Ethen Frey

2022-07-21

Since ${p}^{2}={E}^{2}-{\stackrel{\to }{p}}^{2}={m}^{2}$ and
$E=h\nu =\frac{hc}{\lambda }$ and
$|\stackrel{\to }{p}|=\frac{h}{\lambda }$
we have that
${p}^{2}=\frac{{h}^{2}{c}^{2}}{{\lambda }^{2}}-\frac{{h}^{2}}{{\lambda }^{2}}$
If I go to Planck-units ($c=1,h=1$), this becomes zero. Is this a correct thing to do?

wintern90

Expert

You're right that, if the result of a calculation depends on the units you use, there's something wrong with it. In fact, even more than that, you can't even get any result out of ${p}^{2}=\frac{{h}^{2}{c}^{2}}{{\lambda }^{2}}-\frac{{h}^{2}}{{\lambda }^{2}}$ in SI units, because you'd be taking a difference between two quantities with different units.
In this case, if you're using SI units, the energy-momentum relation is ${p}^{2}=\frac{{E}^{2}}{{c}^{2}}-|\stackrel{\to }{p}{|}^{2}$; that extra factor of ${c}^{2}$ is necessary to keep the units

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