Worked examples — General relativity — equivalence principle, curved spacetime (overview)

Before any symbol appears, here is the toolbox — every quantity we will use, in plain words:

The two master results from the parent, restated so we can point at each piece:

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The scenario matrix

Every problem in this topic is one (or a combination) of these cells:

Cell	Case class	What changes	Example
A	$h>0$ (clock higher up)	GR: higher clock faster, light redshifts going up	Ex 1
B	$h<0$ (clock lower down)	GR: lower clock slower, light blueshifts going down	Ex 2
C	$h=0$ (same height)	Degenerate: no GR shift	Ex 3
D	Pure velocity, no height	SR only: moving clock slower	Ex 3
E	Both effects competing (GPS)	GR $+$, SR $-$, which wins?	Ex 4
F	Strong-field / exact potential	Use $U=-GM/r$, not $gh$	Ex 5
G	Limiting behaviour ($h\to 0$, $h\to \infty$, weak field $\to$ SR)	Sanity checks on the formula's edges	Ex 6
H	Real-world word problem	Mountain-vs-valley aging	Ex 7
I	Exam twist — light bending, not clocks	Free-fall / deflection reasoning	Ex 8

The eight examples below hit every cell.

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Example 1 — Cell A: a clock carried UP a tower ($h>0$)

Forecast: Guess now — nanoseconds, microseconds, or seconds per day? And does the light get redder or bluer going up?

Steps

1. Fractional rate difference. $\frac{\Delta \tau }{\tau }=+\frac{gh}{c^2}=\frac{9.8\times 100}{(3\times 10^8{)}^2}$. Why this step? $h>0$ (roof is higher) puts us squarely in Cell A, so we use the GR engine with a plus sign — higher means faster.

2. Compute: $\frac{980}{9\times 10^{16}}=1.089\times 10^{-14}$ (pure number). Why this step? This ratio tells us "for every second, how many extra seconds does the roof gain." We keep it dimensionless before multiplying by real time.

3. Multiply by a day: $\Delta \tau =1.089\times 10^{-14}\times 86400=9.4\times 10^{-10}\text{s}\approx 0.94\text{ns/day}$. Why this step? A rate only becomes a duration when multiplied by elapsed time $\tau$.

4. Light direction: light climbing UP loses energy $\Rightarrow f$ drops $\Rightarrow$ redshift ($\Delta f/f=-gh/c^2<0$). Why this step? The photon is "climbing out of the well," paying energy on the way up — exactly the sign of the $\Delta f$ formula.

Verify: Units: $\frac{({\text{m/s}}^2)(\text{m})}{(\text{m/s}{)}^2}=\frac{{\text{m}}^2/{\text{s}}^2}{{\text{m}}^2/{\text{s}}^2}=$ dimensionless ✓. Roughly a nanosecond a day for 100 m — the same order Pound–Rebka measured over 22.5 m scaled up. ✓

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Example 2 — Cell B: a clock lowered DOWN a mine ($h<0$)

Forecast: Does the deep clock run faster or slower than the surface clock?

Steps

1. Plug the negative height: $\frac{\Delta \tau }{\tau }=+\frac{gh}{c^2}=\frac{9.8\times (-300)}{9\times 10^{16}}=-3.27\times 10^{-14}$. Why this step? The formula already carries the sign of $h$; a lower clock ($h<0$) automatically comes out negative = runs slower. This is Cell B and it needs no new physics, just honest bookkeeping of the minus sign.

2. Interpret: negative $\Delta \tau$ ⇒ the mine clock runs slower than the surface clock. Why this step? "Lower in the potential well = slower" is the whole content of gravitational time dilation; the negative $h$ is how the formula encodes "lower."

3. Light sent down: $\Delta f/f=-gh/c^2=-g(-300)/c^2=+$ positive ⇒ $f$ increases ⇒ blueshift. Why this step? A photon falling gains energy (rolling downhill), so its frequency rises — the mirror image of Example 1.

Verify: Magnitude $3.27\times 10^{-14}$ is exactly $3\times$ Example 1's $1.089\times 10^{-14}$ because $|h|$ is $3\times$ larger ($300$ vs $100$) — the formula is linear in $h$. ✓ Sign flipped correctly: up→redshift, down→blueshift. ✓

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Example 3 — Cells C & D: same height ($h=0$) but one clock MOVING

Forecast: With $h=0$, is there any time dilation at all?

Steps

1. (a) GR part: $h=0\Rightarrow \frac{\Delta \tau }{\tau }=\frac{g\cdot 0}{c^2}=0$. Why this step? Cell C is the degenerate case: the $gh$ engine is linear in $h$, so zero height difference gives exactly zero gravitational shift. No trick, no residue.

2. (b) SR part (Cell D): $\frac{\Delta \tau }{\tau }=-\frac{v^2}{2c^2}=-\frac{200^2}{2(3\times 10^8{)}^2}=-2.22\times 10^{-13}$. Why this step? Motion alone slows a clock. We use the velocity engine now because the only difference between the clocks is speed, not height. Why $v^2/2c^2$ and not $v^2/c^2$? For slow speeds we expand the exact factor $\sqrt{1-v^2/c^2}\approx 1-\frac{1}{2}{v}^2/c^2$; the leading correction is half $v^2/c^2$.

3. Per day: $\Delta \tau =-2.22\times 10^{-13}\times 86400=-1.92\times 10^{-8}\text{s}\approx -19.2\text{ns/day}$. Why this step? Convert rate to duration by multiplying by $\tau$, same as before.

Verify: GR shift is exactly $0$ ✓ (correct degenerate behaviour). SR shift is negative ✓ (moving clock slower). The rim clock loses ~19 ns/day purely from motion. ✓

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Example 4 — Cell E: GPS, both effects fighting

Forecast: Does the satellite clock end up ahead or behind the ground clock — and by microseconds or nanoseconds?

Steps

1. GR (height) part, exact potential. Because the height is huge, use $U=-GM/r$ not $gh$: $\frac{\Delta {\tau }_{\text{GR}}}{\tau }=\frac{U_{\text{sat}}-U_{\text{ground}}}{c^2}=\frac{GM}{c^2}\left(\frac{1}{R}-\frac{1}{r}\right).$ Numerically $\frac{3.986\times 10^{14}}{9\times 10^{16}}\left(\frac{1}{6.37\times 10^6}-\frac{1}{2.66\times 10^7}\right)=5.28\times 10^{-10}$. Why this step? Cell F rule: $gh$ only works over small heights where $g$ is constant. Over $20,000$ km, $g$ falls off, so we must integrate the potential — the difference of $-GM/r$ terms. Positive = satellite (higher) runs faster.

2. SR (velocity) part: $\frac{\Delta {\tau }_{\text{SR}}}{\tau }=-\frac{v^2}{2c^2}=-\frac{3874^2}{2(9\times 10^{16})}=-8.34\times 10^{-11}$. Why this step? The satellite also moves, and motion slows clocks. Negative, opposing the GR gain.

3. Add them (Cell E is a superposition): $(5.28-0.834)\times 10^{-10}=4.45\times 10^{-10}$. Why this step? The two small effects simply add to first order. GR ($+$) is bigger than SR ($-$), so the net is positive: the satellite clock runs fast.

4. Per day: $4.45\times 10^{-10}\times 86400=3.84\times 10^{-5}\text{s}\approx 38\mu \text{s/day}$. Why this step? Rate × day. This is the famous $\approx 38\mu$s that GPS must correct or positions drift ~11 km/day.

Verify: GR ($+45$-ish) beats SR ($-7$-ish) in the standard textbook split, net $\approx +38\mu$s/day ✓, and it is positive (satellite ahead) ✓. Sign of each piece matches our convention (higher=faster, moving=slower). ✓ See GPS Corrections.

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Example 5 — Cell F: exact potential vs the $gh$ shortcut

Forecast: Over 100 km, does the simple $gh$ formula still hold — off by 0.1%? 10%? 50%?

Steps

1. Naive: $\frac{gh}{c^2}=\frac{9.8\times 10^5}{9\times 10^{16}}=1.089\times 10^{-11}$. Why this step? This is the flat-field shortcut, valid only while $g$ is roughly constant.

2. Exact: $\frac{GM}{c^2}\left(\frac{1}{R}-\frac{1}{r}\right)=\frac{3.986\times 10^{14}}{9\times 10^{16}}\left(\frac{1}{6.37\times 10^6}-\frac{1}{6.47\times 10^6}\right)=1.076\times 10^{-11}$. Why this step? The real potential difference accounts for $g$ dropping slightly with altitude — this is the honest Cell F calculation.

3. Compare: they differ by about $1.2\%$. Why this step? It shows when the shortcut breaks: over 100 km the error is small but visible; over 20,000 km (Ex 4) it would be enormous, which is exactly why GPS uses the exact form.

Verify: Exact < naive (since real $g$ decreases as you climb, the true potential gain is slightly smaller) ✓. Relative difference $\approx 1.2\%$ ✓.

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Example 6 — Cell G: the three limiting cases

Forecast: Should a good formula give $0$, a known result, and a warning at these three edges?

Steps

1. (a) $h\to 0$: $\Delta \tau /\tau =gh/c^2\to 0$. Why this step? Two clocks at the same place must tick identically — the formula must vanish, and it does (linear in $h$).

2. (b) Weak-field / free-fall limit: in a freely-falling small box, EP says gravity is "off," so only velocity time dilation ($-v^2/2c^2$) survives — you recover Special Relativity. Why this step? GR must reduce to SR locally; the GR term is a gravitational-potential term, and in free-fall the potential difference across a tiny box $\to 0$.

3. (c) Large-$h$ warning: naive $gh/c^2$ with $h=3\times 10^{16}\text{m}$ gives $gh/c^2=9.8\times 3\times 10^{16}/9\times 10^{16}=3.27>1$ — a fractional rate difference exceeding 1 is nonsense. Why this step? It flags that constant-$g$ extrapolation is illegal at large distance; you must switch to $-GM/r$ (Cell F). The formula tells you where it dies.

Verify: (a) exactly $0$ ✓; (c) $3.27>1$, unphysical ✓ — correctly signals breakdown of the linear approximation.

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Example 7 — Cell H: mountain vs valley (word problem)

Forecast: Milliseconds? Seconds? Minutes over a whole lifetime?

Steps

1. Rate: $\frac{\Delta \tau }{\tau }=\frac{gh}{c^2}=\frac{9.8\times 2000}{9\times 10^{16}}=2.178\times 10^{-13}$. Why this step? Cell A again ($h>0$): the higher-up mountain clock (and body) runs faster.

2. Total time: $\tau =80\times 3.156\times 10^7=2.525\times 10^9\text{s}$. Why this step? Convert a lifetime into seconds so the fractional rate becomes an actual duration.

3. Extra ageing: $\Delta \tau =2.178\times 10^{-13}\times 2.525\times 10^9=5.5\times 10^{-4}\text{s}\approx 0.55\text{ms}$. Why this step? Rate × elapsed time — the mountain twin is genuinely older by half a millisecond after 80 years.

Verify: Positive (higher = ages faster) ✓, and tiny (sub-millisecond over a lifetime) ✓ — consistent with the $\sim$ns/day-per-100m of Example 1 scaled up ($20\times$ height, $\sim 2.9\times 10^4$ days).

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Example 8 — Cell I: the exam twist — light bending, not clocks

Forecast: Micrometres, nanometres, or something you could see with a ruler?

Steps

1. Crossing time: $t=L/c=3/(3\times 10^8)=1.0\times 10^{-8}\text{s}$. Why this step? Light travels straight in the inertial frame; we need how long it's in flight to see how far the floor rises under it.

2. Box rises: $\Delta y=\frac{1}{2}g{t}^2=\frac{1}{2}(9.8)(1.0\times 10^{-8}{)}^2=4.9\times 10^{-16}\text{m}$. Why this step? From inside the accelerating box, the box moves up while light goes straight, so light appears to drop by $\Delta y$. By the equivalence principle, the same bend must occur in real gravity.

3. The trap (Cell I twist): naively treating the photon as a Newtonian mass falling gives this $\Delta y$, but the full GR deflection of starlight is twice as big — because space itself is also curved, not just time. Why this step? It kills the "photons have mass" misconception: the EP/accelerating-box argument captures only the time-curvature half; the extra factor of 2 comes from spatial curvature (see Gravitational Lensing).

Verify: $\Delta y=4.9\times 10^{-16}\text{m}$ ✓ — utterly tiny over 3 m (why lab light-bending is unmeasurable and we need starlight grazing the Sun). The "factor of 2" for real deflection is the standard GR-vs-Newton result ✓.

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Recall Self-test across the matrix (hide answers)

Sign of $\Delta \tau$ for a clock lifted higher? ::: Positive — higher clocks run faster. Which effect dominates for GPS, GR or SR? ::: GR (higher = faster) wins; net $\approx +38\mu$s/day. When must you drop $gh$ for $-GM/r$? ::: When height is large enough that $g$ changes appreciably (Cell F). What is $\Delta \tau /\tau$ when $h=0$? ::: Exactly zero (degenerate Cell C). Why is the real light deflection twice the accelerating-box estimate? ::: The box argument only gives time-curvature; spatial curvature adds the other half.

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Connections

• Parent topic — the principles these examples exercise.
• Gravitational Time Dilation — the engine behind Cells A/B/C/F/H.
• GPS Corrections — Cell E in full detail.
• Special Relativity — the $-v^2/2c^2$ velocity engine (Cells D/E, limit G(b)).
• Gravitational Lensing — the "factor of 2" of Example 8.
• Newtonian Gravity — source of the $-GM/r$ potential used in Cells E/F.
• Geodesics & Curvature — why free-fall feels weightless (Example 8's frame).