GR 8677927796770177 | # Login | Register

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
 TimToolMan2018-04-12 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 TimToolMan2018-04-04 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 pranav2012-11-09 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
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 $m=0$.
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 $m=0$
 TimToolMan2018-04-12 16:28:04 Disregard this one, formatting mistake. See my other 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..
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)

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.
 mpdude82012-04-19 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.
 deneb2018-10-10 04:47:41 The difference is that A MUST be true, so you have to pick A. Anything with v = c must be massless.
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
 neon372010-11-03 11:55:06 umm....that looks correct to me. I cant think of an arguement to counter that (atleast yet).
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
2006-10-23 00:41:07
It should be $E=\sqrt{p^2c^2+m^2c^4}$
 spacemanERAU2009-10-18 18:35:28 agreed!

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)
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}\$ $\frac{a}{b}$