3 ms·
Squared four-vectors? I'm only an amateur, but wouldn't that give different results depending on choice of units? I.e, I usually use C=1.
by Filligree 2y ago
Squared four-vectors?
I'm only an amateur, but wouldn't that give different results depending on choice of units? I.e, I usually use C=1.
- dguest 2y agothe math is m^2 c^4 = E^2 - p^2 c^2 where m is mass, E is the total energy in the decay products and p is the 3-vector sum of the momentum. Those units should work out (they certainly do if you set c = 1).
- Filligree 2y agoAh, I see. I was assuming you meant the 4-momentum. Though I'm not sure this doesn't come out to the same thing.
- dguest 2y agowhat I showed is the squared 4 momentum when you use the Minkowski metric [1], assuming "squared" means "self-dot product". The formulation above is just another way to illustrate the Minkowski dot product. [1]: https://en.wikipedia.org/wiki/Minkowski_space#Minkowski_metric https://en.wikipedia.org/wiki/Minkowski_space#Minkowski_metr...
- sixo 2y agoyou use the same units on both sides of the equation, it's fine, it's like counting "meters squared"