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  GR8677 #68
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Verbatim question for GR8677 #68
Special Relativity}Relativistic Energy

The relativistic energy relation is,

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).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
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.Alternate Solution - Unverified
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=0Alternate Solution - Unverified
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)

Alternate Solution - Unverified
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.Alternate Solution - Unverified
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.
Alternate Solution - Unverified
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..		NEC
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)

Alternate Solution - Unverified
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.
NEC
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).
NEC
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 involvedNEC
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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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.

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