Timelike separation: which of these two answers do you prefer?

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In summary: I took differential geometry.In summary, the problem is to prove that for a timelike interval, two events cannot occur simultaneously. The first attempt involves showing that there is no reference frame where this can occur, but this is not true in general. The second attempt is a simpler and more accurate solution, showing that for a timelike interval, there is always a reference frame where ∆s^2<0, and therefore it is impossible for two events to occur simultaneously.
  • #1
daselocution
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Homework Statement


This is the problem as stated in my textbook: Prove that for a timelike interval, two events can never be considered to occur simultaneously.

I thought about two different ways to go about this; both of these ways make sense to me, but I'm not sure if they both make sense to all of you/if one is significantly better/different from the other


Homework Equations


Δs'2=∆s'2

(∆s is invariant)

s^2=x^2 - (ct)^2
s'^2=x'^2 - (ct')^2

For a timelike interval, it is given that: ∆s^2<0



The Attempt at a Solution



The first way I thought of doing the problem: At first I thought that I should be showing that if two events occur at (x1, t1) and (x2, t2) in system K, then there is NO frame K' such that these two events could occur at the same time (simultaneously). If this is a misunderstanding or seems tautological, then please correct me.

I went about proving it as follows:
∆s^2=∆x^2 - (c∆t)^2 = ∆x'^2 - (c∆t')^2 = ∆s'^2

Where, in K', we are assuming that t2'-t1'=0 such that:

∆s^2=∆x^2 - (c∆t)^2 = ∆x'^2 = ∆s'^2

From here, it follows that as ∆s^2=∆x^2 - (c∆t)^2<0 for it to be timelike separation, this entire quantity must be negative in sign. However, the right side of the equation, ∆x'^2, must necessarily be positive. Because it is given that s^2 is invariant, it must also be true that ∆s^2 is invariant from reference frame to reference frame, and so this must be impossible because ∆s^2≠∆s'^2

The second way I tried to solve the problem: At this point I had doubts about whether my first solution made sense, and so I tried to solve it all in the context of ONE inertial frame:

∆s^2=∆x^2 - (c∆t)^2

For the interval to be timelike, ∆s^2=∆x^2 - (c∆t)^2<0;
thus, if we are assuming that t1=t2, then:

∆s^2=∆x^2 - 0 CANNOT be less than 0, as it MUST be a positive number or it must be zero, neither of which allow for a timelike interval.

What do all of you have to say? Are both wrong? One right one wrong? One better?
 
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  • #2
The second one is much cleaner.
I think I would begin with "for a timelike interval, there is a reference frame where ∆s^2<0. As this is invariant, ∆s^2<0 for all reference frames. Assume there is a frame where ∆t=0, ..."

At first I thought that I should be showing that if two events occur at (x1, t1) and (x2, t2) in system K, then there is NO frame K' such that these two events could occur at the same time (simultaneously).
In general, this is not true. As an example, consider (x1,0) and (x2,0), and K'=K. You have to add some constraints on those coordinates, and I don't see the advantage of this.
 
  • #3
Ahhh thank you for clearing that up. I don't think I actually fully understood what I was getting into with the first attempt...
 

1. What is timelike separation?

Timelike separation refers to the relationship between two events in spacetime where one event can influence the other. It is a term used in the theory of relativity to describe the temporal distance between two events.

2. How is timelike separation different from spacelike separation?

Timelike separation is different from spacelike separation in that it involves a temporal distance between events, while spacelike separation involves a spatial distance between events. In other words, timelike separation represents the time it takes for an event to influence another event, while spacelike separation represents the physical distance between the two events.

3. What is the significance of timelike separation in the theory of relativity?

Timelike separation is significant in the theory of relativity because it is used to determine the causality between events. It helps us understand how one event can affect another event in spacetime and plays a crucial role in understanding the concept of time dilation.

4. Can timelike separation be negative or zero?

No, timelike separation cannot be negative or zero. In the theory of relativity, time is always considered to be a positive quantity, and therefore, timelike separation can only have positive values. Negative or zero values would violate the principles of causality and contradict the theory of relativity.

5. Which answer do you prefer between the two options given?

As a scientist, I do not have a preference between the two options given. Both options can be used to explain the concept of timelike separation, and the choice ultimately depends on the context in which it is being used and the individual's personal understanding and interpretation of the concept.

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