GREPhysics.NET: GR0177 #34

GR0177 #34

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

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.

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.

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.

Reply to this comment

Post A Comment!

You are replying to:

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

Chat Archives | Delete left banner ad | Donate

(Click to view chat.)

Hate being Anonymous? Login or Register

Upgrade to Poser 7 Now

Huge Textbook Selection, Low Prices – Phat Campus.

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone