Solving the Spaceship Paradox: A New Explanation

[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
$$\begin{align*} t_k&=\int_0^{\tau_k} \frac{dt}{d\tau}d\tau\\ x_k &= x_{k0}+\int_0^{\tau_k} \frac{dx}{d\tau}d\tau \end{align*}$$

 

[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.

 

[Mentz114]

This equation which I used above $\tau_2=\tau_1+\gamma\beta(x_{20}-x_{10})$ may not be right. But we can calculate $\tau_2$ given $\tau_1$.

You're correct in your interpretation of the equations for $t_k$ and $x_k$.

I have made the only additional constraint that agrees with what happens in the ship frames. Surely anything else is risky at the least.

I understand what you've said thus - that there are different definitions of separation in the S frame, none of which are incorrect. If this is the case then my definitions (which follow the physics in local frames)do not lead to an apparent paradox, but claiming that the retarded separation is the proper separation ( in the S frame) leads to a paradox.

I know which definitions I would choose in such a case.

 

[Mentz114]

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

Sorry, I missed this post. I think I've given it since.

 

[PeterDonis]

This equation which I used above $\tau_2=\tau_1+\gamma\beta(x_{20}-x_{10})$ may not be right. But we can calculate $\tau_2$ given $\tau_1$.

Once you've decided on a definition of simultaneity, yes (since that's basically what picking out a relationship between $\tau_1$ and $\tau_2$ amounts to). But there are at least three possible definitions of simultaneity: frame S, the instantaneous rest frame of spaceship 1, and the instantaneous rest frame of spaceship 2. Except at $t = 0$, all three are different, so they pick out three different relations between $\tau_1$ and $\tau_2$.

I have made the only additional constraint that agrees with what happens in the ship frames.

But it can't agree with both "ship frames" at once, because they are different (except at $t = 0$, as above). You appear to have picked the instantaneous rest frame of spaceship #1; but that means you have *not* picked the instantaneous rest frame of spaceship #2. You can't pick both at once.

Surely anything else is risky at the least.

Once again, it depends on what you are trying to do. The choice of which frame to use is a practical choice, not a choice that is determined by the laws of physics.

claiming that the retarded separation is the proper separation ( in the S frame) leads to a paradox.

What does "proper separation (in the S frame)" mean? Isn't "proper separation" supposed to specifically mean the separation in one of the ship's instantaneous rest frames?

In any case, I am certainly not trying to claim that what you are calling the "retarded separation" (and which I would just call "the distance between the ships with respect to observers at rest in frame S", since that's the obvious physical interpretation) is the same as the separation in either of the ships' instantaneous rest frames. Obviously it isn't.

 

[Mentz114]

In any case, I am certainly not trying to claim that what you are calling the "retarded separation" (and which I would just call "the distance between the ships with respect to observers at rest in frame S", since that's the obvious physical interpretation) is the same as the separation in either of the ships' instantaneous rest frames. Obviously it isn't.

I'm sorry if that sounded accusatory. I should have said 'If one claims ...'.

You are right, the radar distance between the ships measured from inertial stations does not change. But this is not the proper separation between the ships ( as you say) so the fact that it does not change does not in my understanding lead to a paradox.

I think we agree on the basic physics, but if it comes down to a choice of simultaneity conventions, or definitions of distance, then I choose the ones which do not lead to a paradox.

I don't expect you to agree with me, but this is a loophole I can use to defuse the paradox. That's enough for me.

 

[PeterDonis]

You are right, the radar distance between the ships measured from inertial stations does not change. But this is not the proper separation between the ships ( as you say) so the fact that it does not change does not in my understanding lead to a paradox.

Agreed.

I think we agree on the basic physics

Yes.

if it comes down to a choice of simultaneity conventions, or definitions of distance, then I choose the ones which do not lead to a paradox.

But none of them do. The "paradox" does not arise from choosing the "wrong" simultaneity convention or definition of distance, as though only one gave right answers. All of them give right answers, because right answers depend on invariants, and invariants are the same regardless of which frame you calculate them in. The "paradox" arises from failing to recognize this fact.

In this particular case, the "paradox" arises from failing to recognize that what happens to the string must depend on some invariant describing it, not on "distance" with respect to a particular frame. That invariant is the expansion scalar: the expansion scalar being positive is what causes the string to stretch and ultimately break. But you can calculate that invariant in *any* frame, and it will come out the same; it has to, because it's an invariant.

One may choose to "interpret" that calculation differently depending on which frame you do it in; for example:

* In frame S, one might interpret the positive expansion scalar as showing that the "unstressed length" of the string contracts (due to "length contraction") while its actual length stays the same;

* In the "rest frame" of one of the ships (basically Rindler coordinates in which one of the ships is at a constant spatial location), one might interpret the positive expansion scalar as showing that the actual length of the string increasing.

But these are matters of "interpretation", not physics. The physics is simply: expansion scalar positive -> string streches and ultimately breaks. The second interpretation above might be preferred (you appear to prefer it) because it makes it more "obvious" that the expansion scalar is positive, since you don't have to appeal to concepts like the "unstressed length" of the string which don't have an obvious physical meaning (although the FAQ entry on the Bell Spaceship Paradox suggests a physical meaning for it). But that's a matter of practicality, or perhaps pedagogy, not physics.

 

[Mentz114]

I disagree with almost everything you've said.

It is not the expansion scalar that causes the string to break. It is the increasing separation between the ships that breaks it. The expansion scalar tells us this is true in all frames.

This begs the question - why are we even asking 'why does the string break in the inertial frame ?' when we already know the answer ?

However, there is a more serious issue.

If a measurement is made which contradicts the expansion scalar then there must be something wrong with the measurement. If someone said to me 'OK, the spaceships are always separating, but what about this frame I just found where they aren't ?' I'd have to ask them to check their work. Is that what you are asking me to believe - that in S the spaceships are not separating, in direct contradiction to what you asserted earlier.

And in this case the measurement is wrong because I have shown (with equations even) that the separation between the ships increases in S coordinates, as required by the expansion scalar.

This is why I say that the paradox arises because the (contradictory) retarded measurement is accepted as a valid measurement of the separation.

Regarding the interpretations. What's to interpret ? The only way to interpret 'the elements of the congruence are all separating in all frames' is just that.

I don't see how any interpretation of expansion can ever give rise to contraction. Nor do I see why the 'unstressed length' should be subject to this contraction and not the gap between the ships ( as in the muon half-life explication).

If I've misunderstood what you're saying, then forgive me, I did try but it's pretty hand-wavey

 

[PeterDonis]

It is not the expansion scalar that causes the string to break. It is the increasing separation between the ships that breaks it. The expansion scalar tells us this is true in all frames.

This assumes a particular definition of "separation", which is *not* the definition that is natural to all frames. (For example, it's not the natural definition for frame S.) That's why I put things the other way around: the expansion scalar being positive *is* the invariant way of saying "the separation increases"; you need the expansion scalar to pick out the correct definition of "separation".

This begs the question - why are we even asking 'why does the string break in the inertial frame ?' when we already know the answer ?

Because to most people, the "separation" is not increasing in frame S, so saying "the increasing separation causes the string to break" doesn't make sense from the point of view of frame S.

If a measurement is made which contradicts the expansion scalar then there must be something wrong with the measurement. If someone said to me 'OK, the spaceships are always separating, but what about this frame I just found where they aren't ?' I'd have to ask them to check their work. Is that what you are asking me to believe - that in S the spaceships are not separating, in direct contradiction to what you asserted earlier.

No, I'm just saying that your definition of "separation" does not apply to all frames. That's why the "separation" does not increase in all frames. That's why you need an invariant, the expansion scalar, to pick out the definition of "separation" that *does* increase.

And in this case the measurement is wrong because I have shown (with equations even) that the separation between the ships increases in S coordinates, as required by the expansion scalar.

No, you haven't. You've shown that your particular definition of "separation" picks out a quantity that increases in S coordinates. But that definition is *not* the natural definition of "separation" for frame S.

Regarding the interpretations. What's to interpret ? The only way to interpret 'the elements of the congruence are all separating in all frames' is just that.

No, it isn't. If it were, there would not be so much confusion about the spaceship paradox. The confusion arises because most people do not use the word "separation" the way you are using it.

Once again, I repeat that this is all a question of words, not physics. We agree on the physics. The only thing we appear to disagree about is how the word "separation" should be used. I'm not saying your usage is wrong; I'm saying it's not the way most people use the term, so statements like "the separation is increasing in all frames" lead to confusion. Saying "the expansion scalar is positive" avoids the confusion because it points at an obvious invariant. One can then go on to say that "positive expansion scalar" corresponds to "separation increasing in the rest frame of either ship" if more explanation is needed.

 

[Mentz114]

Ok, I hear what you're saying. I don't agree with it .

One can then go on to say that "positive expansion scalar" corresponds to "separation increasing in the rest frame of either ship" if more explanation is needed.

It also means that there are no coordinates in which the separation is not increasing.

... That's why you need an invariant, the expansion scalar, to pick out the definition of "separation" that *does* increase.

I'm not sure I understand this 'picking'.

I still don't see any paradox to explain. Some good insights have emerged from this analysis so I'll leave you to have the last word.

Thanks for your help.

 

[PeterDonis]

It also means that there are no coordinates in which the separation is not increasing.

No, it means that in the statement you quoted, I was using a particular definition of "separation" which is not the only one that is possible or valid. According to the natural defintion of "separation" for frame S, the "separation" in frame S is not increasing.

(To clarify, when I said "separation increasing in the rest frame of either ship", I meant "separation as defined in the rest frame of either ship". I was not trying to say that that is the only possible definition of the word "separation".)

I'm not sure I understand this 'picking'.

I just mean that, since there are different possible definitions of the word "separation", in order to explain why the definition you are using is the "right" one for this problem, it helps, IMO, to point out that that definition is the one that has an easy, direct correspondence to the expansion scalar, which is the appropriate invariant.

I still don't see any paradox to explain.

I agree that there is no "paradox" (not even an apparent one) *if* you look at things the right way. But many people apparently don't look at things the right way, so they see an apparent paradox. The problem of how to get them to look at things the right way is a problem of pedagogy, not physics. One can disagree about which pedagogical methods work; but I don't think one can disagree that pedagogy is needed.

 

[Mentz114]

Thinking about radar distances to the spaceship measured from a chain of inertial observers, I'm now wavering about whether the result would be the 'retarded' distance or the other one. Earlier I said it was the former, but now I'm not so sure.

The signal sent from the inertial observer has to hit something physical to reflect and give a reading. But there isn't any actual thing at the $x$ that corresponds to the 'retarded' distance.

If this logic is correct then the spaceship scenario analysed in radar coordinates has no apparent contradictions or paradoxes. The spaceships will be separating for every observer.

This also supports the view that the 'retarded' distance is not a measure of any actual distance.

I'm interested to hear what you think if you're not heartily sick of this subject.

 

[PAllen]

Thinking about radar distances to the spaceship measured from a chain of inertial observers, I'm now wavering about whether the result would be the 'retarded' distance or the other one. Earlier I said it was the former, but now I'm not so sure.

The signal sent from the inertial observer has to hit something physical to reflect and give a reading. But there isn't any actual thing at the $x$ that corresponds to the 'retarded' distance.

If this logic is correct then the spaceship scenario analysed in radar coordinates has no apparent contradictions or paradoxes. The spaceships will be separating for every observer.

This also supports the view that the 'retarded' distance is not a measure of any actual distance.

I'm interested to hear what you think if you're not heartily sick of this subject.

In principle, there is something at what you call x. Just have a string of (mutually at rest, inertial) observers watching the ships go by. When the rear one goes right by one, they flip up a mirror. Meanwhile, every observer is also broadcasting their current clock reading with a unique source id also encoded. A series of these observers will also record the time the front ship passes, and the time they receive a reflected signal of theirs whose value is twice as far back as the time they recorded front ship pass. They then compute the separation as c times (time of reception of reflection minus time of font ship passing). Each such distance measuring observer communicates their findings to some lab. The lab finds they all measured the same radar distance between the ships.

[Or you could just imagine the rear ship is appropriately reflective]

 

[Mentz114]

In principle, there is something at what you call x.

Radar signals do not reflect off principles.

Going with the analysis in posts #113 and #116 the signal will reflect off a spaceship at a more advanced position than where you say it is being reflected. In this case an inertial observer in a CIRF of the first ship will measure the distance to the leading ship as increasing, i.e. a later measurement gives a greater distance than an earlier one.

This is because the ships *are* separating and radar only bounces off physical things.

The motions and positions of the ships are determined by the initial conditions, and not every member of the congruence is actually occupied by a ship.

 

[PAllen]

Radar signals do not reflect off principles.

Going with the analysis in posts #113 and #116 the signal will reflect off a spaceship at a more advanced position than where you say it is being reflected. In this case an inertial observer in a CIRF of the first ship will measure the distance to the leading ship as increasing, i.e. a later measurement gives a greater distance than an earlier one.

This is because the ships *are* separating and radar only bounces off physical things.

The motions and positions of the ships are determined by the initial conditions, and not every member of the congruence is actually occupied by a ship.

You put something physical there. Is that profoundly difficult?

All you need to verify what I described is a family of observers at rest in the 'starting inertial frame', and a central 'lab' also at mutual rest with these. You don't even need for these to have synchronized clocks only clocks that measure proper time passage. With a somewhat more complicated procedure than I described, you would need only one observer at rest in the 'starting inertial frame' to measure that the radar distance between the ships (per that observer) does not change, ever. Yet the string breaks.

 

[Mentz114]

The string breaks because the distances measured from S are 'retarded' and do not reflect what is actually happening. No problem there.

I was wrong in my previous post. I finally worked out that the radar distances do remain constant. The reason is that the inertial stations clocks started after the ships clocks by $\gamma\beta\ \delta x$. $\delta x$ being the distance between the measuring station and the ship initial $x$. So the distance measured is the shorter, retarded measurement.

Interesting that the 'retarded' measurement is because of relativity of simultaneity.

My 'wavering' about the radar distance was because I couldn't see why. Now I can see it clearly. Thanks for your responses.

 

[Mentz114]

With a somewhat more complicated procedure than I described, you would need only one observer at rest in the 'starting inertial frame' to measure that the radar distance between the ships (per that observer) does not change, ever.

(my italics)

After a lot of thought I'm not sure the italicised part what you say here is right. I think that there are two ways to make the radar measurement of the separation between two spaceships. One of them is designed to measure the (unchanging) rest separation, and the other gives a sensible result.

The first picture shows two measurements made on the spaceships from an inertial frame. The result will clearly increase with time.

The second picture shows two measurements, made from worldlines O1 and O2 respectively on the trailing and leading ship. Collating those measurements gives the rest-separation. Now, the same result we get from collating the two measurements can be got from O1, provided the measurements are not made at the same time. So the measurement P_0 to A will give the same conclusion as M_1 to C ( when collated with M_0 to B).

So, from O1's frame, there are two ways to measure the separation, one is bound (unwittingly) to give the rest separation, and the other shows an increasing separation.

My point is that the second procedure, where O1 sends carefully timed ranging pulses will always give the rest separation. So what use is it in probing the dynamic separation ?

I assert that the use of an array of inertial measurers is not the correct distance measurement here. It is like a toy thermometer that is stuck on one temperature.

The conventional way of measuring radar distance is correct here and conforms to the physics expressed by the invariants.

 

[PAllen]

After a lot of thought I'm not sure the italicised part what you say here is right. I think that there are two ways to make the radar measurement of the separation between two spaceships. One of them is designed to measure the (unchanging) rest separation, and the other gives a sensible result.

The first picture shows two measurements made on the spaceships from an inertial frame. The result will clearly increase with time.

No, the first picture is not the correct way to apply radar distance measurement. You need pairs of reflections from the measuring world line to each ship, such that each pair is centered on the same time of the measuring world line. This is what gives meaning to the two measurements giving a distance between the ships - you have two distances corresponding to the same time per the measuring world line. Then the difference between these two radar distances, each at the same radar simultaneity, is the radar distance between the ships per the give observer. You will then see that this distance remains constant.

I have no idea what your first picture is trying to show, but it is not radar measurements as universally used in literature.

As for the second picture, the second (green) world line plays no role. Here you have correctly drawn one pair of radar measurements good for the blue vertical world line. More pairs of these will show constant distance between the ships per the blue vertical world line observer. This is the only sense in which radar measurements are made - in astronomy, relativity, or whatever. The measurements must correspond in radar simultaneity for you to compute a distance between objects.

No matter how far the ships get from the blue world line, and how fast they are going relative to it, the blue world line - all by itself - correctly applying radar measurements - computes the same distance between the ships.

The conventional way of measuring radar distance is correct here and conforms to the physics expressed by the invariants.

You don't seem to understand radar measurements. You have it exactly backwards. The only way radar measurements are understood, gives for the starting inertial observer, constant separation between these ships. It doesn't correspond to an invariant because it is coordinate quantity - radar simultaneity and radar distance for an inertial observer in flat spacetime are exactly the same as standard Minkowski coordinates.

 

[Mentz114]

Measuring the expansion

No, the first picture is not the correct way to apply radar distance measurement.

I don't see anything wrong with it practically or in principle.

No matter how far the ships get from the blue world line, and how fast they are going relative to it, the blue world line - all by itself - correctly applying radar measurements - computes the same distance between the ships.

I was trying to show that. If the reflection events have the same $t$ value then the only possible result is the unchanging rest separation. Choosing this procedure guarantees the outcome.I did this calculation based on the attached picture.

In the picture the curved worldlines are in the congruence $\vec{u}=\sqrt{a^2t^2+1}\ \partial_t + at \ \partial_x$, from which the equations of motion are,
$$\begin{align} X_n(t) &=x_n - \frac{1}{a} + \frac{1}{a}\sqrt{a^2t^2+1}=x_n - \frac{1}{a} + \frac{1}{a}\gamma(t),\ \ n=1,2 \end{align}$$
where $x_n$ are the positions when $t=0$. The point $R=(\bar{x}_1, t_1)$ and $S=(\bar{x}_2, t_2)$. So using (1)
$$\begin{align} \bar{x}_2-\bar{x}_1 &= (x_2-x_1) +\frac{1}{a}\left[ \gamma(t_2)-\gamma(t_1)\right]\\ %\Rightarrow a(\bar{x}_2-\bar{x}_1) &= a(x_2-x_1) +\left[ \gamma(t_2)-\gamma(t_1)\right]\\ \Rightarrow a\delta \bar{x} &= a\delta x +\left[ \gamma(t_1+\delta t)-\gamma(t_1)\right] \end{align}$$
Dividing (3) by $\delta t$ and taking limits gives
$$\begin{align} \frac{d\bar{x}}{dt} &= \frac{dx}{dt} + \frac{1}{a}\frac{d\gamma}{dt}\\ &= 2\frac{at}{\sqrt{a^2t^2+1}}\\ &= \frac{2}{a}\theta \end{align}$$
Where $\theta = \partial_a u^a$ is the expansion scalar. ( Obviously the final equality only holds if $a\neq 0$).

So this method gives the physical result.

If the measurement is made using the other method it will alway give the rest separation, and so does not say anything about the later separations. It only tells us that the rest separation does not change, which follows from the deifinition of rest separation.

I don't see why the first method can be ruled out. It gives the correct result for measurements on inertial worldlines.

 

[PAllen]

Radar measurement has a standard definition. If you want to invent something else, call it something else. For me, your first picture in no way explained what the logic of your new procedure, and I still don't see any logic to it:

When things are in motion relative to a given observer, distance, unqualified by time or simultaneity has no meaning. Radar distances are coupled to the concept of radar simultaneity. Comparing non-simultaneous distances for different objects each moving relative to you makes no sense, and has nothing to do with the universally understood meaning of radar distances.

 

[Mentz114]

After more thinking ( so slow ) I think I should have said

\begin{align}
\frac{d\bar{x}}{dt} &= \frac{1}{a}\frac{d\gamma}{dt}\\
&= \frac{at}{\sqrt{a^2t^2+1}}\\
&= \frac{\theta }{a}
\end{align}
because $x_2-x_1$ is constant so there's no differential associated with it.

your first picture in no way explained what the logic of your new procedure, and I still don't see any logic to it

A signal is sent at event A and half reflected at R. The frst reflection arrives back at B. The second reflection is at S and arrives back at C.

The calculation is elementary.

There is one thing you don't seem to be getting ( or maybe I have it backwards)

1. Define rest length (or separation) as a measurement made simultaneously at each end of of said interval.

2. Set up radar measurement with equal $t$ reflection events ( ie simultaneous measurements)

3. Throw away the data because the answer is always the rest length ( distance).

I don't see how this experimental procedure can justify ' ... therefore in S the length/distance ... is constant', because it did not measure the separation, only the rest separation.

Can you give me a reference for the standard definition of radar distance and length measurements, please ?

 

[PAllen]

After more thinking ( so slow ) I think I should have said

\begin{align}
\frac{d\bar{x}}{dt} &= \frac{1}{a}\frac{d\gamma}{dt}\\
&= \frac{at}{\sqrt{a^2t^2+1}}\\
&= \frac{\theta }{a}
\end{align}
because $x_2-x_1$ is constant so there's no differential associated with it.


A signal is sent at event A and half reflected at R. The frst reflection arrives back at B. The second reflection is at S and arrives back at C.

The calculation is elementary.

There is one thing you don't seem to be getting ( or maybe I have it backwards)

1. Define rest length (or separation) as a measurement made simultaneously at each end of of said interval.

2. Set up radar measurement with equal $t$ reflection events ( ie simultaneous measurements)

3. Throw away the data because the answer is always the rest length ( distance).

I don't see how this experimental procedure can justify ' ... therefore in S the length/distance ... is constant', because it did not measure the separation, only the rest separation.

Can you give me a reference for the standard definition of radar distance and length measurements, please ?

The issue is which distances you consider simultaneous, and thus can subtract to get a length of a moving object. Clearly, you cannot subtract distance to one end of an object from distance to the other, if the distances are not simultaneous (by some criterion). The criterion for radar measurements is radar simultaneity. Two distant events are simultaneous for a give observer A making radar measurements if the proper time along A's world line of the midpoint of the signal round trips is the same. Thus if A sends a signal at t=3 (on their world line - no other clock is needed), gets it back at t=5; and there is another signal sent at t=2 and received t=6; then these measurements correspond to simultaneous distances, and the distance between the events is 2 -1 = 1, and this is simultaneous to t=4 on A's world line.

A common reference on this is:

http://arxiv.org/abs/gr-qc/0104077