GREPhysics.NET: GR0177 #34

GR0177 #34

Problem

GREPhysics.NET Official Solution

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Special Relativity $\Rightarrow$ }Spacetime Interval

The spacetime interval for the convention $ds^2 = c^2 dt^2 - dx^2$ has regions where the slope is greater than that of the lightcone line, i.e., $ct=x$ , to be timelike. Slopes that are less than the lightcone slope on the plot of x (horizontal) vs ct (vertical) correspond to spacelike phenomena. In this region, one can have two observers disagree on whether an event happens before the other. Thus, one wants $ct<x$ , as in choice (C).

Alternate Solutions

thebigshow500 2008-10-12 16:10:18	Just Check out Griffiths p.502 - "The invariant interval." Everything is super clear there. Mr. Griffiths derives the result of interval between two events as I = $d^2-c^2t^2$ I < 0 refers to timelike interval. Two events occur simultaneously. I > 0 refers to spacelike interval. Two events occur at the same place. I = 0 refers to lightike interval. Two events are connected by a signal traveling at c! For this question, I > 0, so the term $d^2$ dominates $c^2t^2$ . One can deduct the answer (C)!! Reply to this comment
ivalmian 2008-03-25 19:02:24	You can also show this through a little math. We know that for 2 arbitrary reference frames $R_1$ and $R_2$ , the time intervals between the two events is: $S_1 = S_2$ Plugging in $\Delta x_1$ , $\Delta t_1$ , $\Delta x_2$ , and $\Delta t_2$ we get: ${\Delta x_1}^2-{\Delta t_1}^2 c^2={\Delta x_2}^2-{\Delta t_2}^2 c^2$ If in $R_2$ the event is perceived as simultaneous, then $\Delta t_2=0$ . Thus we get: ${\Delta x_1}^2-{\Delta t_1}^2 c^2={\Delta x_2}^2$ Dividing by $\Delta t_1$ we get: $\frac{{\Delta x_1}^2}{{\Delta t_1}^2 } - c^2=\frac{{\Delta x_2}^2}{{\Delta t_1}^2 }$ Adding $c^2$ to both sides we get: $\frac{{\Delta x_1}^2}{{\Delta t_1}^2 } =\frac{{\Delta x_2}^2}{{\Delta t_1}^2 }+c^2$ And taking square root we get: $\mid\frac{{\Delta x_1}}{{\Delta t_1}}\mid=\sqrt{\frac{{\Delta x_2}^2}{{\Delta t_1}^2 }+c^2}$ And since $\frac{{\Delta x_2}^2}{{\Delta t_1}^2}\g 0$ this means: $\mid\frac{{\Delta x_1}^2}{{\Delta t_1}^2}\mid\g c$ Reply to this comment

Comments

thebigshow500
2008-10-12 16:10:18

Just Check out Griffiths p.502 - "The invariant interval." Everything is super clear there.

Mr. Griffiths derives the result of interval between two events as I = $d^2-c^2t^2$

I < 0 refers to timelike interval. Two events occur simultaneously.
I > 0 refers to spacelike interval. Two events occur at the same place.
I = 0 refers to lightike interval. Two events are connected by a signal traveling at c!

For this question, I > 0, so the term $d^2$ dominates $c^2t^2$ . One can deduct the answer (C)!!

thebigshow500
2008-10-12 16:15:53

Correction:

I < 0 refers to timelike interval. Two events occur at the same place.
I > 0 refers to spacelike interval. Two events occur simultaneously.

thebigshow500
2008-10-12 16:17:15

Correction:

I < 0 refers to timelike interval. Two events occur at the same place.
I > 0 refers to spacelike interval. Two events occur simultaneously.

goldenkid314
2009-04-02 06:37:27

It is actually:

I < 0 refers to timelike interval. Two events occur at the same place and are separated by an interval of time.
I > 0 refers to spacelike interval. Two events occur at the same time and are separated only in space.

The definition you give would lead to answer b) being correct, which it is not.

Reply to this comment

ivalmian
2008-03-25 19:02:24

You can also show this through a little math.

We know that for 2 arbitrary reference frames $R_1$ and $R_2$ , the time intervals between the two events is:

$S_1 = S_2$

Plugging in $\Delta x_1$ , $\Delta t_1$ , $\Delta x_2$ , and $\Delta t_2$ we get:

${\Delta x_1}^2-{\Delta t_1}^2 c^2={\Delta x_2}^2-{\Delta t_2}^2 c^2$

If in $R_2$ the event is perceived as simultaneous, then $\Delta t_2=0$ . Thus we get:

${\Delta x_1}^2-{\Delta t_1}^2 c^2={\Delta x_2}^2$

Dividing by $\Delta t_1$ we get:

$\frac{{\Delta x_1}^2}{{\Delta t_1}^2 } - c^2=\frac{{\Delta x_2}^2}{{\Delta t_1}^2 }$

Adding $c^2$ to both sides we get:

$\frac{{\Delta x_1}^2}{{\Delta t_1}^2 } =\frac{{\Delta x_2}^2}{{\Delta t_1}^2 }+c^2$

And taking square root we get:

$\mid\frac{{\Delta x_1}}{{\Delta t_1}}\mid=\sqrt{\frac{{\Delta x_2}^2}{{\Delta t_1}^2 }+c^2}$

And since $\frac{{\Delta x_2}^2}{{\Delta t_1}^2}\g 0$ this means:

$\mid\frac{{\Delta x_1}^2}{{\Delta t_1}^2}\mid\g c$

joe35
2008-06-27 14:56:42

This is a nice solution. It might help to clarify that $S_1$ and $S_2$ refer to the Lorentz scalar distance, $S^2$ between two events, which is the same in all inertial reference frames.

Prologue
2009-10-23 13:27:30

This last line $\mid\frac{{\Delta x_1}^2}{{\Delta t_1}^2}\mid\g c$ should be $\mid\frac{{\Delta x_1}}{{\Delta t_1}}\mid \g c$ .

Prologue
2009-10-23 13:28:39

This last line $\mid\frac{{\Delta x_1}^2}{{\Delta t_1}^2}\mid\g c$ should be $\mid\frac{{\Delta x_1}}{{\Delta t_1}}\mid \g c$ .

Kentai
2009-11-06 14:34:43

It is slightly easier then that...

you have dx^2 - (c^2 *dt^2) = dx'^2

factorize dt^2 out from the left hand side, the eqn becomes dt^2 ((dx/dt)^2 - c^2) = dx'^2

it is obvious that dx/dt has to be greater then c in order to get a positive dx'^2

(too lazy to follow the latex syntax...)

Reply to this comment

keflavich
2005-11-10 12:22:03

$dt$ should be squared, I think. You can also remember that for an observer to say that two events happened at the same time, they can't have a causal relationship, and therefore their spatial separation has to be greater than $ct$ . i.e. $s^2 > 0$ if $s^2 = d^2-c^2t^2$ (if you use a Minkowski diagram with time on the y-axis, these signs are correct)

yosun
2005-11-10 12:34:06

keflavich: thanks for the typo-alert; the typo is now corrected.

i'm planning on posting diagrams sometimes during winter break.

meanwhile, here's an open call to other users for submitting diagrams... email them as gif/jpg attachments to yosun(at)nusoy(dot)com.

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