Different Reference Frames and Time Dilation (physics question)

Factual Questions
1

Since all motion is relative I’ve run into a paradox I can’t resolve.

Einstein said that for two objects, A & B, there is no way to say which is moving and which isn’t (assume they’re hanging in deep space with no other reference). If they are approaching each other at 99.9999999% light speed each one will see the other as all smooshed up (thin). When they look around at themselves all will seem normal. BOTH can perform measurements proving the other one is moving and both will get consistent results to support their view.

When moving, especially close to light speed, your clocks slow down and your tape measures shrink. So here’s my problem…

We have RocketA and RocketB. RocketA turns on its super-duper rocket motor and accelerates to 99.999999% light speed very quickly and then turns off the engine and coasts towards RocketB. Now, to both of them it appears as if it is the other ship that is moving. However, won’t the time dilation effect RocketA? If I left earth at near light speed for a few months travelling near light speed I would return years later. I’d have aged a few months while everyone on earth aged years. Can’t RocketA and RocketB make a determination on who is moving by using their super-duper telescopes to read a clock through the window of the other spaceship? Otherwise, since RocketA and RocketB both say it is the other one who is moving would they both see the other Rocket moving in slow time? What happens when they meet and match velocities? Who is still young and who is old?

2

The key is that there is no way to escape the rest of the universe. Acceleration with respect to the fixed stars will establish your inertial reference frame.

Try these explanations of the so-called twin paradox with special relativity, and the twin paradox with general relativity.

3

Only Rocket A would feel acceleration. (I’m assuming you’re keeping Rocket B stationary relative to A.) So, Rocket Scientist A would expect time dilation in his frame as observed by Rocket Scientist B. As you stated, both would experience normal rates of time.

If the two observers had sync’d clocks to begin with, then it would be apparent who experienced time dilation relative to the other. Without the measurement, you could not tell how time has passed in the interim for the two reference frames. When they meet, it will be at the same point in time.

4

Nope. In order to try to do that, the person in one rocket (let’s say rocket A, it doesn’t matter) must try to read the clock in Rocket B at two different times and make some calculations based on that result. Since the rockets are moving relative to each other, at least one of those two measurements must involve light travelling between the two rockets. Because all observers see light travelling at the same speed, they see symmetric results in this case.

Let’s do it from A’s point of view.

Say that A and B pass each other infinitesimally close at a point where both of their clocks read 12:00 and each is traveling at half the speed of light relative to the other (c/2). Each has some super-duper telescope and watches the other’s clock. At some point A sees B’s clock read 12:01. What time does A’s clock indicate? Assuming A is stationary, B’s clock is running slow relative to A’s by:

t[sub]B[/sub] = t[sub]A[/sub]*sqrt(1 - (c/2)[sup]2[/sup]/c[sup]2[/sup])

or, approximately:

t[sub]B[/sub] = 0.866 *t[sub]A[/sub]

But A knows that B is travelling at c/2, so A is seeing 12:01 on B’s clock at some time after the light left B’s spaceship. So the time on A’s clock is 12:00 plus the time it took for B’s clock to indicate 12:01 plus the time it took for the light to travel from B to A. We know the time it took for B’s clock to indicate 12:01: B’s clock is running at 0.866 the speed of A’s clock, so B’s clock read 12:01 when A’s clock read 12:00 + 1/0.866 = 12:01.1547 or 12:01:9.28. Now for the travel time of the light. When B’s clock read 12:01, A thinks B had been traveling for 1.1547 minutes (69.282 seconds) at c/2, and we’ll take the speed of light as 310[sup]8[/sup] m/sec, so at that time B was (310[sup]8[/sup]/2)69.282 = 10,392,300,000 m away. It took light 10392300000/310[sup]8[/sup] = 34.64 seconds (by A’s clock) for the light to travel over that distance (and, remember, we are assuming that A is not moving, so A didn’t go anywhere during that travel time), so when A sees B’s clock read 12:01:00 then A’s clock reads:

12:01.1547 + (34.64/60) = 12:01.7321 or 12:01:43.923

Although A can’t directly see B’s clock at that instant, A can predict that at that instant B’s clock reads slower by a factor of 0.866 and should read 12:01.50 or 12:01:30.

Now you can go through the same excercise from B’s point of view (hint; interchange A and B and there’s no math at all). They each see the same thing from their parochial point of view.

To finish it out, you should introduce a third observer, C, moving at some different velocity so C passes B at some time and C passes A at some other time. Note that C must be moving faster than B relative to A, or C must be moving faster than A relative to B. The math gets a little more tedious, but not more complex; if A and B compare clocks when they pass, then B and C compare clocks when they pass, then C and A compare clocks when they pass, the answer will be the same no matter who you assume is moving as long as you make sure you always use velocities relative to the point you have assumed is not moving. This is covered in some more detail in the links RM Mentock provided.

5

As Phobos said, it’s the acceleration that’s relevant. The usual context in which this problem is presented is the so-called Twin Paradox: Abby goes on a rocket ship and flies away very fast, then comes back very fast, while Ann stays home on Earth. Because of the time dilation, when Abby gets back, she’s younger than Ann (until then, you can’t say, because to be fair, you have to make the measurements simultaneously, and simultaneity isn’t defined for spatially separated points). The problem is that, from Abby’s point of view, at any given time, she’s stationary and Ann is moving. The key here is that Ann can consistently say that she never moved, while Abby has to admit that she was moving either on the outward voyage or on the return.

Incidentally, the business of acceleration causing time dilation is the reason why General Relativity predicts gravitational time dilation: The Equivalence Principle posits that a gravitational field is indistinguishable from acceleration.

6

Thanks guys…great answers. RM Mentock’s links helped a lot. While I don’t claim to have it nailed down I at least have a weak grasp on what is happening (kinda subtle).

Next question. The Twin Paradox mentioned in RM Mentock’s links proposes THREE reference frames. Ann, Bob Leaving and Bob Coming Back.

What happens if Bob doesn’t come back? If Bob sends a message back to Ann does the message become the third reference frame? If Bob just leaves and never sees or speaks to Ann again are we left with some kind of Schroedinger Cat deal? (I.e. Nothing can be said to have changed until it is observed to have done so?). While Ann, from previous experiments can say what she thinks probably happened to Bob is it still ultimately speculation?

Just for fun what if Bob sends a message back via a wormhole to Ann (instant messaging)? What happens to reference frames then? (Note: If that is unfair just say so :)).

7

If it’s a light signal, it’s rather hard to treat it as a reference frame, because you get into situations where you’re diving zero by zero. If it is slower than light, uyou certainly can treat it as a reference frame (assuming it does not accelerate). You can treat any velocity at being at rest; you don’t even need there to be an object traveling at that velocity for it to be a reference frame (although it such a case the reference frame usualy isn’t very useful).

Actually, it’s nothing like Schroedinger’s Cat. The cat thought experiment involves two different states that exist simultaneously. When the box is opened, one state disappears and the other becomes realized. In the Twin Paradox, ther sare several “states”, but upon close examination they all lead to the same calculations. If you calculate how much time difference there should be between Alice and Bob using
a) Alice to be at rest the whole time
b) Bob to be at rest at first, with Alice moving away from him, then Bob chasing after Alice.
c) Bob to be first moving faster than Alice, then coming to rest, with Alice moving at a contant speed behind him.
you will get exactly the same answer in each case. If Bob is familiar with relativity, the time difference will be exactly what he expects, even if he did not consider Alice to be at rest. They are really the same “state”.
In the cat case, something changes in the observation. In the Twin Paradox, nothing changes. Before Bob returns, all reference frames are valid, and after he returns all reference frames are valid. There is no change to observe.

No, she knows exactly what will happen in her reference frame. Suppose you measure the distance from LA to NYC using miles, and I use kilometers. You will get the exact same answer every time; there is no uncertainty. I will get the exact same answer every time; there is no uncertainty. The fact that our answers are different is not due to any uncertainty, but rather to the fact that we use different measuring systems.

According to relativity, there is no such thing as an absolute meaning to the term “instantaneous”. If someone in one reference frame sees a message as traveling instantaneously, someone in another reference frame will see it as being non-intantaneous, but faster than light, and a third reference frame will see it as going backwards in time. This is part of why people say that relativity implies that nothing can travel faster than light; if something could travel faster than light, then people in different reference frames would disagree about who was the sender and who was the receiver. Thus, your hypothetical would lead to problems with causality, but all reference frames would remain valid.

8

Martin Gardner, as usual, came up with a good explanation of the Twins Paradox, I think its in his book “Relativity for The Million.” I suggest you read that book for a good general explanation of relativity issues.

But anyway, Gardner bases his problem on the frame-of-reference equivalence issue. All frames of references are relative. From one frame, the rocket with a twin accelerates away from the earth until it approaches the speed of light. The twin in the rocket experiences relativistic time compression.

HOWEVER… it is equally valid to say that the rocket stays stationary and the earth accelerates away from the stationary rocket at speeds approaching the speed of light. Thus, the twin on the rocket stays in a “normal” time frame and the twin on earth experiences time dilation. There is no way to determine which frame of reference is “correct.” This is the real twins paradox.

9

There is no real paradox. At least, not as far as Einstein and his theories of relativity are concerned.

10

Well of course. But there is a paradox by Newtonian physics. Its not a paradox in Einsteinian systems. But there’s still no “answer” to the paradox.

11

In what sense? I would say, that there is an answer, in the Einstein relativity context. It’s been worked out neatly, as far as I know.

Unless, you’re talking about the answer. :slight_smile:

12

Well, The answer is, of course, 42, but I don’t see what that has to do with anything. The key is that, in the situation where the astronaut returns to Earth, from the outgoing frame, what you see is a rocket staying still, and the Earth moving quickly to the left (say). Then, at a certain point, the rocket starts moving left even faster, and catches up to the Earth. In this frame, we see that the rocket experiences more time dilation, just like before.
Now, if the rocket never returns, then the astronaut and the Earthbound twin will disagree on who’s older, because they’ll disagree on when to measure their ages. If we have two twins who stay on Earth, and I say that Alice is 23 in 1999, and Bob is only 20 in 1996, does that mean that Alice is older? Of course not, I have to measure their ages at the same time. The problem is that in relativity, you can’t define “the same time” for spatially separated points unless you specify a reference frame. In the astronaut example, in Alice’s frame, Alice is older, and in Bob’s frame, Bob is older.
My favorite analogy for this: I’m in a train, three cars long. I pass the service yard, where there’s another train three cars long on another track, at an angle to my own. As I look out the window, I see that train forshortened by the rotation, so in the direction I define as “length” for my train, the other train only extends two car lengths. In other words, from my reference frame, the other train is shorter than mine. However, someone else in the other train would say that my train is shorter than theirs, for the same reason. Of course, from my point of view, the other train is also wider than mine, and vice versa, in exactly the right way to compensate: sqrt(l[sup]2[/sup] + w[sup]2[/sup]) = 3 cars. Similarly, in relativity, the “distance” between any two events, defined by sqrt(dx[sup]2[/sup] + dy[sup]2[/sup] + dz[sup]2[/sup] - c[sup]2[/sup]dt[sup]2[/sup]) is invarient of the reference frame-- Motion at different velocities reperesents a “rotation” in space-time.