# B Relativity scenario help

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1. Mar 12, 2017

### MetinErsin

Let's assume there are two observers. Observer B is at rest and observer A is in a spaceship which has transparent walls. A draws line of 1 meter. When A arrives near B, B draws a line of 1 meter. And sees that A's 1 meter is shorter than own 1 meter. before forget i should say that meter is directed along motion. I hope i could explain well.
My question is that how would A see B's line? shorter or longer? If answer is longer, it would be in conradict with relativity. But if answer is shorter, how could this be, it doesn't make sense.

2. Mar 12, 2017

### PeroK

Shorter.

3. Mar 12, 2017

### Staff: Mentor

Why not? From A's point of view, i.e. himself at rest, B is moving.

4. Mar 12, 2017

### Orodruin

Staff Emeritus
It makes perfect sense. Your problem is that you are not accounting for the relativity of simultaneity. This is crucial for the definition of what a length is.

5. Mar 12, 2017

### PeroK

The Earth is orbiting the Sun, because (according to Newton) the Sun exerts a gravitational force on the Earth. The Sun isn't touching the Earth, nor does it have any way to communicate where it is and how massive is it. Yet, the Earth seems to know this and how to respond to this invisible force. How could that be? That doesn't make sense either!

6. Mar 12, 2017

### Staff: Mentor

Aether?

7. Mar 12, 2017

### MetinErsin

Let's say observer A have put his 1 meter stick to wall. Also A have covered with bomb left side of wall from left end of stick to the end of the wall. (Stick's right end is next to cornet of the ship.). And this bombs explode when someone touchs. When A arrives near B, B puts his stick's left end next to corner of the ship. Because A's stick is shorter than itself, right end of the stick of B touch the bombs and A dies. But A's point of view, A also sees that B has shorter stick than itself so the stick of B doesn't reach the left end of the stick of A and so A doesn't die.
How could be this??
I think as Orodruin said i have problems with simultaneity.

8. Mar 12, 2017

### Orodruin

Staff Emeritus
I am sorry, but this sounds quite unintelligible. Regardless, as I said, your problem lies in the relativity of simultaneity. Your statement seems to imply that there is a well defined "when". What "when" is depends on the frame and you therefore cannot state "when someone touches" without specifying what inertial frame that "when" refers to (unless the touching is happening at the same event as the explosion).

9. Mar 12, 2017

### Staff: Mentor

You do. Consider the two events "left end of A's stick touches the wall" and "right end of A's stick touches the wall". Using the frame in which A is at rest, these events have time and space coordinates (x=0,t=0) and (x=1,t=0). Instead of using the length contraction formula (which has a hidden assumption about simultaneity built in) use the Lorentz transformations to find the time and space coordinates of these two events using the frame in which B is at rest.... You will see that the two events no longer happen at the same time, so the spatial distance between where they happen is no longer the length of the stick.

(The length of a stick is the distance between where the two ends are at the same time, which is why I say there is a simultaneity assumption hidden in the length contraction formula).

Last edited: Mar 12, 2017
10. Mar 12, 2017

### Mister T

Study the Ladder in the Barn Paradox (also called the Barn-Pole Paradox). Just Google it. What you are proposing is almost certainly a version of that paradox, but you've worded it in such a way that I'm having trouble following the specifics. Otherwise I'd give a specific response.

11. Mar 14, 2017

### MetinErsin

I try to explain well my question. And i'll solveand if you can say my falses, i'll appreciate,
lets say v=0.5 c=1 from this
γ=1.15
x' = 1.15(x - 0.5t)
t' = 1.15(t - 0.5x) and ofcourse observer at the ship can say i am at rest
x = 1.15(x' + 0.5t')
t = 1.15(t' + 0.5x')
From A's frame of reference: (A is moving)
There is a stone at x'=0 at all times, and there is another stone at x'=2 at all times. And there are bombs x'>2. These bombs explode someone put a stone on it.

At reference of B, B sees at t=0, there is a stone (A's stone) at x=0. And sees There is another stone (A's stone too) at t=0 at x=1.74. So B sees there are bombs at t = 0, at x>1.74.
at t=0, B puts a stone at x=2 and touch the bombs. And both die.

From reference of A
A sees B have put a stone to x'= 2.3 at t'=0.15. And both die.
No problem but as if A, at t'=0, put a stone to x'= 2, because of bombs both would die,
But from b's reference; sees A's action happened at t = 3.15 at x=2.3 so according to B at t=3.15 both die. But until t = 3.15, (i wrote 5 line above)at t=0 B had put a stone and had kill both. So my question is when did observers die to own reference?
I think a died at t'=0 and b died at t=0 is it true, or is it more complicated?

12. Mar 14, 2017

### pixel

While I know what you mean, to be precise there is no such thing as absolute rest, just rest relative to something else.

13. Mar 14, 2017

### pixel

What specifically about this doesn't make sense to you?

14. Mar 14, 2017

### Jeronimus

You have miscalculated something here or so it appears to me.

If A observes an event at x'=2, t'=0

Then B moving at v=0.5c will observe this event at about:

t = 1.15(t' + 0.5x') = 1.15(0 + 0.5*2) = 1.15 (of whatever units you chose)
x = 1.15(x' + 0.5t') = 1.15(2 + 0.5*0) = 2.3 - you got that one right

edit: I miscalculated it as well on my first attempt. Corrected now to 1.15 instead of 1

As you seemingly understand that events which are simultaneous in one reference frame, will not be simultaneous in another reference frame, i don't think the issue here is you not understanding the relativity of simultaneity.

What seems to be the issue is that you do not know how physicists measure/define the length of an object. The length of an object is measured by measuring two endpoints of an object which are __SIMULTANEOUS__.
If you want to measure a rod for example, you can draw the worldlines of the endpoints of this rod. Then you have to find two endpoints which have the same time location in your diagram.
Hence given two endpoints of this rod with x1, t1 and x2, t2. x1≠x2 obviously BUT t1=t2. As you already know, in another reference frame t1' will be ≠t2' so you won't be able to use those two events in another reference frame to measure the length of the rod. You will have to find two events which are simultaneous on the rod's endpoint worldlines.

Last edited: Mar 14, 2017
15. Mar 14, 2017

### Jeronimus

Here is a drawing which should be depicting what you had in mind.

The purple filled circle on the right diagram, is where A observes the event at x'=2.3 t'=0. In the left diagram, this event is observed by B at x=2.3, t=1

The red lines are the worldlines of your bombs at x=2 and x=0 as observed by B.

For A, the distance between the two bombs is about 1.74 as you already calculated.

The green lines in BOTH diagrams, have endpoints with the same t-location. Their endpoints lie on the bomb's worldlines(red lines). Hence the endpoints are __SIMULTANEOUS__ and can be used to calculate the distance between the two bombs.

Hope that helped.

Last edited: Mar 14, 2017
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