Part III · General Relativity — Chapter 3.1

The Equivalence Principle

Gravity is the one force you can switch off by letting go. What refuses to switch off is the whole of the subject.

Tools you'll need  — Chapter 0.1 §3: the derivative as the coefficient of the best linear approximation. Every appearance of a "tidal" effect below is that sentence applied to a force field. Chapter 0.6 §1 and §6: partial derivatives, the matrix of second partial derivatives (the Hessian), and Taylor's theorem to second order in several variables. Chapter 0.7 §7: the Laplacian $\nabla^{2}$, and the fact that Newtonian gravity obeys $\nabla^{2}\Phi=4\pi G\rho$, which is Poisson's equation, so that $\nabla^{2}\Phi=0$ in empty space. Chapter 1.1 for what is unsatisfactory about forces. Chapter 2.3 §5: proper time as the length of a worldline. Chapter 2.5 §7.2: the longitudinal Doppler shift, of which we need only the first-order form. Chapter 1.2's worked example on geodesics is the picture waiting at the end of this one. We will point at it and not spend it.

1 · Two masses that did not have to be equal

Let's begin with a piece of bookkeeping that Newton himself noticed and could not explain. Write down the two places the word "mass" appears in Newtonian physics, and notice that they are two different words wearing the same name.

The first is in the second law. Push on something and it resists:

The number $m_{I}$ is inertial mass. It is defined by an experiment involving no gravity whatsoever. A spring, a puck on ice, a collision: any of those will do. What it measures is one thing only, which is how reluctant the object is to have its velocity changed.

The second is in the law of gravitation. Put an object near a mass $M$ and it is pulled:

The number $m_{G}$ is gravitational mass. It is defined by an experiment involving no accelerations at all. Hang the object from a balance and read the deflection. What it measures is a completely different thing, which is how strongly the object couples to the gravitational field. It is a charge, the exact analogue of the $q$ in $\vv F=q\vv E$.

There is no reason within Newtonian physics for these two numbers to be related. Electric charge supplies the control case. A proton and an electron have charges equal in magnitude, and inertial masses differing by a factor of $1836$. So charge and inertia are demonstrably independent properties of matter. Nothing in (3.1.1) or (3.1.2) forbids gravitational charge from being independent too.

Now let's combine them. We want the acceleration of a falling body, so substitute (3.1.2) into (3.1.1) and divide both sides by $m_{I}$:

Let's read what that says. The acceleration of a falling body is proportional to its ratio of gravitational to inertial mass. Suppose that ratio varied from substance to substance. Then a lead ball and a wooden ball released together would separate visibly as they fell, and the whole of what follows would be impossible.

Given that result, we may as well choose our unit of gravitational charge so that the ratio is exactly one. From here on there is one mass, written $m$, and it cancels out of (3.1.3) altogether:

Let's pause and look at what happened there. The left-hand side is kinematics, the acceleration of this object. The right-hand side is a field, a property of the point in space and of nothing else. The object has vanished from its own equation of motion.

2 · The path forgets what is falling

That cancellation has a consequence we should draw out at once. Equation (3.1.4) is a second-order differential equation for $\vv r(t)$ with no free parameters describing the body. Chapter 0.8 §1 established what such an equation delivers: given a starting position and a starting velocity, the solution exists and is unique. Therefore:

Nothing about that is true of any other force, and it is worth seeing the contrast immediately. Put a charge in an electromagnetic field and Chapter 2.6 §8 gives

Here the factor $q/m$ stubbornly refuses to cancel. It is $-1.759\times10^{11}\ \mathrm{C\,kg^{-1}}$ for an electron, $9.579\times10^{7}$ for a proton, $4.822\times10^{7}$ for a helium nucleus and exactly zero for a neutron. Release those four particles side by side from rest in the same electric field and they follow four different curves, one of which is a straight line. So there is no such thing as "the trajectory through this field". There is only "the trajectory of this particle through this field."

2.1 · Deleting a uniform field

Here is the first payment universality makes, and it is a large one. Suppose the field is uniform: $\vv g$ is one and the same constant vector everywhere. Define new coordinates by

Notice what that is and what it is not. It is a change of coordinates and nothing more, an instruction for relabelling each event $(t,\vv r)$ with a new spatial address, using a rule that depends on $t$. We want to know how a free body moves in the new labels, so differentiate the definition twice with respect to $t$. The first differentiation gives $\dot{\vv r}^{\,\prime}=\dot{\vv r}-\vv g t$. The second gives

The last step used (3.1.4). In the primed coordinates every freely falling body moves in a straight line at constant speed. The gravitational field is not weak in these coordinates, and it is not approximately absent. It is gone.

Now notice the step that made this work, because it is easy to read past. The substitution (3.1.6) contains no reference to the body at all. No $m$, no $q$, no composition. So that single relabelling removes the field from the equation of motion of every body at once.

Try the same trick on (3.1.5) with $\vv B=0$ and constant $\vv E$. The shift you would need is $\tfrac12 (q/m)\vv E t^{2}$, which is a different relabelling for each species of particle. A coordinate system is not permitted to depend on what you are looking at, so the trick fails. It fails for exactly the reason §1 identified.

Run the argument backwards and it says something equally strong. Start in an inertial frame with no gravity, and change to coordinates $\vv r^{\,\prime\prime}=\vv r+\tfrac12\vv a_{0}t^{2}$. Those are the coordinates an observer accelerating uniformly at $-\vv a_{0}$ would naturally use. Differentiating twice, free particles now obey $\ddot{\vv r}^{\,\prime\prime}=\vv a_{0}$. A uniform gravitational field has appeared out of nothing.

So a uniform field and a uniformly accelerated frame are not merely similar. As far as (3.1.4) is concerned, they are the same statement written in two coordinate systems.

3 · Einstein's enlargement

Everything in §2 concerned trajectories of test bodies. Einstein's move was to assert that the same holds for every experiment, not merely mechanical ones. That is a large step, and it is worth displaying with its clauses separated. Each clause is testable on its own, and the third is where most of the content sits.

Three things must be said about this immediately.

It is a hypothesis, not a theorem. Clause (i) is the experimental result of §1. Clauses (ii) and (iii) are extrapolations from mechanics to optics, electromagnetism, nuclear physics and everything else. They were not derived from anything. They are, however, sharply testable, and §6's redshift is a test of clause (iii). A frequency standard that depended on where it sat in a gravitational field would violate that clause. ⚑ Experiments comparing different atomic clocks at different gravitational potentials constrain such a dependence at the level of a few parts in $10^{6}$ of the predicted effect. Local position invariance therefore holds to that accuracy.

The word "local" carries the entire burden. Both qualifications in the statement are doing real work, the small region and the short time alike, and neither can be dropped. Section 5 turns them into an inequality with numbers in it.

There is a stronger version, and the difference matters. The EEP as stated exempts experiments that themselves involve gravity. Drop that exemption, and insist that even the gravitational binding energy of the apparatus falls at the same rate as everything else. What you then have is the strong equivalence principle.

That is not a formality. A body's gravitational self-energy is a genuine contribution to its mass, so a theory in which self-energy gravitates differently from ordinary mass would show the Earth and the Moon falling toward the Sun at slightly different rates. ⚑ Lunar laser ranging bounds that difference at about one part in $10^{4}$ of the self-energy contribution. General relativity satisfies the strong principle. Most of its competitors do not, which is why the measurement is made.

4 · What free fall cannot remove

This section is the physical content of the chapter. Everything before it says what gravity is not. This section says what is left over when you have taken away everything a change of coordinates can take away. Here is the destination, announced first:

4.1 · The tidal equation, in five lines

The plan is to write down the equation of motion for each of two neighbouring particles and then subtract, so that what is common to both drops out. Let $\vv r_{0}(t)$ be the position of a reference particle in free fall, and let a second free particle sit at $\vv r_{0}(t)+\vv d(t)$, where $\vv d$ is a small separation. Both obey (3.1.4).

Line 1. Start with the reference particle, whose only motion is its own free fall:

Line 2. Now the companion. Its acceleration is the field evaluated at its own location, which is displaced from the reference particle by $\vv d$:

Line 3. We want the motion of one particle relative to the other, so subtract (3.1.8) from (3.1.9) and keep only the separation. Nothing has been approximated yet:

Line 4. The right-hand side is a difference of two nearby values of the same function, so the thing to do is linearise. That is Chapter 0.1 §3 applied in several variables. Writing the $i$-th component and using the multivariable linear approximation of Chapter 0.6 §2,

summed over $j=1,2,3$. The step is legitimate precisely because $\vv d$ is small. Everything below is a statement about nearby particles, and nothing more than that is being claimed.

Line 5. Substitute that expansion into (3.1.10). The zeroth-order terms cancel, and that cancellation is the deletion of the field performed in §2.1. What survives is

Let's read (3.1.12) carefully, because it is the sentence Part III is built on. The field $\vv g$ has disappeared from it. What governs the relative motion of two neighbouring freely falling particles is not the field but the derivative of the field.

Choose coordinates as cleverly as you like and that stays true. The field at a point can always be made zero, since that is three numbers and the shift (3.1.6) has three parameters to spend on them. Its derivative is a different object altogether, and no shift touches it.

4.2 · In terms of the potential

Potentials are easier to differentiate twice than fields are, so let's rewrite the tidal equation in terms of one. Chapter 0.7 §2 established that the Newtonian field is a gradient, $\vv g=-\nabla\Phi$, so $g^{i}=-\partial_{i}\Phi$ and

That array is the Hessian of Chapter 0.6 §6, the matrix of second partial derivatives, and Clairaut's theorem (0.6 §6.1) makes it symmetric. Substituting it, (3.1.12) reads

This is a linear equation whose coefficient matrix is symmetric. By the spectral theorem of Chapter 0.5 §6, it therefore has three real eigenvalues and three mutually perpendicular eigendirections. Along each eigendirection the separation obeys $\ddot d = -\lambda d$, giving exponential growth where $\lambda<0$ and oscillation where $\lambda>0$.

That is worth saying on its own line. Gravity, seen from inside a falling laboratory, is a symmetric $3\times3$ matrix.

4.3 · Computing it for a point mass

An abstract matrix is easier to trust once you have seen a concrete one. So take the potential of a point mass, $\Phi=-GM/r$ with $r=\abs{\vv r}$, and differentiate it twice, one step at a time.

First derivative of $r$. We will need this repeatedly, so get it out of the way. Since $r^{2}=x^{k}x^{k}$, differentiating both sides gives $2r\,\partial_{i}r = 2x^{i}$, so

Those are the components of the unit vector pointing outward from the mass.

First derivative of $\Phi$. With $\partial_{i}r$ in hand, the chain rule applied to $-GM\,r^{-1}$ gives

(Check: $\vv g=-\nabla\Phi$ then points inward with magnitude $GM/r^{2}$, as it must.)

Second derivative. This is the one we actually want. Differentiate (3.1.16) with respect to $x^{j}$, using the product rule on $x^{i}\cdot r^{-3}$. The first factor gives $\partial_{j}x^{i}=\delta_{ij}$. The second gives $-3r^{-4}\partial_{j}r=-3x^{j}/r^{5}$ by (3.1.15). Hence

Before reading anything off, let's give a name to the overall factor, since it is the single number that will appear everywhere below. Note that it has the dimensions of an inverse time squared:

Putting (3.1.17) and that definition into the tidal equation (3.1.14), and expanding the bracket so the two pieces stand apart, we get

Now read off the two cases. Everything hangs on the scalar $n^{j}d^{\,j}=\hat{\vv n}\cdot\vv d$, the component of the separation along the line to the mass.

Radial separation. Put $\vv d = d\,\hat{\vv n}$, one particle directly above the other. Then $\hat{\vv n}\cdot\vv d=d$ and $n^{i}(n^{j}d^{\,j}) = d\,n^{i} = d^{\,i}$, so the two terms of (3.1.19) combine as $-Kd^{i}+3Kd^{i}$:

The sign is positive, so the acceleration points along the separation, away from the reference particle. Radially, the two particles pull apart. The physical reason is plain enough in (3.1.4), since the lower particle is closer to the mass and falls harder. But notice that we did not have to say so. It fell out of the second derivative on its own.

Transverse separation. Put $\vv d\perp\hat{\vv n}$, the two particles side by side at the same height. Then $\hat{\vv n}\cdot\vv d=0$, the second term of (3.1.19) vanishes outright, and

The sign is negative, so the acceleration points back along the separation, toward the reference particle. Transversely, the two particles come together. Again this is as it must be, since both are falling toward the same centre along converging lines.

Grind box — the tidal matrix in general position, and the size of the next term

1 · The two cases of §4.3 were not special. We evaluated (3.1.19) for separations exactly along and exactly across $\hat{\vv n}$. Show that nothing else happens by diagonalising the matrix $M_{ij}\equiv\delta_{ij}-3n^{i}n^{j}$ directly, for an arbitrary unit vector $\hat{\vv n}$.

Along $\hat{\vv n}$. Apply $M$ to $\hat{\vv n}$ itself, using $n^{j}n^{j}=1$:

$$M_{ij}n^{j} \;=\; n^{i} - 3n^{i}\big(n^{j}n^{j}\big) \;=\; n^{i}-3n^{i} \;=\; -2\,n^{i}.$$

So $\hat{\vv n}$ is an eigenvector with eigenvalue $-2$.

Across $\hat{\vv n}$. Take any $\vv w$ with $n^{j}w^{j}=0$. Then the second term dies outright:

$$M_{ij}w^{j} \;=\; w^{i} - 3n^{i}\big(n^{j}w^{j}\big) \;=\; w^{i}.$$

So every vector in the plane perpendicular to $\hat{\vv n}$ is an eigenvector with eigenvalue $+1$. That plane is two-dimensional, and together with $\hat{\vv n}$ it spans all of $\R^{3}$. So we have a complete eigenbasis, and the spectrum is $(-2,+1,+1)$ whatever direction the mass lies in. Multiplying by the overall $K$ and putting in the minus sign of (3.1.14), the relative acceleration has coefficients $(+2K,-K,-K)$. That is one stretch and two squeezes, always.

Two checks fall out for free. The trace is $-2+1+1=0$, which is (3.1.22). The determinant is $(-2)(1)(1)=-2$, so $\det\Phi_{,ij}=-2K^{3}$, which is not zero. The tidal matrix is therefore invertible, and in particular there is no direction along which two free particles stay put. (Eigenvalues, trace and determinant confirmed numerically for randomly chosen $\hat{\vv n}$.)

2 · What (3.1.24) threw away. The closed-form ellipse treated $K$ as constant, and it is not. The reference particle falls, $r$ shrinks and $K$ grows. Our goal here is the next term. Starting from rest, the reference particle's own free fall gives $r_{0}(t)=r-\tfrac12(GM/r^{2})t^{2}+O(t^{4})$, so

$$K(t) \;=\; \frac{GM}{r_{0}^{3}} \;=\; K\left(1-\frac{K t^{2}}{2}\right)^{-3} \;=\; K\left(1+\tfrac32 K t^{2}+O(K^{2}t^{4})\right),$$

by the binomial series of Chapter 0.3 §2. We now want the quartic coefficient, so put $\xi_{z}/\rho = 1 + Kt^{2} + a_{4}K^{2}t^{4}$ into $\ddot\xi_{z}=2K(t)\xi_{z}$ and match the coefficient of $t^{2}$ on both sides. The left side gives $12a_{4}K^{2}$. The right side gives $2K\big(\tfrac32 K + K\big)=5K^{2}$, so $a_{4}=\tfrac{5}{12}$. The same match on $\ddot\xi_{x}=-K(t)\xi_{x}$ gives $-\tfrac1{12}$. Hence

$$\frac{\xi_{z}}{\rho} = 1 + Kt^{2} + \tfrac{5}{12}\big(Kt^{2}\big)^{2}+\ldots, \qquad \frac{\xi_{x}}{\rho} = 1 - \tfrac12 Kt^{2} - \tfrac{1}{12}\big(Kt^{2}\big)^{2}+\ldots$$

This is exactly the quantity the figure below calls error 2, and it explains the numbers there. At $Kt^{2}=0.05$ the estimate $\tfrac{5}{12}(Kt^{2})^{2}=0.104\%$ stands against a measured $0.107\%$. At $Kt^{2}=0.25$ it is $2.60\%$ against $3.06\%$. At $Kt^{2}=0.55$ it is $12.6\%$ against $19.0\%$, where the series has plainly stopped being a good guide and the numerical integration has to be believed instead.

Note what the correction does not depend on, which is the ring's size. It is an error in the expansion in time, not in the expansion in separation. The figure separates the two so that neither can be mistaken for the other.

4.4 · The trace, and what it is secretly telling you

One number extracted from the tidal matrix carries more than its share of meaning, and it is the simplest one available. Add up the diagonal entries of (3.1.17). Contract $i$ with $j$, using $\delta_{ii}=3$ in three dimensions and $n^{i}n^{i}=1$ because $\hat{\vv n}$ is a unit vector:

Zero. Now look at what that zero is. The quantity $\Phi_{,ii}$ is $\nabla^{2}\Phi$, the Laplacian of Chapter 0.7 §7, and Poisson's equation for Newtonian gravity says $\nabla^{2}\Phi=4\pi G\rho$. So (3.1.22) is not an accident of the point-mass potential. It is the statement that we are computing in empty space, where $\rho=0$.

The tidal matrix of any vacuum region is traceless. Stretch along one direction must be paid for by squeeze along the others, in exactly compensating amounts. The eigenvalues $(-2K,+K,+K)$ of (3.1.17) sum to zero for that reason and no other.

Hold onto this. Chapter 3.6 will produce field equations whose vacuum form, $R_{\mu\nu}=0$, is the covariant descendant of (3.1.22), and whose sourced form is the descendant of Poisson's equation.

4.5 · The ring, and the honest arithmetic about its area

Make the shape visible. Release a ring of $N$ particles, all at rest relative to a central reference particle, arranged in the plane containing the radial direction. Put $\hat{\vv z}$ along $\hat{\vv n}$ (radially outward) and $\hat{\vv x}$ transverse, so the ring is

The two components decouple, because $\hat{\vv x}$ and $\hat{\vv z}$ are eigendirections of the tidal matrix. So we can integrate (3.1.20) and (3.1.21) twice from rest, one line each, provided the coefficient $K$ may be treated as constant over the interval. In general it may not, because the reference particle is itself falling and $r$ is shrinking. Keeping only the leading behaviour, which means working to first order in the dimensionless combination $Kt^{2}$,

Those are the parametric equations of an ellipse, stretched along the field line and squeezed across it. That shape is what a tidal field is, and Chapter 3.4 will recover this very ellipse from the curvature tensor.

Now let's say plainly what was assumed, because the figure below measures it. Equation (3.1.24) carries two approximations, and they are logically independent of each other.

• One in the separation, since the tidal equation itself is only first order in $\vv d$.
• One in the time, since only the first two terms of the expansion in $Kt^{2}$ have been kept.

The figure reports them separately so that neither can hide behind the other, and the grind box above computes the leading size of the second.

It is often said that the ring deforms at constant area. Do the arithmetic before believing it. The semi-axes are $a=\rho(1+Kt^{2})$ and $b=\rho(1-\tfrac12 Kt^{2})$, so the enclosed area is

That grows, at the same order in $t$ as the deformation itself. So the slogan is wrong in two dimensions. In three it survives, and the reason is worth seeing. Release a small ball of particles rather than a ring. One direction stretches and both transverse directions squeeze, so the volume ratio is

The $t^{2}$ term has cancelled identically. A small ball of freely falling particles keeps its volume to second order in time while changing its shape.

And the reason it does is (3.1.22). The rate of fractional volume change is the sum of the three fractional stretch rates, which is the trace, which vanishes in vacuum. The volume statement and Laplace's equation are the same fact.

elapsed Kt²

K t² = 0.250

ring size ρ/r

ρ/r = 1.0e-3

hide tidal arrows

closed-form semi-axes: radial 1.2500 ρ transverse 0.8750 ρ

area 1.093750 × initial volume of the ball 0.957031 × initial

error 1 · exact − tidal equation: 0.0476% of ρ

error 2 · tidal equation − closed form: 3.06% of ρ

that is t = 402.7 s at the Earth’s surface, or 0.434 s at 10 000 km from a 10 M☉ black hole

4.6 · The thesis of Part III, stated early

Collect what §2 and §4 have between them established.

• The gravitational field at a point can always be set to zero by a change of coordinates, and can be set to zero for all bodies at once. It is therefore not an invariant feature of the situation. It is a fact about the coordinates, in exactly the sense Chapter 2.4 taught us to be suspicious of.
• The derivative of the field cannot be set to zero. It shows up as the relative acceleration of two freely falling bodies, which is measured by comparing two objects with each other and never by referring to a frame.

So if gravity is going to be described by something coordinate-independent, that something must be built out of second derivatives of the potential, and it must vanish when and only when the tidal effects vanish.

Chapter 3.4 constructs exactly such an object, a tensor with four indices built from second derivatives of the metric. Chapter 3.4 §4 will then derive an equation of the form $\ddot\xi \sim -R\,\xi$, which is (3.1.14) with $-\Phi_{,ij}$ replaced by that tensor. Set the two side by side and the identification is unavoidable: tidal acceleration is curvature. That is the thesis of Part III, and it was decided here.

5 · How big may a falling laboratory be?

The equivalence principle was stated with two qualifications, "sufficiently small" and "sufficiently short". Section 4 supplies everything we need to replace those two words with an inequality. Here is the destination: we want a bound on the size and duration of a region in which the tidal drift stays below the precision of whatever instrument is in the room.

Take the worst case, which is radial separation. From (3.1.20), two particles a distance $\ell$ apart, released at relative rest, have relative acceleration $2K\ell$. Because $\ell$ barely changes over the interval, the acceleration is very nearly constant, and integrating twice from rest gives a relative displacement after time $T$ of

The frame is good enough as long as that drift stays below what the instrument in the room can resolve. Call that resolution $\varepsilon$, which is a length, and demand that the drift stay under it:

Three things are worth extracting from (3.1.28).

The constraint is on a product, not on a length. A local inertial frame is a region of spacetime. You may have a large laboratory for a short time or a small one for a long time, and the trade is $\ell\propto T^{-2}$. This is why general relativity is a differential theory. Special relativity is exact in the infinitesimal and only there.

If your precision is fractional rather than absolute, the size drops out. Dividing (3.1.27) by $\ell$, the fractional drift is $\delta/\ell=KT^{2}$, independent of how big the laboratory is. Requiring $\delta/\ell<\varepsilon_{\text{frac}}$ then gives $T<\sqrt{\varepsilon_{\text{frac}}/K}$, which is a pure statement about duration.

$K$ is the local answer to "how curved is it here". Far from everything, $K\to0$ and the frame may be enormous. Close to a compact mass, $K$ is large and the frame is tiny. And $K$ is a second derivative of the potential, which is §4's object all over again.

5.1 · Numbers

At the Earth's surface, with $M=5.972\times10^{24}\ \mathrm{kg}$ and $r=R_{\oplus}=6.371\times10^{6}\ \mathrm{m}$,

where the middle expression follows because $g=GM/R_{\oplus}^{2}$. Feeding this into (3.1.28):

Laboratory	Resolution $\varepsilon$	Size $\ell$	Time available
Optical bench	$1\ \mu\mathrm{m}$	$1\ \mathrm{m}$	$0.81\ \mathrm{s}$
Optical bench	$1\ \mu\mathrm{m}$	$1\ \mathrm{cm}$	$8.1\ \mathrm{s}$
Atom interferometer	$1\ \mathrm{nm}$	$1\ \mathrm{cm}$	$0.25\ \mathrm{s}$
Gravitational-wave detector	$10^{-18}\ \mathrm{m}$	$4\ \mathrm{km}$	$1.3\times10^{-8}\ \mathrm{s}$

The last row is the interesting one. A gravitational wave is a travelling tidal field, so a detector built specifically to measure tidal effects has a local inertial frame lasting about thirteen nanoseconds. That is not a defect of the instrument. It is the statement that the instrument works.

5.2 · A debt from Chapter 1.1, collected

Chapter 1.1 §1 identified a circularity at the foundation of Newtonian mechanics that no amount of care could remove. An inertial frame was defined as one in which force-free bodies move uniformly. A force-free body was one that moves uniformly in an inertial frame. Newton's escape was to postulate absolute space, an entity he conceded was unobservable. That chapter promised the circle would be broken here, by changing the theory rather than by being more careful.

Here is the break. Freely falling bodies define the inertial frames, and free fall is identifiable by an experiment performed inside, with no reference to anything external. Release two objects and see whether they stay put relative to each other.

Gravity leaves the list of forces entirely. A body under gravity alone is not being pushed. It is the standard of not being pushed. So "force-free" acquires an independent meaning, and the definition stops chasing its own tail.

The price is (3.1.28). The test "do the two objects stay put" succeeds only within the region that inequality allows, so inertial frames now exist locally and not globally.

Whether the local frames can be knitted together into one global frame is a separate question with a definite answer. The obstruction to knitting them is exactly what Chapter 3.4 will call curvature, and Newton's absolute space is the assumption that the knitting always succeeds.

6 · Light climbing out of a well

Now three consequences, each obtained with nothing but the equivalence principle and special relativity. No curvature, no metric, no field equations. Here is the destination for this one: a light signal sent upward through a height $h$ in a gravitational field $g$ arrives with its frequency lowered by the fraction $gh/c^{2}$, and identical clocks at the two ends therefore run at different rates.

6.1 · The cabin

Put a sealed cabin in deep space, far from any mass, and accelerate it with constant proper acceleration $g$ along its own axis, which we call upward. A source is bolted to the floor and a detector to the ceiling, a height $h$ above it. This is a problem in special relativity alone. There is no gravity anywhere in it.

Work in the inertial frame $S$ in which the cabin is momentarily at rest at the instant of emission, and call that instant $t=0$. The source emits light of frequency $\nu_{\text{em}}$ as measured by the source's own clock, which at $t=0$ is a clock at rest in $S$.

The flight time. In $S$ the light travels at $c$ and must cover a distance of $h$ plus whatever the ceiling has moved. So

We are going to work to first order in the small dimensionless quantity $gh/c^{2}$, and the correction in (3.1.30) is itself of that order. Since it multiplies a quantity that already carries one factor of $g$, it contributes at second order and may be dropped. Say this out loud, so it is not mistaken for carelessness: we use $t_{\text{arr}}=h/c$ and the error is $O\!\big((gh/c^{2})^{2}\big)$.

The detector's speed on arrival. Starting from rest in $S$ and accelerating at $g$, the ceiling's velocity when the light reaches it is

(Proper and coordinate acceleration agree at first order in $v/c$, which is the order we are keeping. Chapter 2.5 §6 recorded the difference.)

Doppler. The detector is receding from the emission event at speed $v$, in the same direction the light is travelling. Chapter 2.5 §7.2 derived the longitudinal Doppler formula, so take it from there and expand it to first order in $v/c$:

We know the speed already, so put (3.1.31) into that expansion and write the answer as a fractional shift in frequency:

Light climbing inside an accelerating cabin arrives reddened. That is a result in special relativity and it required nothing else.

6.2 · Handing it to gravity

Now invoke the equivalence principle in the direction established at the end of §2.1: a cabin accelerating at $g$ in empty space and a cabin held stationary in a uniform gravitational field $g$ are the same situation described in two coordinate systems. Every experiment performed inside must give the same answer. Hence, with no further calculation,

6.3 · Why this is a statement about clocks

Equation (3.1.34) is usually read as "light loses energy climbing out". Resist that reading for a moment, because there is a sharper one available and it costs three lines.

Suppose the source emits $N$ wave crests over an interval that its own clock records as $\tau_{\text{em}}$, so the emitted frequency is $\nu_{\text{em}}=N/\tau_{\text{em}}$. Those crests travel up.

Now use the fact that the situation is static. The field does not change with time, the source and the detector do not move, and the region between them is unchanging. Therefore the number of crests in flight at any moment is constant, and no crest is created or destroyed on the way.

So the detector receives exactly $N$ crests, over an interval its own clock records as $\tau_{\text{rec}}$, giving $\nu_{\text{rec}}=N/\tau_{\text{rec}}$. Divide the one frequency by the other and the count $N$ drops out:

Let's look at what that line is actually saying. The same $N$ crests took longer to arrive than to leave, as measured by the two local clocks. Neither clock is moving. Neither is defective, since by the equivalence principle both are ideal. The only consistent reading left is that the lower clock runs slow relative to the upper one, by the fraction $gh/c^{2}$:

6.4 · Non-uniform fields, by chaining cabins

Equation (3.1.34) assumed uniformity, which is only good over a height satisfying §5's bound. Remove the assumption by applying the local result over an infinitesimal rise $\dd h$ and accumulating, which is Chapter 0.2's business. Over $\dd h$ the fractional change in frequency is

Let's fix the sign convention once, out loud, so it does not have to be checked again. Let $h$ measure height upward and let $\Phi$ be the gravitational potential, which by (3.1.16) increases with height. The downward field strength is then $g=+\dd\Phi/\dd h$, so $g\,\dd h=\dd\Phi$. The frequency therefore falls as the potential rises, which is the right direction.

With the sign settled, substitute $\dd\Phi$ into (3.1.37) and integrate all the way from emitter to receiver:

to first order in $\Delta\Phi/c^{2}$. The uniform result is recovered when $\Delta\Phi=gh$. This form is the one used in practice, because real potentials are not uniform.

6.5 · Numbers, and one very large consequence

Situation	$\Delta\Phi/c^{2}$	In seconds per year
Top vs bottom of a $1\ \mathrm{m}$ table	$1.09\times10^{-16}$	$3.4\ \mathrm{ns}$
$33\ \mathrm{cm}$, the Chou et al. optical-clock experiment ⚑	$3.60\times10^{-17}$	$1.1\ \mathrm{ns}$
$22.5\ \mathrm{m}$ Harvard tower, Pound and Rebka ⚑	$2.46\times10^{-15}$	$77\ \mathrm{ns}$
GPS orbit vs the ground ($26\,560\ \mathrm{km}$ vs $6371\ \mathrm{km}$)	$5.29\times10^{-10}$	$16.7\ \mathrm{ms}$

⚑ Pound and Rebka measured the $22.5$-metre shift in 1960 using the Mössbauer effect and confirmed it. Later refinements of the same experiment agree with the prediction to about $1\%$. ⚑ In 2010 a comparison of two aluminium-ion optical clocks resolved the shift over a height difference of $33\ \mathrm{cm}$. Both are quoted as experimental results.

Now the consequence, and it is a large one. Two clocks, at rest with respect to each other, never moving, separated only in height, tick at different rates. In special relativity every rate difference came from relative motion. Here there is no relative motion at all.

So whatever describes spacetime in the presence of a mass, it cannot be $\eta_{\mu\nu}$ everywhere, because $\eta_{\mu\nu}$ assigns the same relation between coordinate time and proper time at every point.

We can go one step further and read off a component of the eventual answer in advance. Chapter 2.3 §5 wrote proper time along a worldline at rest as $\dd\tau^{2}=\dd s^{2}/c^{2}$. Chapter 3.3 will replace $\eta_{\mu\nu}$ by a position-dependent $g_{\mu\nu}(x)$ in that expression, and for a static observer only the $00$ component survives, giving $\dd\tau=\sqrt{g_{00}}\;\dd t$.

Now compare that with (3.1.38), which says the rate ratio is $1+\Phi/c^{2}$ up to a common constant. Matching the two expressions for the rate gives

Hold two things about (3.1.39) in mind. First, it will be derived from the field equations in Chapter 3.6 rather than guessed, and the agreement will be a check. Second, and much more importantly for the next section: this argument has said absolutely nothing about the spatial components of the geometry. Clocks probe $g_{00}$ and nothing else. Keep that grievance in your pocket, because §7 is where it bites.

7 · Light bending, and a debt of exactly one half

The same cabin, turned through a right angle. Here is the destination in advance, including the bad news: we will obtain a deflection of $2GM/bc^{2}$ for a ray passing a mass $M$ at distance $b$, which for a ray grazing the Sun is $0.87$ arcseconds, and the measured value is $1.75$ arcseconds. Our answer will be exactly half. We will say so and identify what is missing.

7.1 · A ray crossing the cabin

Cabin again, accelerating at $g$ upward, but now the light enters horizontally through one wall and crosses a width $L$ to the opposite wall. In the momentarily comoving inertial frame the light travels in a perfectly straight horizontal line, because that frame is inertial and the light is free. It takes time $t=L/c$ to cross, again to first order.

During that time the cabin rises by $\tfrac12 g t^{2}$. So relative to the cabin the light has descended by

More useful than the drop is the change in direction. In the inertial frame the light has zero vertical velocity, while the cabin has acquired vertical velocity $+gt$. So relative to the cabin the light has vertical velocity $-gt$ while retaining horizontal velocity $c$, and the angle by which its path has turned is

Divide by the path length $L$ to get the result in the form we can integrate. Writing $g_{\perp}$ for the component of the gravitational field perpendicular to the ray, and $\ell$ for arc length along it, the equivalence principle converts (3.1.41) into

A ray bends toward a mass at a rate equal to the transverse field divided by $c^{2}$.

7.2 · Integrating past a mass

Send a ray past a point mass $M$ with impact parameter $b$, meaning the closest approach of the undeflected line. Set up coordinates with the ray along the $x$ axis and the mass at $(0,-b)$, so that a point of the ray at abscissa $x$ is at distance $\sqrt{x^{2}+b^{2}}$ from the mass.

The field magnitude there is $GM/(x^{2}+b^{2})$, and its transverse (here, downward) component is obtained by multiplying by the direction cosine $b/\sqrt{x^{2}+b^{2}}$:

Let's name the approximation before making it. The total deflection comes out at order $10^{-6}$ radians, so we may integrate (3.1.42) along the undeflected straight line and take $\dd\ell=\dd x$.

This is first-order perturbation theory. You evaluate a small correction along the uncorrected path, so that the piece you neglect is the correction to the correction. The error here is second order in $GM/bc^{2}$, which for the Sun is one part in $10^{6}$ of an already tiny angle.

With that settled, the total deflection $\alpha$ is the bending rate accumulated along the whole ray, from far before the mass to far after it:

Do the integral by the trigonometric substitution $x=b\tan u$, so that $\dd x=b\sec^{2}u\,\dd u$ and $x^{2}+b^{2}=b^{2}\sec^{2}u$, whence $(x^{2}+b^{2})^{3/2}=b^{3}\sec^{3}u$. The limits $x=\mp\infty$ become $u=\mp\pi/2$:

The integral came out as $2/b$. Restoring the constants that were carried outside it gives the total deflection:

7.3 · The number, and the confession

For a ray grazing the Sun, $b=R_{\odot}=6.957\times10^{8}\ \mathrm{m}$ and $GM_{\odot}=1.327\times10^{20}\ \mathrm{m^{3}\,s^{-2}}$:

The reason is visible if you look at what the cabin argument used. Every step of §7.1 was about time. How long the light takes to cross. How much speed the cabin picks up in that time. The whole derivation is a consequence of the fact established in §6, that the rate of a clock depends on where it sits. That is a fact about $g_{00}$ alone, exactly as (3.1.39) warned. Nothing in the cabin argument constrains how spatial distances are measured near a mass, because a cabin is a rigid box and we assumed its width was $L$ without asking any further.

In the full theory the spatial part of the geometry is also modified, by an amount of the same order. A ray of light covers equal amounts of space and time, since it moves at $c$, so it picks up an equal contribution from that modification. Two equal contributions, hence the factor two.

A slow-moving planet, by contrast, covers a great deal of time and very little space per unit of proper time, so it barely notices the spatial part. That is exactly why the Newtonian limit is recovered from $g_{00}$ alone (Chapter 3.6 §5), and why Newtonian gravity works so well for everything except light.

8 · Why gravity, and only gravity, can be geometry

Everything is now in place for the argument the chapter exists to make. It has three steps.

Step 1. The paths belong to the arena. Section 2 established that through each event, and for each initial velocity, there passes exactly one free-fall trajectory, and that trajectory is the same for every body. So the assignment

is a well-defined rule that mentions nothing about matter. It is therefore permissible to regard it as a property of spacetime rather than of the objects moving through it. Permissible, note, and not yet compulsory.

Step 2. That is exactly what a geometry supplies. Chapter 1.2's second worked example took a rule for the length of a curve, extremised it, and produced from that rule alone a family of "straightest available" curves: straight lines in the plane, great circles on a sphere. The input was the metric and nothing else, with no reference to what was travelling. The output was precisely an assignment of the form (3.1.48), one curve through each point in each direction. So a geometry generates the right kind of object, and it does so for free.

Chapter 2.2 §7 already named which extremum it will have to be, and the answer is the strange one. Because of the minus sign in the interval, the straight worldline between two events carries more proper time than any bent one. So the curve a free body follows is the one of maximal elapsed proper time rather than minimal length.

That chapter's closing remark is the statement we are heading for: that a satellite's orbit is the path maximising the proper time of the clock aboard it. Chapter 3.3 §8 turns it into an equation by feeding $\int\dd\tau$ into the Euler–Lagrange machinery of Chapter 1.2, with $\dd\tau$ now computed from a position-dependent metric. Not a line of that machinery changes. Only $\dd\tau$ does.

Step 3. The geometry must be on spacetime, not on space. The curve in (3.1.48) depends on the initial velocity. A ball thrown gently and a bullet fired hard from the same window follow wildly different arcs through space. So the data selecting a curve is a point plus a direction in four dimensions, not in three, and the arena carrying the geometry has to be spacetime.

This is where §6 rejoins the argument. What the redshift showed is that the rule for the length of a worldline, the proper time of Chapter 2.3 §5, already varies from place to place. A position-dependent rule for the length of a worldline is exactly what Chapter 3.3 will call a metric field.

8.1 · Why the same treatment fails for every other force

The contrast is quantitative, and it is worth doing with numbers rather than in words. Consider four particles released from rest at the same point in the same uniform electric field $E$. By (3.1.5) their accelerations are in the ratio of their charge-to-mass ratios:

Particle	$q/m$ (C kg⁻¹)	Acceleration relative to the proton
Electron	$-1.7588\times10^{11}$	$-1836$
Proton	$+9.5788\times10^{7}$	$+1$
Helium nucleus	$+4.8225\times10^{7}$	$+0.5035$
Neutron	$0$	$0$

Four particles, four different curves through the same point with the same initial velocity, one of them a straight line. No assignment of the form (3.1.48) exists here, because the map is not single-valued. You must know what is moving before you can say where it goes. There is nothing for a geometry to be.

Electromagnetism gets a different treatment. It becomes a connection on an internal space rather than on spacetime itself, which is Part VI, and the reason the treatments differ is the table above.

The same test disqualifies the strong and weak nuclear interactions for the same reason and disqualifies air resistance, which depends on cross-sectional area. Gravity is alone in the class, and it is alone in it because of §1, an experimental fact with error bars.

9 · Worked examples

10 · Your turn