# Noether's theorem and energy in four-momentum

In Newtonian physics, momentum and energy are often treated as distinct entities, which happen to be separately conserved. In relativity, energy is regarded as the "time" component of the four-momentum. Energy and momentum are still separately conserved, but they mix under Lorentz transformations in a way that identifies energy with time and momentum with space.

Meanwhile, there is an analogous relationship energy:time::momentum:space in Newtonian physics, which becomes apparent in the Hamiltonian formulation. The Hamiltonian (energy) is the generator of time translations, whereas momentum is the generator of space translations.

What's the "secret" behind this? As pointed out in the comments, Noether's theorem relates energy and time via time translation symmetry in Newtonian mechanics, so this symmetry is probably also at play in relativity. But how? Is there a clear line of reasoning that goes from Noether's theorem + spacetime translation symmetry in relativity to "energy and momentum combine as a four-vector"?

• The answer to the question in the title is definitely Noether's theorem. But it seems like you are also asking a subtler point that involves the connection between energy and momentum in the body of the post, and how they are "fused" in the relativistic 4-momentum. Am I reading your question correctly? Nov 29 '21 at 7:53
• @MariusLadegårdMeyer Yes. If Noether's theorem is the answer, then my question becomes, "How does time translation symmetry + Noether's theorem explain why it makes sense to put energy in the time component of four-momentum?" Nov 29 '21 at 7:56
• Okay thanks. In my opinion you should update the question/title with this clarification, it might attract better answers. Nov 29 '21 at 7:58
• If energy is the generator of time translation and momentum is the generator of spacial translation, as we put position and time into a four vector in space-time, we are automatically required to put their generators, energy and momentum into a four vector in four-dimensional momentum space. Nov 29 '21 at 9:05
• The question (v4) is still not clear. It seems the answer is "That's how Noether's theorem works in the first place". Nov 29 '21 at 9:09

Consider, for example, the Lagrangian for a free relativistic particle, $$L = \frac{1}{2} U^\mu U_\mu,$$ where $$U^\mu$$ is the four-velocity. If we make a transformation of the coordinates $$x^\mu \to x^\mu + \epsilon\psi^\mu$$, the Lagrangian stays invariant (the coordinates are all cyclic), and hence there is some associated conserved quantity. Picking $$\psi^\nu = \delta^{\nu\mu}$$ for different choices of $$\nu$$ and using Noether's theorem, we find that $$- \sum_{\nu} \frac{\partial L}{\partial U^\nu} \psi^\nu = - \frac{\partial L}{\partial U^\mu} = - m U_\mu$$ are constants of motion. We see we can arrange these quantities in a four vector by defining $$p^\mu = m U^\mu$$ and analysing the explicit expressions for each component we'll find out that these turn out to be precisely the energy (which was indeed associated with time translations in non-relativistic Physics) and the momenta (associated with time translations).