Canonical Energy-Momentum and Improvement

The canonical energy-momentum tensor collects the four Noether currents associated with spacetime translations. It tells us how energy and momentum move locally. Its form depends on the chosen fields and Lagrangian, so it can be asymmetric or gauge dependent. A superpotential improvement redistributes the local density while preserving its divergence; suitable boundary conditions also preserve the total four-momentum. Metric variation supplies the separately defined Hilbert response tensor.

Required background. Classical Fields and Their Actions provides field actions and Euler–Lagrange equations, while Events, Intervals, and Minkowski Geometry fixes the spacetime conventions. Helpful background. Calculus and Linear Algebra for Relativity provides multivariable differentiation; Presymplectic Potentials and Currents develops boundary and current ambiguities.

Translation identity

We work in Cartesian Minkowski coordinates, assume a first-order Lagrangian with no explicit $x$ dependence, and use the physical $(-,+,+,+)$ energy convention. For the active translation $\delta\phi^a=-a^\nu\partial_\nu\phi^a$, define

$$\pi_a^\mu=\frac{\partial\mathcal L}{\partial(\partial_\mu\phi^a)}, \qquad T_{\rm can}^{\mu}{}_{\nu} =\delta^\mu{}_\nu\mathcal L-\pi_a^\mu\partial_\nu\phi^a.$$

Some references reverse the translation generator and hence the overall Noether current. Our choice makes $T^{00}$ the positive scalar-field energy density below.

The index positions matter: with this signature $T^0{}_0=-T^{00}$. A negative mixed time-time component is therefore consistent with positive physical energy density. The conserved four-momentum below uses $T^{0\nu}$ with both indices raised.

Differentiate the canonical tensor and use the chain rule:

$$\partial_\mu T_{\rm can}^{\mu}{}_{\nu} =\left( \frac{\partial\mathcal L}{\partial\phi^a} -\partial_\mu\pi_a^\mu \right)\partial_\nu\phi^a =E_a\partial_\nu\phi^a.$$

Thus translation invariance gives $\partial_\mu T_{\rm can}^{\mu}{}_\nu=0$ only on shell. The four currents are the columns labeled by $\nu$, and

$$P^\nu=\int d^3x\,T_{\rm can}^{0\nu}$$

is constant when the integral converges and boundary flux vanishes. The conventional derivation, with the opposite signature and its corresponding action signs, appears in Tong, §1.3.2, printed pp. 14–16, Eqs. (1.39)–(1.48).

For a vector or spinor field, Lorentz transformations act on internal component indices as well as on the argument. Translation symmetry alone does not require $T_{\rm can}^{\mu\nu}=T_{\rm can}^{\nu\mu}$. In source-free electromagnetism the expression built directly from $A_\mu$ also contains the gauge-dependent potential, even though the improved closed-system four-momentum is gauge invariant under the boundary conditions stated below.

Superpotentials and charge equivalence

Let $B^{\rho\mu\nu}=-B^{\mu\rho\nu}$. Then

$$T'^{\mu\nu} =T_{\rm can}^{\mu\nu}+\partial_\rho B^{\rho\mu\nu}, \qquad \partial_\mu\partial_\rho B^{\rho\mu\nu}=0$$

identically, without equations of motion. This is a superpotential improvement. The charge shift is a boundary term,

$$\Delta P^\nu =\int_{\partial\Sigma}B^{i0\nu}n_i\,dA.$$

Therefore the local density can change while the total charge remains the same only under adequate falloff, periodic cancellation, or explicit boundary counterterms. In $d$ spacetime dimensions, $T$ has dimensions of energy per spatial volume and $B$ has one extra power of length, so $\partial B$ has the correct dimensions.

The Belinfante tensor is one particular improvement constructed from the Lorentz spin current; it becomes symmetric on shell under the hypotheses of Poincaré invariance. An arbitrary superpotential is not automatically Belinfante, and neither object is defined by metric variation. Pons, §§4.1–4.3 and the Appendix establishes the detailed relation for bosonic tensor fields and makes its on-shell qualifications explicit; spinors require the corresponding tetrad treatment.

Scalar-field check

For

$$\mathcal L=-\frac12\partial_\rho\phi\,\partial^\rho\phi-V(\phi),$$

one has $\pi^\mu=-\partial^\mu\phi$ and

$$T_{\rm can}^{\mu\nu} =\partial^\mu\phi\,\partial^\nu\phi +\eta^{\mu\nu}\mathcal L.$$

It is already symmetric. Its energy density and flux are

$$u=T^{00} =\frac12\dot\phi^2+\frac12|\boldsymbol\nabla\phi|^2+V, \qquad S^i=T^{i0}=-\dot\phi\,\partial_i\phi.$$

Using $\ddot\phi-\nabla^2\phi+V'(\phi)=0$ gives $\partial_tu+\boldsymbol\nabla\cdot\mathbf S=0$. This independently checks the sign: a spatially uniform oscillator has $\mathbf S=0$ and positive $u$ when $V\ge0$.

Maxwell check: gauge dependence and improvement

The source-free Maxwell field makes the distinction visible. With

$$\mathcal L=-\frac14F_{\rho\sigma}F^{\rho\sigma}, \qquad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu,$$

the momentum conjugate to $\partial_\mu A_\alpha$ and the canonical tensor are

$$\pi^{\mu\alpha}=-F^{\mu\alpha}, \qquad T_{\rm can}^{\mu\nu} =F^{\mu\alpha}\partial^\nu A_\alpha -\frac14\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}.$$

This tensor depends explicitly on $A_\mu$ and is generally neither gauge invariant nor symmetric. By contrast, metric variation gives

$$T_{\rm H}^{\mu\nu} =F^\mu{}_{\lambda}F^{\nu\lambda} -\frac14\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}.$$

Use $\partial^\nu A_\alpha=F^\nu{}_{\alpha}+\partial_\alpha A^\nu$ and the product rule to keep the field-equation term visible:

$$\boxed{ T_{\rm can}^{\mu\nu} =T_{\rm H}^{\mu\nu} -\partial_\rho\!\left(F^{\rho\mu}A^\nu\right) +\left(\partial_\rho F^{\rho\mu}\right)A^\nu.}$$

For the free field, $\partial_\rho F^{\rho\mu}=0$, so

$$T_{\rm H}^{\mu\nu} =T_{\rm can}^{\mu\nu} +\partial_\rho\!\left(F^{\rho\mu}A^\nu\right).$$

The bracket is antisymmetric in $\rho$ and $\mu$, so this is a superpotential improvement. The integrated four-momenta agree only if $\int_{\partial\Sigma}F^{i0}A^\nu n_i\,dA=0$ or an equivalent boundary accounting is imposed. With an external current, the displayed Euler–Lagrange term does not vanish; one must include the matter sector rather than claim a conserved electromagnetic four-momentum by itself. The gauge-invariant Maxwell response tensor and its energy density are checked independently in Lorentz Force and Electromagnetic Stress-Energy.

Exercise: an identically conserved improvement

Show that

$$\Delta T^{\mu\nu} =\left(\partial^\mu\partial^\nu-\eta^{\mu\nu}\Box\right)F(\phi)$$

is symmetric and identically conserved. When does it leave $P^\nu$ unchanged?

Solution

Symmetry follows because partial derivatives commute. Taking a divergence gives

$$\partial_\mu\Delta T^{\mu\nu} =\Box\partial^\nu F-\partial^\nu\Box F=0,$$

with no field equation. It is a superpotential because

$$\Delta T^{\mu\nu} =\partial_\rho\left[ \eta^{\rho\nu}\partial^\mu F -\eta^{\mu\nu}\partial^\rho F \right],$$

and the bracket is antisymmetric in $\rho,\mu$. Hence $\Delta P^\nu$ is a surface integral. It vanishes for compact support or sufficiently rapid decay of $\partial F$, but not automatically with physical boundaries or slow falloff. This example also shows why local energy density is more ambiguous than the closed-system charge.

Distinctions that must survive

Adding a total divergence to $\mathcal L$ can alter the canonical tensor; field redefinitions can alter its appearance; and external backgrounds can exchange momentum with the fields. Belinfante improvement handles Lorentz spin under stated assumptions, not every boundary ambiguity. Hilbert Stress-Energy from Metric Variation introduces the metric response without identifying these constructions by fiat.

References

• Josep M. Pons, “Noether Symmetries, Energy-Momentum Tensors and Conformal Invariance in Classical Field Theory,” Journal of Mathematical Physics 52, 012904 (2011), doi:10.1063/1.3532941, Open PDF, §§4.1–4.3, printed pp. 11–17, and Appendix, printed pp. 27–29.
• David Tong, Quantum Field Theory, University of Cambridge Part III lecture notes (2006/2007), official PDF, §1.3.2, printed pp. 14–16, Eqs. (1.39)–(1.48).