# Timelike, Spacelike and Null Seperation

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• Aug 5th 2012, 07:01 AM
AA23
Timelike, Spacelike and Null Seperation
I have been given the following problem where I have been asked to find two events that have a timelike, spacelike and null separation. I have obtained an answer for each however am unsure if I have mixed up spacelike and timelike. Could someone tell me if I am correct or not? Thank You

• Aug 6th 2012, 06:11 AM
JohnDMalcolm
Re: Timelike, Spacelike and Null Seperation
You're using the convention ds^2 = dt^2 - dx^2. So if you get ds^2 > 0, this means that the time part, dt^2, of your equation dominates, making the separation 'time-like'. On the otherhand, if ds^2 < 0, then your space part, dx^2, would dominate making the separation 'space-like.' (You weren't getting them mixed up in your solutions).

As a side note, you need to be careful of which convention you're working in. Some texts use ds^2 = -dt^2 + ds^2, which means ds^2 < 0 is timelike. The property of the separation doesn't change, just which regimes you choose to call negative and positive.
• Aug 6th 2012, 06:14 AM
AA23
Re: Timelike, Spacelike and Null Seperation
Thank you JohnDMalcolm for your response. As a follow up could you help me with (iv) and (v), these seem very difficult to me and am unsure where to begin answering them. Thank you once again .
• Aug 6th 2012, 09:47 AM
JohnDMalcolm
Re: Timelike, Spacelike and Null Seperation
For part (iv) you need to (Lorentz) transform the event coordinates to the frame of the new observer, who is travelling at $\displaystyle \beta = \frac{v}{c} = \frac{1}{4}$ relative to the initial observer. Two events are simultaneous in a particular frame if they have the same time coordinate. Similarly, for part (v) (which has a different $\displaystyle \beta$), two events occur at the same location in a particular frame if they have the same spatial coordinate.
• Aug 6th 2012, 10:04 AM
AA23
Re: Timelike, Spacelike and Null Seperation
Taking what you have said on board, is the following correct for (iv) ??
• Aug 6th 2012, 10:20 AM
AA23
Re: Timelike, Spacelike and Null Seperation
Scratch that last post, I have spotted an error. The answer should be E_1 and E_2 since they both produce a t value of 13/SQRt15

Do you agree?
• Aug 6th 2012, 10:28 AM
JohnDMalcolm
Re: Timelike, Spacelike and Null Seperation
Yeah, pal. You got it.
• Aug 6th 2012, 10:37 AM
AA23
Re: Timelike, Spacelike and Null Seperation
Cheers mate (Happy) and just for (v) I get an x value of 0 for E_1 and E_3, which would make them occur at the same location?
• Aug 6th 2012, 11:07 AM
JohnDMalcolm
Re: Timelike, Spacelike and Null Seperation
You betcha.
• Aug 8th 2012, 08:24 AM
AA23
Re: Timelike, Spacelike and Null Seperation
Thank You JohnDMalcolm, I do have a follow up question that has been causing me problems. I have answered the first part but the second part is confusing me. Please see attachment, Thanks again (Nod)
• Aug 8th 2012, 09:54 AM
JohnDMalcolm
Re: Timelike, Spacelike and Null Seperation
Based on the Lorentz transformation, you have expressions for t' and x' in terms of t and x. Plug these into $\displaystyle \left( t' \right)^{2} - \left( x' \right)^{2}$ and show that it equals $\displaystyle t^{2} - x^{2}$.
• Aug 8th 2012, 12:47 PM
AA23
Re: Timelike, Spacelike and Null Seperation
I'm sorry I still don't quite understand, is the following correct? See attachment.
• Aug 8th 2012, 02:53 PM
HallsofIvy
Re: Timelike, Spacelike and Null Seperation
You don't really need to do the squares here (for x and t positive, $x^2> t^2$ if and only if x> t). If it is possible to cover the spacial distance between two events in the given time difference without going over thespeed of light, the interval is "space like". If light could cover the distance in the given time (the distance in lightyears is the same as the time difference in years) it is a "null separation". If it would take more that the given time difference in years to cover the spacial distance in light years, it is "time like". Comparing E1(4,3) with E2(6, 10) we see that the time difference is 6-4= 2 years and the space difference is 10- 3= 7 lightyears.You can't go 7 light years in 2 years- this is "time like". Comparing E1 with E3(8, 6) we see that the time difference is 8- 4= 4 years and the space difference is 6- 3= 3 iightyears. We can go that distance in that time. This is "space like". Comparing E1 and E4(10,4) the time difference is 10- 4= 6 years and the space difference is 4- 3= 1 light year. Yes, we can 1 lightyear in 6 years- this is "space like", etc.
• Aug 8th 2012, 03:29 PM
AA23
Re: Timelike, Spacelike and Null Seperation
Hey HallsofIvy

So based on your post are my original calculations incorrect in #1, because the notes I have been given would suggest my method is correct?
• Aug 9th 2012, 01:08 AM
JohnDMalcolm
Re: Timelike, Spacelike and Null Seperation
Quote:

Originally Posted by AA23
I'm sorry I still don't quite understand, is the following correct? See attachment.

So you have your expressions t'(t,x) and x'(t,x). Just plug these into the left-hand side of your equation. Some manipulations will show that this is equal to the right-hand side. You only need to work on the left side and show that you get the right side.

$\displaystyle LHS = \left( t' \right)^{2} - \left( x' \right)^{2}$
$\displaystyle = \left[ \gamma \left( t - \frac{vx}{c^{2}} \right) \right]^{2} - \left[ \gamma \left( x - vt \right) \right]^{2}$

$\displaystyle ...$

$\displaystyle = t^{2} - x^{2}$
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