Different time coordinates due to different velocities

[Timothy S.]

We have 2+1 spacetime (x,y,t) and objects a) and b). The velocity of the object a) is greater than the velocity of the object b), so, according to special theory of relativity the time passes for the object a) slower than for object b) (t_2 is greater than t_1). So is the pinned picture correct: on the plain t_1 there is only obj. b) and only obj. b) on the t_2 plain?

And if the answer was yes, than in our reality all of the objects with non-zero relative velocities would have different time coordinates but still be able to interact. What does theory of relativity says about that?

Thanks

 

[Orodruin]

No, nothing of this is correct.

 

[Dale]

We have 2+1 spacetime (x,y,t) and objects a) and b). The velocity of the object a) is greater than the velocity of the object b)

You actually need more structure than you have given in order to state that. In addition to the spacetime, you need a coordinate chart or a tetrad, often called a reference frame. I assume you want to use the frame where b) is at rest.

So is the pinned picture correct: on the plain t_1 there is only obj. b) and only obj. b) on the t_2 plain?

No. Objects do not pop in and out of existence. They have continuous worldlines.

Assuming they are neither being destroyed nor created during this scenario then both objects will exist on all “plains”.

The only thing is that a)’s clock will read a different number than the coordinate time.

When you reset your watch, do you suddenly find that you stopped existing during the missing time?

 

[pervect]

I only see one object on your space-time diagram, that is a sideways (time going left-right) space-time diagram of a light clock with three cycles. (Usually space-time diagrams are drawn with the time axis vertical, not horizontal, but I can adapt to that. It's hard to tell from your 2d rendering of a 3d diagram, but it appears that x and y are constant at the center of the clock, while t progresses which is why I say it is a diagram of a stationary light clock, because x=y=constant is a stationary object).

I don't see any moving objects on your diagram at all. If you intended to make a space-time diagram with a moving object, you didn't do it correctly.

I would suggest doing a simpler task first. Draw a space-time diagram with only t and x, and draw the space-time diagram of a stationary object (x = constant), and a moving object (x = vt).

Next up, draw a space-time diagram of a stationary light clock, which should be similar to what you've drawn. It will be easiest if you make the diagram 2d - after you've got a correct 2d diagram (with x and t), you can add y to the picture, but it won't make any material difference.

The last step is to draw the space-time diagram of a moving light clock. I'm guessing that may have been your intention, but all I see is a space-time diagram of a stationary light clock and no hint of any moving objects at all.

[add] I'd also recommend drawing your diagrams at a scale where light always moves at a 45 degree angle from the x and t axis on the diagram.

 

[Mister T]

The velocity of the object a) is greater than the velocity of the object b), so, according to special theory of relativity the time passes for the object a) slower than for object b) (t_2 is greater than t_1).

This sounds like you are thinking of the velocity of an object as a property of that object. It is not. The concept makes sense only when you compare the velocity of an object to something else. In this case a) is faster than b), but that makes sense only if you compare their velocity to a third object.

You can say that a) and b) are in motion relative to each other, but the speed of a) relative to b) equals the speed of b) relative to a).

Moreover, time dilation is symmetrical. You can describe it by saying a) will observe b)'s clocks to be running slow, but b) will observe a)'s clocks to be running slow. An apparent contradiction until you understand how the consideration of relativity of simultaneity makes it possible.