I Twin paradox again

1. Feb 28, 2017

jonjacson

Imagine that we have two metal rods, and both measure 1 meter, a measurement done from the same inertial system when they are at speed = 0. These two rods are equal, the extreme points of the first one are called a and b and its inertial system is called F, the extreme points of the other rod are called c and d and its inertial frame is called F'.

Now imagine there is a third reference system F'' , this system is in the middle of these two rods at point O, the rods are traveling towards O in the same direction but obviously with opposite velocities.

According to F'' the rods at F and F' are coming at speed v and -v.

Everything is disposed so the extremes a and c will coincide at the same space point at some given instant. (Lets imagine they don't collide, it is just they will run very close in a parallel maner, or that they are not metallic simply imaginary).

At some instant in F'' the extremes a and c coincide in the same point O, that means both b and d coincide since the lenghts are the same.
Let's imagine the speeds are close to the speed of light, that means maybe F'' will measure length of rod in F= 0.3 meters, lenght of F'= 0.3 meters.(Nothing strange here)

Now, according to the system F, he is quiet obviously, and the rod at F' is coming with speed= 2v. According to this system the rod in F does not change its size, it measures 1 meter. Also the rod at F' measures 1 meter as observed from F', but this meter compared with the metter of F is much shorter according to F. The conclusion is that points a and c will coincide at the same point, but now seen from F the point b is at a larger distance than the point d, which is closer to c, and from the F rulers is shorter than 1 meter. Is this correct? (In that case points c and d don't coincide and the lenghts are different, How could be this consistent with F'' measurements?)

Now the bizarre stuff is this, What does F' see?

For F' it is F who is traveling, it is the rod at F who will contract the lenghts, and when a and c coincide the point d will be 1 meter away from c, no contraction here, but the distance a - b in these rulers will be shorter than 1 meter. Just the opposite that F measured.

Both cannot be right at the same time.

Do you see the point? Let me know.

UNderstanding that from 1 system another changes its lenght when moving as measured from the "quiet" system looks consistent to me. But if I change to the other moving system I can't see how his measurements could be consistent with the first ones.

2. Feb 28, 2017

Ibix

Look up "relativity of simultaneity". Also, relativistic velocity addition (F' is not moving at 2v as seen from F if both are doing v in opposite directions in F'').

In short, "at the same time" is not a complete sentence in relativity. You can use the Lorentz transforms to determine how each frame will describe your scenario.

3. Feb 28, 2017

phinds

I only got as far in your post as
and stopped reading because right there is likely the problem. To F, the rod in F' is NOT approaching at 2v, he's approaching at something less that c.

Google "Lorentz Transform"

EDIT: I see ibix beat me to it.

4. Feb 28, 2017

jonjacson

I missed that point.

I will calculate the velocity addition using Lorentz and then will see if everything makes sense.

Thanks.

5. Feb 28, 2017

Ibix

You also need to use the Lorentz transforms to show that things that are simultaneous in one frame (e.g. both ends of the rods being coincident in F'') are not simultaneous in other frames. Not realising that is, I think, why you're having trouble imagining how F and F' describe things.

6. Feb 28, 2017

Staff: Mentor

Like @Ibix says, you will also have to use the Lorentz transformations to get the relativity of simultaneity right. There is only one frame in which you can arrange for a and c to be lined up at the same instant that b and d are lined up, and that is F'' (Why? Think about it for a moment). So use that as your initial conditions: in F'' the event "a and c are lined up" has coordinates $(t=0,x=0)$ and the event "b and d are lined up" has coordinates $(t=0,x=L'')$ where $L''$ is the length of the two rods when you do your calculations using the F'' frame.

What are the coordinates of these two events in the frame F? In the frame F'? And as a final sanity check, once you have the coordinates in the frame F, use the Lorentz transformation between F and F' to see if you get the same result for F' as when you transformed from F'' to F'.

7. Feb 28, 2017

jonjacson

I will do it and I will post here the results if they look absurd.

Thanks!

8. Feb 28, 2017

Staff: Mentor

The relativity of simultaneity is very important to understand. Almost all of the "paradoxes" in SR are based on the relativity of simultaneity.

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