Negative squares using the space time interval invariance

1. Sep 17, 2010

ersteller

Hallo
I'm new to this (wonderful) forum, and to SR too...
I've a general question about the space time interval invariance.
Say we have two points A and B, at rest each other, at distance AB.
Now A and B simultaneously in their reference frame emit a flash of light.
The space time interval between these two events is spacelike and equals -(AB)2.
Now a fast spaceship C travels from A toward B and is reached by the two flashes at the middle point between A and B. For C the two flashes are not emitted simultaneously, and I want to use the invariance of the space time interval to compute T, that is the difference in time (measured by C) between the two flashes. The distance between A and B measured by C is shortened by the Lorentz contraction: ABc < AB. The space time interval measured by C should be:
(CT)2 - (ABc)2
and this number should equal -(AB)2. It results
(CT)2 = (ABc)2 - (AB)2 < 0.

So, how can I find T, as a square root of a negative number?

Sorry if the question is stupid....
er

2. Sep 17, 2010

Mentz114

This is a space-time diagram of your scenario as I understand it. The first is from A and B's rest frame, the second is from the ship frame. I think you can read off the time between flashes on the diagram. The ship is travelling at 0.42c in the A,B frame.

I think you must have made a mistake, if you're trying to take the sqrt of a negative number. With this signature (+,-,-,-) a timelike interval is negative, so just ignore the sign, take the square root, the put the sign back.

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3. Sep 17, 2010

ersteller

Thank you

With this signature (+,-,-,-) a SPACElike interval is negative, and this is our case.

And for sure (CT)2 = (ABC)2 - (AB)2 IS negative.

What I find very interesting in your answer is when you say: "just ignore the sign, take the square root, the put the sign back"... Really so simple?

Thank you very much!
er

4. Sep 17, 2010

Mentz114

I don't know if space-like intervals are invariant ( they are, see below) so ignore my advice.

The flashes are at the same time in the A,B frame (tA-tB=0) but clearly not when the diagram is boosted by -0.412c. We're talking about a change in simultaneity here, and I'm not sure how proper length comes into it. I'll try and work it out later.

[later]

Spacelike intervals are invariant, so it seems there's not much point in using one to explain a change in simultaneity. In my diagrams, the interval is sqrt(-100) in both frames. The quantity you want is one side of the triangle that defines the interval.

If two events ( tA,xA), (tB,xB) are boosted by $\gamma$ the transformed time coords are (Y is gamma, b is beta)

t'A = YtA+YbxA
t'B = YtB+YbxB

t'A - t'B = Yb(xA - xB)

which shows that the change in simultaneity depends on the spatial separation, and the relative velocity.

Last edited: Sep 17, 2010
5. Sep 17, 2010

ersteller

Thank you.

My point (now) is not to *find* a number, but only to understand: By using the invariance of the spacetime interval, all quantities in SR kinematics can be found, as I've understood. So my question, that perhaps you have already answered, is: Why have I a negative square in this case?
You said: "just ignore the sign, take the square root, the put the sign back" and perhaps this is the simple answer (but I don't really understand WHY...)

Thanks again
er

6. Sep 17, 2010

Mentz114

The sign of a spacelike interval depends on the signature convention, so there's no harm in changing the sign and then taking the square root and putting sign back.

I still can't see how to get the t'A - t'B in my last post without doing a Lorentz transformation. I'll have another try later.

7. Sep 17, 2010

ersteller

Very good! Now I can do my simple calculation without harm!
Thank you!

Here perhaps I can help you: From the invariance of the spacetime interval

(CT)2 - (ABc)2 = (AB)2

(and without problems arising from the 'negative square'...) one can immediately find T... or not?

Ciao
er

Last edited: Sep 17, 2010
8. Sep 17, 2010

Mentz114

I think that should be

(CT)2 - (ABc)2 = -(AB)2

so

(CT)2= (AB)2( 1/(1-b2) - 1 ) = b2Y2(AB)2

as above. But by using a Lorentz contraction, have you not assumed that an LT connects the frames ?

(it's late so don't assume I've done the math correctly).

9. Sep 17, 2010

ersteller

You're right, I forgot a minus sign...

(CT)2 - (ABc)2 = - (AB)2

that is

(CT)2 = (AB)2 ($$\frac{1}{\gamma^{2}}$$ - 1) (<0 but no more a problem...)

er

10. Sep 17, 2010

Mentz114

Now I am confused ( nothing new there ). This is wrong

[STRIKE](CT)2= (AB)2( 1/(1-b2) - 1 ) = b2Y2(AB)2[/STRIKE]

but is this ?

Now I have doubts the validity of (ABC)=1/Y(AB). This sort of length contraction is because the moving frames perception of simultaneity when measuring the ends of a timelike interval. These events are not causally connected so can length contraction be applied ?

Last edited: Sep 17, 2010
11. Sep 18, 2010

ersteller

I guess YES, but is perhaps better to wait for an explanation....

Now I've less doubts, thank you...
er

12. Sep 19, 2010