GREPhysics.NET: GR8677 #68

GR8677 #68

Problem

GREPhysics.NET Official Solution

Alternate Solutions

Special Relativity $\Rightarrow$ }Relativistic Energy

The relativistic energy relation is,

$E= \sqrt{p^2 c^2 + mc^4.}$

As the speed of a particle $v \rightarrow c$ , $E\rightarrow pc$ , since by the de Broglie relation, $p=h/\lambda=h\nu/c=E/c$ . Plug in the requirement into the energy relation to find that $m\rightarrow 0$ , as in choice (A).

Alternate Solutions

TimToolMan 2018-04-12 16:27:04	We know that $\\beta=v/c=pc/E$ (using relativistic momentum and energy). Since $v=c$ we have $1=pc/E$ which implies $E=pc$ . And, from the relation $E^2 = (pc)^2+(mc^2)^2$ , we can say that $E=pc$ implies $m=0$ . Reply to this comment
TimToolMan 2018-04-04 16:21:50	Another way to think about it:\\\\r\\\\nTake $E^2=(pc)^2+(mc^2)^2$ if $m=0$ then $E=pc$ and then $c=E/p$ which means $c=\\gamma mc^2 / \\gamma mv$ cancelling like terms gives $v=c$ when $m=0$ Reply to this comment
pranav 2012-11-09 13:17:09	One more way of doing this is by recalling the following equation: $m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$ The only way anyone (including Mr. Photon) can go at $v=c$ is by having a rest mass of zero. (For those who doubt this: The above equation can be re-written at v=c as: $m_0=m \sqrt{1-\frac{v^2}{c^2}}=m (1-1)=m (0)=0$ QED) Reply to this comment

Comments

TimToolMan
2018-04-12 16:27:04

We know that $\\beta=v/c=pc/E$ (using relativistic momentum and energy). Since $v=c$ we have $1=pc/E$ which implies $E=pc$ . And, from the relation $E^2 = (pc)^2+(mc^2)^2$ , we can say that $E=pc$ implies $m=0$ .

Reply to this comment

TimToolMan
2018-04-04 16:21:50

Another way to think about it:\\\\r\\\\nTake $E^2=(pc)^2+(mc^2)^2$ if $m=0$ then $E=pc$ and then $c=E/p$ which means $c=\\gamma mc^2 / \\gamma mv$ cancelling like terms gives $v=c$ when $m=0$

TimToolMan
2018-04-12 16:28:04

Disregard this one, formatting mistake. See my other comment.

Reply to this comment

Giubenez
2014-10-17 05:46:11

The easiest way is surely to compare it with the photon (that is known to have a rest mass = 0).
Since
$p=\gamma \cdot m \cdot v$
and we know that photons have finite momentum the only possible explanation is that
$m_\gamma = 0$
( $\gamma \Rightarrow \infty$ ).
Now our particle, having v =c must have m=0 to obtain a finite momentum..

Reply to this comment

pranav
2012-11-09 13:17:09

One more way of doing this is by recalling the following equation:

$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$

The only way anyone (including Mr. Photon) can go at $v=c$ is by having a rest mass of zero.

(For those who doubt this:

The above equation can be re-written at v=c as:

$m_0=m \sqrt{1-\frac{v^2}{c^2}}=m (1-1)=m (0)=0$

QED)

Reply to this comment

Astrophysicist
2011-10-02 11:34:45

One can also remember that photons(which travel at the speed of light) have rest mass = 0. Thus, this particle must have a rest mass = 0 as well.

mpdude8
2012-04-19 15:15:23

You can't really say "particle X will automatically share the same characteristics of a photon" and deduce the answer from there, otherwise answer B also would be valid.

deneb
2018-10-10 04:47:41

The difference is that A MUST be true, so you have to pick A. Anything with v = c must be massless.

Reply to this comment

shak
2010-07-31 21:20:43

Can we make a conclusion from this formula,
m=m(0)/(1-v^2/c^2)^0.5
then,
m(o)=m*(1-v^2/c^2)^0.5

when v approaches c, then m(0) goes to zero
Is this right?
thanks

neon37
2010-11-03 11:55:06

umm....that looks correct to me. I cant think of an arguement to counter that (atleast yet).

Reply to this comment

spacemanERAU
2009-10-20 08:37:41

Also, this problem is really simple if you remember that any particle that has speed of c MUST have rest mass of zero...no math involved

Reply to this comment

radicaltyro
2006-10-23 00:41:07

It should be $E=\sqrt{p^2c^2+m^2c^4}$

spacemanERAU
2009-10-18 18:35:28

agreed!

Reply to this comment

Post A Comment!

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...

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