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.

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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..

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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).

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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

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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!

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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.

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