0
$\begingroup$

A while ago, I asked a question if two events are always simultaneous in some reference frame. I received excellent answers. The point is that if $E_1$ and $E_2$ are time-like separated with time between events $t > 0$ in some frame $S$, then it can easily be shown that for any frame $S'$, these two events cannot be simultaneous in $S'$.

What I'm curious about now, is if someone (I'm a layman) would draw a Minkowski diagram demonstrating this. So, I would like to see a diagram showing that the events $E_1$ and $E_2$ are not on any line where all events on that line are simultaneous.

$\endgroup$
1

2 Answers 2

3
$\begingroup$

You can draw this yourself. Make the vertical axis time and the horizontal axis space. The lines of simultaneity for various observers are all the lines with slopes of less than 45 degrees (in absolute value). So pick any two points that are connected by a line of slope greater than 45 degrees (or equivalently, draw any line with slope greater than 45 degrees and then pick any two distinct points on that line) and you're done.

$\endgroup$
1
$\begingroup$

Spacetime diagram of relativity of simultaneity

I drew the spacetime diagrams for you. On the l.h.s. you may see two simultaneous events in the unprimed (x,t) frame. The axes of a frame going in the positive direction (the primed frame) should be drawn into the unprimed as I have done it. You can find the new space coordinates by drawing a straight line parallel to the new time axis through the event. The point where your line intersects with the new spacial axis, is the new spacial coordinate. The new time coodrdinate is found by drawing a line parallel to the new spatial axis, and finding its intersection with the new time axis. Green lines indicate the trajectory that light would follow, supposing we are working with units so that $c=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.