Q:

if E^2 = (MC^2)^2 + (PC)^2
then for mass less substance it is
E = PC
but then P = momentum = mass times velocity
i.e E = (M.V) C
and this would mean that light is not mass less which, i guess, contradicts the fact that light is mass less
I'm so confused with this thing

- Tanvir (age 15)

NCR, India

- Tanvir (age 15)

NCR, India

A:

That's a nice question, one that most people come across as they start to think about these things. The confusion arises from two distinct meanings used for the symbol "m" by different people in diferent contexts. One meaning is rest mass, and that's the one that appears in E^{2}=p^{2}c^{2}+m^{2}c^{4}. That m is invariant, meaning that it is the same for any inertial observer. The other meaning is inertial mass, the coefficient in the momentum/velocity equation. To avoid confusion let's call that M. The **p**=M**v**. As you point out, we can't have m=M. To make things consistent we need m=M(1-v^{2}/c^{2})^{1/2}, where v is the velocity of the particle in our reference frame. M must depend on the reference frame, since v does and m doesn't. For something that travels at the speed of light but with finite **p**, that equation requires m=0.

Mike W.

*(published on 10/22/2014)*