00:01
In this problem, we're going to be looking at space -like separated events and time -like separated events.
00:07
And we're going to ask, in certain situations, look at, does a frame exist where these two events are simultaneous? could they be at the same point? and when we say, does a frame exist? it's basically saying that this prime frame, as prime frame, it's got a speed v less than c.
00:30
If it doesn't c, it exists.
00:31
If it's greater than equal to c, it doesn't exist.
00:34
So that's how we want to look at it.
00:37
Now, what defines space -like, time -like intervals? well, that's this invariant space -time interval of delta -s square.
00:50
And that's c -delta -t squared minus delta -x square.
00:55
And i should mention before i go on, if you have either professor or a book where the minus -sign is on the delta -t term and a plus in the delta -x, it does not matter.
01:05
It doesn't change one thing.
01:07
As long as you're consistent with whatever choice is made, everything comes out exactly the same.
01:14
So when something is time -like, you get a positive here, a negative there, but you also get a positive on the space -like, as you're going to see.
01:26
We're here to be negative.
01:28
Don't worry about it.
01:29
Everything's the same.
01:31
So if delta s -square is less than zero, that's what we mean by space -like.
01:45
This delta x term, the space term, is larger than the delta t term, the time term.
01:53
Space -like.
01:57
Delta s -squared greater than zero applies time -like.
02:07
And there is a third one, even though we won't be using it here.
02:11
Delta s -squared equal to zero.
02:13
This is what is called light -like.
02:18
In terms of light -comb, light -like, the a -8.
02:24
Think of event a at the origin, light like b is on the light cone.
02:31
In time like, b is inside the cone.
02:35
And space like, b is outside.
02:38
That's how visually works.
02:43
What really this is, this is really a representation of a four -dimensional displacement.
02:50
Really what it is.
02:52
Invariant, this is all, whatever this value is in one frame, it's going to be the same value in another frame.
03:01
That's what we mean by that.
03:04
It's always going to have, it's going to have this form.
03:06
C delta t prime squared minus delta x prime square.
03:10
So i was going to have this form.
03:12
It's always going to have the same value in whatever frame.
03:14
Now, i should mention a lot of times this is written, this is written as c squared, tau squared.
03:22
Why? because say we're in the frame where the two events take place at the same point.
03:29
The clocks are giving me proper time.
03:31
Time.
03:33
In that case, delta x prime is zero.
03:37
So this term is gone in that frame, and the delta t in that frame is the proper time tell.
03:44
So that's what you would, that's, sometimes you'll see that.
03:48
Sometimes you'll see that with the proper time written here, gives you something more concrete deal with.
03:57
Okay.
03:58
So now let's look at the cases.
04:00
One, space like.
04:05
That means delta x, delta x squared greater than c delta t squared or delta x greater than c delta t that's what we have part a we want to know we want to prove that a frame exists where a and b are simultaneous so using lorence transformation so this is going to be this is going to be t b prime minus t a prime so you write the the runs transformation for each event and then subtract.
04:54
So let's give me t -a -b minus v.
04:59
Delta x, ab over c -squared, all over 1 -1 minus v squared over c -square.
05:10
So that's what you get.
05:11
Now, if you want them to be simultaneous, that means delta t -a -b -prime equals zero.
05:20
So the bottom is not going to be zero, but at the top.
05:25
So that gives me delta t, ab, is equal to v, delta x, ab over c squared.
05:39
That gives me delta x, ab, is equal to c over v, c, delta t, ab.
05:55
But we know that delta x is greater than c delta t.
06:01
So this is greater than c, delta t, a, b.
06:08
Okay? well, c -d -t -a -b is cancel out.
06:11
So this gives me that c -v is greater than one, or, a factor the v over c less than one.
06:27
That's what we want.
06:29
So the frame exists where they are simultaneous.
06:36
So we proved that there is a frame where a and b will be simultaneous when their separation is space -like.
06:44
But if that's the case, then a and b can have, a, we're assuming it's acting first, at least in our head, it's initially.
06:55
But if a and b are simultaneous, a cannot have any role in what happens at b.
07:02
There's no connection.
07:07
So remember that.
07:11
It can be no connection.
07:12
Whatever goes on at a can have no effect on b.
07:16
They're happening simultaneously.
07:21
So that's very important.
07:26
If you are not, if you cannot be simultaneous, then that means a can affect b.
07:31
And that's not going to be for space -like.
07:36
Now, b wants to show that no frame exists where a and b take place at the same point in space.
07:47
So we do the same thing, but now with the x around transformation, so delta x, a, b, minus v delta t ab square root or minus v squared over c squared and so we want to see can we find a frame a prime frame where delta x ab prime is equal to zero so delta x ab is equal to v delta x ab is equal to v delta t ab which is greater than c delta t ab still space like a and b or instill in case one, case two will do the time record.
08:44
So, the delta tabs go away.
08:47
This gives me that v over c greater than one.
08:53
That cannot be.
08:57
So no frame exists.
09:06
See, the importance, so the importance of the simultaneous, you know, being simultaneous, that connects with who can affect, can a, whatever's having an a effect, what's going to happen and going, it's going to be at b.
09:21
When you talk about being at the same point, this connects to proper time.
09:28
Remember, proper time.
09:31
The two adventure time, it must take place to the same point in space.
09:37
So without this can be, there can be no, there can be no frame that's going to give you the proper time.
09:46
They cannot be at the same, a and b cannot be at the same point.
09:52
That's a criteria, requirement.
09:55
To be proper time.
10:00
So no frame exists there...