# Shape and even connectedness of accelerating components in SR is frame dependent?

In some inertial frame consider a disk of radius 1 lightyear at rest. Then along the edge of the disk there are some people in spacesuits at rest hovering right above the disk (which has negligible mass I really want it just to paint the scene).

There is a light source at rest in the center with a timer to turn it on and then off so that it will only ever be on for one single minute. When the people see the light they fire thrusters to get up to speed and move in a circle and then after that they fire their thrusters just to maintain uniform circular motion (again as observed by the frame of the disk). And then they hold hands.

When they stop seeing the light they stop holds hands but keep firing their tbrusters the same way. And then after they stop holding hands they fire their thrusters differently to get back to rest relative to the disk.

In the frame of the disk there was a connected rotating object of humans in spacesuits. But since they held hands for less than a minute and the disk is one light year apart (in the frame of the disk) the events of one person holding hands is spacelike separated from all the events where the opposite person held hands. So there are frames where all of the one events happen before all of the other events. And frames where all of the other events happen before all of the one events.

And in fact, if the speed they get up to is high enough, the instantaneously comoving frame of the person originally halfway between them is such a frame.

So whether there ever simultaneously existed a connected object consisting of mutually hand holding people is frame dependent.

There is a frame where this hand holding set of events is connected. And other frames (the frames that are instantaneously comoving with the parts) where the parts are never part of a single object.

I know that an object can be Born rigid if there is a frame where its components are at rest, e.g. from wikipedia

The defining property of Born rigidity is locally constant distance in the co-moving frame for all points of the body in question.

But this collection of components (people) do not have a common comoving inertial frame. An instantaneously comoving frame for one person is never an instantaneously comoving frame for the person initially at the opposite side for instance.

So there is no Born rigid object (except the disk which isn't involved, and the light source and the people but realistically every classical thing has parts that are moving so aren't Born rigid for the same reason, no comoving frame in common).

I fully understand that in one frame if something looks rigid we can call it Born rigid to accept that it might not look rigid to another frame. But that is only reasonable if there is a frame where the object has its parts all at rest. So this example isn't some connected object that is at rest in some frame, so we can't invariantly refer to the frame in which it is at rest because there is no such frame. And every single frame that has a part that is instantaneously at rest sees a component that is disconnected from the rest. Pun (about components) not intended.

So is being connected a frame dependent idea in Special Relativity given that the any instantaneously comoving frame of a component will observe itself as disconnected from other components?

And even if a frame sees some components connected to others, won't it have different shapes based on the spatial geometry of the connected components in the plane of simultaneity of the frame? Specifically that the object will be shorter in the direction the frame is moving relative to the disk compared to directions orthogonal to that motion? Again there isn't a common rest frame for the components to refer to.

When the people see the light they fire thrusters to get up to speed and move in a circle and then after that they fire their thrusters just to maintain uniform circular motion...

OK, we have a disk, and light, and a rotating ring of people. No problem.

So there are frames where all of the one events happen before all of the other events. And frames where all of the other events happen before all of the one events.

These frames are abstract things. They don't actually exist. What does exist is the disk and the light and the people. The light came on, the light moved outwards in all directions, all the people saw it, and they all fired their thrusters and held hands. Some observer might claim that some event preceded another, but his state of motion doesn't change what happens on the disk.

So whether there ever simultaneously existed a connected object consisting of mutually hand holding people is frame dependent.

No it isn't. A frame is little more than a state of motion. When you change your state of motion you change. You don't change the disk or the light or the ring of people.

There is a frame where this hand holding set of events is connected.

The hand-holding event is where the people are connected. Stand in the middle of the disk and you see the light moving outwards and all the people firing their thrusters and holding hands. But you won't see a frame.

The defining property of Born rigidity is locally constant distance in the co-moving frame for all points of the body in question.

I'm not sure it's relevant Timaeus.

So there is no Born rigid object (except the disk which isn't involved

Agreed. You could stage this scenario without the disk.

But that is only reasonable if there is a frame where the object has its parts all at rest.

There is no frame in any objective sense. There's a disk, light, and people, all in motion, in space.

So is being connected a frame dependent idea in Special Relativity?

No. Your relative motion alters how you see it, but it doesn't alter it.

And even if a frame sees some components connected to others, won't it have different shapes...? Specifically that the object will be shorter...

You might see a star looking flattened, but that star didn't change one iota when you fired your thrusters. Instead, your state of motion changed, and so did you.

• A nontraversible wormhole sounds like something people want to say is connected because there is a spacelike surface connecting the end. See my analogy in physics.stackexchange.com/a/201775 so I think it is pedagogically relevant to discuss how people ideas of connectedness are related to frames and spacelike surface rather than traveling to and from. So intuitions about connectivity are like intuitions about simultaneity they pedagogically have to be addressed even of just as you advocate it is to say not to do it at all. It doesn't generalize to relativity is the point. Aug 21, 2015 at 17:29
• @john-duffield You seem intent on the notion that a frame is fictitious and what matters is what happens in the rest frame of the disk (which is not fictitious). But a frame is nothing but a (set of) prescription(s) for measuring coordinates of events relative to an origin that is usually an observer, albeit not necessarily a human observer. As such they are not fictitious at all. Insisting on the rest frame does not solve the problem of what is observed in other frames. Timaeus's question is well-posed.
– udrv
Aug 21, 2015 at 19:55
• @udrv : I consider myself a "relativist", somebody who understands special relativity. And IMHO the key to this is appreciating that when you change your reference frame, what you're really changing is your state of motion. And that whilst this changes you and your observations, it has no effect whatsoever on the thing you're observing. A distant star doesn't change in an instant just because you accelerated towards it. Nor does the ring of people. There's no issue with Timaeus's question. It's interesting and thoughtful. I gave him a +1 too. Aug 21, 2015 at 20:13
• @john-duffield Ok, I agree with your point that in SR an observer's state of motion does not change the observed object (keeping away from quantum issues). But what is discussed is what does the observer observe & measure of the object(s) from said state of motion, given that such measurements are as objective as those in the object's rest frame.
– udrv
Aug 21, 2015 at 20:54
• @udrv : the measurements aren't objective. When you and I are moving in different directions, our measurements of the object and events don't tally. We know from this that we have to compare notes and use our knowledge of relativity to work out that the light flashed and the people did their ring-a-roses dance. They can vouch for that because they completed a circuit and got a mild electric shock or something. Aug 22, 2015 at 10:57

You can look at the problem like this:

Whether or not one person gets to see all the other persons as connected for some period of time depends on

1) the disk size, say radius R

2) the duration $\tau$ of the connection in the disk frame

3) the speed $v$ of individual components on the rim relative to the disk.

Let's denote $T$ the astronauts' period of revolution around the disk rim, as seen in the disk frame: $T = \frac{2\pi R}{v}$. Assuming as you did that the disk is astronomically large and $\tau << T$ (or $v\tau << 2\pi R$), let's approximate the rest frame of one representative person P as inertial relative to the disk D. Within the time frame when this approximation is applicable, let P's local frame move at -v on the line y=-R as seen from D (z=0 throughout), such that P sees D move at velocity v along line y'=R.

If in D's frame P gets connected at time $t_0 = 0$ and disconnects at time $t_1=\tau$, then P must see himself getting connected at time $t'_0 = 0$ and disconnecting at $t'_1 = \tau/\gamma$, with $\gamma = \sqrt{1-\beta^2}$, $\beta = \frac{v}{c}$ as usual. But from P's frame, points of D in different planes x'=const correspond to different times t in D's frame (relativity of simultaneity). At t'=0, events P observes at the rearmost point on the circumference of D, $x'_{rear} = - R/\gamma$, correspond to Ds time $ct_{rear} = \gamma (0 - \beta (-R/\gamma) ) = \beta R$, while events observed at the foremost point on the rim, $x'_{fore} = R/\gamma$, correspond to $ct_{fore} = \gamma (0 - \beta R/\gamma) = - \beta R$. Similarly, events observed by P at different $x'$ between $- R/\gamma$ and $R/\gamma$ correspond to different times $t$ in D, $- \beta R ≤ ct ≤ \beta R$. Scanning the entire range of $x'$ from $- R/\gamma$ to $R/\gamma$ means $ct$ scans backwards the entire range $\beta R$ to $- \beta R$.

(Personal note to Timaeus: I hope you remember our argument on my "space-time cross-section" question. The above is exactly what I meant when I said an "object in motion (here the disk) is observed in space-time cross-section". It spans space and a finite interval of its proper-time. Nothing else. Usually the latter is negligible (nanoseconds), but you found a nice example where it isn't and I like it.)

This means that whatever the connection duration $\tau$ is in D, at $t'=0$ P will see at least some persons in the $x'<0$ half-plane as connected and all persons in the $x'>0$ as disconnected. If $c\tau < \beta R = \frac{\beta^2}{2\pi}cT$ there will also be disconnected persons in the $x'<0$ plane. A similar analysis holds for every $0 ≤ t' ≤ \tau/\gamma$. The range of connected persons will appear to move progressively into the $x'>0$ half-plane, until at $t'=\tau/\gamma$ all persons in the $x'<0$ half-plane will be disconnected.

The problem still is: what does it take to have P see all persons connected during some period of time between $t'=0$ and $t'=\tau/\gamma$? To figure it out, consider the distance P observes at any time $t'$, $0 ≤ t' ≤ \tau/\gamma$, between points corresponding to $t=0$ and $t=\tau$ in Ds frame. By the Lorentz transformations we expect $(x_a, 0) \rightarrow (x'_a, ct')$ and $(x_b, c\tau) \rightarrow (x'_b, ct')$. If the distance $\Delta x' = |x'_b - x'_a|$ is larger than the length contracted span of D as seen by P, then there will be some interval of time $\Delta ct'$ when P will see D entirely between $x'_a$ and $x'_b$. Now, for $x'_a$ we have

$$x'_a = \gamma(x_a + \beta 0) = \gamma x_a \\ ct' = \gamma(0 + \beta x_a) = \beta\gamma x_a \\ x_a = \frac{ct'}{\beta\gamma}\;\; \Rightarrow \;\; x'_a = \frac{ct'}{\beta}$$

while for $x'_b$,

$$x'_b = \gamma(x_b + \beta c\tau) \\ ct' = \gamma(c\tau + \beta x_b) \\ x_b = \frac{1}{\beta}\left( \frac{ct'}{\gamma} - c\tau \right) \;\; \Rightarrow \;\; x'_b = \gamma \left( \frac{1}{\beta}\left( \frac{ct'}{\gamma} - c\tau \right) + \beta c\tau \right) = \frac{ct'}{\beta} - \frac{c\tau}{\beta\gamma}$$

and so

$$\Delta x' = |x'_b - x'_a| = \frac{c\tau}{\beta\gamma}$$

In order for P to see everybody connected for at least a brief moment we need $$\frac{c\tau}{\beta\gamma} > \frac{2R}{\gamma}\\ c\tau > \beta (2R)$$

In terms of the period of revolution around the rim, and taking into account the original condition that $\tau << T$, it comes down to $$\frac{\beta^2}{\pi} < \frac{\tau}{T} << 1$$

Basically this requires low enough $\beta$, $\beta << 1$; for instance $\beta \lesssim 10^{-2}$ will do fine.

In other words, if P's velocity around the rim is so low that $\beta < \frac{c\tau}{2R}$, there will be a period of time in his local frame during which he will see everybody connected. As P's velocity increases and $\beta > \frac{c\tau}{2R}$, he will see a progressively smaller fraction of connected persons, even as his local frame can no longer be approximated as inertial wrt the disk's frame. So yes, temporary connectedness does depend on the frame from which it is observed.