Energy, momentum, and stress

Consider gas in a box. It has energy, pushes on the walls, and can carry energy and momentum from one part of the box to another. An observer moving past the box measures different particle energies and sees a different flow. We need a description that relates all these measurements consistently.

The stress-energy tensor provides that description. Once we understand its entries, we can connect matter to the geometry of the preceding chapters.

11.1 Energy density and momentum flux#

Put an imaginary detector through the gas. It does not block the particles; it records each crossing. A particle carries energy and a momentum arrow. The arrow can point sideways to the detector, so we must specify both which direction is crossed and which momentum component is carried.

First watch a single stream cross the detector. Then choose balanced motion. There is no average motion of the gas, but particles still cross in both directions. Their energies can flow equally both ways while their normal momentum transfers add. This is how pressure can remain when the bulk flow is zero.

MATTER LAB / 01

What crosses the surface?

Each moving particle carries energy and momentum. Put a detector through the box, choose what to count, and watch the crossings build a measurement.

Toward +xToward −xDetector96 particles · no collisions

From counted crossings35.068

From all 96 particles34.772

Crossings toward + / −213 / 0

Latest x-momentum contribution: +1 × (0.659) = +0.659. Crossing sign × carried value.

Readings use mc²/L³. L is the box width; m is one particle’s rest mass. One time unit is the time light takes to cross the box.

Every crossing carries a signed amount into the tally. Divide by the detector area and the elapsed time to get flux. The second reading averages over all 96 particles; finite counting time explains the difference between them.

Why opposite motion can produce pressure

First choose Balanced motion, Across x, and x-momentum. Half of the x-moving particles go right; the other half go left. Their average momentum is zero.

A right-moving particle carries positive x-momentum across the detector in the positive direction, giving a positive contribution. A left-moving particle carries negative x-momentum across it in the negative direction. Negative momentum transported backward also gives a positive contribution. Add both and divide by area and counting time: that is the normal momentum flux. For a gas with equal flux in every direction and no bulk motion, we call it pressure.

A reflecting wall gives the same reading: each collision reverses the particle’s normal momentum, so the wall receives twice that momentum, while only the population moving toward the wall hits it.

Now select Energy. Particle energies are positive on either trip, so the two crossing directions cancel in the long-time energy flux. Zero energy flow and positive pressure are compatible. Finally choose Tilted pairs and measure y-momentum across x: particles can transport momentum sideways to the detector’s normal. This is an off-diagonal flux. It comes from the prepared particle directions here; we have not modeled viscosity.

Build the stress–energy tensor from these measurements

Selecting where to measure chooses a row. Selecting the carried quantity chooses a column. The highlighted entry is the all-particle average corresponding to the detector. The first row records densities in the cube; the remaining rows record fluxes through surfaces.

Columns: energy, x-momentum, y-momentum, z-momentum

In box114.9563.220.000.00

Across x63.2234.770.000.00

Across y0.000.000.000.00

Across z0.000.000.000.00

The time coordinate is $x^0=ct$. To give every entry energy-density units, the time row uses energy density and $c$ times momentum density. The spatial rows use energy flux divided by $c$ and ordinary momentum flux. These factors are already included in the readings.

In these box units, a particle has $p^0=\gamma$ and $p^i=\gamma v^i$, with speed measured in units of $c$. Averaging the particles gives $T^{\mu\nu}=\sum_a p_a^\mu p_a^\nu/p_a^0$ for unit box volume. The formula is symmetric in the two indices. In ordinary units the same expression is $T^{\mu\nu}=(c/V)\sum_a p_a^\mu p_a^\nu/p_a^0$, where $p_a^0=E_a/c$.

The model, units, and counting checks

All 96 particles have the same rest mass and speed. One stream has one shared velocity. Balanced motion contains equal populations along the six directions ±x, ±y, ±z; its average momentum vanishes and its three diagonal momentum fluxes agree. It is a discrete distribution with an isotropic second moment, not a calculation of thermal equilibrium. Tilted pairs contain equal opposing velocities in the xy plane.

The particles move freely. A particle leaving one side re-enters the opposite side, so the box samples an endlessly repeated medium. The detector is an internal plane spanning the cube, of area $L^2$. The drawing is an orthographic view of the three-dimensional positions. Short trails show velocity; a particle brightens briefly after a detector crossing. Fixed pseudorandom initial positions avoid arranging the two velocity directions in separate halves of the box. Changing a setup control recalculates the whole history from those same initial positions.

Every crossing contributes its carried momentum with the sign of its crossing direction. We divide the accumulated sum by area and elapsed time. For energy flux, divide by $c$ as well before entering it in $T$. At zero time the crossing estimate is undefined, shown as a dash. For a short measurement, the crossing and all-particle estimates need not agree. A longer count reduces that sampling difference; no hidden smoothing forces agreement.

Energy is measured in $mc^2$, momentum in $mc$, distance in $L$, and time in $L/c$. Dividing the flux by its corresponding area and time units puts all tensor entries in $mc^2/L^3$. The ensemble construction follows Edmund Bertschinger’s treatment of a gas of noninteracting particles.

A density measures what is present per volume. A flux measures what crosses a surface per area per time. The distinction is physical: counting the particles inside the cube and recording passages through the detector are different experiments. In the lab, selecting “In the box” performs the first; selecting “Across x,” “Across y,” or “Across z” performs the second. The finite crossing estimate fluctuates around the volume average because a short count samples only some of the particles.

Start in a tiny laboratory with orthonormal axes, using $x^{\hat 0}=ct$. Hats distinguish physically calibrated local axes from arbitrary coordinate axes. Let

• $\epsilon$ be energy per volume, in $\mathrm{J/m^3}$;
• $S_i$ be energy flux, in $\mathrm{J/(m^2\,s)}$;
• $\pi_i$ be momentum per volume, in $\mathrm{kg/(m^2\,s)}$;
• $\Pi_{ij}$ be the flux of $j$-momentum through a surface normal to direction $i$, in $\mathrm{N/m^2}$.

The stress–energy tensor packages these quantities as

Every entry has energy-density units. In this display, the second index identifies which energy–momentum component is being tracked; the first identifies the direction through which it flows. The time row records what is present in a spatial volume. The spatial rows record transport through its faces.

For the symmetric stress tensor appearing in ordinary metric general relativity,

This is a local relationship between energy flow and momentum density. It is one of the most useful ways to recognize that energy and momentum belong to one relativistic object. A beam of light has momentum because it transports energy; a moving lump of matter does too.

Check the packaging by taking a divergence in an inertial Cartesian laboratory:

Set $\nu=0$. Since $\partial_0=c^{-1}\partial_t$,

or

Energy disappearing from a little box must flow through its faces. Now set $\nu=j$:

That is the momentum balance law. A pressure gradient changes a fluid’s momentum because one face receives a different momentum flux from the opposite face. Together, these are the four local balance equations: one for energy and one for each component of momentum.

Coordinate components and instrument readings. In spherical coordinates, $T^{\theta\theta}$ is not a pressure read directly from a gauge aligned with an angular ruler. The coordinate basis has its own normalization. Physical readings come from contraction with an observer’s orthonormal frame.

Lowering a time index. In an orthonormal frame with signature $(-,+,+,+)$, lowering a single time index changes its sign. Thus a fluid at rest, with pressure $p$, has

The measured energy density remains $T_{\hat0\hat0}=\epsilon$. The negative mixed component results from lowering just one time index; lowering both gives the positive energy-density component.

11.2 Measurements made by a chosen observer#

A particle’s energy depends on who measures it. So does a field’s energy density. The covariant object is $T_{\mu\nu}$; “energy density” is one observer’s projection of it.

Let an observer have four-velocity $w^\mu$, with $w_\mu w^\mu=-c^2$, and define the dimensionless unit timelike vector

The observer’s spatial projector is

Why the plus sign? Apply it to $n^\nu$:

It removes the time direction, leaving precisely the observer’s local three-dimensional rest space. The observer measures

Here $j^\mu$ is spatial and has energy-density units: the physical energy-flux vector is $c j^\mu$, and momentum density is $j^\mu/c$. The tensor $P^{\mu\nu}$ is the spatial stress seen by that observer. Together they reconstruct the whole tensor:

Check these definitions in the observer’s own orthonormal frame. There $n^\mu=(1,0,0,0)$, $n_\mu=(-1,0,0,0)$, and $H^\mu{}_{\nu}=\operatorname{diag}(0,1,1,1)$. Thus $\epsilon_{(n)}=T_{00}$, $j^i=T^{i0}=S_i/c$, and $P^{ij}=T^{ij}$. The minus sign in the definition of $j$ compensates for the negative lowered time component. We recover exactly the quantities in §11.1.

For a numerical observer change, take a sample with negligible pressure and no energy flow in its own frame, so its only nonzero component is $T_{00}=\epsilon$. An observer moving along $x$ has $n^\mu=\gamma(1,\beta,0,0)$ and measures

At $v=0.6c$, $\gamma=1.25$, giving $1.5625\epsilon$. One factor of $\gamma$ reflects the increased energy of each particle; the other reflects the increased number of particles per volume measured in the moving frame. The tensor describes both observers’ results.

Some matter admits a local rest frame with zero energy flux. A single beam of light does not: catching up with the beam would require a timelike observer to become null. Never assume that every stress tensor can be put into a perfect-fluid rest-frame form.

11.3 Dust: particles with negligible pressure#

The pressureless sample just considered is called dust. Within a small fluid element, its particles share one velocity; their random motion and pressure are neglected. This is useful for sufficiently cold, dilute matter on scales where those approximations hold.

In the common rest frame of a small dust element, only the energy-density component is nonzero. If its four-velocity is $u^\mu$, the covariant expression reproducing this is

In the rest frame $u^{\hat\mu}=(c,0,0,0)$, so $T^{\hat0\hat0}=\rho c^2=\epsilon$. A moving observer sees energy flux and spatial momentum flux automatically, because $u^i$ is then nonzero.

The curved-spacetime conservation equation is

Contract this with $u_\nu$. The second term vanishes because

The first term therefore gives

Insert that result back into the uncontracted equation. Wherever $\rho\ne0$,

The same conservation law has produced both continuity of dust mass and geodesic motion. For this pressureless model, continuity and free fall are two consequences of the same equation.

The qualification matters. Charged matter exchanging momentum with an electromagnetic field need not have a separately conserved matter stress tensor. A pressured fluid accelerates because neighboring fluid elements push it. Extended spinning bodies can have curvature-dependent corrections to simple geodesic motion. Dust is a controlled idealization, not a universal description of matter.

11.4 Perfect fluids, derived from isotropy#

The particle experiment showed momentum crossing a surface because particles traveled through it. Real fluids also transmit tangential forces between neighboring layers. Slide a plate over a layer of oil: the oil beside the plate follows it and drags the oil farther away. The force required depends on how quickly velocity changes across the layers. The material coefficient relating that velocity gradient to force per area is its viscosity.

In the next experiment, the blue dots are dye following the average fluid motion. The gold arrows show the forces between layers. A faster flow need not have a larger shear force: inspect the middle of the pressure-driven channel, where the fluid moves fastest but the velocity profile has zero slope.

MATTER LAB / 02

How one layer pulls another.

Move the upper plate. The fluid beside it moves too, pulling the layers below. Slide the gold cut through the flow to measure the force between neighboring layers.

Velocity and dyeShear forceDye follows the fluid’s average motion.

H = 2 cm · viewing window 12 cm long · height expanded to show the layers

Speed at the cut0.210 m/s

Force/area on lower fluid3.60 Pa

x-momentum flux toward +y-3.60 Pa

The faster upper layers exert a rightward force on the lower fluid. The upper fluid feels the opposite force. Increase viscosity at a fixed plate speed: the velocity profile stays the same, but the required force grows.

Why force and momentum flux have opposite signs here

Take x to the right and y upward. Focus on the fluid below the gold cut. If the fluid just above it moves faster to the right, it pulls the lower fluid to the right. The lower fluid exerts an equal leftward force on the upper fluid. These are the two gold arrows.

A Newtonian fluid’s shear force per area is proportional to the rate at which its velocity changes across layers:

$$\tau_{yx}=\mu\frac{dv_x}{dy}$$

Here $\mu$ is the dynamic viscosity, measured in pascal-seconds. One pascal is one newton per square metre. The first index names the cut’s normal direction; the second names the force direction. Positive $\tau_{yx}$ means the upper fluid pushes the lower fluid toward +x.

That rightward momentum is entering the lower region downward through its top face. Its flux measured toward +y therefore has the opposite sign:

$$\Pi_{yx}=-\tau_{yx}$$

No average fluid motion crosses this horizontal cut: $v_y=0$. The dye stays in its layer. The shear force nevertheless transfers momentum between layers through microscopic interactions. Those interactions are represented by the viscosity law; the dye dots are not a molecular simulation.

For a general slow Newtonian fluid, the spatial momentum flux separates into transport by bulk flow, pressure, and viscous stress:

$$\Pi_{ij}=\rho v_i v_j+p\delta_{ij}-\tau_{ij}$$

The mass density is $\rho$. The first term carries momentum with moving material. The pressure term acts normally across a cut. The last term is the negative of the viscous force per area. Equivalently, the mechanical Cauchy stress is $\sigma_{ij}=-p\delta_{ij}+\tau_{ij}$. Compressive pressure has a negative mechanical-stress sign and a positive normal-momentum-flux sign.

Derive the two velocity profiles

The flow is steady and moves only along x. Neighboring layers are labeled by height y between plates at $y=0$ and $y=H$. A thin layer receives a pressure force per volume $-dp/dx$. The difference between the shear forces on its top and bottom contributes $d\tau_{yx}/dy$. Newton’s law with zero acceleration gives

$$0=-\frac{dp}{dx}+\mu\frac{d^2v_x}{dy^2}$$

Moving plate. The pressure is constant. Thus $d^2v_x/dy^2=0$, so the profile is a straight line. No slip means the fluid matches each plate’s velocity: zero below and $U$ above. Therefore

$$v_x(y)=U\frac yH,\qquad \tau_{yx}=\frac{\mu U}{H}$$

This is Couette flow. Fixing the two plate speeds fixes the velocity profile. Increasing viscosity changes the force required to maintain it.

Pressure difference. Both plates are stationary. Define $G=-dp/dx$, so positive G drives flow toward +x. Integrating the force-balance equation twice and imposing zero velocity at both walls gives

$$v_x(y)=\frac{G}{2\mu}y(H-y),\qquad \tau_{yx}=\frac G2(H-2y)$$

This is plane Poiseuille flow. The speed is greatest at the middle, where the slope and shear vanish. Increasing viscosity at fixed pressure gradient reduces the speed. The shear profile remains fixed by the imposed pressure force. A constant pressure gradient is balanced by the change in shear from one layer to the next.

What the animation represents

This is an exact steady solution for an incompressible Newtonian fluid with constant viscosity, straight parallel plates, and no slip. Changing a control selects another steady solution; it does not simulate the transient while a flow starts. The plate separation is 2 cm and the visible window is 12 cm long. The drawing expands the height to make the layers readable.

Blue dots are passive dye markers following the calculated velocity. A repeating train of markers enters and leaves the viewing window. Short trails and blue arrows show velocity; the linked profile uses the same velocity calculation. The gold cut selects the same height in both views. Gold arrows show forces on the two adjacent fluid regions, not velocities.

Time is physical seconds, played at one tenth speed. The pressure-driven example uses a pressure of 100 Pa above an arbitrary constant background at the window’s center. These are gauge pressures: adding the same constant to every pressure does not change the pressure gradient or this classical steady flow. Its left and right pressures change consistently with the pressure gradient. The fluid model includes viscosity; a perfect-fluid stress tensor deliberately leaves that effect out.

Independent checks solve the steady force-balance equation on a grid with the stated wall velocities. Integrated viscous dissipation is checked against the work supplied by the moving wall or pressure gradient. The same classical solutions appear in David Tong’s fluid-mechanics notes and Cambridge’s Fluid Dynamics II notes.

We now make a simpler approximation, setting aside those viscous shear forces. It is useful when their effect on the motion is negligible over the distances and times being studied. The remaining rest-frame stress is pressure, acting equally in every direction.

A perfect fluid can have pressure from microscopic particle motion, but its local rest frame has no preferred spatial direction and no net heat flow. The model neglects viscosity: the additional momentum transfer associated with neighboring layers moving differently. Rotating the rest-frame axes must leave its stress unchanged. This requires zero off-diagonal entries and equal diagonal entries, giving $p\delta^{ij}$, where $p$ is pressure.

Let

be the projector onto the fluid’s rest space. Its energy contribution must be $\epsilon u^\mu u^\nu/c^2$; its isotropic spatial stress must be $pH^{\mu\nu}$. Adding them gives

The combination $\epsilon+p$ follows from the observer’s spatial projector. Pressure initially entered as a purely spatial stress. Expressing “spatial” covariantly requires a projector containing $u^\mu u^\nu$, so pressure also joins the coefficient of the velocity term.

For an observer moving at relative speed $v$ with respect to the fluid,

This follows by contracting $T_{\mu\nu}$ with the observer’s unit time vector and using $u\cdot w=-c^2\gamma$. Even the energy density measured by a moving observer knows about pressure.

To extract the fluid equations, first apply the product rule to the full tensor:

Contract with $u_\nu$. Normalization gives $u_\nu u^\nu=-c^2$ and $u_\nu a^\nu=0$. The two derivatives of $p$ cancel, leaving

Here $\theta$ measures fractional expansion of a small comoving volume. Write its proper volume as $V$ so that $dV/d\tau=\theta V$. Multiplying the equation by $V$ yields

This is the pressure–volume work law. To see the mechanical meaning, imagine a piston with area $A$ moving outward by $d\ell$. Pressure exerts force $pA$, doing work $pA\,d\ell=p\,dV$. The fluid’s energy decreases by that amount when there is no heat transfer.

Now apply the spatial projector $H^\alpha{}_{\nu}$ to the expanded equation. It removes the term proportional to $u^\nu$ and leaves $a^\alpha$ unchanged because $u\cdot a=0$. The result is

The pressure gradient supplies the force density; $(\epsilon+p)/c^2$ supplies the relativistic inertial coefficient. In a cold, slow fluid, $p\ll\epsilon\simeq\rho c^2$, recovering $\rho\mathbf a=-\boldsymbol\nabla p$ locally.

Pressure and tension conventions. Relativity’s $T^{ij}$ is momentum flux. Some engineering conventions define Cauchy stress as positive in tension, giving a stationary fluid the mechanical stress $-p\delta^{ij}$. When comparing formulas, check whether the tensor describes momentum flux or mechanical traction with this tension convention.

11.5 Radiation pressure and the mass of a box#

The field equation developed next implies, in an orthonormal frame comoving with an isotropic fluid,

The factor $3p$ is the sum of the stresses in three spatial directions. It describes a particular Ricci-curvature projection relevant to focusing timelike geodesics. It is not a universal instruction to replace all mass densities everywhere by $(\epsilon+3p)/c^2$.

For isotropic radiation, $p=\epsilon/3$. Here is the momentum-flux calculation. A photon of energy $E$ traveling along a unit direction $n$ carries momentum $(E/c)n_j$ in direction $j$. Its crossing rate through a face normal to $i$ contains the velocity component $cn_i$. Thus a beam with energy density $\epsilon_{\rm beam}$ contributes $\epsilon_{\rm beam}n_in_j$ to the momentum flux.

Average over an isotropic distribution of directions. The three averages $\langle n_x^2\rangle$, $\langle n_y^2\rangle$, and $\langle n_z^2\rangle$ are equal and sum to one, so each is $1/3$. The off-diagonal averages vanish by symmetry. Each diagonal stress is therefore $\epsilon/3$.

Substituting gives $\epsilon+3p=2\epsilon$. Taken alone, this appears to assign radiation twice the gravitational effect expected from its energy. To determine the mass of a box containing radiation, however, we must include the box’s stresses too.

The missing member of that calculation is the box. Radiation pushes outward; walls develop stresses to hold it in. An isolated static system must include its supports in its total stress tensor. Wall tensions compensate the extra integrated radiation-pressure contribution in the regime of negligible internal self-gravity. This is the resolution studied explicitly by Misner and Putnam in “Active Gravitational Mass”.

We can expose the mechanism with a short independent calculation. For a localized stationary system in approximately flat spacetime, total momentum conservation says

Multiply by $x^i$ and use the product rule:

Integrate over a volume enclosing the system. The divergence theorem turns the integral of a divergence into outward flux through its boundary. It follows by assembling the small boxes from Chapter 0: neighboring boxes share a face with opposite outward normals, so their interior-face contributions cancel. Only the outer faces remain. Here the relevant flux is $x^iT^{kj}$. Take the enclosing boundary to infinity; if the complete system’s stresses decay sufficiently fast, its surface contribution vanishes. We obtain

The positive pressure stresses of the contents cannot be the whole story: other stresses must balance their integral. Consequently the leading static weak-field mass contribution

reduces to total energy divided by $c^2$ for the complete system under these assumptions. If you supply additional energy $E$ without otherwise changing the total energy accounting, the leading mass increase is $E/c^2$. Gravitational binding corrections require the appropriate relativistic total-energy definition.

Pressure contributes to the field equation. In this stationary box, the wall tensions also contribute. Including both reconciles the local pressure term with the total mass inferred far from the complete system.

11.6 Two field examples: scalar waves and electromagnetism#

PREPARATION FOR THIS SECTION

A field has motion and stored energy at every point

Where is energy while a wave crosses an otherwise empty region?

Builds onWork and energy Flux and divergence

See the idea

A classical scalar field assigns one number to each event: $\phi(t,\mathbf x)$. A stretched string is a useful one-dimensional model. Its displacement changes with time, and neighboring points can have different displacements. Motion stores kinetic energy; spatial variation stores elastic energy. A field replaces a finite list of particle coordinates by a value at each position. In the following calculation, subscripts denote partial derivatives: $\phi_t=\partial\phi/\partial t$ and $\phi_x=\partial\phi/\partial x$.

Work it out

1. 01
   Turn a string into a field model

   For small transverse slopes on a string with constant tension $\mathcal T$ and mass per length $\mu$, a segment has kinetic energy per length $\mu\phi_t^2/2$. Expanding its length $\sqrt{1+\phi_x^2}\,dx$ gives extra length $\phi_x^2dx/2$ and elastic energy per length $\mathcal T\phi_x^2/2$. The vertical tension components at the segment’s ends differ by $\mathcal T[\phi_x(x+dx)-\phi_x(x)]\simeq\mathcal T\phi_{xx}dx$. Equating this force to mass $\mu\,dx$ times acceleration $\phi_{tt}$ gives $\mu\phi_{tt}=\mathcal T\phi_{xx}$.

   $$e=\frac\mu2\phi_t^2+\frac{\mathcal T}2\phi_x^2,\qquad c_s^2=\mathcal T/\mu.$$

   Why this step works The approximation requires small slopes; this wave speed is a property of the string.

2. 02
   Derive the energy current

   Differentiate $e$ with respect to time and substitute the string equation. The two terms combine into an x derivative. Define flux $j=-\mathcal T\phi_t\phi_x$ so local energy conservation reads $\partial_te+\partial_xj=0$.

   $$\partial_te=\mathcal T\partial_x(\phi_t\phi_x),\qquad j=-\mathcal T\phi_t\phi_x.$$

   Why this step works Energy lost from a short segment leaves through its ends; local conservation is a flux balance.

3. 03
   Meet electric and magnetic fields

   An electric field $\mathbf E$ supplies force per charge on a stationary test charge $q$. A magnetic field $\mathbf B$ contributes $q\mathbf v\times\mathbf B$ to a moving charge’s force. The cross product $\times$ gives a vector perpendicular to its inputs, with magnitude equal to the product of their lengths times the sine of their angle; its direction follows the right-hand rule. Maxwell’s theory supplies a separate physical model for these fields. In SI vacuum it assigns energy density $u=\epsilon_0 E^2/2+B^2/(2\mu_0)$ and energy-flux vector $\mathbf S=\mathbf E\times\mathbf B/\mu_0$. Here $\epsilon_0,\mu_0$ are the vacuum electromagnetic constants and $c^2=1/(\epsilon_0\mu_0)$.

   $$\partial_tu+\nabla\cdot\mathbf S=-\mathbf J\cdot\mathbf E.$$

   Why this step works The charge-current density J measures charge crossing unit area per time, in amperes per square metre. The right side is minus the power per volume delivered to this current; energy transferred to matter leaves the field.

Go deeper

For a plane electromagnetic wave in vacuum, $\mathbf E\perp\mathbf B$, $B=E/c$, and both fields are perpendicular to the propagation direction. Substituting gives $u=\epsilon_0E^2$ and $|\mathbf S|=cu$. The two equal terms in $u$ are electric and magnetic contributions. In a local orthonormal frame using $x^0=ct$, $T^{00}=u$ and $T^{0i}=S_i/c$. Chapter 13 develops the field action and its variations. The Maxwell action below is a preview of that construction; the energy/flux interpretation now has a direct definition. The string’s $c_s$ need not equal the invariant light speed $c$.

Further readingDavid Tong · Electromagnetism

Test the idea

FIRST, PREDICT

Why does a traveling string wave transport energy?

Because every wave in every medium travels at c.The local energy change is balanced by flux through neighboring points.Because the same piece of string travels the whole distance.

Compare the reasoning

Because every wave in every medium travels at c.

The string wave speed depends on tension and mass per length.

The local energy change is balanced by flux through neighboring points.

The wave equation gives an explicit continuity equation and energy current.

Because the same piece of string travels the whole distance.

Material points oscillate; the disturbance and energy propagate along the string.

A hint

Distinguish the motion of the disturbance from the motion of the material.

NOW CHANGE THE EXAMPLE

A vacuum plane electromagnetic wave has electric energy density $2\,\mathrm{J/m^3}$. Find its total instantaneous energy density.

A hint

Include the equal magnetic contribution.

Work through the solution

$u=u_E+u_B=2+2=4\,\mathrm{J/m^3}$.

A stress tensor includes energy density and energy flow; empty of particles does not mean empty of fields.

For this scalar-field subsection only, choose time and length units with $c=1$. A real scalar field assigns one real number $\phi$ to each event. Choose its normalization so that squared derivatives have energy-density units, and let $V(\phi)$ be its potential energy density. A Lagrangian density is the local integrand used to build a field action, analogous to the particle Lagrangian of Chapter 5. The model considered here uses

This is a specified model for matter, an additional physical input to GR. Its kinetic sign is chosen for signature $(-,+,+,+)$, so time-dependent excitations have positive kinetic energy. For now we will evaluate its energy and pressure; Chapter 13 derives the field and stress equations by variation.

The metric-variation method of Chapter 13 gives

where $(\partial\phi)^2=g^{\alpha\beta}\partial_\alpha\phi\partial_\beta\phi$. In a local inertial frame,

Time variation, spatial gradients, and potential energy all contribute. For a homogeneous field in its comoving frame,

Kinetic energy gives positive pressure; potential energy gives negative pressure. This prepares the intuition for scalar-field cosmology without requiring a field to be imagined as a literal fluid of tiny balls.

Return to SI units and $x^0=ct$. Before packaging electromagnetism into a tensor, give its two fields a measuring procedure. A small test body’s electric charge $q$ determines its electromagnetic response; charge is measured in coulombs ©, and its sign can be positive or negative. The electric field $\mathbf E$ is the force per unit positive charge on a test body momentarily at rest. It is measured in newtons per coulomb. The magnetic field $\mathbf B$ describes an additional force on a moving charge. Together they enter the experimentally established Lorentz-force law:

Here $\mathbf p$ is the particle’s ordinary three-momentum and $\mathbf v$ its velocity in the chosen local inertial frame. The magnetic field has units $\mathrm{N\,s/(C\,m)}$, called teslas. The cross product $\mathbf a\times\mathbf b$ is perpendicular to both arrows, with magnitude $|\mathbf a||\mathbf b|\sin\theta$ when the angle between them is $\theta$. Its direction follows the right-hand rule: curl the fingers from $\mathbf a$ toward $\mathbf b$; the thumb points along the product. In right-handed Cartesian axes its components are

For example, velocity along +x and magnetic field along +y give a magnetic force on a positive charge along +z. The magnetic force is perpendicular to the velocity, so it changes the momentum’s direction without doing work on that particle. The electric contribution supplies power $q\mathbf E\cdot\mathbf v$.

Electric and magnetic fields depend on the observer. The electromagnetic field-strength tensor $F_{\mu\nu}$ packages their six components so another observer can transform them together. It is antisymmetric: $F_{\nu\mu}=-F_{\mu\nu}$. With our metric signature, choose

in a local orthonormal frame. Here $\varepsilon_{ijk}$ is the three-dimensional antisymmetric symbol: $\varepsilon_{123}=+1$, exchanging two indices reverses the sign, and repeating an index gives zero. It is unrelated to energy density $\epsilon$.

The factor $1/c$ gives the temporal and spatial entries of $F$ the same units. It also makes the covariant force law $dp^\mu/d\tau=qF^\mu{}_{\nu}u^\nu$ reproduce the three-force above: insert $u^\nu=\gamma(c,\mathbf v)$ and divide by $dt/d\tau=\gamma$. The temporal contribution becomes $q\mathbf E$ and the spatial contribution becomes $q\mathbf v\times\mathbf B$.

Two vacuum constants set the SI normalization: the permittivity $\epsilon_0$ and permeability $\mu_0$. They relate the strengths of the electric and magnetic fields to charge, electric current (charge flowing past a point per unit time), and stored field energy. Their units are $\mathrm{C^2/(N\,m^2)}$ and $\mathrm{N\,s^2/C^2}$ respectively, and $\epsilon_0\mu_0c^2=1$. Thus $\epsilon_0E^2$ and $B^2/\mu_0$ both have units of energy per volume.

The electromagnetic Lagrangian energy density and stress tensor are

These conventions produce

The energy-flux vector $\mathbf S$ is called the Poynting vector. Its direction is the direction of energy transport; its magnitude gives energy crossing a perpendicular unit area per unit time. The corresponding momentum density is $\mathbf S/c^2$, as in §11.1. The field-action formulas here preview the variational construction developed in Chapter 13. The same tensor includes electric and magnetic stresses, so a magnetic field can affect geometry even in a region containing no material particles.

A light wave as a check. A plane wave has the same field across each plane perpendicular to its direction of travel. At one event in such a wave traveling along +z, suppose $\mathbf E=(E,0,0)$ and $\mathbf B=(0,E/c,0)$. The two contributions to the energy density are equal because $1/(\mu_0c^2)=\epsilon_0$. Therefore

The field energy moves at $c$ and carries forward momentum. Here $E$ is the instantaneous field value, which can vary along the wave. With other local field configurations, stored energy need not imply a nonzero net flux: if $\mathbf E$ and $\mathbf B$ are parallel, their cross product vanishes even when the energy density is positive.

Its classical trace in four spacetime dimensions vanishes:

Contract the displayed tensor: the first term gives $F_{\mu\alpha}F^{\mu\alpha}$, and the second gives $4\times\frac14$ times that same quantity. They cancel. Zero trace does not mean zero energy, zero stress, or zero gravitational influence.