From Free Fall to Curved Spacetime
From Light Signals to Spacetime constructed the rule relating observations made by inertial laboratories. We now begin with a laboratory near Earth and ask:
Can one rule predict how freely released objects move, how clocks at different heights compare, and how light travels?
We will first solve the falling-object problem without geometry or gravitational mass. We will then ask whether the measurements admit one gravitational rule that works for different test objects, and introduce curved spacetime only when a single observational model must describe falling objects, clocks, and light.
1. What does a released object do near Earth?
Choose a vertical position coordinate \(z\) that increases upward. Release a small object in an evacuated chamber and record its position with the chamber’s clocks. Over a sufficiently small height and time, the observations are described by
\[z(t)=z_0+u_0t-\frac12gt^2,\]where \(z_0\) is its position at \(t=0\), \(u_0\) is its initial upward velocity, and \(g\) is a measured positive constant. Differentiating twice gives
\[\frac{d^2z}{dt^2}=-g.\]This rule already predicts the position of that object. If it is released from rest at height \(h\) above the floor, then \(z_0=h\) and \(u_0=0\). It reaches the floor when
\[0=h-\frac12gt_{\mathrm{fall}}^2,\]so
\[t_{\mathrm{fall}}=\sqrt{\frac{2h}{g}}.\]Now repeat the evacuated-chamber experiment with objects made from different materials. Give them the same initial position and velocity:
| Released object | Measured vertical acceleration |
|---|---|
| Aluminum sphere | \(-g\) |
| Glass sphere | \(-g\) |
| Dense metal sphere | \(-g\) |
Within experimental precision, the same value of \(g\) describes every tested object. Objects released with the same initial conditions therefore follow the same trajectory, regardless of their material or mass. This conclusion applies to sufficiently small test objects whose own influence on Earth can be neglected.
The original drop problem is solved for every such object. We did not need either inertial mass or gravitational mass. Now ask for a different prediction:
How strong must a spring or floor be to support different objects at different locations?
2. How strong must the support be?
The earlier mechanics construction defined relative inertial mass by comparing the accelerations of two objects under the same controlled interaction:
\[\frac{m_{\mathrm{i},A}}{m_{\mathrm{i},B}} :=\frac{a_B}{a_A}.\]The subscript \(\mathrm{i}\) labels the inertial mass already calibrated in that separate experiment. Once those masses were calibrated, the same construction defined force by
\[F:=m_{\mathrm{i}}a.\]Suppose those experiments assign object \(A\) an inertial mass of \(1\ \mathrm{kg}\) and object \(B\) an inertial mass of \(2\ \mathrm{kg}\).
Now construct the support experiment. Fix one end of a calibrated spring to the ceiling and attach a test object \(X\) to its lower end. Wait until the object is at rest. Let \(x_X\) be the spring’s downward extension from its natural length.
The drop rule tells us how \(X\) moves after release, but it does not predict the extension \(x_X\) while the spring holds it. Suppose the spring’s calibrated stiffness is \(k=100\ \mathrm{N/m}\). It pulls upward with magnitude \(kx_X\). Because the supported object has zero acceleration, the total force on it is zero. The downward gravitational force must therefore have the same magnitude:
\[W_X:=kx_X.\]Here \(W_X\) is the object’s weight at that location. This is an operational measurement of gravitational force: the calibrated spring force balances it.
Now repeat the measurement at two locations, \(L_1\) and \(L_2\). The following values illustrate the observed pattern:
| Object | Inertial mass | \(x\) at \(L_1\) | Weight at \(L_1\) | \(x\) at \(L_2\) | Weight at \(L_2\) |
|---|---|---|---|---|---|
| \(A\) | \(1\ \mathrm{kg}\) | \(0.098\ \mathrm{m}\) | \(9.8\ \mathrm{N}\) | \(0.095\ \mathrm{m}\) | \(9.5\ \mathrm{N}\) |
| \(B\) | \(2\ \mathrm{kg}\) | \(0.196\ \mathrm{m}\) | \(19.6\ \mathrm{N}\) | \(0.190\ \mathrm{m}\) | \(19.0\ \mathrm{N}\) |
There is not one numerical gravitational force shared by both objects. At either location, object \(B\) has twice the weight of object \(A\). But the ratio belonging to the pair of objects is stable:
\[\frac{W_A(L_1)}{W_B(L_1)} =\frac{W_A(L_2)}{W_B(L_2)} =\frac12.\]Experiments with other calibrated springs check that this ratio does not depend on the supporting apparatus. Experiments at other locations check that it does not depend on the local gravitational environment. When those checks succeed, we can assign the stable factor to the objects. Define relative passive gravitational mass by
\[\frac{m_{\mathrm{g},A}}{m_{\mathrm{g},B}} :=\frac{W_A(L)}{W_B(L)}\]when both weights are measured at the same location \(L\). Choosing one reference object fixes the overall gravitational-mass unit. “Passive” means that this quantity describes how strongly the test object responds to an existing gravitational environment. It does not yet describe how strongly the object would produce gravity if used as a source.
An object’s gravitational mass stays fixed when we move it to another location, while its weight can change.
Once the object factors have been calibrated, divide them out of the measured weights. At location \(L\), define
\[\gamma(L):=\frac{W_X(L)}{m_{\mathrm{g},X}}.\]The measured stability of the weight ratios makes \(\gamma(L)\) independent of which test object \(X\) is used. It describes the location and source, while \(m_{\mathrm{g},X}\) describes the test object. Every separate weight measurement is now summarized by one reusable rule:
\[W_X(L)=m_{\mathrm{g},X}\gamma(L).\]This factorization does predictive work. After measuring \(\gamma(L)\) with one reference object, we can predict the extension of a spring of stiffness \(k\) supporting any object whose passive gravitational mass is known:
\[x_X(L)=\frac{m_{\mathrm{g},X}\gamma(L)}{k}.\]A floor supporting the same object must provide the upward force \(m_{\mathrm{g},X}\gamma(L)\). Thus the construction answers the question that free-fall acceleration alone could not answer.
What happens when the support is removed?
Now release \(A\) and \(B\) in the evacuated chamber at \(L_1\). Their measured accelerations are
| Released object | Inertial mass | Passive gravitational mass | Measured vertical acceleration |
|---|---|---|---|
| \(A\) | \(1\ \mathrm{kg}\) | \(1\) gravitational-mass unit | \(-9.8\ \mathrm{m/s^2}\) |
| \(B\) | \(2\ \mathrm{kg}\) | \(2\) gravitational-mass units | \(-9.8\ \mathrm{m/s^2}\) |
To connect the weight measurements to the earlier spring calibration, imagine removing gravity while keeping the spring attached and initially stretched by \(x_X\): this returns us to the spring experiment in Rediscovering Mechanics and Energy from Questions. Let \(a_X\) be the initial upward acceleration in that spring-only experiment. The spring calibration gives \(m_{\mathrm{i},X}a_X=kx_X\), which by our definitions of weight and \(\gamma\) is also \(m_{\mathrm{g},X}\gamma\):
\[m_{\mathrm{i},X}a_X =kx_X=W_X =m_{\mathrm{g},X}\gamma,\]and hence
\[a_X=\gamma\frac{m_{\mathrm{g},X}}{m_{\mathrm{i},X}}.\]For \(A\) and \(B\), the measured weights and independently calibrated inertial masses give \(a_A=9.8/1=9.8\ \mathrm{m/s^2}\) and \(a_B=19.6/2=9.8\ \mathrm{m/s^2}\). Compare these spring-calibrated accelerations with the separate release measurements in the table: each object’s downward free-fall acceleration has the same magnitude as its upward spring-only acceleration. This agreement must be checked experimentally; it does not follow from the definition of \(\gamma\).
These equal spring-only accelerations imply equal ratios \(m_{\mathrm{g}}/m_{\mathrm{i}}\) at the same location. Repeating both experiments with other objects tests whether the agreement and the mass proportionality persist. Within experimental precision, they hold for every tested object.
Choose the gravitational-mass unit so that one reference object has the same numerical inertial and passive gravitational mass. The universal proportionality then gives
\[m_{\mathrm{g}}=m_{\mathrm{i}},\]and \(\gamma(L_1)=9.8\ \mathrm{m/s^2}=g\). The larger gravitational force on \(B\) does not give it a larger acceleration because its inertial mass is larger by the same factor. After establishing this result, either mass symbol alone is sufficient for calculations; keeping them separate allowed the equality to be discovered rather than assumed.
3. What does an accelerometer measure while falling?
The balance experiment used a supported spring to measure weight. Now attach the same kind of spring to a box’s ceiling and hang the test mass from it. This physical box, together with its clocks, is what we mean by a laboratory.
We want to predict the spring’s extension while the box is supported and while it falls.
Take downward as positive for this spring calculation. Let \(x_{\mathrm{mass}}\) and \(x_{\mathrm{ceiling}}\) be the positions of the mass and the spring’s ceiling attachment, measured from the same fixed ground origin. The extension is
\[x=x_{\mathrm{mass}}-x_{\mathrm{ceiling}}-\ell_0,\]where \(\ell_0\) is the spring’s constant natural length. Differentiating twice with respect to time gives
\[\frac{d^2x}{dt^2}=a_{\mathrm{mass}}-a_{\mathrm{ceiling}},\]Using the earlier force rules, the mass has downward gravitational force \(m\gamma\) and upward spring force \(kx\), so
\[\begin{aligned} m a_{\mathrm{mass}}&=m\gamma-kx,\\ m\frac{d^2x}{dt^2}&=m(\gamma-a_{\mathrm{ceiling}})-kx. \end{aligned}\]Here \(m\) is the test mass, with inertial and gravitational mass equal in our chosen units, and \(\gamma=g\) locally. Assume gravity is approximately uniform across the apparatus and the box is much heavier than the test mass, so the spring’s pull has a negligible effect on the box’s motion.
While the box rests on the ground, the ground holds the box up, and the stretched spring supports the test mass. Mark the spring’s extension in this state. This is the earlier support experiment: \(a_{\mathrm{ceiling}}=0\), and the settled mass has \(d^2x/dt^2=0\). Therefore
\[0=m\gamma-kx, \qquad x=\frac{m\gamma}{k}.\]Now release the entire box. Its ceiling accelerates downward at approximately \(a_{\mathrm{ceiling}}=\gamma\). Substituting into the same relative-motion equation gives
\[m\frac{d^2x}{dt^2}=-kx.\]This is the spring equation already solved in Rediscovering Mechanics and Energy from Questions. Set \(t=0\) at release. The mass initially has zero velocity relative to the box and extension \(x_0=m\gamma/k\), so the earlier solution gives
\[x(t)=x_0\cos\!\left(\sqrt{\frac{k}{m}}\,t\right).\]We want the instrument’s reading after it has settled, so we need to ask what happens as \(t\to\infty\). This solution oscillates forever and has no such limit.
To model settling, include a damping force \(-b\,dx/dt\) that opposes the mass’s motion relative to the box. Here \(b>0\) is a calibrated damping coefficient, in \(\mathrm{kg/s}\). This is an additional force model, whose applicability must be checked experimentally. The equation becomes
\[m\frac{d^2x}{dt^2}+b\frac{dx}{dt}+kx=0.\]For positive \(m\), \(k\), and \(b\), every solution of this damped model decays to zero, so
\[\lim_{t\to\infty}x(t)=0, \qquad \lim_{t\to\infty}\frac{kx(t)}{m}=0.\]We could display the extension \(x\) directly. However, different springs and test masses would give different extensions under the same conditions: in the supported experiment, doubling the stiffness halves \(x\), while doubling the test mass doubles it. Using the earlier calibration, we display \(kx/m\) so that these instruments agree once settled. This normalization expresses the result in acceleration units, giving our accelerometer the displayed reading
\[\text{reading}=\frac{kx}{m},\]with a positive reading displayed as upward. This conversion applies whether the box is stationary or moving.
This gives a settled reading of \(\gamma=g\) upward while the box is supported and zero during free fall. In a finite drop, the instrument needs time to settle before that zero reading is a good approximation. We have now predicted both readings; the instrument reports acceleration relative to free fall. We must distinguish two meanings of acceleration:
-
Coordinate acceleration is calculated by differentiating position twice with respect to time. For our box, it is \(a_{\mathrm{ceiling}}=d^2x_{\mathrm{ceiling}}/dt^2\) relative to the ground, with downward positive.
-
Proper acceleration is reported by an accelerometer attached to the object. For our box, its vertical component is the settled reading \(kx/m\), with upward positive.
The direction conventions belong to this calculation; either type of acceleration can point upward or downward. Both quantities above were calculated using the same ground reference frame. A zero accelerometer reading identifies free fall, while a nonzero reading measures how strongly the box is being pushed away from free fall.
| Motion of the box | Accelerometer reading |
|---|---|
| Resting on the ground | Approximately \(g\) upward |
| Falling freely | Approximately \(0\) |
4. Can an accelerating rocket reproduce the falling observations?
Consider a rocket far from gravitating bodies. Its engines make the floor accelerate upward with constant proper acceleration \(a\). Place our laboratory box inside it, initially held at rest above the rocket floor, then release the whole box. The box follows an inertial path while the rocket floor accelerates toward it. Take the rocket floor as the origin, with \(z\) increasing upward and \(t\) measured by a clock at the floor. Over a small enough region and duration after release, the box’s position obeys
\[\frac{d^2z}{dt^2}=-a.\]Set \(a=g\). This is the same local update measured for our released laboratory box near Earth, with height measured above the ground:
\[\frac{d^2z}{dt^2}=-g.\]An accelerometer fixed to the rocket floor reads upward acceleration \(g\), just as one supported on the ground does. The released box has zero proper acceleration; relative to the rocket floor, it appears to fall. These local observations do not reveal whether the surrounding room is supported near Earth or accelerating in empty space.
Our spring calculation predicts the same behavior in the two freely moving laboratories. We extend this agreement, as a hypothesis, to other local experiments, such as collisions and light propagation. This is the equivalence principle:
In a sufficiently small freely falling laboratory, these experiments follow the same rules as in a freely moving laboratory far from gravity.
Here the laboratory far from gravity is the released box inside the rocket. “Sufficiently small” matters because Earth’s gravitational acceleration changes with position and direction. A single falling laboratory can remove the common acceleration at one place, but it cannot remove differences across an arbitrarily large region.
5. What does the rocket predict for clocks at different heights?
Place a light source at the rocket floor and a receiver a fixed ruler distance \(h>0\) above it. Both remain attached to the rocket. The source emits \(\nu_{\mathrm{lower}}\) wave cycles per second of its own clock. A crest is a peak of the wave; following successive crests lets us compare the instruments’ clock readings.
How many cycles arrive per second of the upper clock, and what does this tell us about the two clocks?
For this calculation, \(t\) and \(z\) are coordinates of one inertial frame in which the rocket is initially at rest, with \(z\) increasing upward. At emission, \(t=0\), the source is at \(z=0\) and the receiver at \(z=h\). The emitted crest follows \(z_{\mathrm{light}}=ct\).
When does the crest arrive?
Let \(a_{\mathrm{upper}}>0\) be the receiver’s constant proper acceleration, in \(\mathrm{m/s^2}\); \(a\) remains the floor’s. Keeping the rocket’s ruler length fixed requires a different acceleration at each height. We first need the receiver’s trajectory in our inertial frame.
Write \(v(t)=dz_{\mathrm{upper}}/dt\) for its velocity. At any chosen instant, set \(w\) equal to that velocity and choose a second inertial frame moving at the fixed velocity \(w\) relative to the first, with the same upward axis. Place its spatial origin at the receiver at that instant and set the clock there to zero. The receiver is momentarily at rest at this origin; afterward, the origin continues at velocity \(w\) while the receiver accelerates.
Write \(F_w(v(t))\) for the receiver’s velocity in this second frame. Section 10 of From Light Signals to Spacetime gives
\[F_w(v(t))=\frac{v(t)-w}{1-v(t)w/c^2},\qquad \lvert v(t)\rvert,\lvert w\rvert<c.\]Let \(\tau_{\mathrm{upper}}(t)\) be the reading of the clock attached to the upper receiver at inertial time \(t\). Its rate is positive, so a derivative with respect to this reading means \(d/d\tau_{\mathrm{upper}}:=(d\tau_{\mathrm{upper}}/dt)^{-1}d/dt\) on records along the receiver’s path. At the chosen instant, the receiver is at rest in the second frame. Its proper acceleration is therefore
\[a_{\mathrm{upper}} =\left.\frac{dF_w(v(t))}{d\tau_{\mathrm{upper}}}\right|_{w=v(t)}.\]Apply the chain rule with \(w\) held fixed during differentiation:
\[a_{\mathrm{upper}} =\left. \frac{dF_w}{dv}(v(t))\, \frac{dv}{dt}\, \frac{dt}{d\tau_{\mathrm{upper}}} \right|_{w=v(t)}.\]We need the velocity-conversion derivative and the clock-rate factor. Writing \(v\) for \(v(t)\), differentiate the velocity formula and use the ideal-clock rule from Section 8 of the preceding post:
\[\begin{aligned} \frac{dF_w}{dv} &=\frac{(1-vw/c^2)+(v-w)w/c^2}{(1-vw/c^2)^2} =\frac{1-w^2/c^2}{(1-vw/c^2)^2},\\ \frac{d\tau_{\mathrm{upper}}}{dt} &=\sqrt{1-v^2/c^2},\qquad \frac{dt}{d\tau_{\mathrm{upper}}} =\frac{1}{\sqrt{1-v^2/c^2}}. \end{aligned}\]Substitute these factors, then evaluate at \(w=v(t)\):
\[\begin{aligned} a_{\mathrm{upper}} &=\left. \frac{1-w^2/c^2}{(1-vw/c^2)^2}\, \frac{dv}{dt}\, \frac{1}{\sqrt{1-v^2/c^2}} \right|_{w=v(t)}\\ &=\frac{1}{(1-v^2/c^2)^{3/2}}\frac{dv}{dt} =\frac{d}{dt}\left(\frac{v}{\sqrt{1-v^2/c^2}}\right). \end{aligned}\]Integrate both sides from emission at \(t=0\), when \(v(0)=0\), to a later inertial-frame time \(t\). Use \(s\) for the integration time:
\[\begin{aligned} \int_0^t a_{\mathrm{upper}}\,ds &=\int_0^t\frac{d}{ds} \left(\frac{v(s)}{\sqrt{1-v(s)^2/c^2}}\right)ds,\\ a_{\mathrm{upper}}t &=\frac{v(t)}{\sqrt{1-v(t)^2/c^2}} -\underbrace{\frac{v(0)}{\sqrt{1-v(0)^2/c^2}}}_{0}. \end{aligned}\]Solve for \(v(t)\), then integrate the velocity over the same time interval, starting at \(z_{\mathrm{upper}}(0)=h\):
\[\begin{aligned} v(t)&=\frac{a_{\mathrm{upper}}t}{\sqrt{1+(a_{\mathrm{upper}}t/c)^2}},\\ z_{\mathrm{upper}}(t) &=h+\int_0^t v(s)\,ds\\ &=h+\frac{c^2}{a_{\mathrm{upper}}} \left[\sqrt{1+(a_{\mathrm{upper}}t/c)^2}-1\right]. \end{aligned}\]For \(a_{\mathrm{upper}}t/c\ll1\), expand the square root to obtain
\[z_{\mathrm{upper}}(t) =h+\frac12a_{\mathrm{upper}}t^2 +O\!\left(\frac{a_{\mathrm{upper}}^3t^4}{c^2}\right).\]The notation \(O(q^n)\) denotes a remainder bounded by a constant times \(\lvert q\rvert^n\) as \(q\to0\), with other parameters fixed.
Let \(T\) be the reception time in the original inertial frame. The crest and receiver meet then, so their positions are equal. Using the light path \(z_{\mathrm{light}}(t)=ct\) gives
\[z_{\mathrm{light}}(T)=z_{\mathrm{upper}}(T) \quad\Longrightarrow\quad cT=z_{\mathrm{upper}}(T).\]Substitute the receiver’s approximate trajectory into this equation and solve for \(T\):
\[\begin{aligned} &cT\approx h+\frac12a_{\mathrm{upper}}T^2,\\[0.5em] &\frac12a_{\mathrm{upper}}T^2-cT+h\approx0. \end{aligned}\]Define \(\epsilon:=a_{\mathrm{upper}}h/c^2\ll1\). Of the two quadratic roots, choose the smaller:
\[T\approx\frac{c-\sqrt{c^2-2a_{\mathrm{upper}}h}}{a_{\mathrm{upper}}} =\frac{c}{a_{\mathrm{upper}}}\left[1-\sqrt{1-2\epsilon}\right].\]The larger root has \(a_{\mathrm{upper}}T/c\approx2\), outside the regime of our trajectory approximation. Expand the smaller root using \(\sqrt{1-2\epsilon}=1-\epsilon-\tfrac12\epsilon^2+O(\epsilon^3)\):
\[\begin{aligned} T&\approx\frac{c}{a_{\mathrm{upper}}} \left[\epsilon+\frac12\epsilon^2+O(\epsilon^3)\right]\\ &=\frac{h}{c}\left[1+\frac12\epsilon+O(\epsilon^2)\right]. \end{aligned}\]Substitute this time into the velocity formula to find the arrival speed:
\[\frac{v(T)}{c} =\frac{a_{\mathrm{upper}}T/c}{\sqrt{1+(a_{\mathrm{upper}}T/c)^2}} =\epsilon+O(\epsilon^2).\]This also checks the condition \(a_{\mathrm{upper}}T/c\ll1\) used in the trajectory expansion.
How far apart are nearby arrivals?
Let \(t_{\mathrm e}\) be a crest’s emission time and \(t_{\mathrm r}(t_{\mathrm e})\) its reception time, both in the same inertial frame. For the crest considered above, \(t_{\mathrm e}=0\) and \(t_{\mathrm r}(0)=T\). Write \(z_{\mathrm{lower}}(t)\) for the source’s trajectory. Each crest travels at speed \(c\), so
\[c(t_{\mathrm r}-t_{\mathrm e}) =z_{\mathrm{upper}}(t_{\mathrm r})-z_{\mathrm{lower}}(t_{\mathrm e}).\]
Differentiate with respect to \(t_{\mathrm e}\), using the chain rule for \(t_{\mathrm r}(t_{\mathrm e})\):
\[c\left(\frac{dt_{\mathrm r}}{dt_{\mathrm e}}-1\right) =\frac{dz_{\mathrm{upper}}}{dt_{\mathrm r}}\frac{dt_{\mathrm r}}{dt_{\mathrm e}} -\frac{dz_{\mathrm{lower}}}{dt_{\mathrm e}}.\]At \(t_{\mathrm e}=0\), the source’s velocity is zero and the receiver’s velocity at \(t_{\mathrm r}(0)=T\) is \(v(T)\). Substituting these velocities gives
\[\begin{aligned} c\left(\left.\frac{dt_{\mathrm r}}{dt_{\mathrm e}}\right|_{t_{\mathrm e}=0}-1\right) &=v(T)\left.\frac{dt_{\mathrm r}}{dt_{\mathrm e}}\right|_{t_{\mathrm e}=0},\\ \left.\frac{dt_{\mathrm e}}{dt_{\mathrm r}}\right|_{t_{\mathrm r}=T} &=1-\frac{v(T)}{c}. \end{aligned}\]This derivative is the limiting ratio of nearby emission and reception intervals.
What do the instruments’ own clocks record?
Let \(N\) count wave cycles, including fractions of a cycle. The instantaneous frequency recorded by an instrument is \(\nu=dN/d\tau\), where \(\tau\) is its own clock reading. A measurement over a short interval approximates this rate.
Here \(\tau_{\mathrm{lower}}\) is the reading of the clock attached to the source at the rocket floor, and \(\tau_{\mathrm{upper}}\) is the reading of the clock attached to the receiver above it.
At emission \(t_{\mathrm e}=0\) the source is at rest; at reception \(t_{\mathrm r}=T\) the receiver has velocity \(v(T)\). The ideal-clock rule gives
\[\begin{aligned} \left.\frac{d\tau_{\mathrm{lower}}}{dt_{\mathrm e}}\right|_{t_{\mathrm e}=0} &=\sqrt{1-0^2/c^2}=1,\\ \left.\frac{d\tau_{\mathrm{upper}}}{dt_{\mathrm r}}\right|_{t_{\mathrm r}=T} &=\sqrt{1-v(T)^2/c^2}. \end{aligned}\]For each reception reading \(\tau_{\mathrm{upper}}\), let \(\tau_{\mathrm{lower}}(\tau_{\mathrm{upper}})\) be the source-clock reading at the corresponding emission, just as we paired \(t_{\mathrm e}\) and \(t_{\mathrm r}\). Write \(N_{\mathrm{lower}}\) and \(N_{\mathrm{upper}}\) for the cycle counts as functions of the respective clock readings, with matching labels for the same cycles. Then
\[N_{\mathrm{upper}}(\tau_{\mathrm{upper}}) =N_{\mathrm{lower}}\bigl(\tau_{\mathrm{lower}}(\tau_{\mathrm{upper}})\bigr).\]Differentiate with respect to \(\tau_{\mathrm{upper}}\) using the chain rule:
\[\begin{aligned} \frac{dN_{\mathrm{upper}}}{d\tau_{\mathrm{upper}}} &=\frac{dN_{\mathrm{lower}}}{d\tau_{\mathrm{lower}}} \frac{d\tau_{\mathrm{lower}}}{d\tau_{\mathrm{upper}}},\\ \nu_{\mathrm{received}} &=\nu_{\mathrm{lower}} \frac{d\tau_{\mathrm{lower}}}{d\tau_{\mathrm{upper}}}. \end{aligned}\]The source frequency is evaluated at the corresponding emission. Divide by it, then express the clock-reading derivative through the inertial emission and reception times. For the crest emitted at \(t_{\mathrm e}=0\) and received at \(t_{\mathrm r}=T\), substituting the clock rates and interval ratio gives
\[\begin{aligned} \frac{\nu_{\mathrm{received}}}{\nu_{\mathrm{lower}}} &=\frac{d\tau_{\mathrm{lower}}}{d\tau_{\mathrm{upper}}}\\ &=\left.\frac{d\tau_{\mathrm{lower}}}{dt_{\mathrm e}}\right|_{t_{\mathrm e}=0} \left.\frac{dt_{\mathrm e}}{dt_{\mathrm r}}\right|_{t_{\mathrm r}=T} \left.\frac{dt_{\mathrm r}}{d\tau_{\mathrm{upper}}}\right|_{t_{\mathrm r}=T}\\ &=\frac{1\cdot\bigl(1-v(T)/c\bigr)}{\sqrt{1-v(T)^2/c^2}}\\ &=1-\epsilon+O(\epsilon^2). \end{aligned}\]The square-root factor differs from one only at order \(\epsilon^2\). To first order, the reduction comes from the receiver’s motion during the flight.
Transfer this prediction to a small supported laboratory in a steady gravitational field using the equivalence principle. Set \(a_{\mathrm{upper}}=g\), with \(g\) measured at the receiver, and require negligible field variation across height \(h\). Our earlier definition \(\epsilon=a_{\mathrm{upper}}h/c^2\) then gives \(\epsilon=gh/c^2\). Keeping terms through first order gives
\[\frac{\nu_{\mathrm{received}}}{\nu_{\mathrm{lower}}} \approx1-\epsilon =1-\frac{gh}{c^2}.\]The clock-interval ratio is the reciprocal of the frequency ratio, so
\[\frac{d\tau_{\mathrm{upper}}}{d\tau_{\mathrm{lower}}} \approx\frac{1}{1-gh/c^2} \approx1+\frac{gh}{c^2}.\]This answers the clock question to first order in \(gh/c^2\): the upper clock records more elapsed time between receptions than the lower clock records between the corresponding emissions. The frequency reduction is called gravitational redshift. It vanishes when \(g=0\) or \(h=0\).
6. What does the rocket predict for light paths?
Now emit a horizontal pulse across width \(L>0\), toward a mark on the opposite wall initially level with the source. Both start at floor height and have proper acceleration \(a\).
How far below the mark does the pulse arrive?
Again use an inertial frame in which the rocket is at rest at emission. Put the source at \((x,z)=(0,0)\) and the mark at \((L,0)\). The light follows \(x_{\mathrm{light}}=ct\) and \(z_{\mathrm{light}}=0\).
The wall remains at \(x=L\), so the pulse arrives at \(T=L/c\). Substituting this time into Section 5’s trajectory, with initial height zero and acceleration \(a\), gives
\[\begin{aligned} \Delta z:=z_{\mathrm{mark}}(T) &=\frac{c^2}{a}\left[\sqrt{1+(aL/c^2)^2}-1\right]\\ &=\frac{aL^2}{2c^2}\left[1+O\!\left((aL/c^2)^2\right)\right]. \end{aligned}\]Thus \(aL/c^2\ll1\) makes the leading gap \(\Delta z\approx aL^2/(2c^2)\) reliable. The mark’s arrival speed is approximately \(aL/c\), so its Lorentz factor differs from one only at order \((aL/c^2)^2\). Converting the gap to a wall-ruler distance therefore preserves the leading result.
Repeating the calculation at intermediate horizontal positions gives the path recorded with the rocket’s rulers, to this accuracy:
\[z_{\mathrm{relative}}(x)\approx-\frac{ax^2}{2c^2},\qquad 0\leq x\leq L.\]Setting \(a=g\) predicts the leading deflection in a sufficiently small supported laboratory near Earth. This answers the horizontal-pulse question. Both rocket predictions were obtained in flat spacetime; a bent path in accelerating coordinates does not establish curvature.
What observation distinguishes Earth’s gravitational field from the effects of accelerating a laboratory in flat spacetime?
7. Why can gravity be removed only locally?
Earlier we treated gravity as giving every released object the same downward acceleration \(g\) throughout a small laboratory. Under that approximation, two objects released from rest keep the same distance apart.
Return to Earth and release two objects from rest side by side. Each accelerates approximately toward Earth’s center, so their paths converge and the distance between them decreases. Now release one object above the other. The lower object is closer to Earth’s center and accelerates downward more strongly, so the distance between them increases. These are the small effects we neglected when we used the same downward acceleration throughout the laboratory.
The earlier result about mass and material still applies: objects given the same position and initial velocity follow the same trajectory. Here the two objects start at different positions. A freely falling observer can measure the changing distance between them using light signals.
To calculate the leading effect, use the ground laboratory’s time \(t\) and approximately Cartesian positions \(\mathbf x_1(t)\) and \(\mathbf x_2(t)\). Let \(\mathbf g(\mathbf x)\) be the smooth, steady acceleration field measured by release experiments. The separation \(\boldsymbol\xi:=\mathbf x_2-\mathbf x_1\) obeys
\[\begin{aligned} \frac{d^2\xi^i}{dt^2} &=g^i(\mathbf x_1+\boldsymbol\xi)-g^i(\mathbf x_1)\\ &\approx\frac{\partial g^i}{\partial x^j}(\mathbf x_1)\,\xi^j. \end{aligned}\]The second line is the first-order Taylor expansion for a small separation. Spatial indices \(i,j\) run from \(1\) to \(3\); a repeated index is summed. The derivative measures how one acceleration component changes with one position coordinate while the other coordinates remain fixed.
These relative accelerations are called tidal effects. Subtracting the same laboratory acceleration from both objects leaves their difference unchanged. In the rocket, freely released objects have constant relative velocities in an inertial frame. Near Earth, the measured tidal acceleration remains even in a freely falling laboratory.
8. What geometry predicts clocks and free paths together?
The small-laboratory clock and light questions are answered. We now seek a description that can also accommodate the tidal measurements across a larger region.
How can we extend the special-relativistic interval while preserving the clock and light rules in each freely falling laboratory?
Return to the supported laboratory near Earth from Section 5. Fix a light source and its clock to the floor, and hold a receiver and its clock at height \(h\) above it. Send light upward through the evacuated space between them. Assume the apparatus stays in place and Earth’s gravitational field does not change during the comparison.
We now use coordinates fixed to this ground laboratory, replacing the inertial coordinates used for the rocket calculation. Set \(z=0\) at the floor and \(z=h\) at the receiver, with \(z\) increasing upward. Choose the coordinate time \(t\) in seconds to agree with the floor clock there. Each signal repeats the same journey under unchanged conditions, so it has the same coordinate travel time \(D\). If \(t_{\mathrm e}\) and \(t_{\mathrm r}\) now denote emission and reception times in these ground coordinates, then
\[t_{\mathrm r}(t_{\mathrm e})=t_{\mathrm e}+D, \qquad \Delta t_{\mathrm r}=\Delta t_{\mathrm e}.\]Thus signals emitted a coordinate-time interval apart arrive that same coordinate-time interval apart. The clocks’ own elapsed times can differ. Write the rate of a clock held at height \(z\) as \(d\tau/dt=A(z)\), where \(A>0\) is dimensionless and \(A(0)=1\) because \(t\) agrees with the floor clock. Using the equal coordinate-time intervals, Section 5’s comparison of the clocks gives
\[\frac{d\tau_{\mathrm{upper}}}{d\tau_{\mathrm{lower}}} =\frac{A(h)}{A(0)} =A(h) \approx1+\frac{gh}{c^2}.\]
Repeating this comparison at other small heights gives \(A(z)\approx1+gz/c^2\). To express the clock-rate rule using one-forms, we use the pullback construction from Section 8 of From Light Signals to Spacetime.
Let \(\zeta:I\to M\) describe one clock held at a fixed height \(z_0\) in the spacetime region \(M\). Use \(\lambda\in I\subset\mathbb R\) for its parameter, chosen to equal the ground coordinate time in seconds, so \(t(\zeta(\lambda))=\lambda\) and \(z(\zeta(\lambda))=z_0\). Let \(\tau:I\to\mathbb R\) give that clock’s reading in seconds. The measured rate is then \(d\tau/d\lambda=A(z_0)\).
At \(p=\zeta(\lambda)\), the coordinate one-form \(dt_p\) acts on \(T_pM\), while \(d\tau_\lambda\) acts on \(T_\lambda I\). Pulling \(dt\) back along \(\zeta\) puts it on the clock’s parameter interval:
\[\zeta^*(dt)=d(t\circ\zeta)=d\lambda.\]The differential of the clock-reading function therefore satisfies
\[\begin{aligned} d\tau &=\frac{d\tau}{d\lambda}\,d\lambda\\ &=A(z_0)\,d\lambda\\ &=A(z_0)\,\zeta^*(dt). \end{aligned}\]At each \(\lambda\), both sides are covectors in \(T_\lambda^*I\). Following the preceding post’s ideal-clock rule, we require the line element along this path to satisfy
\[ds^2=-c^2\,d\tau\otimes d\tau =-A(z_0)^2c^2\,d\lambda\otimes d\lambda.\]This is a bilinear form on \(I\) and determines the time coefficient that a spacetime metric must reproduce along the stationary path. Describing moving clocks and light also requires spatial and mixed coefficients. We adopt the same bilinear structure used in special relativity, allowing its coefficients to depend smoothly on the event.
At an event \(p\in M\), the metric is a symmetric bilinear map
\[g_p:T_pM\times T_pM\longrightarrow\mathbb R.\]As in the preceding post, \(T_pM\) contains tangents to paths through \(p\). Require the metric to be nondegenerate, so its matrix is invertible, with signature \((-,+,+,+)\): in a suitable basis it has one negative and three positive squared directions. This preserves the local distinction between clock, light, and spatial directions.
Choose coordinates \(x^\mu=(ct,x,y,z)\), all in length units. Greek indices run from \(0\) to \(3\) and repeated indices are summed. Write \(\partial_\mu:=\partial/\partial x^\mu\) for the coordinate tangent basis and \(dx^\mu\) for its dual covectors. They satisfy \(dx^\mu(\partial_\nu)=\delta^\mu{}_{\nu}\), where \(\delta^\mu{}_{\nu}\) is one for equal indices and zero otherwise. The metric components and their assembled map are
\[g_{\mu\nu}(p):=g_p(\partial_\mu,\partial_\nu),\qquad g=g_{\mu\nu}\,dx^\mu\otimes dx^\nu.\]Here \(g_{\mu\nu}\) are dimensionless metric components; the scalar \(g\) in the earlier drop rule still denotes acceleration. The line element along a path is \(ds^2:=\zeta^*g\). For our stationary clock, \(\zeta^*(dx^0)=c\,d\lambda\) and the three spatial coordinate one-forms pull back to zero. Thus \(\zeta^*g=g_{00}(\zeta(\lambda))c^2\,d\lambda\otimes d\lambda\). Comparing with the clock result fixes \(g_{00}=-A(z)^2\) at each height. The components depend on coordinates, so their variation alone does not diagnose curvature.
What does the metric predict along a recorded path?
Now allow the clock to move. Keep \(\lambda\) equal to laboratory time along its path \(\zeta:I\to M\). The differential of the path, also called its pushforward, is the linear map
\[d\zeta_\lambda:T_\lambda I\longrightarrow T_{\zeta(\lambda)}M.\]It sends the parameter’s unit tangent to the clock’s spacetime tangent:
\[V(\lambda):=d\zeta_\lambda\!\left(\left.\frac{\partial}{\partial\lambda}\right|_\lambda\right) \in T_{\zeta(\lambda)}M.\]Each coordinate one-form \(dx^\mu_{\zeta(\lambda)}:T_{\zeta(\lambda)}M\to\mathbb R\) then takes this spacetime tangent to a real number. These numbers are the components of \(V\), and the chain rule identifies them with the coordinate rates:
\[V^\mu(\lambda) :=dx^\mu_{\zeta(\lambda)}\bigl(V(\lambda)\bigr) =\frac{d(x^\mu\circ\zeta)}{d\lambda}(\lambda).\]Write \(g(V,V)\) for the scalar function \(\lambda\mapsto g_{\zeta(\lambda)}(V(\lambda),V(\lambda))\). Evaluating the metric yields
\[g(V,V)=g_{\mu\nu}(\zeta(\lambda)) \frac{d(x^\mu\circ\zeta)}{d\lambda} \frac{d(x^\nu\circ\zeta)}{d\lambda}.\]We extend the ideal-clock rule from special relativity by requiring, for a clock path with \(g(V,V)<0\),
\[\frac{d\tau}{d\lambda}=\frac1c\sqrt{-g(V,V)},\qquad ds^2=g(V,V)\,d\lambda\otimes d\lambda.\]For the stationary clock, \(V^\mu=(c,0,0,0)\), recovering \(d\tau/d\lambda=\sqrt{-g_{00}}=A(z_0)\). Integrating the rate predicts the reading along any specified clock path.
For light, we retain the local special-relativistic condition: a nonzero tangent \(K\in T_pM\) must satisfy \(g_p(K,K)=0\). Such a direction is called null. These clock and light rules are physical assumptions of the metric model.
For a smooth metric of this signature, coordinates can be chosen so that \(g_{\mu\nu}=\operatorname{diag}(-1,1,1,1)\) and its first coordinate derivatives vanish at a chosen event. Identifying these with the coordinates of a local freely falling laboratory implements the equivalence principle. Light then has speed \(c\) measured with that observer’s clocks and rulers, although its speed calculated from coordinates covering an extended region can differ from \(c\).
9. How does the metric predict free fall?
Knowing the metric now lets us calculate the time along a given path and identify possible light directions. We still need a rule that selects the path of a released object.
How do we express continued inertial motion when the tangent spaces at successive events are different?
Keep the path \(\zeta:I\to M\) and its clock-reading function \(\tau:I\to\mathbb R\) from Section 8. The tangent \(V\) measures change in the clock’s spacetime position per second of laboratory time. To express that change per second recorded by the moving clock, we first choose the corresponding tangent on \(I\). The clock’s rate \(\tau'(\lambda):=d\tau/d\lambda\) is positive, so \(\tau\) is a smooth coordinate on \(I\). Write \(\partial_\tau:=\partial/\partial\tau\) for its coordinate tangent field, characterized by \(d\tau_\lambda(\left.\partial_\tau\right\rvert_\lambda)=1\). Since \(d\tau=\tau'\,d\lambda\), this gives
\[\left.\partial_\tau\right|_\lambda =\frac{1}{\tau'(\lambda)} \left.\frac{\partial}{\partial\lambda}\right|_\lambda \in T_\lambda I.\]The differential of the path maps this tangent into spacetime:
\[\begin{aligned} U(\lambda) &:=d\zeta_\lambda\!\left(\left.\partial_\tau\right|_\lambda\right)\\ &=\frac{1}{\tau'(\lambda)}\, d\zeta_\lambda\!\left(\left.\frac{\partial}{\partial\lambda}\right|_\lambda\right)\\ &=\frac{V(\lambda)}{\tau'(\lambda)} \in T_{\zeta(\lambda)}M. \end{aligned}\]For any smooth scalar record \(q:I\to\mathbb R\), its rate per moving-clock second is the action of this coordinate tangent on \(q\):
\[(\partial_\tau q)(\lambda) =dq_\lambda\!\left(\left.\partial_\tau\right|_\lambda\right) =\frac{1}{\tau'(\lambda)}\frac{dq}{d\lambda}(\lambda).\]In particular, let \(q^\mu:=x^\mu\circ\zeta\) be the four coordinate records. The chain rule relates these rates to the components of \(U\):
\[U^\mu(\lambda) =dx^\mu_{\zeta(\lambda)}\bigl(U(\lambda)\bigr) =d(x^\mu\circ\zeta)_\lambda\!\left(\left.\partial_\tau\right|_\lambda\right) =(\partial_\tau q^\mu)(\lambda).\]The clock rule from Section 8 gives the normalization
\[g(U,U)=\frac{g(V,V)}{(\tau')^2} =\frac{-c^2(\tau')^2}{(\tau')^2} =-c^2.\]How do we differentiate a vector along the path?
We can differentiate each scalar component \(U^\mu\), but tangents at different events belong to different tangent spaces. The metric compares two vectors at the same event; a derivative also needs a rule for comparing vectors along a path. To specify such a rule, we first describe how it acts on vector fields in the surrounding region.
A smooth vector field assigns a tangent vector to each event. Use the coordinates and dual bases from Section 8 to write two such fields:
\[X=X^\mu\partial_\mu,\qquad Y=Y^\nu\partial_\nu.\]We want to differentiate \(Y\) along \(X\). First consider each component \(Y^\nu\), which is a smooth real-valued function within the coordinate chart. Its differential is the one-form field
\[d(Y^\nu)=(\partial_\mu Y^\nu)\,dx^\mu, \qquad \partial_\mu Y^\nu:=\frac{\partial Y^\nu}{\partial x^\mu}.\]To find the component’s rate of change along \(X\), apply this one-form to \(X\). At an event \(p\), the dual-basis identity \(dx^\mu_p(\left.\partial_\alpha\right\rvert_p)=\delta^\mu{}_\alpha\) gives
\[\begin{aligned} d(Y^\nu)_p\bigl(X(p)\bigr) &=(\partial_\mu Y^\nu)(p)\, dx^\mu_p\!\left(X^\alpha(p)\left.\partial_\alpha\right\rvert_p\right)\\ &=(\partial_\mu Y^\nu)(p)\,X^\alpha(p)\delta^\mu{}_\alpha\\ &=X^\mu(p)(\partial_\mu Y^\nu)(p). \end{aligned}\]The input is a tangent vector and the output is a real number. Doing this at each event defines the scalar function \(d(Y^\nu)(X)=X^\mu\partial_\mu Y^\nu\). Assemble these component rates using the coordinate basis at that event:
\[d(Y^\nu)(X)\,\partial_\nu =X^\mu(\partial_\mu Y^\nu)\,\partial_\nu.\]This expression differentiates only the component functions. A derivative of \(Y=Y^\nu\partial_\nu\) must also specify how to differentiate the basis vectors. Different coordinate choices change the division between these two contributions.
For any smooth scalar function \(f\), the notation \(df(X)\) has the same meaning: the scalar function whose value at \(p\) is \(df_p(X(p))\). Multiplication of a vector field by a scalar function is defined at each point by \((fX)(p):=f(p)X(p)\).
For vector fields \(X\) and \(Y\), write \(\nabla_XY\) for the covariant derivative of \(Y\) in direction \(X\). Its value at an event \(p\) must be a vector in \(T_pM\), independent of the coordinates used to calculate it. We require this derivative to be additive in both inputs, linear over smooth functions in its direction input, and to obey the product rule in the field being differentiated. For smooth real-valued functions \(f,h\) and another vector field \(Z\), the last two requirements mean
\[\begin{aligned} \nabla_{fX+hZ}Y&=f\nabla_XY+h\nabla_ZY,\\ \nabla_X(fY)&=df(X)Y+f\nabla_XY. \end{aligned}\]A rule \(\nabla\) with these properties is a connection on the tangent bundle, the collection of tangent spaces over \(M\). These are requirements defining the kind of derivative we seek; they do not yet choose one connection.
Once a connection is chosen, each \(\nabla_{\partial_\mu}\partial_\nu\) is a vector field and therefore has an expansion in the coordinate basis. Define its coefficient functions \(\Gamma^\rho{}_{\mu\nu}\) by
\[\nabla_{\partial_\mu}\partial_\nu =\Gamma^\rho{}_{\mu\nu}\partial_\rho.\]Here \(\mu\) specifies the direction of differentiation, \(\nu\) specifies the basis vector being differentiated, and \(\rho\) specifies the output component. These functions are the connection coefficients. Linearity in the direction input now gives the basis derivative along \(X\):
\[\nabla_X\partial_\nu =X^\mu\nabla_{\partial_\mu}\partial_\nu =X^\mu\Gamma^\rho{}_{\mu\nu}\partial_\rho.\]Apply the product rule to \(Y=Y^\nu\partial_\nu\), then substitute this basis derivative:
\[\begin{aligned} \nabla_XY &=\nabla_X(Y^\nu\partial_\nu)\\ &=d(Y^\nu)(X)\,\partial_\nu +Y^\nu\nabla_X\partial_\nu\\ &=d(Y^\nu)(X)\,\partial_\nu +X^\mu Y^\nu\Gamma^\rho{}_{\mu\nu}\partial_\rho. \end{aligned}\]The first term is exactly the component rate we calculated above; the second is the contribution from differentiating the basis. Relabel the summed index \(\nu\) as \(\rho\) in the first term and use \(d(Y^\rho)(X)=X^\mu\partial_\mu Y^\rho\) to combine them:
\[\begin{aligned} \nabla_XY &=\left[d(Y^\rho)(X) +\Gamma^\rho{}_{\mu\nu}X^\mu Y^\nu\right]\partial_\rho\\ &=\left[X^\mu\partial_\mu Y^\rho +\Gamma^\rho{}_{\mu\nu}X^\mu Y^\nu\right]\partial_\rho. \end{aligned}\]The coefficients \(\Gamma^\rho{}_{\mu\nu}\) themselves are not tensor components: a coordinate change also differentiates the functions relating the two bases. Their transformation makes the complete derivative independent of the coordinate choice.
Return to our clock path. Let \(W(\lambda)=W^\mu(\lambda)\left.\partial_\mu\right\rvert_{\zeta(\lambda)}\) be a smooth assignment of vectors along it. Write \(DW/D\lambda\) for the path derivative induced by the connection in the parameter direction \(\partial/\partial\lambda\). The notation \(\partial_\beta\circ\zeta\) restricts a coordinate basis field to this path. Since its tangent has components \(V^\alpha\), the induced derivative of the basis is
\[\frac{D(\partial_\beta\circ\zeta)}{D\lambda} =V^\alpha(\Gamma^\mu{}_{\alpha\beta}\circ\zeta) (\partial_\mu\circ\zeta).\]The component functions \(W^\beta\) live on \(I\), where their differentials satisfy \(d(W^\beta)(\partial/\partial\lambda)=dW^\beta/d\lambda\). Applying the same product rule to \(W=W^\beta(\partial_\beta\circ\zeta)\) gives
\[\begin{aligned} \frac{DW}{D\lambda} &=d(W^\beta)\!\left(\frac{\partial}{\partial\lambda}\right) (\partial_\beta\circ\zeta) +W^\beta\frac{D(\partial_\beta\circ\zeta)}{D\lambda}\\ &=\frac{dW^\beta}{d\lambda}(\partial_\beta\circ\zeta) +W^\beta\frac{D(\partial_\beta\circ\zeta)}{D\lambda}\\ &=\left(\frac{dW^\mu}{d\lambda} +\Gamma^\mu{}_{\alpha\beta}(\zeta(\lambda))V^\alpha W^\beta\right) \left.\partial_\mu\right|_{\zeta(\lambda)},\\ \frac{DW}{D\tau} &:=\frac{1}{\tau'}\frac{DW}{D\lambda} =\left(\partial_\tau W^\mu +\Gamma^\mu{}_{\alpha\beta}(\zeta(\lambda))U^\alpha W^\beta\right) \left.\partial_\mu\right|_{\zeta(\lambda)}. \end{aligned}\]The second path derivative uses \(\partial_\tau=(\tau')^{-1}\partial/\partial\lambda\) on \(I\). Setting \(W=U\) gives the clock’s acceleration vector \(DU/D\tau\). A vector assignment with zero path derivative is called parallel along the path; solving that equation from an initial vector defines parallel transport.
Which connection does the metric select?
We have defined the derivative once its coefficients are supplied. To calculate those coefficients from the metric, impose two further conditions.
First, require the metric comparison of two parallel vectors to remain constant along a path. For vector fields \(Y,Z\), the notation \(g(Y,Z)\) denotes the scalar function \(p\mapsto g_p(Y(p),Z(p))\). Its differential \(d\bigl(g(Y,Z)\bigr)\) is a one-form field; applying it to \(X\) gives the rate of change of the pairing along \(X\). Metric preservation requires that this derivative account for the change of each vector:
\[d\bigl(g(Y,Z)\bigr)(X) =g(\nabla_XY,Z)+g(Y,\nabla_XZ).\]This condition is called metric compatibility. Substitute \(X=\partial_\alpha\), \(Y=\partial_\beta\), and \(Z=\partial_\nu\). Using \(g(\partial_\beta,\partial_\nu)=g_{\beta\nu}\) and the definition of the connection coefficients gives
\[\begin{aligned} \partial_\alpha g_{\beta\nu} &=g(\Gamma^\rho{}_{\alpha\beta}\partial_\rho,\partial_\nu) +g(\partial_\beta,\Gamma^\rho{}_{\alpha\nu}\partial_\rho)\\ &=\Gamma^\rho{}_{\alpha\beta}g_{\rho\nu} +\Gamma^\rho{}_{\alpha\nu}g_{\beta\rho}. \end{aligned}\]Second, require \(\nabla_{\partial_\alpha}\partial_\beta=\nabla_{\partial_\beta}\partial_\alpha\) in a coordinate basis, or equivalently \(\Gamma^\rho{}_{\alpha\beta}=\Gamma^\rho{}_{\beta\alpha}\). This is the zero-torsion condition in coordinates. It is an additional choice, not a consequence of metric compatibility. Standard general relativity uses both conditions.
To solve for the coefficients, write metric compatibility with three arrangements of the indices:
\[\begin{aligned} \partial_\alpha g_{\beta\nu} &=\Gamma^\rho{}_{\alpha\beta}g_{\rho\nu} +\Gamma^\rho{}_{\alpha\nu}g_{\beta\rho},\\ \partial_\beta g_{\alpha\nu} &=\Gamma^\rho{}_{\beta\alpha}g_{\rho\nu} +\Gamma^\rho{}_{\beta\nu}g_{\alpha\rho},\\ \partial_\nu g_{\alpha\beta} &=\Gamma^\rho{}_{\nu\alpha}g_{\rho\beta} +\Gamma^\rho{}_{\nu\beta}g_{\alpha\rho}. \end{aligned}\]Add the first two equations and subtract the third. Symmetry of \(g\) and of the lower connection indices cancels the terms containing \(\Gamma^\rho{}_{\alpha\nu}\) and \(\Gamma^\rho{}_{\beta\nu}\), leaving
\[\partial_\alpha g_{\beta\nu} +\partial_\beta g_{\alpha\nu} -\partial_\nu g_{\alpha\beta} =2g_{\rho\nu}\Gamma^\rho{}_{\alpha\beta}.\]Let \(g^{\mu\nu}\) be the inverse matrix, defined by \(g^{\mu\rho}g_{\rho\nu}=\delta^\mu{}_{\nu}\). Multiplying by \(\tfrac12g^{\mu\nu}\) gives
\[\Gamma^\mu{}_{\alpha\beta} =\frac12g^{\mu\nu} \left(\partial_\alpha g_{\beta\nu} +\partial_\beta g_{\alpha\nu} -\partial_\nu g_{\alpha\beta}\right).\]Substitution verifies that these coefficients satisfy both conditions; the calculation shows that any connection satisfying them must have this formula. The resulting unique connection is the Levi-Civita connection of \(g\). Its coordinate coefficients are called Christoffel symbols. They have units of inverse length because the metric components are dimensionless and the coordinates have length units. If those components are constant in a coordinate chart, the formula gives \(\Gamma^\mu{}_{\alpha\beta}=0\) there, and the covariant derivative reduces to differentiating vector components in that basis.
Which paths represent free motion?
We postulate that an ideal test object with no nongravitational forces has \(DU/D\tau=0\). To express this condition in coordinates, first set \(W=U\) in the path derivative defined above, then substitute \(U^\mu=\partial_\tau q^\mu\):
\[\begin{aligned} \frac{DU}{D\tau} &=\left[\partial_\tau U^\mu +\Gamma^\mu{}_{\alpha\beta}(\zeta(\lambda))U^\alpha U^\beta\right] \left.\partial_\mu\right\rvert_{\zeta(\lambda)}\\ &=\left[\partial_\tau(\partial_\tau q^\mu) +\Gamma^\mu{}_{\alpha\beta}(\zeta(\lambda)) (\partial_\tau q^\alpha)(\partial_\tau q^\beta)\right] \left.\partial_\mu\right\rvert_{\zeta(\lambda)}. \end{aligned}\]The coordinate basis vectors at \(\zeta(\lambda)\) are linearly independent. Therefore \(DU/D\tau=0\) holds exactly when each coefficient vanishes, giving one equation for each \(\mu=0,1,2,3\):
\[\partial_\tau(\partial_\tau q^\mu) +\Gamma^\mu{}_{\alpha\beta}(\zeta(\lambda)) (\partial_\tau q^\alpha)(\partial_\tau q^\beta)=0.\]These paths are called geodesics. In the freely falling coordinates of Section 8, the first metric derivatives and hence \(\Gamma\) vanish at the chosen event. The equation reduces there to the inertial rule \(\partial_\tau(\partial_\tau q^\mu)=0\). For a supported clock, \(DU/D\tau\) instead gives its nonzero proper-acceleration vector; its spatial components in the clock’s rest frame are accelerometer readings.
Light follows null geodesics in the geometric-optics approximation, where its wavelength is small compared with the scales on which the geometry varies. A null ray has no increasing proper-time reading to use as a coordinate on its parameter interval. Give its path a separate name \(\chi:J_{\mathrm{light}}\to M\), where \(J_{\mathrm{light}}\subset\mathbb R\) is an interval with a dimensionless, increasing parameter \(r\). Its coordinate records are \(q_{\mathrm{light}}^\mu:=x^\mu\circ\chi\). Choose \(r\) so that the covariant derivative of its tangent along the ray vanishes:
\[\begin{aligned} \frac{d^2q_{\mathrm{light}}^\mu}{dr^2} +\Gamma^\mu{}_{\alpha\beta}(\chi(r)) \frac{dq_{\mathrm{light}}^\alpha}{dr}\frac{dq_{\mathrm{light}}^\beta}{dr}&=0,\\ g_{\mu\nu}(\chi(r))\frac{dq_{\mathrm{light}}^\mu}{dr}\frac{dq_{\mathrm{light}}^\nu}{dr}&=0. \end{aligned}\]A parameter for which the first equation has this form is called affine; a constant positive rescaling and shift preserve it. It need not equal laboratory coordinate time. The first equation continues the light direction along the ray, while the second imposes the local light-speed condition.
10. How does the familiar gravitational field appear?
Return to the weak, steady field measured in the ground laboratory. We want to check whether the geodesic rule reproduces the drop acceleration as well as the clock comparison. Use approximately Cartesian spatial coordinates, with the floor at \(\mathbf x=0\) and \(t\) calibrated to its clock.
Assume the measured field \(\mathbf g(\mathbf x)\) has path-independent work per unit mass. Work here in the approximately Euclidean space \(\Omega\) of ground-fixed positions. Choose any smooth spatial path \(\boldsymbol\ell:[0,1]\to\Omega\) from the floor to \(\mathbf x\). As in the conservative-energy construction, define the potential from release measurements:
\[\Phi(\mathbf x):=-\int_0^1 \delta_{ij}g^i(\boldsymbol\ell(u))\frac{d\ell^j}{du}\,du, \qquad \Phi(\mathbf 0)=0.\]Here \(u\) is a dimensionless path parameter, \(\ell^j\) are the path’s Cartesian components, and \(\delta_{ij}\) supplies their Euclidean scalar product. In this spatial calculation, \(x^i\) and \(dx^i\) denote coordinates and one-forms on \(\Omega\). The potential has units \(\mathrm{m^2/s^2}\). Path independence gives \(d\Phi=-\delta_{ij}g^i\,dx^j\) on \(\Omega\), or \(g^i=-\delta^{ij}\partial_j\Phi\) in components. Near the floor, \(\mathbf g=(0,0,-g)\), so \(\Phi(z)=gz\).
Allow the stationary clock factor \(A:\Omega\to(0,\infty)\) to depend on all three spatial coordinates. For a tilted light path, Section 5’s arrival-spacing calculation uses the velocity component along the ray. Applying that comparison between nearby supported clocks gives, to first order,
\[A^{-1}dA\approx-\frac{\delta_{ij}g^i\,dx^j}{c^2} =\frac{d\Phi}{c^2}.\]All three expressions are one-forms on the same spatial domain \(\Omega\); \(A^{-1}dA\) means multiplication by the scalar \(1/A\). The chain rule gives \(d(\ln A)=A^{-1}dA\). In the spacetime metric, \(A\) and \(\Phi\) mean these spatial functions composed with the projection \((t,\mathbf x)\mapsto\mathbf x\). Integrating along a spatial path from the floor, where \(A=1\), and expanding for \(\lvert\Phi\rvert/c^2\ll1\) gives
\[\ln A\approx\frac{\Phi}{c^2},\qquad A\approx1+\frac{\Phi}{c^2},\qquad g_{00}=-A^2\approx-\left(1+\frac{2\Phi}{c^2}\right).\]We have fixed this coefficient using the clock result. Now assume a static metric with \(g_{0i}=0\) and spatial components close to \(\delta_{ij}\). Its time derivatives vanish, so Section 9’s connection formula gives, to leading order,
\[\begin{aligned} \Gamma^i{}_{00} &=-\frac12g^{ij}\partial_j g_{00}\\ &\approx-\frac12\delta^{ij}\partial_j \left(-1-\frac{2\Phi}{c^2}\right) =\frac{\partial_i\Phi}{c^2}. \end{aligned}\]Keep the falling object’s path parameter \(\lambda\) equal to ground coordinate time, so its coordinate records satisfy \(q^0(\lambda)=c\lambda\). For slow motion in the weak field, the clock rule gives \(\tau'\approx1\). The exact conversion for each spatial record is
\[\partial_\tau(\partial_\tau q^i) =\frac{1}{\tau'}\frac{d}{d\lambda} \left(\frac{1}{\tau'}\frac{dq^i}{d\lambda}\right) =\frac{1}{(\tau')^2}\frac{d^2q^i}{d\lambda^2} -\frac{\tau''}{(\tau')^3}\frac{dq^i}{d\lambda}.\]Multiplying the spatial geodesic equation by \((\tau')^2\) therefore gives
\[\frac{d^2q^i}{d\lambda^2} =-\Gamma^i{}_{\alpha\beta}\frac{dq^\alpha}{d\lambda}\frac{dq^\beta}{d\lambda} +\frac{\tau''}{\tau'}\frac{dq^i}{d\lambda}.\]All coefficients here are evaluated at \(\zeta(\lambda)\). For the static metric, \(\Gamma^i{}_{0j}=0\). The terms with two spatial velocities and the clock-rate correction are higher order in speed divided by \(c\): the time component of the geodesic equation gives \(\tau''/\tau'=2\Gamma^0{}_{0j}\,dq^j/d\lambda\). Keeping the two time inputs, with \(dq^0/d\lambda=c\), yields
\[\frac{d^2q^i}{d\lambda^2} \approx-c^2\Gamma^i{}_{00} =-\partial_i\Phi=g^i.\]Since \(q^3=z\circ\zeta\) and \(\lambda\) is ground time, the vertical equation at the floor is precisely the original drop rule \(d^2z(t)/dt^2\approx-g\) for the recorded height. Agreement with the clock formula used to construct \(g_{00}\) is built in; recovering the drop rule also uses the geodesic postulate. Spatial metric corrections are unnecessary at this order for slow objects, but they matter for extended light-path predictions.
11. What remains after the free-fall rule is known?
The metric and geodesic rule now predict clocks and motion for a specified geometry. We still need to express the tidal measurement in that geometry and determine which metric surrounds a given source.
How do the components of one direction change along another?
Section 7 compared neighboring drops. We now want to calculate their relative acceleration using the connection. This will require differentiating both along a clock’s path and across the family of nearby paths. To understand how the two directions enter the calculation, first examine the component changes of one vector field along another.
For smooth vector fields \(X=X^\mu\partial_\mu\) and \(Y=Y^\nu\partial_\nu\), Section 9 derived the one-form \(d(Y^\nu)\) and its scalar evaluation \(d(Y^\nu)(X)\). Reuse the resulting component rates:
\[d(Y^\nu)(X)\,\partial_\nu =X^\mu(\partial_\mu Y^\nu)\,\partial_\nu.\]This is the first term in our calculation of \(\nabla_XY\), before adding \(Y^\nu\nabla_X\partial_\nu\). By itself it depends on the coordinate choice.
Exchanging the roles of the two fields gives the component changes of \(X\) along \(Y\):
\[d(X^\nu)(Y)\,\partial_\nu =Y^\mu(\partial_\mu X^\nu)\,\partial_\nu.\]Their difference defines the Lie bracket in these coordinates:
\[\begin{aligned} [X,Y] &:=\bigl(d(Y^\nu)(X)-d(X^\nu)(Y)\bigr)\partial_\nu\\ &=\bigl(X^\mu\partial_\mu Y^\nu-Y^\mu\partial_\mu X^\nu\bigr)\partial_\nu. \end{aligned}\]This construction takes two vector fields and produces a vector field in the chosen chart. We still need to check that their difference gives the same vector field when we change coordinates, even though either term separately need not.
To check this, apply the differential of an arbitrary smooth scalar function \(f\) to the result. Since \(df(\partial_\nu)=\partial_\nu f\), we obtain
\[df([X,Y]) =\bigl(d(Y^\nu)(X)-d(X^\nu)(Y)\bigr)\partial_\nu f.\]We can recover this expression by differentiating \(f\) in two orders. The first rates are \(df(Y)=Y^\nu\partial_\nu f\) and \(df(X)=X^\mu\partial_\mu f\). Differentiate each rate along the other field and use the product rule:
\[\begin{aligned} d\bigl(df(Y)\bigr)(X) &=d(Y^\nu)(X)\,\partial_\nu f +Y^\nu X^\mu\partial_\mu\partial_\nu f,\\ d\bigl(df(X)\bigr)(Y) &=d(X^\mu)(Y)\,\partial_\mu f +X^\mu Y^\nu\partial_\nu\partial_\mu f. \end{aligned}\]Subtract the second equation from the first. Relabel the summed index \(\mu\) as \(\nu\) in its first term; the second derivatives cancel because coordinate partial derivatives commute. The result is
\[\begin{aligned} &d\bigl(df(Y)\bigr)(X)-d\bigl(df(X)\bigr)(Y)\\ &\quad=\bigl(d(Y^\nu)(X)-d(X^\nu)(Y)\bigr)\partial_\nu f\\ &\quad=df([X,Y]). \end{aligned}\]The left side is defined using differentials acting on vector fields, independently of any coordinates. A vector is determined by these evaluations: taking the coordinate functions as \(f\) recovers each of its components. Since the identity holds for every smooth \(f\), it determines \([X,Y]\) independently of the chosen chart. The bracket therefore measures the difference between the two orders of directional differentiation, without a metric or connection. It depends on derivatives of the input fields as well as their values at the event. For coordinate basis fields, the components are constant, so \([\partial_\mu,\partial_\nu]=0\).
Which part of this difference measures curvature?
We can now compare the bracket with the covariant derivatives using the same component rates. Subtract the two versions of Section 9’s formula and exchange the summed indices in the second connection term:
\[\begin{aligned} \nabla_XY-\nabla_YX &=\bigl(d(Y^\rho)(X)-d(X^\rho)(Y)\bigr)\partial_\rho\\ &\quad+X^\mu Y^\nu \bigl(\Gamma^\rho{}_{\mu\nu}-\Gamma^\rho{}_{\nu\mu}\bigr)\partial_\rho\\ &=[X,Y]. \end{aligned}\]The last equality uses the symmetry of the lower indices of the Levi-Civita coefficients. Thus each covariant derivative contains a basis contribution, but those contributions cancel in their difference.
Now compare the two orders of covariant differentiation of a vector field \(Z\). Expanding \(X=X^\mu\partial_\mu\) and \(Y=Y^\nu\partial_\nu\) with the connection’s product rule gives
\[\begin{aligned} \nabla_X(\nabla_Y Z) &=X^\mu\nabla_{\partial_\mu} \bigl(Y^\nu\nabla_{\partial_\nu}Z\bigr)\\ &=X^\mu(\partial_\mu Y^\nu)\nabla_{\partial_\nu}Z +X^\mu Y^\nu\nabla_{\partial_\mu}(\nabla_{\partial_\nu}Z). \end{aligned}\]Subtract the expression with \(X\) and \(Y\) exchanged. The terms differentiating their components give \(\nabla_{[X,Y]}Z\) by the bracket formula just derived:
\[\begin{aligned} &\nabla_X(\nabla_YZ)-\nabla_Y(\nabla_XZ)\\ &\quad=\nabla_{[X,Y]}Z\\ &\qquad+X^\mu Y^\nu\left[ \nabla_{\partial_\mu}(\nabla_{\partial_\nu}Z) -\nabla_{\partial_\nu}(\nabla_{\partial_\mu}Z)\right]. \end{aligned}\]To isolate the remaining difference, define the connection’s curvature by
\[\mathcal R(X,Y)Z :=\nabla_X(\nabla_YZ)-\nabla_Y(\nabla_XZ)-\nabla_{[X,Y]}Z.\]Here \(X,Y\) specify the two directions of differentiation and \(Z\) is the vector field being differentiated. This equation fixes our sign convention.
To calculate the result, use the coordinate formula from Section 9 twice. For \(Z=Z^\rho\partial_\rho\), it gives
\[\nabla_{\partial_\nu}Z =(\partial_\nu Z^\rho+\Gamma^\rho{}_{\nu\sigma}Z^\sigma)\partial_\rho.\]Differentiating this field along \(\partial_\mu\) gives
\[\begin{aligned} \nabla_{\partial_\mu}(\nabla_{\partial_\nu}Z) &=\Bigl[ \partial_\mu(\partial_\nu Z^\rho+\Gamma^\rho{}_{\nu\sigma}Z^\sigma)\\ &\qquad+\Gamma^\rho{}_{\mu\kappa} (\partial_\nu Z^\kappa+\Gamma^\kappa{}_{\nu\sigma}Z^\sigma) \Bigr]\partial_\rho\\ &=\Bigl[ \partial_\mu\partial_\nu Z^\rho +(\partial_\mu\Gamma^\rho{}_{\nu\sigma})Z^\sigma +\Gamma^\rho{}_{\nu\sigma}\partial_\mu Z^\sigma\\ &\qquad+\Gamma^\rho{}_{\mu\kappa}\partial_\nu Z^\kappa +\Gamma^\rho{}_{\mu\kappa}\Gamma^\kappa{}_{\nu\sigma}Z^\sigma \Bigr]\partial_\rho. \end{aligned}\]Subtract the expression with \(\mu\) and \(\nu\) exchanged. The second derivatives of \(Z^\rho\) cancel, and the two terms containing first derivatives of \(Z\) cancel against their counterparts. Since the coordinate directions have zero bracket, the remaining terms give
\[\begin{aligned} \mathcal R(\partial_\mu,\partial_\nu)Z &=Z^\sigma R^\rho{}_{\sigma\mu\nu}\partial_\rho,\\ R^\rho{}_{\sigma\mu\nu} &:=\partial_\mu\Gamma^\rho{}_{\nu\sigma} -\partial_\nu\Gamma^\rho{}_{\mu\sigma}\\ &\quad+\Gamma^\rho{}_{\mu\kappa}\Gamma^\kappa{}_{\nu\sigma} -\Gamma^\rho{}_{\nu\kappa}\Gamma^\kappa{}_{\mu\sigma}. \end{aligned}\]Setting \(Z=\partial_\sigma\) identifies these as the curvature components: \(\mathcal R(\partial_\mu,\partial_\nu)\partial_\sigma=R^\rho{}_{\sigma\mu\nu}\partial_\rho\). The indices \(\mu,\nu\) label the two directions, \(\sigma\) labels the vector being differentiated, and \(\rho\) labels the output component. Substituting into the expression for general directions gives
\[\mathcal R(X,Y)Z =X^\mu Y^\nu Z^\sigma R^\rho{}_{\sigma\mu\nu}\partial_\rho.\]Only the values of \(X,Y,Z\) at each event remain. The result is linear in each of those three vectors, so it defines a tensor at \(p\):
\[\begin{gathered} \mathcal R_p:T_pM\times T_pM\times T_pM\longrightarrow T_pM,\\ \mathcal R_p\bigl(X(p),Y(p)\bigr)Z(p) =\bigl(\mathcal R(X,Y)Z\bigr)(p). \end{gathered}\]For the Levi-Civita connection this is the Riemann curvature tensor of the metric. Its components have units of inverse length squared. A flat metric in inertial coordinates has \(\Gamma=0\) throughout the chart, giving \(\mathcal R=0\). Setting \(\Gamma=0\) only at one event leaves its derivatives in the curvature formula.
How does curvature predict the separation of free paths?
To compare neighboring drops, label the clocks by a dimensionless number \(b\) in an interval \(B\) containing zero, with \(b=0\) for the reference clock. Set their initial zero readings at simultaneous events in the reference clock’s local rest frame. Define a smooth map \(F:J\times B\to M\), where \(J\) is a common interval of clock readings in seconds: \(F(s,b)\) is the event where clock \(b\) reads \(s\). Here \(s\) is the clock reading itself; this family is specified directly by those readings rather than by laboratory time.
The differential of \(F\) gives two vectors at that same event:
\[U(s,b):=dF_{(s,b)}(\partial_s),\qquad \xi(s,b):=dF_{(s,b)}(\partial_b).\]The first follows a clock, while the second varies which clock we follow at a fixed reading. It is a tangent vector, not a subtraction of spacetime points. Let \(D_s\) and \(D_b\) denote the connection’s derivatives along these two parameter directions, defined as for a single path in Section 9. The parameter directions commute, so zero torsion gives \(D_s\xi=D_bU\). The curvature definition gives \((D_sD_b-D_bD_s)U=\mathcal R(U,\xi)U\). Since each clock falls freely, \(D_sU=0\), and therefore
\[\begin{aligned} D_s^2\xi &=D_sD_bU\\ &=D_bD_sU+\mathcal R(U,\xi)U\\ &=\mathcal R(U,\xi)U. \end{aligned}\]This is the geodesic-deviation rule: curvature predicts the relative acceleration of neighboring free paths. For a small label difference \(\Delta b\), the spatial components of \(\Delta b\,\xi(s,0)\) in the reference clock’s rest frame give their separation to first order. Unlike Section 7’s Euclidean difference \(\mathbf x_2-\mathbf x_1\), \(\xi\) is a vector at the reference event; it provides the infinitesimal separation needed in curved spacetime.
A coordinate change can set the connection coefficients to zero at one event. It cannot make a nonzero curvature tensor vanish there. The accelerating rocket has zero spacetime curvature; Earth’s tidal measurements require a nonzero value.
What selects the metric around Earth?
How do the source’s measured energy, momentum, and stress determine the geometry?
Organize these source measurements in a symmetric bilinear map \(T_p:T_pM\times T_pM\to\mathbb R\), called the stress-energy tensor. At each event choose the unit time and spatial axes of a freely falling laboratory. Hats label components in this orthonormal basis, where the metric is \(\operatorname{diag}(-1,1,1,1)\):
| Components | Local measurement |
|---|---|
| \(T^{\hat0\hat0}\) | Energy per unit volume |
| \(T^{\hat0\hat i}=T^{\hat i\hat0}\) | Energy flux in direction \(i\) divided by \(c\); also momentum density in direction \(i\) times \(c\) |
| \(T^{\hat i\hat j}\) | Flux of \(i\)-momentum across a surface normal to \(j\); diagonal entries are pressures |
All entries have energy-density units and can be calibrated through transfers to local detectors. In any coordinate basis, \(T_{\mu\nu}=T(\partial_\mu,\partial_\nu)\); upper indices mean \(T^{\mu\nu}=g^{\mu\alpha}g^{\nu\beta}T_{\alpha\beta}\). The same raising rule applies to the hatted components using the inverse orthonormal metric.
To form a two-index equation with this source, contract the curvature components just calculated. Summing one output index with one input gives the Ricci tensor, and a further contraction gives the scalar curvature:
\[R_{\mu\nu}:=R^\rho{}_{\mu\rho\nu},\qquad R:=g^{\mu\nu}R_{\mu\nu}.\]The source’s local conservation rule involves a derivative of its tensor. Extend the connection to tensor inputs using the product rule, then contract the derivative index with the first tensor index using the inverse metric:
\[\begin{aligned} (\nabla_\alpha T)_{\mu\nu} &:=\partial_\alpha T_{\mu\nu} -\Gamma^\rho{}_{\alpha\mu}T_{\rho\nu} -\Gamma^\rho{}_{\alpha\nu}T_{\mu\rho},\\ \nabla^\mu T_{\mu\nu} &:=g^{\mu\alpha}(\nabla_\alpha T)_{\mu\nu}. \end{aligned}\]Here \(\nabla_\alpha\) means \(\nabla_{\partial_\alpha}\). Every factor is evaluated at the same event. The resulting components define a covector field, the covariant divergence of \(T\).
We therefore seek a curvature expression with zero covariant divergence. The combination \(\mathcal G_{\mu\nu}:=R_{\mu\nu}-\tfrac12Rg_{\mu\nu}\), called the Einstein tensor, satisfies the geometric identity \(\nabla^\mu\mathcal G_{\mu\nu}=0\). It can match a locally conserved source.
The additional physical postulate is the Einstein field equation:
\[\boxed{\mathcal G_{\mu\nu} =\frac{8\pi G_{\mathrm N}}{c^4}T_{\mu\nu}.}\]Newton’s constant \(G_{\mathrm N}>0\) has units \(\mathrm{m^3/(kg\,s^2)}\) and is calibrated by gravitational measurements with known source masses and distances. The coefficient matches the Newtonian source law in the weak, slow limit. Both sides then have units of inverse length squared. The geometric identity requires \(\nabla^\mu T_{\mu\nu}=0\), the local energy-momentum conservation equation.
This source equation does not follow from the equivalence principle or the clock experiment. Its predictions require experimental tests. We have neglected a possible cosmological term \(\Lambda g_{\mu\nu}\) on the left, where \(\Lambda\) has units of inverse length squared; its effects are negligible for the laboratory problems here. Solving for a metric also requires appropriate initial or boundary data. Even where \(T_{\mu\nu}=0\), the full curvature can be nonzero, as outside Earth or in a gravitational wave.
12. How are special and general relativity related?
Special relativity supplies the local clock, light, and inertial-motion rules. General relativity uses a metric to relate those measurements across a region and a field equation to predict that metric from sources and initial or boundary data. Where curvature is nonzero, one inertial coordinate system cannot describe the entire region.
The main question is answered within this model: one metric determines ideal-clock readings, and its connection determines the free paths of test objects and light. Curvature describes their tidal separation. Together with the source equation, these rules form general relativity. The rocket calculations explain the leading clock shift and light deflection; the weak, steady, slow-motion limit recovers the measured drop acceleration.