$). In any other frame, because of relativity of simultaneity, the ships start accelerating at different times.

Well, a proper model will take all this into account.

 

[PeterDonis]

The spaceships have synchronised starts but the strings are being towed so they have a lag.

Strictly speaking, yes, all the pieces of the string won't start moving at the same time (in A's original rest frame) as the ships do. The assumption of constant proper acceleration for all the pieces of the string is an idealization. However, dropping the idealization doesn't change the conclusion: it just makes the math more complicated (because you have to show that the expansion scalar I computed is the invariant that describes the *average* motion of the string).

Preachy. I've been trying to stick to invariants and covariant calculations throughout.

I didn't mean to be preachy; but your question about time dependence that I was responding to doesn't make sense if you're only looking at invariant/covariant quantities, since the "time dependence" you were describing is dependence on the coordinate time in a particular frame. The only invariant/covariant time dependence in the problem is dependence on proper time along one of the given worldlines (the worldline of one of the spaceships or a piece of the string).

 

[WannabeNewton]

Peter, the length issue could be a cruncher, so it could be back to the pencil and note book for me.

You keep trying to do calculations but you aren't thinking about the physics. Put down the pencil and note and set aside the blind calculations. Just think about what's going on conceptually in the inertial frame. It's really simple. Length contraction causes the equilibrium (natural) length of the string to continuously decrease in the inertial frame but the length of the string itself in the inertial frame is held constant by being fastened between the two instantaneously equal velocity spaceships. Clearly there comes a point when the stresses are too much to sustain the difference between fixed length in the inertial frame and equilibrium length in the inertial frame and the string breaks.

 

[Mentz114]

You keep trying to do calculations but you aren't thinking about the physics. Put down the pencil and note and set aside the blind calculations. Just think about what's going on conceptually in the inertial frame. It's really simple. Length contraction causes the equilibrium (natural) length of the string to continuously decrease in the inertial frame but the length of the string itself in the inertial frame is held constant by being fastened between the two instantaneously equal velocity spaceships. Clearly there comes a point when the stresses are too much to sustain the difference between fixed length in the inertial frame and equilibrium length in the inertial frame and the string breaks.

I only believe fully covariant calculations. I would prefer being guided to the right way to do the calculation myself.

If I'm not able to play the acceleration gradient card, then there must be a kinematic explanation, viz. one which deends only on the velocities, not accelerations. Perhaps I will end up in the right place.

 

[PAllen]

I only believe fully covariant calculations. I would prefer being guided to the right way to do the calculation myself.

If I'm not able to play the acceleration gradient card, then there must be a kinematic explanation, viz. one which deends only on the velocities, not accelerations. Perhaps I will end up in the right place.

It seems to me that you have one covariant explanation providing minimum 'explanation', and a choice of additional frame dependent explanations - each providing more 'motivation' but at the cost of being frame dependent.

1) The covariant description is simply that by virtue of how the congruence is specified, its expansion tensor is nonzero, thus the string is under growing tension or separation. The force causing this (assuming the string is passive) is the rockets pulling on it. Nothing can be said about distances because that requires specification of simultaneity. Nothing can be said about why the expansion tensor is nonzero except: that is how the congruence was defined. Different congruence, representing a different physical set up, would have no expansion.

2) In (any) inertial frame, the distance between the rockets is constant (once they are both accelerating), and the additional frame dependent explanation is that the reason the rockets cause tension and a nonzero expansion tensor is that the equilibrium length of the string contracts, but the rockets are forcing it to stay the same length.

3) In either rocket frame, realized, for example, as Fermi-Normal coordinates, the additional explanation is that the distance between the rockets grows, causing the tension and expansion of the string.

This is similar to any number of other situations where there is an invariant/covariant description, and multiple frame dependent explanations.

 

[Mentz114]

I have a calculation which may do the trick.

The expansion tensor is defined as $\theta_{\mu\nu}= \Theta h_{\mu\nu}$ where $h=g_{\mu\nu}+U_\mu U_\nu$. For $U_\mu=-\sinh(at) dt + \cosh(at) dx$ I get the components of $\theta_{\mu\nu}$

$\left[\begin{array}{cccc} a\,cosh\left( a\,t\right) \,\left( {sinh\left( a\,t\right) }^{2}-1\right) & -a\,{cosh\left( a\,t\right) }^{2}\,sinh\left( a\,t\right) & 0 & 0\\\ -a\,{cosh\left( a\,t\right) }^{2}\,sinh\left( a\,t\right) & a\,cosh\left( a\,t\right) \,\left( {cosh\left( a\,t\right) }^{2}+1\right) & 0 & 0\\\ 0 & 0 & a\,cosh\left( a\,t\right) & 0\\\ 0 & 0 & 0 & a\,cosh\left( a\,t\right) \end{array}\right]$

and if we boost this tensor by $\beta=at$ the components are.

$\left[\begin{array}{cccc} -a\,cosh\left( a\,t\right) & 0 & 0 & 0\\\ 0 & 2\,a\,cosh\left( a\,t\right) & 0 & 0\\\ 0 & 0 & a\,cosh\left( a\,t\right) & 0\\\ 0 & 0 & 0 & a\,cosh\left( a\,t\right) \end{array}\right]$

I think the answer is right there in the way those components change between the two frames. The $\theta_{tx}$ is crucial.

Sorry about the matrix explosion, it's hot off the presses.

 

[WannabeNewton]

This is similar to any number of other situations where there is an invariant/covariant description, and multiple frame dependent explanations.

Which is exactly what I've been trying to say over and over again in this thread. And yet there has been no progress since the first page of this thread because blind calculations are being used to substitute for conceptual understanding, to no avail unsurprisingly.

I feel that some advice from Feynman is in order:

https://www.youtube.com/watch?v=obCjODeoLVw

tl;dr no calculation you do is useful if you don't know how to actually interpret it in terms of physics.

 

[Mentz114]

tl;dr no calculation you do is useful if you don't know how to actually interpret it in terms of physics.

Please stop the anodyne advice and disparaging remarks about my lack of understanding of physics.

The calculation in my last post explains the situation exactly without videos or handwaving. The expansion tensor shows how time enters the expansion in the inertial frame. Do you understand what that calculation is saying ? I do.