# F.O.R. and Acceleration/Deceleration Question

1. Jul 29, 2009

### TrashmanL

Here's the impossible scenario I THINK I understand, from the train/platform thought experiments and the "twin paradox"... I prefer to use spaceships:

We are in Spaceship A - this shall be our Frame of Reference (F.O.R.).
At some time prior in our F.O.R., Spaceship A and B are synchronized.
Spaceship B flies by us at relativistic speed.
In our F.O.R. clocks on Spacehip B run slower.
In the F.O.R. of the crew of Spacehip B, our clocks are running slower.
If someone in Spacehip B could suddenly jump back into our F.O.R. after they pass us by, their clock would show an earlier time than ours.
If we could suddently jump into their F.O.R. after they pass us by, our clocks would show an earlier time than theirs.

Hopefully, there's nothing wrong with the above statements (the impossibility of shifting F.O.R. instantly aside - and yes, I know even the term "instantly" is ambiguous in this context - let's take it from a "common sense" perspective). If there are other issues I'm sure someone will let me know.

When I try to get specific and make it more "realistic" by adding acceleration and deceleration into the mix is when I have a total failure to understand.

Say Spaceship A and Spaceship B are next to each other in space and we synchronize our clocks. Lets even say we calculate how much time will pass for them as they venture to some specific point to give them time to accelerate.. Spaceship B takes off and accelerates, reaching a speed of 0.986c (this gives a Lorentz factor of 6, if I'm not mistaken) just as it passes us and it maintains that speed for the rest of its journey. It ventures out 4.5 ly, turns around and comes back (I don't even want to consider the consequences of that turnaround right now - lets ignore that for right now and just consider the acceleration/deceleration at the end and I'll ask about the turnaround later). At the moment when it passes us by in our F.O.R. 9.13 years have passed by for us (3331.6 days - ignoring the turnaround time), but only 1.52 years (555.3 days) have passed for the crew of Spaceship B.

OK, so far I think I've got it. Now here's where it gets weird for me:

If, instead of merely waiting for Spaceship B to pass by, we do some calculations to plot an intercept course to accelerate to .986c and rendezvous w/Spaceship B (which would have to happen at some point after Spaceship B passes our initial position) and then we compare times - who's clock says what?

As we accelerate towards their F.O.R. the differences in our clock speeds would get smaller (they would also be decelerating towards us from our F.O.R. in fact, we'd be decelerating towards them in their F.O.R. as well).

To us, their clocks would always be slower than ours, so I would think the difference would be similar to to the 9.13 years vs. 1.52 years above, with a slight difference.

To them, OUR clocks would always be slower than theirs. they only see 1.52 years pass by before our F.O.R.s begin to meet, so we should have less than 1.52 years pass on our clocks.

(This all assumes we can accelerate to .986c in a trivial amount of time (days?) and the numbers are ignoring this time... is this my wrong assumption?)

BUT... more than 9 years have passed to us in our F.O.R. before we try to accelerate to the same F.O.R. as Spaceship B - so our clocks would have to have progressed at least that much - we wouldn't suddenly start going backward in time in our own F.O.R. even if our F.O.R. is changing.

Does the changing time dilation between the two F.O.R.s during acceleration make up for the difference (just using SR alone, or including the gravity equivalence of GR)? Did we have to travel much farther and take longer (space dialation?) to accelerate from their F.O.R. than ours? Or is there something else I'm completely missing (I'm pretty sure this is the case)? How do I resolve these two scenarios? What happens as we accelerate and change F.O.R.s? And, what the heck WOULD our clocks say? Depending on the answers to these questions, I have more questions - but this is already a long post so I'll stop it here.

2. Jul 29, 2009

### Staff: Mentor

Yes, you are neglecting the relativity of simultaneity. Because of simultaneity issues there is no standard rest frame for an accelerating objece and you get bad results by naievely "frame jumping" as you suggest.

3. Jul 30, 2009

### TrashmanL

I am familiar with the relativity of simultaneity and the relativistic car in the garage (I guess this is more commonly a "ladder") analogy. Apparently, only the sheer basics (or not even) of it, as I fail to understand how it applies to this situation.

I realize in the first scenario I painted that I was impossibly "frame jumping" as I made a gross simplification to try to make a point. I did not realize that in continuously accelerating towards a relativistic speed I was also guilty of "frame jumping." Certainly, it is possible to do this, so what happens as one accelerates from a non-relativistic speed to a relativistic speed?

This question actually arises from a thought experiment I was making involving a simultaneous event apparently independent of F.O.Rs and the complications presented by the relativity of simultaneity. I'm sure this experiment sounds ridiculously stupid to you right now, and obviously I have a large hurdle of understanding to jump over in regards to the relativity of simultaneity. This question I thought was going to answer something else, but I guess my brain was going on a more circuitous route - when this question actually goes right to the heart of the matter. I will do some more research, but if anyone here can help me understand more about the relativity of simultaneity and how it applies to this question please do.

Last edited: Jul 30, 2009
4. Jul 30, 2009

### Staff: Mentor

OK, here is the basic problem with "frame jumping". A frame of reference is a coordinate system, it has no physical existence and is only used as a mathematical tool for analyzing a physics problem. Objects do not jump in and out of reference frames (since a coordinate system extends out to infinity). When you choose a reference frame in which to do an analysis you need to stick with that reference frame throughout the entire analysis, if you switch midway or if you attempt to use a non-inertial coordinate system then (unless you do your math very carefully - which very few people do) you will get nonsense.

Now, there is no reason that you need to have any object at rest in your chosen coordinate system. You can arbitrarily pick any inertial coordinate system. That is the whole point of relativity. But once you have made that choice stick with it through the entire problem. So go ahead and pick any inertial frame and do the analysis all the way through. If you are bored, then go back to the beginning pick a different inertial frame and do the analysis all the way through again. By the principle of relativity you are guaranteed to get the same result. But don't switch frames in the middle of the analysis! If you do, you are effectively using a non-inertial coordinate system and most likely you are doing it wrong.

5. Jul 31, 2009

### TrashmanL

I think I might have it. The first paragraph here using your suggestion -

Use my calculations of 9.13 and 1.52 -> Let's add a 3rd ship, A'. A' never undergoes the acceleration that A does, it remains in close proximity to A's initial position the entire time. A' sees ship B travel 9 ly A' knows the Lorentz factor of 6, and thus B has only aged 1.52 years after making the round trip, even though 9.13 years has passed to A'. A has aged 9.13 years while waiting however, then makes its jump to .986c. So A and B are together with a new age difference from this F.O.R., like in the Twin paradox, any further aging would happen at the same rate, however.

I could understand it from that F.O.R.

That doesn't really answer my question, though. I want to know what the experience of this acceleration and deceleration is. I want to know what it is like from B's perspective on the entire trip, likewise what it is from A's perspective on the entire trip. They do, unfortunately "jump frames" but their frames come together and meet and they have experiences to share.

I looked more into the Twin Paradox which I thought I understood pretty well, but now realize I didn't.

Still ignoring the time it takes to accelerate and decelerate (which I will assume is not entirely insignficant but much smaller than the scale of years I'm using and will only throw my numbers off slightly) here's what I have:

To ship B, the space in front of it became shorter and it only travelled 1.5 ly and that's why it only took them 1.52 years (though at the turn around point they could look back and say "hey, we are 4.5 light years from where we started and only 0.76 years have passed!" then once at 0.986c in the other direction, the return trip becomes 0.75 ly, just as the trip there was). Again, only 1.52 years has passed for them total. But this really uses two inertial reference frames, and what I couldn't wrap my head around...A did what relative to B.... In B's first inertial reference frame, I can say A was moving away from B at 0.986c. But, when B is decelerating/accelerating I can't say B was staying at rest and A was decelerating/accelerating relative to B. This took me a little while to wrap my head around - I wanted to quantize B's speed and say at each moment B is in a new inertial reference frame. I guess a force on B cannot be shown to be an opposite force on the rest of the universe as opposed to B, so B is uniquely experiencing its acceleration/deceleration. Prior to this turnaround, however, A only appeared to age 0.76 years according to B. During the turnaround, B somehow "sees" A rapidly age from 0.76 years to 4.565 years. One explanation of this is to say that B's "plane" of simultaneity (really should be "space" of simultaneity shouldn't it?) is rotating. At the point when B is at rest relative to A, A and B are both in the same frame of reference and all their calculations agree (I know, if they take relativity into account they should ALWAYS agree... but now they're both seeing it the same way). Then, upon accelerating back towards A, A appears to have suddently aged another 3.805 years - when B reaches 0.986c again (in this case, would a good analogy be that time is like a rubber band snapping back?) A now appears 8.37 years old and will age another 0.76 years on the trip back to B, just as B will age to A. The crew of B all being spatially close to eachother do not see a significant time difference between eachother (though at these speeds, it is possible that someone at the bow of the ship might age by a small, but noticeable difference from someone at the stern - but I guess that would depend on the ships axis of rotation or if it even has one...lets not debate this point in this thread - if I get interested and not shot down in this thread I'll make a new thread on that topic).

That is all without A having a rendezvous with B - so up to this point it's pretty much the twin "paradox". I know it's not completely accurate, but is my summation reasonable up to this point? If it is, I think I finally understand a Minkowski diagram! It seems kind of easy in a way...

Lets push back a little now and go back to the original question. A is accelerating up to reach B - by using proper equations (which I don't know) it is "timed" so that A can reach B without exceeding 0.986c (A will have to leave before B reaches the position it first achieved 0.986c). Again neglecting some of this acceleration time - now, as A accelerates towards B, its plane of simultaneity is shifting - but they are somewhat spatially close together, so the difference of that is again minimal. When A reaches B, A has "aged" approximately 9.13 years and B has aged approximately 1.52 years and everyone both A and B agree - nothing additional changed according to either of them - except the rest of the universe,As A' and the rest of the universe will have aged significantly whenever they come back to our F.O.R. The calculations in their F.o.R agree with the calculation in the F.O.R. of A' and thus the universe is happy.

OK, now that I think I know it all, what did I get wrong? If this is right, it makes my previous thought experiment extremely interesting.

6. Jul 31, 2009

### Staff: Mentor

Good, that is all special relativity says (that all inertial reference frames are equivalent).

What do you mean by "from B's perspective" and "from A's perspective"? They are non-inertial objects, so there is no standard definition of "their perspective". You need to completely define the coordinate system that you are using and the transformations from those coordinate systems to an inertial coordinate system like A'. If you really want to go down this road then you must do your math completely rigorously at each step.

Here is an Arxiv paper that uses an approach that I like: http://arxiv.org/abs/gr-qc/0104077

Read it carefully, if you are not completely comfortable with the math used there then I urge you to stick to inertial frames. Also, since I believe that they present something like 4 different non-inertial coordinate systems, hopefully it will drive home the point that there is no standard meaning to a non-inertial object's perspective. You have to be really explicit.

No, it is not meaningful. The frames do not "come together" and it is not possible to jump in or out or between reference frames.

Again, reference frames are coordinate systems. They extend to infinity, they cover all space and all time. You cannot ever enter or leave or jump between reference frames. Sitting at your desk right now you are in all possible reference frames. In some reference frames you are stationary, in others you are moving with v close to c.

The only thing that changes is the position and its derivatives in a given reference frame. But you do not go in or out of the reference frame by going far away or fast.

Last edited: Jul 31, 2009
7. Aug 3, 2009

### TrashmanL

OK, you're hammering home the point that an inertial reference frame is a contruct of math and not something a person is physically in.

But, if I am a person on a ship traveling at near a constant velocity v, would not all my measurements be relative to that inertial reference frame? If I was on ship A and took measurements before it attempted a rendezvous with ship B, wouldn't my meaurements pertain to that inertial reference frame? Then, if I took measurements after accelearating to the speed of ship B wouldn't those measurements now correspond to that inertial reference frame? And heck, what would the measurements be like during acceleration/deceleration?

Though I'd find numbers helpful, what I'm asking is what would the experience be like for a person on ship A or ship B? I use numbers when I can just to get some grounding in something quantifiable. They help gain perspective, but it's really the qualitative description that I'm after.

When I have the time to go through, I'll check out that link. I imagine the math probably is beyond my knowledge.

8. Aug 3, 2009

### TrashmanL

I reread your post after making mine. I suppose I'm still miusing the term "reference frame" since, as you say, we are in all reference frames at all times. So saying that A or B's measurements would correspond to "that frame" is nonsensical.

Still, if A' were to take a measurement of the diameter of a nearby star, and B were to take a measurement of that same star, and then they compared notes, would they not get different answers? What terminology would be appropriate to use when comparing those two measurements if not to say "from frame A'' and "from frame B".

9. Aug 3, 2009

### vin300

If you attempt a rendezvous with ship B being in ship A first, your ref frame turns non-inertial making up for the time difference

10. Aug 3, 2009

### Staff: Mentor

Precisely!

OK, why don't we spend a little time on this point. The first thing is to note that relativity postulates that the laws of physics are the same in all reference frames, and therefore that all reference frames will agree on the result of any given experiment. So, if you use a ruler to measure the length of a piece of paper and find that it is 11" long then all reference frames must agree that your ruler will give a measurement of 11". We can disagree about whether or not the paper is "really" 11" long, but we must agree on the result of your measurement. In that sense, the raw measurements are absolute, not relative. Does that make sense so far?

Now, how do you go about measuring the length of a moving piece of paper? Well, we can take our ruler and, as the piece of paper goes by, just measure where the far end of the paper is at the same time that the near end of the paper reaches the 0 mark on the ruler. Notice the key phrase "at the same time". In order to make this kind of measurement we need to choose a synchronization convention. For inertial reference frames the standard convention is the Einstein synchronization convention, but there is no standard convention for non-inertial reference frames. Once you have chosen the synchronization convention then you can make this kind of measurement. However, note that this is actually a compound measurement, you are really taking two measurements and combining them using your simultaneity convention to get your measurement of length. Other frames will agree about your two raw measurements, but they will disagree that your simultaneity convention is correct, so they will disagree that what you measured was actually the length of the paper. This is how you get "measurements" that are relative to a reference frame. Essentially, it is due to choosing different synchronization conventions in different reference frames, and that is how measurements become "relative" to a reference frame.

So, talking about a frame's perspective or how things pertain relative to that frame is only unambiguous if the frame is inertial. The reason is that only for inertial frames can you immediately know the synchronization convention. For non-inertial frames the synchronization convention is completely open, and until it is defined it doesn't make sense to talk about the frame's perspective or measurements relative to it.

Excellent, I think you will find it worthwhile.

"Relative to". That makes it clear that the same situation can be described relative to multiple frames and that no object is "entering" or "leaving" the coordinate system, just changing their velocity or position relative to the coordinate system.

11. Aug 4, 2009

### TrashmanL

Thank you, DaleSpam, that was very helpful. I'm sure I'll have questions about the link at some point and I'll ask them here, but that'll probably be a while.