# SR and Accelerations

1. Jan 30, 2009

### MikeLizzi

An inertial observer in reference frame S makes the following determinations about you and me. (By that I mean all values below are with respect to the inertial observer.)

You:
Position (x, y, z) = ( 0, 0, 0)
Velocity (vx, vy, vz) = ( 0, 0, 0)
Acceleration (ax, ay, az) = (999, 0, 0)

Me:
Position (x, y, z) = (999, 0, 0)
Velocity (vx, vy, vz) = ( 0, 0, 0)
Acceleration (ax, ay, az) = (999, 0, 0)

Question: As far as Special Relativity is concerned, do we both have the same Proper Time as the observer?

I’m sure I will follow up with a request for an explanation later, but could I just get one of the following answers for now?

1. Yes
2. No
3. Indeterminate because the question is so poorly stated.

2. Jan 30, 2009

### JesseM

I'd say #3. Proper time normally refers to the time interval between two different events on an observer's worldline as measured by the observer's own clock--it doesn't really make sense to talk about proper time at a particular moment unless you specify at what earlier moment the observer set their clock to 0. Also, different frames define simultaneity differently (the relativity of simultaneity), so there is no frame-independent way to decide if two observers at different locations are showing the same time on their clocks, you can only ask whether their clocks show the same time at a particular moment in one particular frame.

3. Jan 30, 2009

### tiny-tim

Hi MikeLizzi!

Instantaneously, yes, because your velocities are 0 …

but they are changing, so your proper times will change rate also …

your proper time goes slower (as seen by the observer) by the factor 1/√(1 - v2/c2).

4. Jan 30, 2009

### JesseM

But how can you say what their "proper time" is without knowing at what point on their worldlines they set their clocks to zero? We can certainly talk about the rate their proper times advance in the frame of the observer as you did, but that wasn't what MikeLizzi actually asked.

5. Jan 30, 2009

### MikeLizzi

Thanks tiny-tim. I was hoping everyone would say that.

Thank you also JesseM. In response to your comment I tried rewriting the question. I hope the new version doesn’t end up making tiny-tim change his mind.

An inertial observer in reference frame S (t, x, y, z) = (0, 0, 0, 0) makes the following determinations about me. (By that I mean all values below are with respect to the inertial observer.)

Me:
Space Time (t, x, y, z) = (0, 999, 0, 0)
Velocity (vx, vy, vz) = ( 0, 0, 0)
Acceleration (ax, ay, az) = (999, 0, 0)

Assume I am in a non-inertial reference frame that is coincident with S at time t = 0.

Question: If I transform those space time coordinates with respect to the observer into space time coordinates with respect to me what will I get for the time coordinate?

1. Zero
2. Not Zero
3. Indeterminate because the question is still too poorly stated.

6. Jan 30, 2009

### JesseM

Are you asking what the coordinates would be in the inertial reference frame where you are instantaneously at rest at this moment (though obviously you won't stay at rest in this frame since you're accelerating), or are you asking what they'd be in a non-inertial coordinate system where you're at rest at all times? There isn't really any single way to construct a non-inertial coordinate system, you can come up with different coordinate systems where the same non-inertial observer is at rest, so if it's the latter I'd still say 3. And if the former, at this moment you are at rest in the frame of the observer, so your instantaneous inertial rest frame is exactly the same as the inertial rest frame of the observer.

7. Jan 30, 2009

### MikeLizzi

I am seeking the time coordinate with respect to the non-inertial reference frame in which I am at rest. So I guess you are still replying that the answer is indeterminate.

8. Jan 30, 2009

### JesseM

Yes, there are an infinite number of different possible non-inertial coordinate systems where you'd be at rest throughout the trip, to answer the question you'd need to specify the details of which one you're talking about (either by describing physically how it'd be constructed or just giving a coordinate transformation between it and an inertial coordinate system).

9. Jan 30, 2009

### tiny-tim

Hi MikeLizzi!

Sorry, but your new question doesn't make any sense …

well, you can set it to be anything, can't you?

(plus what JesseM says about non-inertial observers )

10. Jan 30, 2009

### Staff: Mentor

I agree with tiny-tim. You can set any arbitrarily moving clock to any value you want at any single event on its worldline.

JesseM is also right, there is no "standard" convention for simultaneity in non-inertial frames. I like this one, but it is just an arbitrary convention.

11. Jan 30, 2009

### MikeLizzi

Ok. I am going to try another statement of the problem tomorrow. I thank you all for for your patience. There is a reward for patience, you know. Someplace. Don't know where.....

12. Jan 30, 2009

### MikeLizzi

Third try. Please disregard all my previous posts.

First, I will summarize what I already know. Errr… What I think I already know.

Given two inertial reference frames S and S’ that coincide at t = t’ = 0.
If I know the space time coordinates (t, x, y, z) for an event in ref frame S, I can calculate the space time coordinates (t’, x’, y’, z’) for the same event in S’ using the Lorentz Transformation and the relative velocity (vx, vy, vz) between S and S’.

Now to the part I am trying to figure out.

Using Special Relativity, not General Relativity, if S’ is non-inertial how do I calculate (t’, x’ , y’, z’) ?

Lets say S’ is undergoing a constant acceleration with respect to S. I want to maintain the setup that the reference frames coincide at t = t’ = 0. I have a value for the instantaneous relative velocity between reference frames at that time.

Have I set up the problem with sufficient precision to determine (t’, x’ , y’, z’) ?

1. Yes
2. No
3. Only if the Easter Bunny was going to do the calculations.

13. Jan 30, 2009

### Fredrik

Staff Emeritus
Still 3. Oops, no wait, you asked a different question this time. It's 2. You didn't completely specify what frame S' is.

There are some problems with the statement "S’ is undergoing a constant acceleration with respect to S". It can at best specify which events the t' axis consists of, but it doesn't quite do that either, because you didn't say if you meant a constant coordinate acceleration or a constant proper acceleration. A constant coordinate acceleration can't go on for very long since the speed would eventually exceed c, so I'm guessing that you would prefer a constant proper acceleration. (That means that the coordinate acceleration is the same constant "a" in all the co-moving inertial frames).

Now that the t' axis has been specified, you need to specify, for each event on the t' axis, which events are simultaneous (in S') with it. The article that DaleSpam linked to suggests one way to do it. This one uses another.

14. Jan 31, 2009

### MikeLizzi

Fourth try.
Please disregard all my previous posts.

I think I’m making progress. Unfortunately, when I looked at the references suggested by Fredrik and DaleSpam they were over my head. I have no idea what a hypersurface is. Up till now, I have not needed to understand that concept to understand Special Relativity. I assume it is part of the knowledge base for General Relativity. I would like to solve this problem using Special Relativity.

With regard to Fredrik’s last post, I had intended that the accelerating reference frame would have a constant coordinate acceleration. But I will buy Fredrik’s suggestion and switch to an accelerating reference frame that has a constant proper acceleration. Easier to deal with. Don’t have to calculate a new value if the observer changes between inertial reference frames.

So here is the new statement of the problem.

Given :
One inertial reference frame, S and one accelerating reference frame, S’ that coincide at t = t’ = 0.
I know the instantaneous velocity of S’ with respect to S at t = t’ = 0.
I know the constant proper acceleration of S’.
I know the space time coordinates (t, x, y, z) for an event in ref frame S.

Goal:
I want to calculate the space time coordinates (t’, x’, y’, z’) for the same event in S’.

I’m thinking for sure I have all the information I need. But Fredrik’s last paragraph stopped me cold. He wrote,

“Now that the t' axis has been specified, you need to specify, for each event on the t' axis, which events are simultaneous (in S') with it.” Reads like I have yet to determine the x’ axis on a Minkowski diagram.

Wouldn’t the Lorentz Transformation tell me which events are simultaneous in S’? Or does that queasy feeling in my stomach mean the Lorentz Transformation doesn’t apply to accelerating reference frames?

Last edited: Jan 31, 2009
15. Jan 31, 2009

### JesseM

No, it's still part of SR, no spacetime curvature is being assumed. A spacelike hypersurface is just a way of taking a continuous 3D slice through 4D spacetime (which on a 2D Minkowski diagram with 2 spatial dimensions dropped would just look like a 1D curve), such that no point on the surface is in the past or future light cone of any other point on the surface--an example would be a surface of constant t in some inertial frame, which in a Minkowski diagram will look like a straight line whose slope is closer to horizontal than 45 degrees (the angle of light beams in Minkowski diagrams), but you could also imagine drawing some curvy line on a Minkowski diagram, and as long as its slope never gets closer to vertical then 45 degrees, then this surface should be spacelike everywhere. Part of the reason there's an infinite variety of non-inertial coordinate systems is that any spacelike surface can be a surface of constant coordinate time in some coordinate system.
The second. The Lorentz transformation just relates the coordinates of different inertial frames in motion relative to one another, you need a different coordinate transformation for a non-inertial (accelerating) coordinate system.

16. Jan 31, 2009

### Staff: Mentor

I admire your persistence! Would now be a good or bad time to tell you that it took me about 7 years to "get" SR (and that was without worrying about accelerating reference frames)

If you take a 3-dimensional space and pick a specific 2-dimensional subset then that is called a surface. If you take a 4-dimensional space and pick a specific 3-dimensional subset then that is called a hypersurface. So, if you start with spacetime (ct,x,y,z) and then choose the subset of all points that are simultaneous with t=0 you get (0,x,y,z), which is a hypersurface (3D subset of a 4D space) of simultaneity.

Your instincts are right. The Lorentz transform specifically relates only inertial reference frames.

Well, luckily this particular answer was already given in Frederik's link to the Rindler coordinates (where the capitalized coordinates are the inertial ones and the lower-case coordinates are the accelerating ones and T and t are in units where c=1).

$$t = \operatorname{arctanh}(T/X), \; x= \sqrt{X^2-T^2}, \; y = Y, \; z = Z$$

with the inverse transformation

$$T = x \, \sinh(t), \; X = x \, \cosh(t), \; Y = y, \; Z = z$$

As JesseM mentioned, there is no "standard" convention for defining arbitrary non-inertial coordinate systems. But for uniform proper acceleration the Rindler coordinates are very well accepted. In fact, one of the reasons that I like the method that I linked to is that it is more general but automatically reduces to Rindler coordinates for an observer undergoing uniform proper acceleration and Einstein synchronization for an inertial observer.

Last edited: Jan 31, 2009
17. Jan 31, 2009

### Fredrik

Staff Emeritus
The basic idea in DaleSpam's reference is that if the accelerating observer sends out a light signal in the positive x' direction at t'=-T, x'=0, and it gets reflected and comes back to x'=0 at t'=T, then the we assign the coordinates t'=0, x'=T/2 to the reflection event. (Actually it's cT/2, but I'm using units such that c=1). This procedure can be extended to assign coordinates to all events in spacetime.

The basic idea in the Wikipedia article that I linked to is to (for each point on the t' axis) do a Lorentz transformation to the co-moving inertial frame (let's call it S'') and define the x' axis to be the same set of points as the x'' axis, and the assignment of spatial coordinates to points on it to be identical to the assignment done by S''. (Note however that S'' is a different frame for each point on the t' axis).

Technically it's an (n-1)-dimensional submanifold of an n-dimensional manifold. But in this case you can think of it as an infinitely long curve that cuts the x-t plane in two disjoint halves (or a plane that cuts the x-y-t space in two disjoint pieces). This "hypersurface" is said to be "spacelike" if the angle it makes with the t axis is > 45° everywhere.

Note that a hypersurface of constant time coordinate doesn't have to be spacelike for the coordinate system to be valid (since a coordinate system is just a function that assigns four numbers to each event), but if it isn't, it doesn't make much sense to think of it as representing "space, at time t".

Now I realize that even that isn't enough. We also have to specify the scale on both the t' axis and on each hypersurface of constant t'. It's easy to do it on the t' axis. Just define the t' coordinate of a point on the t' axis to be the proper time along the t' axis from the origin to that point. It's a bit harder to specify the x' coordinate of each point on the x' axis, but both of the procedures described above take care of that (in different ways, and they also disagree about which set of events the x' axis is. (Edit: I see now that DaleSpam said that the two methods agree in the case of constant proper acceleration. He's probably right. I didn't really think this part through)).

As Jesse said, it's the second alternative. But as I said above, you can Lorentz transform to the co-moving inertial frame and use that frame's notion of simultaneity. You should however be aware that that's not the only option. The other option discussed above is at least as "natural" (because it's the method used for inertial frames) even though it has the weird property that what events are simultaneous with you right now depends on how you're going to move in the future.

Last edited: Jan 31, 2009
18. Jan 31, 2009

### Staff: Mentor

Yes, it is derived on page 6 starting with the second paragraph after figure 6. (hmm, I can't figure out how to get it to jump to the right page when you follow the link)

19. Jan 31, 2009

### robphy

20. Jan 31, 2009

### MikeLizzi

Well, I'm totally lost. I can remember about a year ago a moderator on this forum stating that Special Relativity could be understood with merely high school algebra and geometry. Perhaps he was exagerating.

If I get motivated again, I might sign up for a couple of linear algebra courses at the local university.