# My first ST diagram

1. Dec 3, 2008

### Saw

I've drawn the attached sort of ST diagram to test my undersanding on how SR works. It intends to reflect a situation where:

(i) B ship goes through A ship.
(ii) Relative velocity is 0.5 c.
(iii) When the left edge of both ships meet, both observers situated there send light signals to the other edge.
(iv) Both signals reflect back on the other edge simultaneously.

I would very much appreciate any spotting of mistakes in the calculations or any criticism on the way it is drawn. By the way, is this the Minkowsky or the Epstein-like style of diagram? And what would be here the spacetime interval on which both frames agree? Thanks in advance!

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2. Dec 4, 2008

### JesseM

Looks good to me. It's definitely a typical Minkowski diagram, Epstein diagrams have some weird properties like the fact that two worldlines which meet at a single point in space and time may be shown as not meeting at all in an Epstein diagram (I've never seen Epstein diagrams discussed outside of the context of discussions of Epstein's own book, so unless you're interested it's not something you really need to learn about). Often Minkowski diagrams also a line or lines of simultaneity for the moving frame (you might draw a line connecting points on the front and back end of the moving rod that are simultaneous in the rod's own rest frame) but it's not necessary. As for spacetime intervals that they both agree on, you can pick any pair of events at all (like the back end of rod B reading a certain time and the front end of rod A reading a certain time), and if you calculate the spatial distance $$\Delta x$$ between them in each frame, and the time difference $$\Delta t$$ in each frame, then both frames will agree on the value of $$\Delta s^2 = \Delta x^2 - c^2 * \Delta t^2$$ when calculated using their own respective values for $$\Delta x$$ and $$\Delta t$$. Note that the spacetime interval $$\Delta s$$ between two events is actually just c * the proper time $$\Delta \tau$$ between those events for an inertial clock that has both events on its worldline: $$\Delta \tau^2 = \Delta t^2 - \frac{1}{c^2} * \Delta x^2$$, or $$\Delta \tau = \sqrt{\Delta t^2 - \frac{1}{c^2} * \Delta x^2}$$.

3. Dec 5, 2008

### Saw

Thanks a lot. I am wondering whether, if I placed the legends expressing the clock readings of the "foreign" clock (eg, those of B frame in A frame ST diagram) at the height where the caption of the relevant legend coincides with the time coordinate of the local diagram, this would result into the line connecting those legends being perpendicular to the line reflecting the slope (v) of the foreign frame. It looks so, but I'm not sure and wonder if that's worth the effort. Should that be true, would it have any meaning?

Last edited: Dec 5, 2008
4. Dec 5, 2008

### JesseM

Hmm, I'll see if it works in another example where the math is a bit simpler...imagine we have a rod 10 l.s. long in its own rest frame, moving at 0.6c in our frame (this is a nice speed to use in numerical examples because gamma = 1/0.8 = 1.25). In our frame, clocks at either end of the rod which are synchronized in the rod's frame will be out of sync by (0.6c)(10 l.s.)/c^2 = 6 seconds. So if the back end of the rod is at x=0 at t=0 in our frame, and the clock there reads t'=0, then the clock at the front end (position in our frame: x = 8 l.s.) will read t'=-6 seconds. 10 seconds later in our frame, the clock at the front end will be at position x = 8 + 6 = 14 l.s., and its time will have advanced forward by 8 seconds, so it'll read t'=-2 seconds.

Now, when you talk about relocating clock-readings so "the relevant legend coincides with the time coordinate of the local diagram", do you mean moving the legend of the first event--the clock at the front reading t'=-6 at coordinates x=8 and t=0--to new coordinates x=8 and t=-6? And do you mean moving the legend of the second event--the clock at the front reading t'=2 at coordinates x=14 and t=10--to new coordinates x=14 and t=2? If so, then the line connecting these relocated legends will be given by t(x) = (4/3)*x - 50/3. This is not at all close to being at right angles to the slope of the clock's worldline, so either I'm misunderstanding what you meant or it's just a quirk of the numbers you chose that it looks close to a right angle.

5. Dec 7, 2008

### Saw

Yes. I meant so.

OK, thanks, this spares me the effort of trying to draw that line. It was just a curiosity to be explored, if it had worked.

Incidentally, another question has come to my mind. This sort of examples usually start with the two rods being at rest with one another and being attributed a certain length. Then it is said that one of them accelerates and the SR effects arise: reciprocal measurement of TD, LC and RS. I was trying to construct my example with this introduction and then this puzzling doubt arose: once in inertial relative motion, with my initial numbers, in A frame, B rod is 0.86 l.s. long; instead, in B frame, B rod is 1 l.s. long; but before there is relative motion, which numbers should I choose? I see several possibilities:

- A’s version: When the two rods were at rest with each other, B rod was 0.86 l.s. long; after relative motion, poor B has started measuring her own rod as longer (1 l.s.).
- B’s version: When the two rods were at rest with each other, B rod was 1 l.s. long; after relative motion, that crazy A has started measuring my rod as shorter (0.86 l.s.).
- None of the others: some intermediate solution, more or less biased towards one side or the other.

The same doubt would apply to rod A.

This must have an obvious answer, but I can’t find it. This looks to me like the twin paradox, but applied to length and located at the beginning of the exercise, instead of at the end. Given this, can the answer be dependent on who has accelerated? For example, if it was B who accelerated, we would opt for A's version...

Last edited: Dec 8, 2008
6. Dec 8, 2008

### JesseM

Acceleration is complicated in SR because pushing on one end of an object doesn't instantaneously influence the other end (if it did this would give you a method of FTL communication)--I'll quote my comment on this thread:
But let's say we have an object like a spring that, after some oscillating when it's accelerated, will eventually return to the same rest length it had prior to the acceleration. In this case, the answer would basically be your second possibility: in frame A before the B spring is accelerated, it will have a length of 1 l.s.; then after it's accelerated and the oscillations stop, the B spring's new length in the A frame will be 0.866 l.s. (while in the B spring's new rest frame it'll again have a length of 1 l.s.)

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