Timelike separated events cannot occur at the same time

• daselocution
In summary: Why is ∆s^2=(x1-x2)^2-c^2*(t1-t2)^2? My book kind of led me to believe that it is possible to find s^2 for one event, which would look like s^2= (x1)^2 - (ct1)^2, and so I kind of assumed from this that ∆s^2 would have to have the form I listed earlier. Did I just straight misunderstand my book? Is s^2 a quantity that relates TWO events within one system (do I need to have both (x1, t1) AND (x2, t2) in order to get a
daselocution

Homework Statement

Prove that for a timelike interval, two events can never be considered to occur simultaneously.

Homework Equations

Δs'2=∆s'2

(∆s is invariant)

s2=x2 - (ct)2
s'2=x'2 - (ct')2

The Attempt at a Solution

I first imagined a reference frame K in which two events happened at (x1, t1) and (x2, t2)

Here, ∆s2= x22 - x12 - c2t22 + c2t12 = s'2=∆x'2 - ∆(ct')2 where ∆(ct')2 = 0 (if I'm trying to prove that this CAN'T happen I thought it would be best to show what would happen if it were indeed zero)

...which can be rewritten as:

∆s2= x22 - x12 - c2t22 + c2t12 = s'2=∆x'2 = x2'2 - x1'2< 0 (my book says that for events with timelike separation that ∆s2<0)

I guess I'm not really sure where to head from here. Should I substitute in x'=x-vt...(the lorentz transformations for velocity) and try and solve for some value of v such that (ideally) it might be v>c and thus false? I tried doing that but saw that the algebra seemed pretty long and so I thought that I might not be on the right track...I'm not exactly sure what I'm looking for here, I'm quite new to proof-based math (or physics or really anything). Any guidance would be appreciated.

daselocution said:

Homework Statement

Prove that for a timelike interval, two events can never be considered to occur simultaneously.

Homework Equations

Δs'2=∆s'2

(∆s is invariant)

s2=x2 - (ct)2
s'2=x'2 - (ct')2

The Attempt at a Solution

I first imagined a reference frame K in which two events happened at (x1, t1) and (x2, t2)

Here, ∆s2= x22 - x12 - c2t22 + c2t12 = s'2=∆x'2 - ∆(ct')2 where ∆(ct')2 = 0 (if I'm trying to prove that this CAN'T happen I thought it would be best to show what would happen if it were indeed zero)

...which can be rewritten as:

∆s2= x22 - x12 - c2t22 + c2t12 = s'2=∆x'2 = x2'2 - x1'2< 0 (my book says that for events with timelike separation that ∆s2<0)

I guess I'm not really sure where to head from here. Should I substitute in x'=x-vt...(the lorentz transformations for velocity) and try and solve for some value of v such that (ideally) it might be v>c and thus false? I tried doing that but saw that the algebra seemed pretty long and so I thought that I might not be on the right track...I'm not exactly sure what I'm looking for here, I'm quite new to proof-based math (or physics or really anything). Any guidance would be appreciated.

You are thinking way too complicated. If the separation between two events is timelike then (∆s)^2<0. If the two events can be seen as happening at the same time in some frame then what's the sign of (∆s)^2? Now remember it is supposed to be invariant.

Ah I see what you're saying--the sign would have to be ∆s^2>0 and the sign flip would be inconsistent with ∆s^2 also being invariant. Would it be feasible for me to show that the sign would have to be positive? Should I just state that it is known? I guess I'm not really sure how to prove that s^2 has to be positive in this case...

daselocution said:
Ah I see what you're saying--the sign would have to be ∆s^2>0 and the sign flip would be inconsistent with ∆s^2 also being invariant. Would it be feasible for me to show that the sign would have to be positive? Should I just state that it is known? I guess I'm not really sure how to prove that s^2 has to be positive in this case...

You'd better show it because you've got some misunderstanding about what ∆s^2 means. If (x1,t1) and (x2,t2) are your two events then ∆s^2=(x1-x2)^2-c^2*(t1-t2)^2. Not (x1^2-c^2*t1^2)-(x2^2-c^2*t2^2) or whatever you seem to think it is. Now suppose t1=t2?

Dick said:
You'd better show it because you've got some misunderstanding about what ∆s^2 means. If (x1,t1) and (x2,t2) are your two events then ∆s^2=(x1-x2)^2-c^2*(t1-t2)^2. Not (x1^2-c^2*t1^2)-(x2^2-c^2*t2^2) or whatever you seem to think it is. Now suppose t1=t2?
Why is ∆s^2=(x1-x2)^2-c^2*(t1-t2)^2? My book kind of led me to believe that it is possible to find s^2 for one event, which would look like s^2= (x1)^2 - (ct1)^2, and so I kind of assumed from this that ∆s^2 would have to have the form I listed earlier. Did I just straight misunderstand my book? Is s^2 a quantity that relates TWO events within one system (do I need to have both (x1, t1) AND (x2, t2) in order to get a value for s^2??)

daselocution said:
Why is ∆s^2=(x1-x2)^2-c^2*(t1-t2)^2? My book kind of led me to believe that it is possible to find s^2 for one event, which would look like s^2= (x1)^2 - (ct1)^2, and so I kind of assumed from this that ∆s^2 would have to have the form I listed earlier. Did I just straight misunderstand my book? Is s^2 a quantity that relates TWO events within one system (do I need to have both (x1, t1) AND (x2, t2) in order to get a value for s^2??)

Yes, you are misunderstanding or the book in unclear. The interval ∆s^2 is measured between two events. Just like when you say 'distance' it implies there are two points. If you write ∆s^2= (x1)^2 - (ct1)^2, that's the interval between (x1,t1) and (0,0). The interval is usually written ∆s^2=(∆x)^2-(c∆x)^2, with the ∆'s indicating a difference between the two points. Once you figure this out it will be easy to figure out the sign if t1=t2.

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Can someone clarify this for me? If we are in a timelike interval, X^2-(ct)^2>0, but why exactly can't events occur at the same time? Say they are in the same time and same place?

kornha said:
Can someone clarify this for me? If we are in a timelike interval, X^2-(ct)^2>0, but why exactly can't events occur at the same time? Say they are in the same time and same place?

Timelike means x^2-(ct)^2<0. If they occur at the same time then t=0. x^2 can't be less than zero. If they occur at the same time AND place then the interval is 0. It's not timelike. It's null.

1. What does it mean for two events to be timelike separated?

Timelike separated events are two events in spacetime that cannot be connected by any path that travels at or below the speed of light. This means that there is a definite time interval between the two events and they cannot occur simultaneously.

2. Can two events that are timelike separated occur at the same time?

No, according to the theory of relativity, it is not possible for two events that are timelike separated to occur at the same time. This is because the concept of simultaneity is relative and depends on the observer's frame of reference.

3. How is the concept of causality related to timelike separated events?

Causality refers to the idea that an event cannot occur before its cause. In the case of timelike separated events, there is a definite time interval between the two events, indicating that one event could not have caused the other.

4. Are there any exceptions to the rule that timelike separated events cannot occur at the same time?

There are no known exceptions to this rule. It is a fundamental principle of the theory of relativity and has been supported by numerous experiments and observations.

5. How does the concept of timelike separation affect our understanding of time and space?

The concept of timelike separation is an important aspect of the theory of relativity, which revolutionized our understanding of time and space. It shows that the relationship between time and space is not absolute and can vary depending on the observer's frame of reference.

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