GR8677 #21



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Comments 
blacksheep 20170920 11:36:53  Although this site doesn\'t see a lot of activity any more, maybe someone will see my question: every source I consult says the spacetime interval is (deltas)^2, NOT deltas. Yet clearly it\'s the latter quantity ETS is after here. It\'s not a problem with respect to answering the question, since \"4\" is not among the answer choices, but what gives?
misael24 20171019 07:04:11 
Hello, yeah not many people are using this site but it is proving to be very helpful in studying for the GRE.\r\n\r\n Everywhere that I look I seem to find s^2 = [delta r]^2  c^2[delta t]^2. I did find other notations but the other was simply s = [r^2 (c^2)(t^2)]^(1/2).\r\n

  mpdude8 20120415 20:29:10  As the original poster points out, this boils down to finding the distance between two points (sqrt((x1  x2)^2 + (y1y2)^2 + (z1z2)^2)  (t1t2)^2). The only part is remembering to subtract the change in time, rather than add.   FortranMan 20081017 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.
So how on Earth do you manage to get rid of the ? I've tried redefining dt as
or using the length contraction eq.
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 20081018 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 m=s

  rajsareen 20061201 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 20061201 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.

  dicerandom 20060909 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 20071005 13:56:18 
I agree, although one should arrive at the correct answer after Lorentztransforming the events, it would just be a big waste of time. Which is just the kind of evil trick ETS likes.

lattes 20081013 08:11:29 
agree with both of you. The key fact here is to remember that spacetime intervals are invariant.

alemsalem 20100920 06:26:00 
the key fact is that ETS likes evil tricks :)

neon37 20101101 08:52:54 
I dont think this is an evil trick, on the contrary, it is a fundamental concept that the metric is invariant for all inertial reference frames.

 




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