Symmetry and conservation

We have reached the equation, but understanding it requires knowing what follows from its structure. Why is stress–energy conserved? What kind of conservation is that? Does the universe have one total energy? And why does an apparently harmless constant in a matter Lagrangian suddenly matter to gravity?

We will derive the local conservation law first, then identify the symmetry that turns it into a conserved integrated quantity.

15.1 Smooth maps and changes of fields#

A smooth map $F:M\to N$ takes points of one manifold to points of another. “Smooth” means that its coordinate expression has continuous derivatives of every order in overlapping charts. Section 4.1’s transition rules ensure this property does not depend on which compatible charts we choose. A diffeomorphism is a smooth map with a smooth inverse. It preserves the smooth structure; it need not preserve a chosen metric’s distances. A diffeomorphism that does preserve the metric is an isometry.

Before using a map on an entire field, work out what it does to one arrow. A curve $x^\mu(\lambda)$ has tangent $V^\mu$ at a point $p$. Its image curve has coordinates $y^a(\lambda)=F^a(x(\lambda))$. The chain rule gives

This is the pushforward: apply the map to the curve, then take its tangent. The Jacobian carries the tangent from $T_pM$ to $T_{F(p)}N$. An arbitrary smooth map can squash a direction to zero; a diffeomorphism cannot, because its inverse Jacobian undoes the operation. The formula resembles a coordinate transformation, but the interpretation can differ: a chart transition relabels the same point, whereas a map of the manifold may send a point elsewhere.

A covector is a linear measuring rule. We can use the map to define such a rule at its starting point. If $\omega$ measures vectors at $F(p)$, define a question at $p$ by first pushing a vector forward and then asking $\omega$:

This is the pullback. It brings a measuring rule back to the starting point. The star placement keeps track of the direction: $F_*$ carries tangents forward, while $F^*$ brings covectors back. These constructions use the smooth map and the chain rule; no metric or parallel-transport connection is required.

For a scalar field, pullback is just composition, $F^*f=f\circ F$. For a metric it applies the same measuring procedure to both vector inputs:

Now make the map into a continuous motion. A flow $F_s$ follows the integral curves of a vector field $\xi$: each starting point moves with coordinate velocity $\xi^\mu$. For a sufficiently small parameter interval, the motion can be reversed, so these maps are local diffeomorphisms. At first order,

To compare a field before and after that motion, pull it back to the same starting points before subtracting. Its Lie derivative is the first-order change:

For a one-dimensional check, take $F_s(x)=e^sx$ and $f(x)=x^2$. Then $F_s^*f=e^{2s}x^2$, so $\mathcal L_\xi f=2x^2$, with $\xi=x\,\partial_x$. For the ordinary line metric $g=dx\otimes dx$, the map multiplies each vector input by $e^s$, hence $F_s^*g=e^{2s}g$ and $\mathcal L_\xi g=2g$. Dilation is a diffeomorphism but not an isometry of that fixed metric. This example contains the whole comparison rule without any curved-spacetime algebra.

For a general metric, differentiating its pullback gives three terms: one for moving to a new field value, and one for changing each vector input:

To rewrite this using the connection, lower $\xi$ and expand:

The Christoffel formula makes the bracket equal to $\partial_\rho g_{\mu\nu}$, reproducing the three terms above. This step uses the metric-compatible, torsion-free connection. The Lie derivative itself does not require a connection: it compares fields through the specified flow. In contrast, the covariant derivative $\nabla_\xi$ uses parallel transport to compare them.

For the metric,

For the inverse metric,

For a scalar field,

The sign convention here uses the pullback generated by $\xi$. One can formulate passive coordinate changes with the opposite infinitesimal sign; all terms must be changed consistently.

The essential idea is that the geometry and matter fields are transformed together. Changing only the matter field while leaving geometry fixed is generally a physical alteration, not the gauge redundancy being discussed.

The distinction can be checked on the dilation example. Pulling back both the metric and all matter fields describes the configuration consistently through the same map. Pulling back only a matter field while keeping the metric fixed can change measured lengths and energies. Whether a transformation counts as gauge freedom or as a physical boundary symmetry also depends on the boundary conditions.

15.2 Symmetry and local conservation#

PREPARATION FOR THIS SECTION

Earn a conserved quantity from a symmetry

What must stay unchanged before you are allowed to claim a conservation law?

Builds onVariations Connection

See the idea

A symmetry changes a description or configuration in a way that leaves the relevant physical action unchanged, with suitable boundary conditions. Start with one particle and one continuous translation. The cancellation will show which quantity stays constant.

Work it out

1. 01
   Translate a free direction

   For $L=m(\dot x^2+\dot y^2)/2-U(y)$, a constant shift $x\to x+\epsilon$ changes neither the velocities nor the potential. The x Euler–Lagrange equation is therefore $d(\partial L/\partial\dot x)/dt=0$.

   $$p_x=m\dot x=\text{constant}.$$

   Why this step works The missing coordinate dependence expresses a translation symmetry.

2. 02
   Check time translation separately

   For a general particle Lagrangian, define $E=\dot q^i(\partial L/\partial\dot q^i)-L$. Differentiate with the product rule and substitute the Euler–Lagrange equations. The remaining term is $-\partial L/\partial t$.

   $$\frac{dE}{dt}=-\frac{\partial L}{\partial t}.$$

   Why this step works An explicitly time-dependent external apparatus can exchange energy with the chosen particle system.

3. 03
   Carry the cancellation into spacetime

   For an affinely parameterized geodesic tangent $u^a$, differentiate $\xi_a u^a$. The term containing $u^b\nabla_bu^a$ vanishes. The other term vanishes when $\nabla_{(a}\xi_{b)}=0$, the Killing equation. By the metric Lie derivative in §15.1, such a vector generates a symmetry that leaves the metric unchanged.

   $$\frac{d}{d\lambda}(\xi_a u^a)=u^au^b\nabla_{(a}\xi_{b)}=0.$$

   Why this step works A symmetric tangent product contracts only the symmetric part of the derivative.

Go deeper

For a symmetric conserved stress tensor and Killing field, $J^a=T^{ab}\xi_b$ has $\nabla_aJ^a=0$. Integrating this identity over a spacetime region relates charges on two slices to flux through the side boundary. Equality of slice charges requires that flux to vanish or be included in the accounting. A local divergence identity without a suitable comparison field, slicing, and boundary prescription is not automatically a global gravitational energy. Gauge symmetries also lead to identities among equations of motion; the next text distinguishes this from the finite-dimensional first-theorem example.

Further readingDavid Tong · General relativity

Test the idea

FIRST, PREDICT

A particle moves in a potential U(x,t) controlled by an external apparatus. Must its mechanical energy stay constant?

No. Explicit time dependence can transfer energy to or from the particle.Yes. Noether’s theorem makes every energy constant.No conservation law can ever include the apparatus.

Compare the reasoning

No. Explicit time dependence can transfer energy to or from the particle.

The differentiated energy satisfies dE/dt = −∂L/∂t = ∂U/∂t for this mechanical Lagrangian.

Yes. Noether’s theorem makes every energy constant.

A conservation statement requires its symmetry and a clearly specified system.

No conservation law can ever include the apparatus.

An enlarged closed system can include the apparatus and its energy transfer.

A hint

Check explicit time dependence before invoking time-translation symmetry.

NOW CHANGE THE EXAMPLE

For $L=\dot x^2-3y^2+\dot y^2$ in a consistent unit convention and $\dot x=4$, calculate the conserved x momentum.

A hint

Differentiate L with respect to $\dot x$.

Work through the solution

$p_x=2\dot x=8$.

Specify the symmetry, system, and boundary flux before naming a conserved charge.

In mechanics, a continuous global symmetry produces a conserved quantity. Diffeomorphism invariance has an arbitrary spacetime-dependent vector field as its generator. This local freedom leads to differential identities among equations of motion—an instance of Noether’s second theorem.

First consider the matter action. Assume its matter fields satisfy their equations of motion, so their first-order action variation vanishes apart from boundary terms. Choose $\xi^\mu$ with compact support to remove the boundary terms. Diffeomorphism invariance and the stress-tensor definition give

Insert the inverse-metric Lie derivative. Symmetry of $T_{\mu\nu}$ makes its two terms equal:

Integrate by parts:

Since the interior vector field $\xi_\nu$ was arbitrary,

This derivation used the matter equations, but not Einstein’s equation. Matter on a prescribed curved background can satisfy this conservation law if its action and equations have the stated covariance and no unaccounted external exchanges.

The phrase on shell means that the relevant equations of motion are satisfied. Off shell means we consider arbitrary field configurations. Conservation of the matter stress tensor is generally an on-shell statement. For the canonical scalar field in natural units, the more informative off-shell identity is

You can check it directly by differentiating the scalar stress tensor, using the product rule and commuting derivatives on a scalar. Its divergence vanishes once the scalar equation holds. Conversely, stress conservation alone need not imply every matter equation: if $\nabla^\nu\phi$ vanishes, that product can be zero without enforcing the scalar equation.

Apply the same diffeomorphism argument to the gravitational action. Its metric variation is proportional to $G_{\mu\nu}+\Lambda g_{\mu\nu}$. Because this is a pure geometric symmetry, the resulting identity

holds off shell for constant $\Lambda$. This reproduces the contracted Bianchi identity from the action’s symmetry. Einstein’s equation joins an identically compatible geometric tensor to a matter tensor conserved when matter evolves consistently.

This is not circular reasoning. One route derives a geometric identity from curvature; another derives matter conservation from matter dynamics; the action explains why the two structures can be coupled.

15.3 From local balance to an integrated charge#

Expand the covariant divergence:

For a vector current $J^\mu$, there is no last free-index connection term:

That difference matters. A divergence-free vector current can be integrated using the divergence theorem to obtain a scalar charge balance. A stress tensor still carries a free vector index, and vectors at different points cannot generally be added without specifying how their frames are related. Curvature makes that comparison path dependent.

A useful mixed-index version is

To derive it, the lower-index correction in $\nabla_\mu T^\mu{}_{\nu}$ moves to the right as $\Gamma^\lambda{}_{\mu\nu}T^\mu{}_{\lambda}$. Lower the first connection index and substitute the Christoffel formula:

The first and third terms cancel after exchanging the summed indices $\mu,\alpha$, since $T^{\mu\alpha}$ is symmetric. The middle term is the displayed metric derivative. If the metric depends on the time coordinate, the corresponding coordinate energy equation has a geometric term on its right-hand side. It is not generally the ordinary flat-spacetime continuity equation for one global energy.

In a sufficiently small freely falling laboratory, the connection vanishes at a chosen event and the law reduces there to the familiar local conservation equations. Across a finite curved region, comparing different local laboratories is part of the problem.

15.4 Conserved charges from Killing vectors#

A Killing vector satisfies

It generates a symmetry of the metric itself. For a symmetric conserved stress tensor define

Its divergence is

The first term vanishes by matter conservation. Since $T^{\mu\nu}$ is symmetric, only the symmetric part of $\nabla_\mu\xi_\nu$ contributes to the second term. The Killing equation makes that symmetric part zero. Therefore

A timelike Killing vector supplies an energy symmetry; a rotational Killing vector supplies angular-momentum symmetry. The same construction works for the corresponding charges, with their conventional normalizations.

To make the energy case concrete, normalize a timelike $\xi^\mu$ so its components are dimensionless and it approaches a unit time direction in an appropriate asymptotically flat region. On a spacelike hypersurface $\Sigma$—a three-dimensional slice with a timelike normal—choose its future-directed unit normal $n^\mu$ and define the matter Killing energy

In Minkowski spacetime, with $n^\mu=\xi^\mu=(1,0,0,0)$, this is $\int\epsilon\,d^3x$. Apply the divergence theorem to a region between two such hypersurfaces: the change of $E_\xi$ equals minus the outward flux through the intervening side boundary. If there is no side flux, the charge is conserved.

The result requires the Killing symmetry, the field equations, and suitable boundaries or falloff. It is a matter Killing charge; it is not by itself a universal formula for the full gravitating system’s total energy.

For a static chart, the metric is time independent and has $g_{0i}=0$. Take $\xi=\partial_{x^0}$ and let $N=\sqrt{-\xi^\mu\xi_\mu}=\sqrt{-g_{00}}$. Observers at fixed spatial coordinates have unit time direction $n^\mu=\xi^\mu/N$. A photon with four-momentum $p^\mu$ has symmetry energy $E_\xi=-c\,p_\mu\xi^\mu$; its local measured energy is $E_{\rm local}=-c\,p_\mu n^\mu$. Substitution gives

The quantity $p\cdot\xi$ is conserved along the free ray: its derivative contracts the symmetric product of the ray tangent and momentum with $\nabla_{(\mu}\xi_{\nu)}=0$. Two static observers can therefore measure different local photon energies while agreeing on $E_\xi$. Gravitational redshift is consistent with energy conservation when the correct conserved quantity and observer normalization are used.

15.5 Energy of gravitational fields#

Gravitational waves transfer energy to detectors, and black holes have measurable mass. Yet general relativity provides no universal, unique, exact local gravitational stress tensor playing the same role as matter’s $T_{\mu\nu}$ in every spacetime.

The equivalence principle offers a useful warning: expressions constructed as a supposed gravitational energy density from connection coefficients can be changed dramatically by changing coordinates, including setting the connection to zero at one event. This warning is not by itself a complete mathematical proof of every nonexistence statement. Curvature tensors certainly exist. The full issue is finding an exact local object with all the desired covariance, conservation, normalization, and physical-energy properties without extra structure.

Several well-defined constructions answer more specific questions:

Each construction specifies its boundary conditions or approximation before defining an energy. Brown and York’s quasilocal construction, for example, derives a boundary stress tensor from variation of a gravitational action with respect to the boundary metric; in the appropriate asymptotically flat limit it recovers ADM quantities. See their original “Quasilocal Energy and Conserved Charges Derived from the Gravitational Action”.

A generic expanding cosmology has no global timelike Killing vector. Photons redshift, and matter still obeys local covariant conservation. Asking where every lost photon joule “went” presupposes a globally conserved energy account that the spacetime may not possess.

The comoving relation $d(\epsilon V)=-p\,dV$ remains useful and exact for a homogeneous perfect-fluid element under its assumptions. It does not require inventing an external reservoir into which “space stores the missing energy.” Nor is “the total energy of every universe is zero” a general theorem of Einstein’s equation.

15.6 The cosmological constant as vacuum stress–energy#

Move the cosmological term to the right-hand side:

with

For any unit timelike observer $n^\mu$,

Every such observer measures the same energy density and zero energy flux. Spatial projection gives isotropic pressure

This is precisely the perfect-fluid form with $\epsilon+p=0$: its velocity-dependent part disappears. Vacuum stress of this form does not select a preferred local rest frame.

Negative pressure is not merely a verbal synonym for repulsion. Its role follows from the equations. In the timelike Ricci projection,

For positive $\Lambda$, that contribution has the opposite sign from positive-density, nonnegative-pressure matter. The resulting expansion or focusing behavior also depends on the spacetime and family of observers being studied; Chapter 19 develops the cosmological case.

There is also a thermodynamic check. If $\epsilon_\Lambda$ is constant while a comoving volume changes, $E=\epsilon_\Lambda V$ implies $dE=\epsilon_\Lambda dV$. Comparing with $dE=-p\,dV$ gives $p=-\epsilon_\Lambda$. The energy in the volume grows with the volume; the local work relation is satisfied by negative pressure.

15.7 Adding a constant to the matter Lagrangian#

In nongravitational mechanics, adding a constant to $L$ changes the action by a fixed amount for a fixed time interval and leaves the equations of motion unchanged.

In gravity, replace the matter Lagrangian energy density by

where $C$ is a constant energy density. The matter field equations remain unchanged, but the metric varies the volume multiplying $C$. The stress tensor shifts by

Consequently the effective cosmological constant becomes

That is the conceptual doorway to the cosmological constant problem discussed in Chapter 23: gravity responds to the metric dependence of a vacuum-energy term even when nongravitational equations are insensitive to a constant offset.

A constant $\Lambda$ is consistent with the Bianchi identity because $\nabla g=0$. If someone simply replaces it by an arbitrary function $\Lambda(x)$ while keeping $\nabla^\mu T_{\mu\nu}=0$ and leaving the rest of the equation unchanged, taking the divergence forces

A varying dark-energy model therefore needs additional consistent dynamics or energy exchange. Changing a parameter into a function is not a free modification of a constrained field theory.

15.8 Coupling a matter model to curved spacetime#

A common way to place a nongravitational field theory on curved spacetime is to begin with its special-relativistic action and make the replacements

where the last replacement is appropriate to the kind of field involved. Scalars have $\nabla_\mu\phi=\partial_\mu\phi$. For the electromagnetic field,

because the two symmetric Levi-Civita connection terms cancel. Spinors require a local frame and spin connection, introduced in Chapter 21.

This procedure is called minimal coupling. Applying it at the action level is especially useful: it produces matter equations and a stress tensor with mutually consistent variational origins.

It is not a uniqueness theorem. In natural units, a term such as

is also generally covariant, with dimensionless coupling $\zeta$ in four dimensions, and vanishes in exactly flat spacetime. A flat-spacetime theory alone cannot tell you its coefficient. Moreover, expressions related by commuting derivatives in flat spacetime need not remain equivalent after covariantization, because covariant derivatives on tensors do not generally commute.

The equation’s assumptions should now be visible: a Lorentzian metric, a specified connection structure, chosen gravitational and matter actions, appropriate boundary data, and a consistent coupling between them. With those in place, covariance, curvature identities, local energy–momentum balance, and dynamics reinforce one another.

15.9 Mach’s question: what fixes an inertial frame?#

Imagine a rotating bucket of water. The surface becomes concave. Rotating relative to what? Relative to the bucket cannot be the whole answer: once the water co-rotates with the bucket, the concavity remains. Mach’s critique of absolute space encouraged the idea that inertia might ultimately be tied to the matter elsewhere in the universe. Einstein found this line of thought influential.

General relativity does not reduce to the claim that distant stars uniquely determine every local inertial frame. A simple counterexample is empty Minkowski spacetime: it contains no matter but has a well-defined family of inertial worldlines. A metric and suitable initial or boundary data remain part of the theory. Frame dragging shows that rotating matter does influence local inertial directions; it does not establish every stronger formulation of “Mach’s principle.” The name covers several related proposals, not one universally accepted theorem. Einstein’s own 1918 discussion of the principles of GR.

This distinction is another application of the book’s discipline. An idea can motivate a theory without becoming a theorem of the final theory. We test what the actual field equation predicts.

15.10 A tiny density that does not dilute#

Use $\Lambda=1.1\times10^{-52}\,\mathrm{m^{-2}}$ as an illustrative cosmological scale, not a newly measured parameter estimate. The effective vacuum mass-equivalent density is

Dividing by the mass of a hydrogen atom, about $1.67\times10^{-27}\,\mathrm{kg}$, gives the mass equivalent of roughly 3.5 hydrogen atoms per cubic metre. The vacuum component is not literally a gas of hydrogen. The comparison translates an unfamiliar density into a familiar mass scale.

For a cosmological constant, the energy density $\epsilon_\Lambda=\rho_\Lambda c^2$ remains constant while a comoving region expands, and its pressure is $p_\Lambda=-\epsilon_\Lambda$. For comparison, let $a$ denote a common scale factor multiplying all spatial lengths in a uniformly expanding region, so a comoving volume is proportional to $a^3$. Dust keeps the same rest energy in that volume and therefore has $\epsilon\propto a^{-3}$. For radiation, substitute $p=\epsilon/3$ into $d(\epsilon a^3)=-p\,d(a^3)$. Expanding gives $a^3d\epsilon+4\epsilon a^2da=0$, hence $\epsilon\propto a^{-4}$. A constant density can therefore become larger than these declining densities even if it starts smaller. That conclusion assumes the component really is constant and the cosmological solution evolves into that regime; it is not a general forecast for every possible dark-energy model.