# Origin of momentum. Noether's theorem

My professor talked about Noether's theorem and how translation is the origin of momentum conservation. But why is it not velocity that is conserved but mass times velocity. And on the same note why isn't it mass squared times velocity.

• Speed of what?
– Danu
Dec 9, 2014 at 0:43
• @Danu - Since momentum conservation implies the vector sum of momentum vectors of each part of an isolated composite system will be constant (at least that's what it means in classical physics), maybe Melle Sterk is asking why it's not the vector sum of the velocity vectors of each part of a system that's constant. I doubt there is any simple conceptual explanation though, if there was I would think this important result would have been discovered before Emmy Noether published it in 1918. Dec 9, 2014 at 1:00
• Important remark : translation is not origin - origin is INVARIANCE to something. And this something can be - translation, rotation, shifting in time, something else. Dec 9, 2014 at 1:20

The laws of physics are discovered through a mixture of heuristics, modelling and inference. In case of momentum, the story goes like this:

It is possible to 'transfer motion' from one body to another. However, experiment shows that it is not velocity that is conserved during such transfers, but another 'quantity of motion'. We give that quantity the name 'momentum'.

It turns out that velocity and momentum are (approximately) proportional, and mass is just the name of the constant of proportionality. By 'coincidence', this constant appears to be the same quantity we can measure using scales.

Once we have a model, we can then come up with stories why the world should behave the way it does.

One story that gets told is that momentum should be conserved because physics is translation invariant. Or is it the other way around? Is physics translation invariant because momentum is conserved?

Eg in Hamiltonian mechanics, symmetries and conservation laws pretty much stand on equal footing as $$\mathcal L_{X_H} p = 0 \Leftrightarrow \mathcal L_{X_p} H = 0$$ ie momentum is conserved under time evolution generated by the Hamiltonian iff the Hamiltonian is conserved under translations generated by momentum.

• I was confused because working with free particles this proportionality constant shows up. Will it become clearer the moment you add interaction terms to a Hamiltonian. Dec 9, 2014 at 11:08