Understanding Photons: The Role of Energy and Momentum in their Existence

[Goodver]

i have read that existence of photons as a mass less particles came from the energy momentum equation.

E^2 = (mc^2)^2 + (pc)^2

and that since when m = 0, there is still an energy = pc

but, sunce momentum defined as m*v, and mass is absolute quantity, then why that m = 0 which we were using to eliminatethe first term does not apply for momentum

 

[Matterwave]

Momentum is only equal to m*v in the low velocity limit. It does not work for particles approaching the speed of light, and completely breaks down for light itself. the proper definition of momentum is in the equation you stated, which is valid for all velocities, including light speed.

For light, the momentum is p=E/c.

 

[Goodver]

Thank you Matterwave, however I am confused.

if p = E / c

then what is a relarivistic momentum p = m*v / sqrt( 1 - v^2/c^2 ) ?

and how come from energy momentum equation one can derive p=mv for low velocities?

thank you

 

[jtbell]

if p = E / c

This applies only to massless particles like photons

then what is a relarivistic momentum p = m*v / sqrt( 1 - v^2/c^2 ) ?

This applies only to massive particles.

and how come from energy momentum equation one can derive p=mv for low velocities?

p = mv applies only to massive particles, so you can safely derive it from your second equation above. Write it as $p = mv (1-v^2/c^2)^{-1/2}$ and use the binomial approximation: $(1-x)^n \approx 1-nx$ when x << 1.

You can also get it from $E^2 = (pc)^2 + (mc^2)^2$, but you also need to use an equation that gives you the velocity, namely $v = pc^2/E$ or the way I like to remember it, $v/c = pc/E$. And you have to use the binomial approximation.

 

[Goodver]

Great, thanks