Peskin and Schroeder’s Problem 2.1 is really about a subtle question…

If Maxwell’s equations follow from a perfectly Lorentz- and gauge-invariant Lagrangian, why does the straightforward Noether energy–momentum tensor look neither symmetric nor gauge invariant?

The solution has two stages: derive Maxwell’s equations, then “repair” the canonical energy–momentum tensor without changing the physical energy or momentum. This matches the indexed descriptions of parts (a) and (b).

I will use the metric $\eta_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$.

1. The electromagnetic field as a field theory

The dynamical variable is the four-potential$$A_\mu(x),$$

and the electromagnetic field tensor is$$F_{\mu\nu} = \partial_\mu A_\nu-\partial_\nu A_\mu.$$

The free electromagnetic Lagrangian density is$$\boxed{ \mathcal L=-\frac14F_{\mu\nu}F^{\mu\nu} }$$

The factor $1/4$ compensates for the fact that the antisymmetric tensor $F_{\mu\nu}$ counts every independent component twice.

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2. Part (a): Recover Maxwell’s equations

The field Euler–Lagrange equation is$$\frac{\partial\mathcal L}{\partial A_\nu} – \partial_\mu \left( \frac{\partial\mathcal L} {\partial(\partial_\mu A_\nu)} \right)=0.$$

Because $\mathcal L$ contains $A_\nu$ only through its derivatives,$$\frac{\partial\mathcal L}{\partial A_\nu}=0.$$

Now vary the Lagrangian:$$\delta\mathcal L = -\frac12F^{\rho\sigma}\delta F_{\rho\sigma}.$$

Since$$\delta F_{\rho\sigma} = \partial_\rho\delta A_\sigma – \partial_\sigma\delta A_\rho,$$

antisymmetry of $F^{\rho\sigma}$ makes the two terms equal:$$\delta\mathcal L = -F^{\rho\sigma}\partial_\rho\delta A_\sigma.$$

Therefore,$$\frac{\partial\mathcal L} {\partial(\partial_\mu A_\nu)} = -F^{\mu\nu}.$$

The Euler–Lagrange equation becomes$$-\partial_\mu(-F^{\mu\nu})=0,$$

or$$\boxed{\partial_\mu F^{\mu\nu}=0.}$$

These are the source-free inhomogeneous Maxwell equations:$$\nabla\cdot\mathbf E=0, \qquad \nabla\times\mathbf B-\frac{\partial\mathbf E}{\partial t}=0.$$

The other two Maxwell equations,$$\nabla\cdot\mathbf B=0, \qquad \nabla\times\mathbf E+\frac{\partial\mathbf B}{\partial t}=0,$$

follow automatically from the definition $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$. Covariantly,$$\partial_\lambda F_{\mu\nu} +\partial_\mu F_{\nu\lambda} +\partial_\nu F_{\lambda\mu}=0.$$

So the important distinction is:

• Two Maxwell equations are equations of motion.
• The other two are identities resulting from writing $F=dA$.

With a current, one adds$$\mathcal L_{\text{source}}=-j_\mu A^\mu,$$

and obtains$$\boxed{\partial_\mu F^{\mu\nu}=j^\nu.}$$

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3. Part (b): The apparent problem with the stress-energy tensor

Translation symmetry gives the canonical Noether tensor$$T_{\text{can}}^{\mu\nu} = \frac{\partial\mathcal L} {\partial(\partial_\mu A_\lambda)} \partial^\nu A_\lambda -\eta^{\mu\nu}\mathcal L.$$

Substituting$$\frac{\partial\mathcal L} {\partial(\partial_\mu A_\lambda)} =-F^{\mu\lambda},$$

gives$$\boxed{ T_{\text{can}}^{\mu\nu} = -F^{\mu\lambda}\partial^\nu A_\lambda +\frac14\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma} }$$

This tensor is conserved:$$\partial_\mu T_{\text{can}}^{\mu\nu}=0.$$

But it has two unattractive features.

It is not manifestly gauge invariant

Under$$A_\mu\rightarrow A_\mu+\partial_\mu\alpha,$$

the potential-dependent term changes.

It is not symmetric

In general,$$T_{\text{can}}^{\mu\nu}\neq T_{\text{can}}^{\nu\mu}.$$

That is awkward because the physical electromagnetic stress-energy tensor should be symmetric, particularly for angular momentum conservation and coupling the theory to gravity.

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4. The key freedom: conserved tensors are not unique

If $T^{\mu\nu}$ is conserved, we may add a term of the form$$\Delta T^{\mu\nu} = \partial_\lambda X^{\lambda\mu\nu},$$

where$$X^{\lambda\mu\nu}=-X^{\mu\lambda\nu}.$$

Then$$\partial_\mu\Delta T^{\mu\nu} = \partial_\mu\partial_\lambda X^{\lambda\mu\nu}=0,$$

because the derivatives are symmetric under $\mu\leftrightarrow\lambda$, while $X$ is antisymmetric.

For electromagnetism, choose$$\Delta T^{\mu\nu} = \partial_\lambda \left(F^{\mu\lambda}A^\nu\right).$$

Using the source-free equation$$\partial_\lambda F^{\mu\lambda}=0,$$

this becomes$$\Delta T^{\mu\nu} = F^{\mu\lambda}\partial_\lambda A^\nu.$$

Add it to the canonical tensor:$$T^{\mu\nu} = -F^{\mu\lambda}\partial^\nu A_\lambda + F^{\mu\lambda}\partial_\lambda A^\nu + \frac14\eta^{\mu\nu}F^2.$$

Combine the first two terms:$$F^{\mu\lambda} \left( \partial_\lambda A^\nu-\partial^\nu A_\lambda \right) = -F^{\mu\lambda}F^\nu{}_{\lambda}.$$

Thus the improved tensor is$$\boxed{ T_{\mathrm{EM}}^{\mu\nu} = -F^{\mu\lambda}F^\nu{}_{\lambda} + \frac14\eta^{\mu\nu} F_{\rho\sigma}F^{\rho\sigma} }$$

This is the standard electromagnetic stress-energy tensor.

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5. Why this is the correct physical answer

The improved tensor has all the desired properties:$$T_{\mathrm{EM}}^{\mu\nu}=T_{\mathrm{EM}}^{\nu\mu},$$

it is gauge invariant because it depends only on $F_{\mu\nu}$, and it remains conserved:$$\partial_\mu T_{\mathrm{EM}}^{\mu\nu}=0.$$

Its components have familiar physical meanings:$$T^{00} = \frac12\left(\mathbf E^2+\mathbf B^2\right)$$

is the electromagnetic energy density, while$$T^{0i}=(\mathbf E\times\mathbf B)^i$$

is the energy flux or momentum density—the Poynting vector. The spatial components are$$T^{ij} = -E_iE_j-B_iB_j + \frac12\delta^{ij} \left(\mathbf E^2+\mathbf B^2\right),$$

which describe electromagnetic pressure and shear stress.

The added divergence does not change the total four-momentum$$P^\nu=\int d^3x\,T^{0\nu},$$

provided the fields vanish sufficiently rapidly at spatial infinity. It changes the local bookkeeping, not the total conserved quantities.

The real lesson of Problem 2.1

This problem is teaching three foundational QFT ideas:

1. Fields are obtained by applying the Euler–Lagrange principle to a Lagrangian density, just as particle equations follow from an ordinary Lagrangian.
2. Noether currents are not unique. You may add identically conserved “improvement terms.”
3. The naïve Noether tensor is not always the most physically useful representative. For gauge fields, it must be improved to make gauge invariance and symmetry manifest.

In one line:$$\boxed{ \text{Canonical Noether tensor} + \text{harmless total divergence} = \text{physical electromagnetic stress-energy tensor}. }$$