# Does the Twin Die?

1. Oct 22, 2006

### jmenke

Twins Variation

[Mentor's note: I split this off from the "twins paradox confusion" thread where it was originally posted because that thread is pretty long already and this is not a continuation of the scenario being discussed there. It would be confusing if M1keh comes back to the other thread and we end up discussing both his scenario and yours simultaneously!]
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I wonder if someone might be able to offer an explanation of the following somewhat related twins paradox I just thought up. I've attached a crude drawing to better explain it.

Twin's names are A and B.

A and B travel @ .9c in a straight line, A following B, away from Earth.
A shoots a laser at B.

A sees the laser slowly catching up to B but B turns before it hits arriving safetly back at Earth.

B sees the laser traveling towards him @ c, crashing into him before he makes the turn and killing him.

Can A and B go for coffee?

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Last edited by a moderator: Oct 22, 2006
2. Oct 22, 2006

### JesseM

This scenario isn't very specific, and I don't understand your diagram (what do the three arrows represent?)...are they both travelling at 0.9c relative to the earth, in the same direction? If so then they both share the same rest frame, so they wouldn't disagree about simultaneity or the distance between them or anything else. If one is moving at 0.9c relative to the other, then you need to specify stuff like how far apart they are at the moment A fires the laser (in A's frame or in B's frame), and how long between the moment A fires the laser and the moment B is planning to make the turn (again, in your choice of frame, since the time-interval will be different in the two frames). Once you specify these sorts of details, it should be relatively easy to show that both frames will agree on whether the laser kills B before he can make the turn.

3. Oct 22, 2006

### pess5

Could you explain how you figure that B will experience the laser striking him if A does not?

pess

4. Oct 22, 2006

### jmenke

Sorry for the confusion with the earlier example, I forgot to mention that either A stops after firing or that there is a 3rd party on the Earth. With the original A certainly would see the same as B.

The arrows represent their planned course. A and B do share rest frame.
The distances at the time of firing could be:
Earth->A: 10ls
A->B: 20ls
B-> turn: 80ls
Both A and B travel at .9c

For clarity I'll use Earth as the stationary observer and have A simply continue to follow B instead of halting (to avoid A changing FOR relative to B).
A fires the laser at B at t.

From A and B's perspective: The laser travels from A to B @ c colliding at t+20seconds when B has travelled 18ls(0.9c*20ls) since t and is 62ls(80-18) or 68.888(62/0.9) seconds from the turn.
B dies.

Earth sibling sees: A fires at B and the laser travels @ c from A to B but because B is moving away from the Earth at .9c the distance between the laser and B from Earth's perspective decreases at a rate of only .1ls/s(1-0.9).
80ls to B's turn at .9c=88.888(80/0.9) seconds before the turn. Distance between laser and B at t is 20ls / .1ls/s = 200seconds before laser hits B. B reaches the turn ~112seconds before the laser would have caught up to him.
B lives.

I hope this is more clear.

Last edited: Oct 22, 2006
5. Oct 22, 2006

### Hurkyl

Staff Emeritus
The distance between A and B as measured in Earth's rest frame is not the same as the distance between A and B as measured in B's rest frame.

6. Oct 22, 2006

### pervect

Staff Emeritus
B sees the laser coming towards him at 'c', but B also sees the distance before the turn as being .1 of what A sees the distance as being.

So both A and B will agree that either B makes it, or he doesn't. One could say this in advance - in relativity,at least, events either happen, or they don't happen. Everyone will have different perceptions of the events due to light speed delays, but they will all agree on what happened.

So either the laser strikes B, or he turns in time. These are facts, and both A and B will agree on them.

7. Oct 22, 2006

### JesseM

Oh OK, you didn't specify that A would stop in the Earth's frame.
Again, can you specify how far B is from A (in the Earth's frame, say) when A fires the laser, and how far B must be from A before he can turn? (Or else how much time must pass between the event of A firing and the moment when B can turn, if you prefer.) For example, if B is 0.5 light-years from A at the moment the laser is fired in the Earth's frame, then in the Earth's frame B's distance from A as a function of the time t since the laser was fired will be (0.9c)*t + 0.5, while the laser's distance from A will be (1c)*t. So, the time needed for the laser to catch up with B can be found by solving the equation (1c)*t = (0.9c)*t + 0.5, which can be solved to give t = 5 years. During this time B will have travelled an additional (5 years)*(0.9c) = 4.5 light-years in the Earth's frame, and since in the Earth's frame B started out 0.5 light-years from A at the moment the laser was fired, this would be at a distance of 4.5 + 0.5 = 5 light years from A in the Earth's frame. So in the Earth's frame, as long as the distance between A and the "turnaround point" for B is less than 5 light years, B will reach the turnaround point before the laser hits it. So if you pick a distance to the turnaround point--say, 4.9 light-years from A as seen in the Earth's frame--then you can do the math and see that in B's frame, B will still reach the turnaround point before the laser hits him, so there is no disagreement between frames. I can do the math for the figures I gave above, or you can pick your own numbers if you would prefer, but you need some set of detailed numbers in order to analyze the problem.

8. Oct 22, 2006

### jmenke

You guys are too quick, I edited my second post :P
The distances can be found there.

Last edited: Oct 22, 2006
9. Oct 22, 2006

### JesseM

Distances in which frame? You understand that measures of distance differ from one frame to another, right?
You mean the distance between B and the turn at the moment A fired the laser, in the Earth's frame?

If so, in the Earth's frame B turns before being hit--the laser takes 100 seconds to reach the point of the turn, but B only takes (80 ls)/(0.9c) = 88.89 seconds to reach the turn.

If you imagine there's a space-buoy which is at rest relative to the Earth, and B makes the turn right when it passes the buoy, then in the Earth's frame the buoy must be 20 + 80 = 100 ls from A at the moment A fires the laser. If you imagine the buoy is attached to one end of a ruler which is 100 ls long in the earth's frame, then A would be passing the other end of the ruler at the same moment it fired the laser. But because of length contraction (given by the formula $$L = L_0 * \sqrt{1 - v^2/c^2}$$, where $$L_0$$ is the length of an object in its own rest frame and L is its length in a frame where the object is moving at speed v), in the rest frame of A and B the ruler will only be (100 ls)*(1 - 0.9^2) = 43.589 ls long, which is the distance between A and the buoy at the moment A fires the laser in the A-B rest frame. On the other hand, since length contraction causes the distance between two objects which are at rest with respect to each other to appear shorter in other frames than in their own rest frame, we know the distance between A and B in their rest frame must be longer than the contracted distance of 20 ls seen in the Earth's frame--we can figure out the distance in their rest frame by rearranging the length contraction formula to read $$L_0 = \frac{L}{\sqrt{1 - v^2/c^2}}$$, which tells us that the distance between A and B in their rest frame must be (20 ls)/sqrt(1 - 0.9^2) = 45.883 ls. So in the A-B rest frame, the distance between A and the turnaround point at the moment A fires the laser, or 43.589 ls, is actually shorter than the distance between A and B! This means that in the A and B rest frame, B actually turns around before A even fires the laser (since the difference in distances is 45.883 - 43.589 = 2.294 ls, B must have turned around 2.294/0.9 = 2.55 seconds before A fired the laser in this frame), so there is no danger that B will get fried. This is just a consequence of the relativity of simultaneity, which says that different frames can disagree on the order of events which have a "spacelike" separation.

You could also get the same conclusion by finding the coordinates of the laser-firing and B's turnaround in the Earth's frame, then using the Lorentz transformation to find their coordinates in the A-B rest frame. If x=0 and t=0 are defined as the position and time that A fires the laser in the Earth-frame's coordinate system, then the coordinates of B reaching the turnaround are x=100 ls, t=88.89 s. Now if we use the coordinates x' and t' to represent the position and time of different events in the A-B frame's coordinate system, with x'=0 and t'=0 also defined as the position and time of A firing the laser, the Lorentz transform tells us that the two coordinate systems relate like this:

$$x' = \gamma (x - vt)$$
$$t' = \gamma (t - vx/c^2)$$
with $$\gamma = 1/(1 - v^2/c^2)$$

So with v=0.9c, in this case $$\gamma$$ is 2.294. So if the coordinates of B reaching the turnaround are x=100 ls, t=88.89 s in the Earth frame, in the A-B frame they'd be:

x' = 2.294*(100 - 0.9*88.89) = 45.88 ls
t' = 2.294(88.89 - 0.9*100) = -2.55 s

So again we see that the event of B's turnaround happens 2.55 seconds before the event of A firing the laser, in the A-B frame.

Last edited: Oct 22, 2006
10. Oct 22, 2006

### jmenke

JesseM: Thanks for the reply. I indeed did not take length contraction of the distances in the AB frame into account and I think I understand the solution you posted. Thanks to everyone for the help!