2026-09-21
Noether's Theorem and Spacetime Symmetries
This note is basically based on David Tong’s lecture notes and Schwartz’s textbook. I use $g_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$ and $c=\hbar=1$. Repeated indices are summed. All transformations are kept to first order, and $\delta\phi_a$ denotes the variation at fixed $x$.
Noether’s theorem
This theorem reads:
Every continuous global symmetry of the action gives rise to a conserved current:
$\boxed{\partial_\mu j^\mu=0}$
when the equations of motion hold.
Proof:
For a set of fields $\phi_a(x)$, the action is
$$
S=\int d^4x\,\mathcal L(\phi_a,\partial_\mu\phi_a).
$$
Consider an infinitesimal transformation
$$
\phi_a\longrightarrow\phi_a+\delta\phi_a,
\qquad
\delta\phi_a=\epsilon X_a,
\qquad
\partial_\mu\epsilon=0.
$$
Using the chain rule,
$$
\delta\mathcal L
=
\frac{\partial\mathcal L}{\partial\phi_a}\delta\phi_a
+
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
\partial_\mu(\delta\phi_a),
$$
where $\delta$ and $\partial_\mu$ commute.
Applying the product rule,
$$
\begin{aligned}
\delta\mathcal L
={}&
\left[
\frac{\partial\mathcal L}{\partial\phi_a}
-
\partial_\mu
\left(
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
\right)
\right]\delta\phi_a\\
&+
\partial_\mu
\left(
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
\delta\phi_a
\right).
\end{aligned}
$$
The fields satisfy the Euler–Lagrange
equations:
$$
\frac{\partial\mathcal L}{\partial\phi_a}
-
\partial_\mu
\left(
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
\right)
=0.
$$
Therefore, on solutions,
$$
\delta\mathcal L
=
\partial_\mu
\left(
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
\delta\phi_a
\right).
$$
Independently, if the transformation is a symmetry, the Lagrangian density may change by a total divergence:
$$
\boxed{\delta\mathcal L=\partial_\mu F^\mu.}
$$
This symmetry condition is checked without using the equations of motion. It is allowed because
$$
\delta S
=
\int d^4x\,\partial_\mu F^\mu
=
\int_{\partial\Omega}d\Sigma_\mu\,F^\mu
$$
is only a boundary term. If the boundary integral vanishes, $\delta S=0$. Strict invariance of $\mathcal L$ corresponds to choosing $F^\mu=0$.
We now have two expressions for the same variation:
$$
\delta\mathcal L
=
\underbrace{
\partial_\mu
\left(
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
\delta\phi_a
\right)
}_{\text{Euler–Lagrange equations}}
=
\underbrace{\partial_\mu F^\mu}_{\text{symmetry}}.
$$
Subtracting,
$$
\partial_\mu
\left[
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
\delta\phi_a-F^\mu
\right]=0.
$$
Define the Noether current:
$$
j^\mu
\equiv
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
\delta\phi_a-F^\mu.
$$
Then
$$
\partial_\mu j^\mu=0.
$$
Vanishing $\epsilon$
Since $\delta\phi_a=\epsilon X_a$, write $F^\mu=\epsilon K^\mu$. Thus,
$$
j^\mu
=
\epsilon
\left(
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
X_a-K^\mu
\right),
$$
and
$$
\epsilon\,\partial_\mu
\left(
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
X_a-K^\mu
\right)=0.
$$
Because $\epsilon$ is an arbitrary constant,
$$
\partial_\mu
\left(
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
X_a-K^\mu
\right)=0.
$$
So we can ignore $\epsilon$, $j^\mu$ includes the infinitesimal transformation parameters.
Translations
Consider the infinitesimal translation
$$
x'^\nu=x^\nu-\epsilon^\nu,
\qquad
\partial_\mu\epsilon^\nu=0.
$$
Since $\phi’_a(x’)=\phi_a(x)$,
$$
\phi'_a(x)
=
\phi_a(x+\epsilon)
=
\phi_a(x)+\epsilon^\nu\partial_\nu\phi_a(x).
$$
Therefore,
$$
\delta\phi_a=\epsilon^\nu\partial_\nu\phi_a.
$$
If $\mathcal L$ has no explicit dependence on $x$, similarly,
$$
\delta\mathcal L
=
\epsilon^\nu\partial_\nu\mathcal L.
$$
To express this as a divergence with index $\mu$,
$$
\begin{aligned}
\delta\mathcal L
&=
\epsilon^\nu\delta^\mu{}_\nu\partial_\mu\mathcal L\\
&=
\partial_\mu
\left(
\epsilon^\nu\delta^\mu{}_\nu\mathcal L
\right).
\end{aligned}
$$
We choose
$$
F^\mu=\epsilon^\nu\delta^\mu{}_\nu\mathcal L.
$$
Substitute into the Noether current:
$$
\begin{aligned}
j^\mu
&=
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
\delta\phi_a-F^\mu\\
&=
\epsilon^\nu
\left(
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
\partial_\nu\phi_a
-\delta^\mu{}_\nu\mathcal L
\right).
\end{aligned}
$$
Define the canonical energy–momentum tensor:
$$
T^\mu{}_\nu
\equiv
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
\partial_\nu\phi_a
-\delta^\mu{}_\nu\mathcal L.
$$
Then
$$
j^\mu=\epsilon^\nu T^\mu{}_\nu,
$$
so
$$
0=\partial_\mu j^\mu
=\epsilon^\nu\partial_\mu T^\mu{}_\nu.
$$
The four components of $\epsilon^\nu$ are independent. Their coefficients must vanish separately:
$$
\partial_\mu T^\mu{}_\nu=0,
\qquad \nu=0,1,2,3.
$$
We are comparing coefficients, not dividing by a vector. The index $\nu$ labels the four translation currents.
Raising the second index,
$$
T^{\mu\nu}
=
g^{\nu\rho}T^\mu{}_\rho
=
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi_a)}
\partial^\nu\phi_a
-g^{\mu\nu}\mathcal L,
$$
and
$$
\partial_\mu T^{\mu\nu}=0.
$$
The conserved four-momentum is
$$
P^\nu=\int d^3x\,T^{0\nu}.
$$
Indeed,
$\frac{dP^\nu}{dt}
=
-\int d^3x\,\partial_iT^{i\nu}
=
-\oint dS_i\,T^{i\nu}
=0,$
assuming that the boundary flux vanishes.
In particular,
$$
E=P^0=\int d^3x\,T^{00},
\qquad
P^i=\int d^3x\,T^{0i}.
$$
Example: the real Klein–Gordon field
The Lagrangian density is
$$
\mathcal L
=
\frac12g^{\alpha\beta}
\partial_\alpha\phi\,\partial_\beta\phi
-\frac12m^2\phi^2.
$$
Then
$$
\begin{aligned}
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}
&=
\frac12g^{\alpha\beta}
\left(
\delta^\mu{}_\alpha\partial_\beta\phi
+
\partial_\alpha\phi\,\delta^\mu{}_\beta
\right)\\
&=g^{\mu\nu}\partial_\nu\phi
=\partial^\mu\phi.
\end{aligned}
$$
Hence,
$$
T^{\mu\nu}
=
\partial^\mu\phi\,\partial^\nu\phi
-g^{\mu\nu}\mathcal L.
$$
For this field,
$T^{\mu\nu}=T^{\nu\mu}$, and
$$
T^{00}
=
\frac12\dot\phi^{\,2}
+\frac12(\nabla\phi)^2
+\frac12m^2\phi^2.
$$
Here we consider a scalar field with a Lorentz-invariant Lagrangian, such as the Klein–Gordon field.
An infinitesimal Lorentz transformation is
$$
x'^\mu=\Lambda^\mu{}_\nu x^\nu,
\qquad
\Lambda^\mu{}_\nu=\delta^\mu{}_\nu+\omega^\mu{}_\nu,
$$
where $\omega^\mu{}_\nu$ is infinitesimal and constant.
The Lorentz condition is
$$
\Lambda^\mu{}_\rho g^{\rho\sigma}
\Lambda^\nu{}_\sigma=g^{\mu\nu}.
$$
Expanding to first order,
$$
g^{\mu\nu}+\omega^{\mu\nu}+\omega^{\nu\mu}
=g^{\mu\nu}.
$$
Thus,
$$
\omega^{\mu\nu}=-\omega^{\nu\mu},
\qquad
\omega_{\mu\nu}=-\omega_{\nu\mu}.
$$
In particular,
$$
\omega^\mu{}_\mu=g^{\mu\rho}\omega_{\rho\mu}=0.
$$
For a scalar field,
$$
\phi'(x)=\phi(\Lambda^{-1}x).
$$
Since $\Lambda^{-1}=I-\omega$ to first order,
$$
\phi'(x)
=
\phi(x)-\omega^\rho{}_\sigma x^\sigma\partial_\rho\phi(x).
$$
Therefore,
$$
\delta\phi=-\omega^\rho{}_\sigma x^\sigma\partial_\rho\phi.
$$
Similarly,
$$
\delta\mathcal L
=
-\omega^\mu{}_\nu x^\nu\partial_\mu\mathcal L.
$$
Using the product rule,
$$
\begin{aligned}
\delta\mathcal L
&=
-\partial_\mu
\left(\omega^\mu{}_\nu x^\nu\mathcal L\right)
+\omega^\mu{}_\nu(\partial_\mu x^\nu)\mathcal L\\
&=
-\partial_\mu
\left(\omega^\mu{}_\nu x^\nu\mathcal L\right)
+\omega^\mu{}_\mu\mathcal L\\
&=
-\partial_\mu
\left(\omega^\mu{}_\nu x^\nu\mathcal L\right).
\end{aligned}
$$
Thus, we may choose
$$
F^\mu=-\omega^\mu{}_\nu x^\nu\mathcal L.
$$
Substitute into the Noether current:
$$
\begin{aligned}
j^\mu
&=
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}
\delta\phi-F^\mu\\
&=
-\omega^\rho{}_\sigma x^\sigma
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}
\partial_\rho\phi
+\omega^\mu{}_\sigma x^\sigma\mathcal L\\
&=
-\omega^\rho{}_\sigma x^\sigma
\left(
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}
\partial_\rho\phi
-\delta^\mu{}_\rho\mathcal L
\right)\\
&=
-\omega^\rho{}_\sigma x^\sigma T^\mu{}_\rho\\
&=
-\omega_{\rho\sigma}x^\sigma T^{\mu\rho}.
\end{aligned}
$$
Interchanging the dummy indices and using antisymmetry,
$$
-\omega_{\rho\sigma}x^\sigma T^{\mu\rho}
=
\omega_{\rho\sigma}x^\rho T^{\mu\sigma}.
$$
Averaging these two forms gives
$$
j^\mu
=
\frac12\omega_{\rho\sigma}
\left(
x^\rho T^{\mu\sigma}-x^\sigma T^{\mu\rho}
\right).
$$
Define
$$
J^{\mu\rho\sigma}
\equiv
x^\rho T^{\mu\sigma}-x^\sigma T^{\mu\rho}.
$$
Then
$j^\mu=\frac12\omega_{\rho\sigma}J^{\mu\rho\sigma},
\qquad
J^{\mu\rho\sigma}=-J^{\mu\sigma\rho}.$
The factor $1/2$ avoids counting each antisymmetric pair twice:
$$
j^\mu
=
\sum_{\rho<\sigma}\omega_{\rho\sigma}J^{\mu\rho\sigma}.
$$
Since $\partial_\mu j^\mu=0$,
$$
\sum_{\rho<\sigma}
\omega_{\rho\sigma}\partial_\mu J^{\mu\rho\sigma}=0.
$$
There are six independent parameters,
$$
\omega_{01},\ \omega_{02},\ \omega_{03},\
\omega_{12},\ \omega_{13},\ \omega_{23}.
$$
Their coefficients vanish separately:
$$
\partial_\mu J^{\mu\rho\sigma}=0,
\qquad \rho<\sigma.
$$
The index $\mu$ is the current index, while the pair $(\rho,\sigma)$ labels the six Lorentz currents. Antisymmetry leaves
$$
4\times 6=24
$$
component slots: six four-component currents. Contracting $\mu$ in the divergence gives six conservation equations.
The conserved charges are
$$
Q^{\rho\sigma}
=
\int d^3x\,J^{0\rho\sigma}
=
\int d^3x
\left(
x^\rho T^{0\sigma}-x^\sigma T^{0\rho}
\right),
$$
again assuming vanishing boundary fluxes.
For spatial rotations,
$$
Q^{ij}
=
\int d^3x
\left(
x^iT^{0j}-x^jT^{0i}
\right).
$$
These are the angular-momentum components:
$$
L^1=Q^{23},
\qquad
L^2=Q^{31},
\qquad
L^3=Q^{12}.
$$
The other three charges correspond to boosts:
$$
Q^{0i}
=
tP^i-\int d^3x\,x^iT^{00}.
$$
For the Klein–Gordon field, we can check conservation directly:
$$
\begin{aligned}
\partial_\mu J^{\mu\rho\sigma}
&=
T^{\rho\sigma}-T^{\sigma\rho}
+x^\rho\partial_\mu T^{\mu\sigma}
-x^\sigma\partial_\mu T^{\mu\rho}\\
&=0,
\end{aligned}
$$
using $T^{\rho\sigma}=T^{\sigma\rho}$ and $\partial_\mu T^{\mu\nu}=0$.
For vector or spinor fields, the canonical Lorentz current generally also includes a spin term. The expression above is the scalar-field result.