GREPhysics.NET: GR8677 #21

GR8677 #21

Problem

GREPhysics.NET Official Solution

Special Relativity $\Rightarrow$ }Metric

Recall that the metric $dS^2=-dt^2+dr^2$ (equivalently, $dS^2=dt^2-dr^2$ )

$dr^2 = dx_i dx_i= (dx)^2+(dy)^2+(dz)^2=4+0+4=8$ and $dt^2=4$ . Thus $dS^2=4$ , and $S=2$ , as in choice (C).

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Comments

FortranMan
2008-10-17 13:28:48

So this is all about how the interval ds is invariant because c is the same in all inertial systems in relative motion, right? So the relations are given as follows.

$\Delta s^{2} = \Delta r^{2} - c^{2} \Delta t^{2} = 0$

$\Delta {s'}^{2} = \Delta {r'}^{2} - c^{2} \Delta {t'}^{2} = 0$

$\Delta {s}^{2} = \Delta {s'}^{2}$

So how on Earth do you manage to get rid of the $c^{2}$ ? I've tried redefining dt as

$\Delta t = \frac{\Delta r}{v}$

or using the length contraction eq.

$L = \frac{L_{0}}{\gamma}$

Both get me closer to 2 compared to the other given answers, but nothing I can think of can give me an exact answer of two. As for the answer explanation here, I am just fairly uncomfortable with the idea that adding two different units together can get the correct answer.

mangree
2008-10-18 07:03:30

The problem tells you to use units such that the speed of light is 1.This is how you "get rid" of c^2 (couldn't manage the syntax for that).

As for the different units,if c=1 then (m/s)=1 $\Rightarrow$ m=s

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rajsareen
2006-12-01 04:32:27

If you use the equation for the interval in paranthesis (equivalenty ... ) you get -4, upon which the length becomes imaginary. Maybe only the first equation is appropriate?

VanishingHitchwriter
2006-12-01 14:01:01

Note that imaginary numbers are just another tool to use when doing relativity calculations. You can either use a metric or use imaginary numbers. Both views are valid.

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dicerandom
2006-09-09 12:06:07

The other thing you need to recall in order to do this problem quickly is that the spacetime interval between two events is invariant, i.e. it does not depend on the reference frame. By asking what the interval is in the O' frame I think the problem is trying to trick you into transforming the events into that frame before calculating the interval.

bkardon
2007-10-05 13:56:18

I agree, although one should arrive at the correct answer after Lorentz-transforming the events, it would just be a big waste of time. Which is just the kind of evil trick ETS likes.

lattes
2008-10-13 08:11:29

agree with both of you. The key fact here is to remember that space-time intervals are invariant.

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