Confusion surrounding Noether's Theorem

As an example of Noether's Theorem, my QFT textbook gave the example of how the conservation of momentum and energy arises from symmetry in space-time translations. The book arrives at the conclusion that the Noether Current is the Energy-Momentum Tensor and that conserved charges from this current can be written as: $$P^{a}=\int_{ }^{ }d^{3}x\ \ T^{0a}$$ (Where $$T$$ is the energy momentum Tensor), My first Question is: How did they arrive at this conclusion, and why is the conserved quantity called a 'charge'? The book then demonstrates how Energy is conserved for when $$a=0$$, however when $$a$$, is non-zero, the book says that the conserved 'charge' is momentum, stating: $$P^{a}=\int_{ }^{ }d^{3}x\ \ T^{0a}=\int_{ }^{ }d^{3}x\ \ Π^{0}∂^{a}ϕ$$ Which is said to be the momentum. My second Question is: How is the above expression equal to the momentum?

I am not studying QFT professionally, so there may be some gaps in my knowledge!

• – user288901
Mar 1, 2021 at 1:12
• I am asking for the additional intuition about the explicit version. Mar 1, 2021 at 2:11
• Perhaps someone with time will write an answer to this or link to another similar question on this site (it seems likely to have been asked before), but if not I suggest checking out chapter 2 of Di Francesco et al.'s book on conformal field theory which gives a nice derivation of Noether's theorem and also discusses the relation between the Noether charges associated to spacetime symmetries and the stress tensor. Mar 1, 2021 at 3:18
• The best way to gain intuition about a new formula is to apply it to examples. For example, are you familiar with the idea of the classical electromagnetic field having momentum? Mar 1, 2021 at 4:27
• Yes, it is the momentum carried by an excitation in the electromagnetic field (i.e.: a photon). Mar 1, 2021 at 5:55

The continuity equation for electric charge can be written $$\frac{\partial \rho}{\partial t} + \nabla \cdot \vec J = 0$$

where $$\rho$$ is the electric charge density and $$\mathbf J$$ is the electric current density. In integral form, this expresses the idea that if the amount of charge in some arbitrary volume changes, it is because some charge passed through the boundary; in other words, charge does not simply appear or disappear.

If we define the current 4-vector as $$\mathbf J = (c\rho, \vec J)$$ and the position coordinates $$x^\mu = (ct,\vec r)$$, this can be expressed as

$$\frac{\partial }{\partial x^\mu} J^\mu = 0$$

where the Einstein summation convention is used. In particular, note that $$J^0$$ is identified with the charge (density) and the spatial components $$J^a$$, $$a=1,2,3$$ are identified with the current (density). In some volume $$V$$, the total charge is given by $$Q = \int_V \mathrm d^3 x\ J^0$$.

From Noether's theorem, we find that spacetime translation invariance yields $$\frac{\partial}{\partial x^\mu} T^{\mu\nu} = 0$$. That is, translation invariance in the $$x^0$$ direction (i.e. a time translation) yields the continuity equation $$\frac{\partial}{\partial x^\mu} T^{\mu 0}= 0$$, with total "charge" given by $$\int \mathrm d^3x \ T^{00}$$. Similarly, translation invariance in the $$x^3$$ direction yields $$\frac{\partial}{\partial x^\mu} T^{\mu 3} = 0$$ which has total charge $$\int \mathrm d^3x \ T^{03}$$.

Noether's theorem takes a continuous symmetry as an input and spits out a continuity equation of the form $$\partial_\mu J^\mu = 0$$, where $$J^0$$ is the conserved Noether charge density (and $$\int \mathrm d^3 x \ J^0$$ is the total charge). In this framework, the definition of energy is the Noether charge corresponding to time translations, while momentum is the Noether charge corresponding to spatial translations.

• Thankyou for your well-written answer! My only problem now is, how would you know that by integrating the energy-momentum tensor for when the second index is non-zero, that you would get momentum? (I know that for when the second index is zero you get energy, because of Hamilton's equation). I.e.: How would you know that in this framework, this definition of momentum holds? Mar 2, 2021 at 9:06
• @Matrix001 What does it mean for a definition to hold? You’d need some more fundamental definition of momentum to compare it to, so what do you propose? Mar 2, 2021 at 13:52
• Say you where trying to prove the theorem, how would you know that it implies that spatial symmetry corresponds to momentum, without referring to it? Mar 2, 2021 at 22:45
• @Matrix001 I think you’re still misunderstanding me - momentum is - by definition - the conserved quantity which correspond to spatial translation invariance. From that POV, the answer to that last question is “by definition.” Mar 3, 2021 at 0:44