- #1

- 580

- 21

Given this, one then introduces the notion of 4-momentum, defining it as $$p^{\mu}=m\frac{dx^{\mu}}{d\tau}=\gamma m\frac{dx^{\mu}}{dt}$$ where ##x^{\mu}## is the position 4-vector, ##\tau## is the proper time, ##t## is the coordinate time and ##\gamma =\frac{1}{\sqrt{1-v^{2}/c^{2}}}## is the Lorentz factor (with ##\mathbf{v}## the relative velocity between two inertial frames).

I'm trying to reason why the zeroth component of 4-momentum of a particle, i.e. ##p^{0}## is proportional to its energy ##E##, $$p^{0}=\frac{E}{c}$$ where ##c## is the speed of light.

Can one do this as follows.

We first note that, by definition, ##p^{0}=\gamma mc=\frac{mc}{\sqrt{1-v^{2}/c^{2}}}##, and so for ##v<<c## we have that $$ p^{0}=\gamma mc=\frac{mc}{\sqrt{1-v^{2}/c^{2}}}\approx mc(1+\frac{v^{2}}{2c^{2}})=\frac{1}{c}(mc^{2}+\frac{mv^{2}}{2})$$ We recognise ##\frac{mv^{2}}{2}## as the kinetic energy of a non-relativistic particle and hence interpret ##mc^{2}+\frac{mv^{2}}{2}## as the total energy of the particle in the non-relativistic limit. Thus we are motivated to interpret ##p^{0}## as being proportional to the energy of the particle, i.e. ##p^{0}=\frac{E}{c}##.

Given this we can then proceed to derive the relativistic energy-momentum relation: $$ (p^{0})^{2}=\frac{E^{2}}{c^{2}}=\gamma^{2}m^{2}c^{2}= \frac{m^{2}c^{2}}{1-v^{2}/c^{2}}= \frac{m^{2}c^{2}+m^{2}v^{2}-m^{2}v^{2}}{1-v^{2}/c^{2}}\\ =\frac{m^{2}v^{2}}{1-v^{2}/c^{2}}+m^{2}\frac{c^{2}-v^{2}}{1-v^{2}/c^{2}}=p^{2}+m^{2}c^{2}$$ and hence $$E^{2}=m^{2}c^{4}+p^{2}c^{2}.$$

Have I understood this concept correctly? Is the derivation correct? I'd really appreciate any feedback.