4
$\begingroup$

I have a question regarding the physical significance of the canonical energy momentum tensor $T_\nu ^\mu$ in the context of classical field theory. It is defined as

$T_\nu ^\mu = \frac{\partial \mathcal{L}}{\partial ( \partial_\mu \Phi^I)} \partial_\nu \Phi^I - \delta_\nu ^\mu \mathcal{L} $, where $\Phi^I$ is the set of all relevant scalar, vector, or tensor fields in the Lagrangian, and $I$ is the corresponding indices. For example, if we consider the Lagrangian for the free gauge field in electrodynamics $\mathcal{L}_A = - \frac{1}{4} F_{\mu \nu} F^{\mu \nu}$, where $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$, we have that $\Phi^I = A_\mu$. Accordingly, the (covariant) canonical energy momentum tensor is

$T_{\mu \nu} = \frac{1}{4}\eta_{\mu \nu} F_{\alpha \beta} F^{\alpha \beta} - F_{\mu \alpha} \partial _ \nu A^ \alpha$.

It satisfies that $\partial _\mu T_\nu ^\mu = 0$ provided that the system, which it describes, is independent of translations in spacetime. This is a consequence of Noethers theorem.

My problem: On the one hand, it is used to define the canonical 4-momentum by

$P_\nu = \int T_{\nu}^0 d^3 x $,

i.e. it is used to define some physical observables of the system. Also, the canonical Hamiltonian $\mathcal{H}$ density is defined by $\mathcal{H}=T_{0}^0$

On the other hand, $T_\nu ^\mu$ is not in general gauge invariant - but physical observables must be gauge invariant. If the gauge field is time independent, i.e. $\partial_0 A^\alpha = 0$ for each $\alpha$, then $T_0 ^0$ (and thus also the energy) can be gauge independent, as is the case for $T_0 ^0$ for free electrodynamics defined above.

However, in general for time-dependent gauge field $T_\nu ^\mu$ is not gauge-invariant, which must imply that the canonical Hamiltonian density is not in general gauge independent and thus not observable.

My question: Is it possible to show that the 4-momentum is always gauge invariant and thus observable despite the fact that $T_\nu ^\mu$ is not?

If this is not possible, what even is the significance (and relevance) of $T_\nu ^\mu$ and the canonical 4-momentum?

$\endgroup$
1

2 Answers 2

1
$\begingroup$

First §: perhaps mention that you are talking about electromagnetism and show or mention which lagrangian you assume. Belinfante, I assume. Otherwise, well put.

Second §: the Noether tensor derived from the Belinfante tensor is not gauge invariant and asymmetric. Could you explain what you mean by "in general"? I don't know any Noether energy-momentum density that is gauge invariant. There is none.

Now your question. In order for the four momentum to be gauge invariant, the underlying density must be. The Noether expression is not. The Belinfante expression is needed. However, the latter is a physically not completely justifiable, ad hoc modification of the former. If you by all means want a gauge invariant theory you have to live with this. If not, take a look at my paper at https://arxiv.org/abs/physics/0106078.

$\endgroup$
5
  • $\begingroup$ I have tried to address your comments. I want to add that I am not talking about the Belifante tensor (which is in fact symmetric and gauge invariant). I am talking about the Noether energy momentum tensor. I have provided an example of a gauge invariant Noether energy momentum tensor. Good point! The gauge invariant Belifante tensor is defined from the Noether tensor such that the energies match for trivial topology - and this only makes sense if the energy defined by the Noether tensor is in fact observable. That is what is bugging me here, because the Belifante tensor is widely used. $\endgroup$
    – Simon G.
    Commented Aug 9, 2019 at 17:54
  • $\begingroup$ Correction: "I have provided an example of a gauge invariant Noether energy momentum density". $\endgroup$
    – Simon G.
    Commented Aug 9, 2019 at 18:14
  • $\begingroup$ The tensor you give is not gauge invariant and is asymmetric. $\endgroup$
    – my2cts
    Commented Aug 9, 2019 at 18:29
  • $\begingroup$ You asked for a "Nother energy momentum density", which I took to mean the Hamiltonian density. In static free electrodynamics, it is gauge invariant, which allows for the energy to be as well. If you meant the Noether tensor, you might be right that it is never gauge invariant except for fields constant in time and space. Can you answer why we even consider canonical momentum as defined by the Noether tensor if it is not an observable quantity? $\endgroup$
    – Simon G.
    Commented Aug 9, 2019 at 18:39
  • $\begingroup$ The Noether theorem is sound. Once you pick a hamiltonian it gives you the conserved currents straight away. That is why Noether currents are of essential importance. In the case of electromagnetism physicists are not happy with the results and modify them. They consider this permissible because the physical consequences are very small. $\endgroup$
    – my2cts
    Commented Aug 9, 2019 at 21:52
1
$\begingroup$

The stress-energy tensor you derived above by applying isometry to Noether's theorem is called "the Canonical Energy-Momentum Tensor", which is typically not gauge-invariant.

There is another form of stress-energy tensor, called "the Hilbert Energy-Momentum Tensor", defined as

$$\mathfrak{T}^{\mu\nu}=-\frac{2}{\sqrt{|g|}}\frac{\delta S[g]}{\delta g_{\mu\nu}},$$

where $S[g]$ is the action of matter fields that depends on the metric tensor $g$. This version of strss-energy tensor is both symmetric and gauge-invariant. It's (on- and off-shell) conservation law

$$\nabla_{\mu}\mathfrak{T}^{\mu\nu}=0$$

cannot be derived from Noether's (first) theorem, but is a consequence of diffeomorphism invariance of $S[g]$. For instance, in GR this is equivalent to the Bianchi identity. In electrodynammics, the Hilbert stress-energy tensor is given by

$$\mathfrak{T}^{\mu\nu}=g_{\alpha\beta}F^{\mu\alpha}F^{\nu\beta}-\frac{1}{4}g^{\mu\nu}F_{\delta\gamma}F^{\delta\gamma},$$

which is both symmetric and gauge-invariant.

When talking about symmetries of the action, one must distinguish two types of "symmetries": Physical Symmetrie (dynamical symmetries) and Gauge Redundancies (non-dynamical symmetries).

The former an is inherent property of matter fields living in spacetime. Examples are global (internal) symmetries of fields leading to charge observables that are conserved on-shell. Such conservation laws are further constraints imposed on the solutions of the equations of motion. In quantized field theory, matter fields must fulfill projective representations of the physical symmetry group in the Hilbert space. The physical symmetry group acts on a physical state in the Hilbert space by transforming it into other physical states.

The latter arises from one's freedom of making choices of the way he may formulate the action. Examples are diffeomorphism (re-parameterization invariance) in GR and in string theory, and $U(1)$-gauge in QED. For example, the Einstain-Hilbert action must be a scalar whether or not the action is extremized. Such redundancies lead to mathematical identities that are independent of the equations of motion. In quantized field theory, the action of gauge group on any physical state in the Hilbert space leaves it unchanged, since physics should be invariant under gauge transformations. In other words, physical states should form a trivial representation of the gauge group. More specifically, the Gauss's law is promoted into an quantum equation. For example, in classical electromagnetism the primary constraint

$$\nabla\cdot\vec{E}-\rho=0$$

is promoted into

$$\left(\nabla\cdot\vec{E}-\rho\right)|\psi\rangle=0,$$

in QED, for any physical state $|\psi\rangle$.

Isometries (metric components being independent of a coordinates along such directions) are also physical symmetries, because under isometric transformations of coordinates, matter fields must transform accordingly in field representations of the isometry group to maintain the invariance of the Lagrangian. They lead to conservation of energy and momentum. However, the metric coupled with matter fields play a role of a Lagrangian multiplier (similar to $A_{0}$ component in QED) since it is non-dynamical in the matter Lagrangian. Consequantly, directly applying Noether's first theorem to isometries would naturally lead to non-gauge invariant quantities.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.