TimToolMan 20180412 16:27:04  We know that (using relativistic momentum and energy). Since we have which implies . And, from the relation , we can say that implies .  
TimToolMan 20180404 16:21:50  Another way to think about it:\\\\r\\\\nTake if then and then which means cancelling like terms gives when
TimToolMan 20180412 16:28:04 
Disregard this one, formatting mistake. See my other comment.

 
Giubenez 20141017 05:46:11  The easiest way is surely to compare it with the photon (that is known to have a rest mass = 0).
Since
and we know that photons have finite momentum the only possible explanation is that
().
Now our particle, having v =c must have m=0 to obtain a finite momentum..  
pranav 20121109 13:17:09  One more way of doing this is by recalling the following equation:
The only way anyone (including Mr. Photon) can go at is by having a rest mass of zero.
(For those who doubt this:
The above equation can be rewritten at v=c as:
QED)
 
Astrophysicist 20111002 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 20120419 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 20181010 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 20100731 21:20:43  Can we make a conclusion from this formula,
m=m(0)/(1v^2/c^2)^0.5
then,
m(o)=m*(1v^2/c^2)^0.5
when v approaches c, then m(0) goes to zero
Is this right?
thanks
neon37 20101103 11:55:06 
umm....that looks correct to me. I cant think of an arguement to counter that (atleast yet).

 
spacemanERAU 20091020 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  
radicaltyro 20061023 00:41:07  It should be
spacemanERAU 20091018 18:35:28 
agreed!

 