# Accelerated rest frame

1. Feb 14, 2014

### analyst5

I was reading some posts on one of my previous threads (great discussion there btw) and I red something that I really didn't understand.

So the basic premise was that if we have 2 clocks that undergo acceleration on the same level, to that they are mutually at rest, that they will see their time rates differently. Can somebody explain this? I've always thought that an observer that is accelarating with an object will see the time pass on that object by the same rate as in his own frame. What are the rules here? How will the time flow relative to each other's frame (both frames are mutually at rest and accelerating at the sam rate)?

2. Feb 14, 2014

### stevendaryl

Staff Emeritus
Here's an intuitive argument that explains time dilation on board an accelerating rocket.

Imagine that you have a rocket that is initially at rest in some inertial reference frame. There is a clock at the rear of the rocket, and another clock at the front of the rocket. Suppose that every $T$ seconds, the rear clock sends a light signal toward the front clock. If the rocket is at rest, then the first signal will arrive at time $t=\frac{L}{c}$, where $L$ is the length of the rocket, and $c$ is the speed of light. Thereafter, signals from the rear clock will arrive every $T$ seconds.

Now, suppose the rocket is accelerating at rate $a$. Then, using nonrelativistic physics (which we can do, because the rocket will take a while before it gets going fast enough for relativity to be important), we can estimate how long the signals will take to travel from the rear of the rocket to the front. The first signal will take approximately time $\frac{L}{c}$ as before. The second signal will take a little longer, because by then, the front of the rocket will be moving away from the rear, and it will take a little longer for the light signal to catch up. If we assume that $a T$ is much smaller than $c$, and that $a L$ is much smaller than $c^2$, then we find that the time between arrival of signals coming from the rear clock is approximately:

$T' = T(1+\frac{a L}{c^2})$

If the people aboard the rocket interpret their situation as if they were at rest in a fictitious gravitational field, what they would conclude is that the rear clock is running slower than the front clock, by a factor of $1+\frac{a L}{c^2}$.

3. Feb 14, 2014

### WannabeNewton

4. Feb 14, 2014

### 94JZA80

i think i completely understand your explanation, but i also think one needs to be very careful with the wording when discussing motion, relativistic or not. to say that the front of the rocket will have moved away from the rear implies that the rocket is in fact stretching in length. if anything, the rear of the rocket will move toward the front of the rocket (at the instant the engine is fired) before the front of the rocket starts to move too (b/c there is no such thing as a perfectly rigid body, and the impulse of the thrust generated by the engine takes time to reach the front of the rocket). now, if we ignore the fact that nothing in this universe is perfectly rigid, and that impulse travels at the speed of light (and not instantaneously), then the relationship between the front of the rocket and rear of the rocket is no different than the scenario the OP suggested in the first place...that is, regardless of whether the rocket is at rest, undergoing uniform motion, or undergoing accelerated motion, the front of the rocket and the rear of the rocket are simply mutually at rest with each other the entire time.

thus, it would probably be more appropriate to say that the second signal will take a bit longer to arrive at the front of the rocket because at T > 0, the front of the rocket will have moved some distance from its original position. hence, the front of the rocket will have also moved some distance from the position at which the rear of the rocket resided at T = 0 (specifically, the front of the rocket will have moved directly AWAY from the rear of the rocket's previous position, creating a distance longer than L). thus the signal takes longer to traverse the length of the rocket not because the rocket stretched its length, but because the distance from the rear of the rocket at T = 0 to the front of the rocket at T > 0 is greater than L (the distance from the rear of the rocket at T = 0 to the front of the rocket at T = 0).

5. Feb 14, 2014

### WannabeNewton

You're thinking of Born rigid acceleration of the rocket. While it's true that in Born rigid acceleration the front and rear ends of the rocket will accelerate at different proper rates in order to remain relatively at rest and furthermore clocks at rest at the respective locations will tick at different rates relative to one another, even in the Newtonian approximation wherein all parts of the rocket have the same proper acceleration whilst maintaining rigidity one will still be able to derive time dilation of resting clocks at different locations in the rocket so what you're proposing as the reason isn't quite correct.

What Steven said is in fact perfectly fine.

6. Feb 14, 2014

### stevendaryl

Staff Emeritus
No, from the point of view of the initial rest frame, the position of the front of the rocket is given, for small values of $t$, by

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

The rear of the rocket will have a position given by:

$x= \frac{a t^2}{2}$

A light signal sent from the rear after $T$ seconds will have a position given by:
$x = \frac{a T^2}{2} + c (t - T)$

So, the light signal will catch up with the front when

$\frac{a T^2}{2} + c (t - T) = L + \frac{a t^2}{2}$

We can rearrange this to get:
$(\frac{a (T+t)}{2} + c) (t-T) = L$

which has the approximate solution

$t = T (1+\frac{aL}{c^2})$

That is certainly true, but pretty much irrelevant. If you assume that rear clock is sending out signals at a rate of one every $T$ seconds, and assume that $c T$ is much greater than $L$, then the front of the rocket will be moving by the time the second signal gets to it.

The notion of two accelerating objects "being mutually at rest" is a frame-dependent notion. They may be at rest according to one frame, but not according to another.

You're the one who said that it had something to do with the rocket stretching, I didn't.

7. Feb 15, 2014

### analyst5

I think you didn't understand my question. I was refering to the situation where we have a collection of observers that are at rest but accelerating at the same rate. In inertial frames, if we are at rest with some object, time will flow with the same rate in in as in our system. What is the difference for accelerated objects mutually at rest? That's what I don't understand.

8. Feb 15, 2014

### Bill_K

In the initial frame, all the rockets have the same velocity at the same time t. But events that appear simultaneous in one frame are not simultaneous in another. An observer on one of the rockets will have his own definition of simultaneous, and in his frame at t' = const the other rockets will be at different stages of their acceleration. Those ahead of him will have traveled longer and gained more velocity, while those behind him will have just begun, or not even started yet.

9. Feb 15, 2014

### stevendaryl

Staff Emeritus
There is no such thing as accelerating observers accelerating identically. It's one of the quirks of special relativity that makes this impossible. If, according to the initially "launch" frame, two observers are accelerating identically, then the observers themselves will see the forward observer move steadily away. They would not see themselves as "at rest" relative to one another.

Alternatively, if the two observers try to remain "at rest" relative to one another, it would require that the forward observer accelerate less strongly than the rear observer. So the forward observer would (from the point of view of the initial rest frame) be traveling slower than the rear observer, and would experience less time dilation. His clock would run faster than the one of the rear observer.

10. Feb 15, 2014

### analyst5

How do we measure proper time elapsed during acceleration, since a clock at rest with an accelerated object would tick differently?

11. Feb 15, 2014

### WannabeNewton

It looks like you completely missed Steven and Bill's important and related points. Have you worked with Rindler space-time before? Do you know how to compute invariants of time-like congruences? Do you know the relationship between clock synchronization and lack of Born rigidity of a time-like congruence (e.g. compressibility of a fluid)?

If you do then it will be really easy to explicitly see why two observers accelerating at the same rate along the line joining them cannot be at rest relative to one another in either of their rest frames and therefore will notice their clocks ticking at different rates of proper time due to kinematical time dilation. On the other hand if you have two observers accelerating at different rates along the line joining them in exactly the manner needed to maintain Born rigidity then the same formalism of Rindler space-time and the kinematical decomposition of time-like congruences will make it easy to see that the observers remain at rest relative to one another in their rest frames but will notice their clocks ticking at different rates of proper time due to "gravitational" time dilation.

12. Feb 17, 2014

### johnny_bohnny

So can an object accelerate in a manner that all of its parts accelerate simultaneously viewed from its own rest frame? This is really confusing. What would happen if an inertial, launch frame, saw the object accelerating part by part?

13. Feb 17, 2014

### WannabeNewton

Well for starters this assumes the extended object has a rest frame to begin with. If the constituents of the object are moving radially relative to one another, such as in an arbitrary fluid, then it doesn't make much sense to talk about the rest frame of the object itself, right? In fact in general the best we can do is talk about the individual rest frames of the constituents of the objects. In a fluid these individual rest frames would correspond to those of the individual fluid elements. When can we talk about the global rest frame of an extended object itself? Precisely when the object is rigid.

But the definition of rigidity in relativity is quite different from the definition of rigidity in Newtonian mechanics. The precise definition requires some mathematical machinery but intuitively rigidity in relativity, often termed Born rigidity, means that given any constituent of the extended object, the spatial distances in the rest frame of this constituent to all neighboring constituents of the object remain constant i.e. there are no relative radial velocities between the neighboring constituents of the object. Note that this definition still allows a rigidly rotating object to have an extended rest frame.

Now coming back to your question, imagine we have a rod at rest in an inertial frame and say we want to accelerate the rod along its length. For the reasons explained above, if we even want to talk about the rest frame of the entire rod itself during the acceleration phase, we better make sure the acceleration is done in a Born rigid manner i.e. the rod must remain rigid during the acceleration phase. Then we can indeed talk about the global rest frame of the rod i.e. an extended frame in which all points of the rod are at rest. In order to do this, each point of the rod must receive a different proper acceleration-in fact we must impart a proper acceleration to each point of the rod inversely proportional to the fixed spatial location of the point in the rest frame of the rod. Now obviously in this case if all points of the rod are accelerated simultaneously in the inertial frame in which the rod is initially at rest then all points of the rod will accelerate simultaneously in the rest frame of the rod by construction*

On the other hand, say we don't accelerate the rod Born rigidly. Say we impart to all points of the rod the same proper acceleration and do so simultaneously in the inertial frame in which the rod is initially at rest. Then in fact it no longer makes sense to talk about the extended rest frame of the rod itself because in the individual rest frames of the points of the rod, the neighboring points will have radial velocities. Furthermore, now if we go to the rest frame of any given point of the rod, then relative to this frame the other points of the rod will have actually accelerated at different times i.e. even though all points were accelerated simultaneously in the initial inertial frame, because we gave all points the same proper acceleration we find that in the rest frame of any given point of the rod the other points will not have accelerated simultaneously. This is in fact the content of the famous Bell spaceship paradox: http://en.wikipedia.org/wiki/Bell_spaceship_paradox

*We have implicitly assumed here that we can actually talk about global Einstein simultaneity in the extended rest frame of the rod. Rigidity is actually not enough to guarantee this. The reason we can make this assumption is due to the irrotationality of the rod motion.

14. Feb 18, 2014

### johnny_bohnny

I think I get it Wannabe Newton, thanks for your answer and the work you put into it.

So we may consider an object to be our frame of reference when it is Born rigid accelerated so that defines its own rest frame? I didn't knew this was possible. Interesting.

15. Feb 19, 2014

### pervect

Staff Emeritus
"Rest frame" is a confusing topic. But I think that whatever significance you ascribe to a "rest frame", I believe can be understood by the less ambiguous idea of a coordinate system.

The rules you are asking for are then are just a mathematical expression for what a clock measures (proper time) given the details of the coordinate system. This is just

for timelike separations:
($\Delta$ proper time) ^2 = $\sum_{ij} g^{ij} \Delta x_i \Delta x_j$

If the RHS is negataive we have a spacelike separation
($\Delta$ proper distance) ^2 = -$\sum_{ij} g^{ij} \Delta x_i \Delta x_j$

Here $g^{ij}$ are the metric coefficients associated with your particular choice of coordinates.

(note: there's a sign convention here that is sometimes variable, it won't matter in this post but it might if you compare to other posts or textbooks).

We've specified the change in a clock reading between two nearby points, (or the change in distance if the separation is a space-like one). To get the change in a clock reading between two distant points you need to plot a particular path or curve between the two distant points and add up (integrate) the change in clock reading between nearby points on the connecting curve. You use the same integration process to find the distance between two points given a connecting curve.

I'm sure you have been told this before, but it doesn't seem to be getting through to you. I'm not sure if just repeating this is going to help, but I'm not sure what about the explanation you may not be getting, so I'll try.

The choice of coordinates to use is arbitrary, in your example in one case you choose inertial coordinates and in the other case you choose "accelerated coordinates".

The numbers associated to events change when you change coordinate systems. The readings of actual clocks do NOT change when you change coordinate system - whatever coordinate system you choose, the clocks report the same amount of time. Note that actual clocks follow some specific path through space-time, you need that path to integrate along.

The notion of clocks "speeding up" or "slowing down", unlike the actual readings of the clocks, DOES depend on the coordinate system you choose. The observer-independent (coordinate independent) notion of what is going on is the actual readings of actual clocks. The observer dependent notion of what is going on is what coordinates you assign to specific clock readings.

If you want more about how the metric coefficients are determined given your choice of frame or coordinates, we can perhaps give a few examples. The simplest thing to note is that inertial observers always have a metric of diag(-1,1,1,1) - (I hope that notation makes sense?).

Before discussing further the details of the metric coefficients, I want to see if the preliminaries make sense.

I rather get the impression you are viewing things backwards - giving observer independence and/or mental priority to the notion of "frames" - which I think is likely the fundamental issue that's confusing you. What's really observer independent are the actual readings of clocks. The "frame" is a mental construct you use to compactly describe all of these observer-independent things, and different observers have different mental constructs.

16. Feb 19, 2014

### johnny_bohnny

So how do we measure the proper time of an accelerating object when all points are accelerated at different rates?

17. Feb 19, 2014

### PAllen

If you imagine a radioactive mass of some size, accelerated for a long time, then the forward (direction of the acceleration) parts of the object will have decayed more than the rear. Different parts of the object experience different proper time elapsed. This is no different from the fact that with current super accurate clocks, we can measure that more time passes for your head than your feet (assuming you are mostly upright).

18. Feb 19, 2014

### johnny_bohnny

So in accelerated bodies there is no possibility that different points on the worltube experience same time dilation, like they do in inertially moving bodies?

19. Feb 19, 2014

### WannabeNewton

Sure there is but again it depends entirely on the acceleration profile i.e. on the 4-acceleration field of the world-tube.

As a simple example take a thin ring in free space rotating about its symmetry axis with a constant angular velocity relative to an inertial observer at the center. All points of the ring experience the same time dilation relative to this observer.

20. Feb 19, 2014

### johnny_bohnny

What about the acceleration that occurs after or before coming into inertial motion? Like in the twins paradox scenario (where the space twin starts his trip). Or for example when somebody starts his car?

21. Feb 19, 2014

### WannabeNewton

You can do it but the motion might not be Born rigid. You can accelerate a rod along its length in such a way that all points of the rod receive the same proper acceleration and hence have the same velocity and time dilation factor relative to the inertial frame the rod was initially at rest in but in the rest frame of any one point of the rod, the proper length of the rod will be increasing, meaning neighboring points will have radial velocities relative to this rest frame, until the stresses break the rod.

Fun fact: you cannot, even in principle, put a thin disk into uniform rotation in a Born rigid manner-the tangential acceleration applied will necessarily violate Born rigidity.

22. Feb 19, 2014

### PAllen

However, for realistic acceleration profiles (and turns), you would expect different parts of you and your car to undergo slightly different aging in the course of a trip.

23. Feb 19, 2014

### johnny_bohnny

This is basically the scenario you mentioned when you said that different points will disagree in simultaneity, the 'forward' ones will already be in different stages of their acceleration, and the 'rear' ones will be perhaps even not starting their acceleration. If I'm understanding you correctly?

24. Feb 19, 2014

### WannabeNewton

Well which points start accelerating when depends on which point's rest frame you choose to boost to but yes the idea is the acceleration of all the points won't be simultaneous in the rest frame of any given point of the rod even if it was simultaneous in the initial inertial frame. It's the exact same set of calculations as in the Bell spaceship paradox.