Timelike separated events cannot occur at the same time

[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.

 

[Dick]

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.

 

[daselocution]

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...

 

[Dick]

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?

 

[daselocution]

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??)

 

[Dick]

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.

 

[kornha]

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?

 

[Dick]

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.