Since p 2 = E 2 − p → 2 = m 2 and E...

Ethen Frey

Ethen Frey

Answered

2022-07-21

Since p 2 = E 2 p 2 = m 2 and
E = h ν = h c λ and
| p | = h λ
we have that
p 2 = h 2 c 2 λ 2 h 2 λ 2
If I go to Planck-units ( c = 1 , h = 1), this becomes zero. Is this a correct thing to do?

Answer & Explanation

wintern90

wintern90

Expert

2022-07-22Added 12 answers

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 = h 2 c 2 λ 2 h 2 λ 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 = E 2 c 2 | p | 2 ; that extra factor of c 2 is necessary to keep the units

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