- #1
Parcival
- 6
- 1
I'm in the process of learning special relativity (SR), and I'm a bit confused as to why the relativistic energy dispersion relation ##E^{2}=m^{2}c^{4}+p^{2}c^{2}## gives the energy for a free particle? I get that it is the sum of (relativistic) kinetic energy plus the rest mass term (a constant), but where in the derivation of this expression does one assume that the particle is free?
From what I've read, one way that one can deduce that ##E^{2}=m^{2}c^{4}+p^{2}c^{2}## is from the relativistic 3-momentum of a particle, ##\mathbf{p}=\gamma m\mathbf{v}## (where ##\gamma## is the Lorentz factor, ##m## the rest mass of the particle, and ##\mathbf{v}=\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}##, with ##t## the coordinate time), and the line element, ##-c^{2}\mathrm{d}\tau^{2}=-c^{2}\mathrm{d}t^{2}+\mathrm{d}\mathbf{x}^{2}## (where ##\tau## is the proper time). Indeed, from the latter, we can imply that ##(mc)^{2}=(\gamma mc)^{2}-p^{2}##. Taylor expanding ##(\gamma mc)^{2}##, we have $$(\gamma mc)^{2}=\frac{1}{c^{2}}\left(mc^{2}+\frac{1}{2}mv^{2}+\cdots\right)^{2}$$ where the dots denote higher order terms in ##v/c##. Recognising ##\frac{1}{2}mv^{2}## as the non-relativistic kinetic energy of a particle, we identify the expression in brackets with the energy of the particle, such that $$\frac{E^{2}}{c^{2}}=m^{2}c^{2}+p^{2}$$ I realize that one can also arrive at this expression, by noting that ##\gamma mc## is the zeroth element of the 4-momentum vector. The thing is, in neither of these ways does one assume that the particle is free, so is the fact that it describes the energy of a free particle something that is determined a posteriori?
From what I've read, one way that one can deduce that ##E^{2}=m^{2}c^{4}+p^{2}c^{2}## is from the relativistic 3-momentum of a particle, ##\mathbf{p}=\gamma m\mathbf{v}## (where ##\gamma## is the Lorentz factor, ##m## the rest mass of the particle, and ##\mathbf{v}=\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}##, with ##t## the coordinate time), and the line element, ##-c^{2}\mathrm{d}\tau^{2}=-c^{2}\mathrm{d}t^{2}+\mathrm{d}\mathbf{x}^{2}## (where ##\tau## is the proper time). Indeed, from the latter, we can imply that ##(mc)^{2}=(\gamma mc)^{2}-p^{2}##. Taylor expanding ##(\gamma mc)^{2}##, we have $$(\gamma mc)^{2}=\frac{1}{c^{2}}\left(mc^{2}+\frac{1}{2}mv^{2}+\cdots\right)^{2}$$ where the dots denote higher order terms in ##v/c##. Recognising ##\frac{1}{2}mv^{2}## as the non-relativistic kinetic energy of a particle, we identify the expression in brackets with the energy of the particle, such that $$\frac{E^{2}}{c^{2}}=m^{2}c^{2}+p^{2}$$ I realize that one can also arrive at this expression, by noting that ##\gamma mc## is the zeroth element of the 4-momentum vector. The thing is, in neither of these ways does one assume that the particle is free, so is the fact that it describes the energy of a free particle something that is determined a posteriori?