Can we observe two different events occuring at the same point?

Question

I was wondering a situation. Suppose a train is moving with a certain acceleration. It turns on its front light at a certain time, and it's back light say t s after. But to a inertial, or non inertial reference frame, can the two events appear to occur at the same time?

Answer 1

The situation is quite possible.As you have not mentioned any Exact value of acceleration or time between the two lights opening it is difficult to tell exact values which will satisfy your event.But I have made a similar thing which may help you understand.

Imagine AE is the train and B,C,D are points on it such that AB=BC=CD=DE.Now AE is moving with a constant acceleration 'a'.Imagine a stationary body ,X, at D.Now when D and E coincide the front light is switched on at E.As the wave fronts from E reach X the train has moved to position C.Meanwhile when the body X is midpoint between D and C The back light at A is switched on.The events happen in a way such that when body X reach position C the wave fronts from A and E reach C.So to X the events are simultaneous. P.S.-Wave fronts from A take lesser time to reach C than E because the train is acceleration from A to C. Hope this helps!

Answer 2

The answer is, "as long as the light from the front of the train has not hit the back of the train first, then yes, there exists some reference frame which thinks of both as simultaneous."

In special relativity, there is a number which everyone agrees on: take any two events that are separated by a vector (in your coordinates) $\vec r$ and time interval $t$, then you can calculate what's called their spacetime interval, $$I = c^2 t^2 - \vec r \cdot \vec r = c^2 t^2 - x^2 - y^2 - z^2.$$It turns out that everyone who calculates this will get the same answer for those two events.

As you can see, the number can be positive or negative. We say that positive intervals are "timelike separated." Everyone agrees that they are timelike separated, that the one happened before the other. If you think of the "bubbles" of light that expand outward at speed $c$, announcing the two events as fast as anything can, one bubble is objectively "inside" the other one and everyone will agree on this fact.

Negative intervals are "spacelike separated". Everyone agrees that they are spacelike separated, that the two happened at different places. If you think of the "bubbles" of light again, these are overlapping bubbles. Neither one is inside the other.

If something is objectively timelike separated then it is not objectively spacelike separated. This is actually a lot easier than it sounds. The space inside of one of these "light bubbles" is the set of points which an observer, who was at that event, can visit in a spaceship moving at some constant speed. You can't go faster than light, out of the bubble, but you can go at any speed slower than light, inside the bubble. And if one bubble begins inside another one, that means that some potential observer, in an inertial reference frame, was "at" both events. Where did they happen? For that observer, they both happened "right here". So they have no objective spacelike separation.

The reverse is also true, but it may be a little harder to see. Lorentz transforms map those light-bubbles onto other light-bubbles, preserving their geometry (which is inside of which other one) but not their relative sizes at any particular point in time. One special case: consider an observer whose spaceship is, at some time just on the intersection between the two bubbles, so they are seeing both events "simultaneously." The way Lorentz transforms work is, based on how fast and which direction she is going, she will measure different distances to those two events: but she thinks that the light still goes at the same speed $c$. There is always a choice of her speed such that she will measure the distances to both events to be the same, and therefore will think that they happened simultaneously with each other. So objectively spacelike separated things are not objectively timelike separated: there are reference frames who think that A happened before B, or B happened before A, or both are identical.

And the threshold between them, one bubble beginning inside-of or outside-of the other, corresponds exactly to the light from the one event reaching or not-reaching the event, respectively.

Answer 3

Let me describe a geometrical way to approach your question. We start by drawing a spacetime diagram with distance along the horizontal axis and time on the vertical axis. We'll also take the speed of light to be $1$, so on our diagram a light ray travels along the line $x = t$ i.e. at 45º.

If we have some other observer moving at a speed $v$ relative to us they will have their own spacetime diagram with $x'$ and $t'$ axes. If we draw their axes on our graph what we get is:

So the moving observer's $t$ and $x$ axes are rotated towards the line $x = t$ by an angle $\theta$ that is related to the relative speed of the moving observer. The angle $\theta$ is given by:

$$ \tan\theta = \frac{v}{c} $$

where $v$ is the velocity of the moving observer. If the moving observer has a relative velocity of $c$ then $\tan\theta = 1$ and therefore $\theta$ is 45º. So as we increase the relative velocity the moving observer's $x'$ axis rotates upwards and at $v = c$ merges with the path of the light ray.

The point of all this is that we can now easily answer your question. Let's draw the two spacetime points corresponding to turning the front then the rear lights. For convenience we'll put the first point at the origin so the second point lies somewhere to the upper right of it:

Actually I've drawn two pairs of points, one pair in green and one point in red. Now your question is:

But to a inertial, or non inertial reference frame, can the two events appear to occur at the same time?

and what that means is for the moving observer the two points have to lie on the $x'$ axis, because all points on the $x'$ axis have the same value of $t'$ and are therefore simultaneous.

If you consider the two red points then as the moving observer speeds up the $x'$ axis rotates upwards and somewhere it will lie along both red points. So there is some speed where both red points lie on the $x'$ axis and the two red points can appear simultaneous to a moving observer.

But consider the two green points. These can never both lie on the $x'$ axis because the $x'$ axis can never rotate upwards by more than 45º so it can never reach the second green point. That means the two green points cannot appear to be simultaneous to a moving observer.

So your question can be answered simply by looking to see where the two points appear on the spacetime diagram. In fact there's a very simple way to see whether the can appear simultaneous or not. If the time between the two points is $\Delta t$ and the distance between them is $\Delta x$ then to travel between the two points means travelling at a speed:

$$ v = \frac{\Delta x}{\Delta t} $$

If this speed is less than $c$ (green points) then they can't appear simultaneous to a moving observer while if $v$ is greater than $c$ they can.

Answer 4

Events can never univerally only be said to happen in a particular order if they are space-like events, i.e. $(\Delta r)^2 > c^2(\Delta t)^2$. If this is true, the there is no universal agreement about the order in which the events happen.

If, however, $(\Delta r)^2 ≤ c^2(\Delta t)^2$ holds true, then every observer will agree on the order.