What is the proper time interval for simultaneous events?

Question

Let us take a rod of length L and the two ends A and B have two photon guns.

                        A----------------------B

Let the end A fire a pulse (event 1) and end B fire another pulse (event 2) at the same time as observed by someone who is at rest equidistant from the ends with respect to the rod. It is evident that the observer will infer the events being simultaneous.

Can we take the time interval between the two events 1 and 2 as observed by the mentioned observer as proper time interval? If yes, then the proper time interval is zero in this case. This leads to the conclusion that any frame moving with a relative velocity with this observer will also find the time interval as zero (as zero multiplied with the Lorentz factor is also zero) and the two events simultaneous, which is absurd. What's wrong?

Answer 1

"Can we take the time interval between the two events 1 and 2 as observed by the mentioned observer as proper time interval?"

No, if two events are simultaneous in any frame, the interval between them is a space-like one, not a time-like one. If you're not familiar with the idea of the three categories of intervals, time-like, space-like and light-like, see the "Spacetime intervals in flat space" section of the wikipedia spacetime article. If you know the coordinates of two events in some inertial coordinate system, you can figure out the coordinate differences $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ between the events (i.e. if the first event had x-coordinate $x_1$ and the second had $x_2$, then $\Delta x = x_2 - x_1$), and then the spacetime interval can be defined as $s^2 = \Delta x^2 + \Delta y^2 \Delta z^2 - c^2\Delta t^2$. If $s^2$ is negative the interval is time-like (and the absolute value of $s$ is the proper time between the two events along the world line of an inertial object that passes through both, so this would be the actual time measured on this object's clock between two events which both occurred right next to the object), if $s^2$ is positive the interval is space-like (and $s$ is then the proper distance along a straight line path joining the two events--see my answer here for a little more on what "proper distance" means physically), and if $s^2$ is zero the interval is light-like (meaning only something moving at the speed of light could have both events on its world line).

Also, note that this interval is frame-invariant, which means that if you switched to a different inertial coordinate systems, then even though the individual values of $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ could change, $s^2$ would be unchanged. This is analogous to the geometric case where if you describe two points in a 2D Euclidean plane using two Cartesian coordinate systems with differently-oriented $x$ and $y$ axes, the values of $\Delta x$ and $\Delta y$ may differ depending on the coordinate system, but the distance $d$ between the points, determined by the Pythagorean theorem to be $d^2 = \Delta x^2 + \Delta y^2$, would be the same in both coordinate systems.

Since the interval is frame-invariant, then if you calculate two events to have a space-like separation in the coordinates of one inertial frame, they will have the same space-like separation in all inertial frames, and thus it's meaningless to ask what the proper time between these two events is. And if the events are simultaneous in at least one frame, then in that frame $\Delta t = 0$, so naturally $s^2$ will be positive and the interval must be space-like.

Answer 2

This is really a comment on Hypnosifl's answer but it got a bit long to fit in the comment box.

Be cautious about interpreting the word time in proper time too literally. We use the phrase proper time because it is the time measured by an unaccelerated observer travelling between the two points. However no such observer can exist in this case because that observer would have to travel at an infinite speed. That means the proper time has no physical interpretation.

We tend to use the terms proper distance, $s$, and proper time, $\tau$, interchangably, and they are defined as:

$$ ds^2 = -c^2d\tau^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$

In your example $d\tau^2$ is going to come out negative meaning $d\tau$ will be imaginary. This is an indication that the interval between the spacetime points is spacelike i.e. you would have to travel faster than light to go from one point to the other. In these circumstances you might find it more convenient to work with the proper distance instead.

Note that the proper time and distance are invariants, so if an interval is spacelike in one frame it is spacelike in all frames.

Answer 3

The expression for time dilation which we come across states the fact that the "time interval between two event is minimum when observed from a frame where two events occure at same place".. In your case the events did not occur at the same place.. so you can instead use the lorentz transformation equation to determine the time interval between event A and B which simply gives

t2-t1= v(x2'-x1')/c^2(lorentz factor) ..in this case x2' is not equal to x1' because two events did not occur at same place in the frame(proper)..

Answer 4

vL/c^2 is the difference (for a stationary observer) between two clocks at the two ends of a rod of rest (or proper) length L traveling at velocity v.

The clock at the forward-end (in the direction of v) of the rod trails by this amount (as perceived by the stationary observer).

The two clocks are synchronized in the rest frame of the rod, but for the stationary observer they show a difference of vL/c^2 .