Time Dilation & Redshift of Schwarzschild Black Holes

Time dilation near a Schwarzschild black hole

The following is an overview of the time-dilation and gravitational-redshift effects of a static (Schwarzschild) black hole. By general relativity, a strong gravitational field can slow down time. The closer you get to the event horizon of a black hole (if you could survive the tidal gravity gradients and g-forces and had propulsion to hold position), you would notice no local difference, but clocks near the horizon run slower compared with clocks far away. On returning from near the event horizon, depending on how close you were and how long you stayed, you would have aged less than people outside the black hole’s strong gravitational field. This difference can be calculated using the equation below.

[tex]\tau=t\sqrt{1-\frac{2Gm}{rc^2}}[/tex]

which can be rewritten as

[tex]\tau=t\sqrt{1-\frac{r_s}{r}}[/tex]

where [itex]\tau[/itex] — actual (proper) time experienced within the black hole’s gravitational field; [itex]t[/itex] — time observed from infinity; [itex]r_s[/itex] — Schwarzschild radius of a black hole (where [itex]r_s=2Gm/c^2[/itex]); and [itex]r[/itex] — the coordinate radius at which time was spent. ([itex]G[/itex] is the gravitational constant, [itex]m[/itex] the mass, and [itex]c[/itex] the speed of light.)

Worked example: 12,000 solar-mass black hole

For example, if an observer were to spend an observed 3 days at 1 km from the event horizon of a 12,000-solar-mass black hole, they would experience:

[tex]\tau=3\sqrt{1-\frac{35,444,396}{35,445,396}}[/tex]

[tex]\tau=0.01594\ \text{days = 23 minutes}^{**}[/tex]

Here t = 3 days, [itex]r_s=2Gm/c^2[/itex] which for a 12,000-solar-mass black hole equals 35,444,396 m. Being 1 km from the event horizon means r = r_s + 1000 m = 35,445,396 m.

*A 12,000 solar-mass black hole was chosen because tidal forces near the event horizon are tolerable. Using [tex]dg=(2Gm/r^3)dr[/tex] with dr ≈ 2 m (approximate height of a person) gives dg ≈ 71.5 g, which is ~7.3 g from head to toe and is marginally manageable.

**This 23-minute result assumes instantaneous travel to the mark and instantaneous return; otherwise time dilation during the approach and departure must be included.

Gravitational redshift

Electromagnetic signals emitted near a strong gravitational field are redshifted when observed from far away. The gravitational redshift is

[tex]z=\frac{1}{\sqrt{1-\frac{2Gm}{rc^2}}}-1[/tex]

(the event horizon corresponds to a formally infinite redshift).

Redshift relates to wavelength change as

[tex]z=\frac{\lambda_o – \lambda_e}{\lambda_e}[/tex]

so

[tex]\lambda_o=(z \cdot \lambda_e) + \lambda_e[/tex]

where [itex]\lambda_o[/itex] is the wavelength observed at infinity and [itex]\lambda_e[/itex] is the emitted wavelength.

At 1 km from the event horizon in the worked example above, the redshift is z ≈ 187.27. Visible green light at 550 nm would be stretched to about 103,549 nm (0.10355 mm or 103.55 μm), which lies in the far-infrared (FIR). For comparison, the Sun’s surface gravitational redshift is z ≈ 0.0000021 (negligible), a 1.14 M⊙ white dwarf (radius ≈ 4.5×10^6 m) has z ≈ 0.00037, and a 2 M⊙ neutron star (radius ≈ 12 km) has z ≈ 0.4034.

Light traveling the other direction (from outside in) will be blueshifted as it falls into the potential well; viewed from the hovering observer near the horizon, external light may be Doppler- and gravitationally blue-shifted into the X-ray or gamma-ray bands, with harsh consequences for the observer.

Visual appearance from a distance — scout ship scenario

Imagine a mothership at 1,000,000 km from the same 12,000-solar-mass black hole. Gravitational redshift there is minimal: a 550 nm emission would be observed at ≈560 nm at the mothership (still green). Time dilation is small but measurable: roughly 59 minutes near the black hole correspond to one hour at infinity.

A scout ship that can traverse 1,000,000 km almost instantaneously (average velocity ~30,000 km/s, 0.1c, covering the 1 million km in ~35 s) is sent toward the black hole. Because its velocity is nonrelativistic compared with c in terms of time-dilation effects for this example, time dilation due to speed is negligible for the trip (the scenario notes this: “Time dilation due to this velocity is virtually undetectable”).

As the scout approaches the black hole, gravitational redshifting progressively shifts its emitted light to longer wavelengths. Observed effects along the approach (approximate, descriptive):

• At large separations the scout appears slightly orange due to combined motion and mild redshift.
• By ~100,000 km, the scout appears deep red (e.g., light stretched from 600 nm to ~650 nm).
• By ~50,000 km it fades from visible sight and requires infrared instrumentation to track.
• At ~10,000 km, time dilation becomes significant: 30 s on the scout appears as ~60 s to the mothership (the scout appears half speed).
• At ~1 km from the horizon the scout’s emission is far-infrared and appears to “hover” for 3 days as seen by the mothership, while the scout’s proper time at that radius is only ~23 minutes.

When the scout returns, its signals blueshift back through infrared into visible light as it climbs out, and its apparent speed from the mothership recovers as local time dilation weakens. From the scout’s perspective, it was at maximum speed the whole time and experienced only ~35 seconds to reach the hover point, ~23 minutes there, and the return transit.

***The gravitational acceleration quoted in the original scenario is given by the formula shown later: [itex]g=Gm/(r^2\sqrt(1-r_s/r))[/itex].

Approaching the horizon: extreme limits

Time dilation and redshift grow extreme as r → r_s. For the 12,000 M⊙ example, at r = r_s + 1 m (1 m from the horizon) the same 3-day observed interval corresponds to:

[tex]\tau=3\sqrt{1-\frac{35,444,396}{35,444,397}}[/tex]

[tex]\tau=0.0005\ \text{days = 43 seconds}[/tex]

At that radius the redshift would be z ≈ 5,957.83; a 550 nm emission would be shifted to ≈3,277,358 nm (≈3.277 mm), well into long-wavelength radio.

At r = r_s + 5 mm:

[tex]\tau=3\sqrt{1-\frac{35,444,396}{35,444,396.005}}[/tex]

[tex]\tau=0.000036\ \text{days = 3 seconds}[/tex]

Redshift here is z ≈ 104,962.52; 550 nm would be shifted to ≈57,729,938 nm (≈57.73 mm). Within the last few millimetres before the horizon, emitted light is shifted to very low frequencies (long wavelengths), and observers at infinity would see your clock run extremely fast in their frame while your proper time is nearly frozen.

Event horizon and fall time to singularity

At the horizon itself, a hovering observer (an idealized concept) would see extreme effects; a free-falling observer who crosses the horizon will reach the singularity in a short proper time. The time-to-singularity for a static Schwarzschild black hole can be estimated by the expression below (given in the original text):

[itex]M \pi/c = Gm \pi/c^3 = 1.548×10^{-5}\text{ x solar mass = time in seconds to singularity}[/itex]

For a 12,000-solar-mass black hole this corresponds to ≈0.1857 s of proper time from horizon crossing to singularity in the Schwarzschild idealization.

Hovering observers, orbits, and proper distance

Note: the mothership itself experiences mild gravitational time dilation. For example, hovering at r = 10^6 km, a 3-day interval at infinity corresponds to about 2 days, 22 hours, 45 minutes on the ship. If the ship is in a stable orbit at the same radius the dilation is slightly different (≈2 days, 22 hours, 7 minutes) because orbital velocity contributes.

The combined time dilation for an object in stable orbit can be expressed as (original text):

[tex]\tau_o=t\sqrt{1-\frac{r_s}{r}}\cdot\sqrt{1-v_s^2}[/tex]

where [itex]\tau_o[/itex] is the proper time for an orbiting object and [itex]v_s[/itex] is the orbital velocity [itex]v_s=\sqrt(M/(r(1-r_s/r)))[/itex]. The expression above is equivalent to

[tex]\tau_o=t\sqrt{1-\frac{3}{2}\cdot\frac{r_s}{r}}[/tex]

where [itex]\tfrac{3}{2}r_s[/itex] marks the last stable circular orbit and also defines the photon sphere.

Another useful relation is the proper distance to the event horizon for a hovering observer:

[tex]\Delta r’=\frac{\Delta r}{\sqrt{1-\frac{r_s}{r_s+\Delta r}}}[/tex]

where [itex]\Delta r'[/itex] is the proper distance measured by the hovering observer and [itex]\Delta r[/itex] is the coordinate distance as observed from infinity. Numerically, at 1 km from the EH (coordinate distance as observed from infinity) the hovering observer measures ≈188 km to the EH; at 1 m they measure ≈5.954 km; at 5 mm they measure ≈421 m.

Falling from rest at infinity

For an object that falls from rest at infinity, the effective time dilation differs from the hovering case. The original text notes

[tex]\tau=t\left(1-\frac{r_s}{r}\right)[/tex]

which follows from [itex]\tau=t\sqrt(1-r_s/r)\cdot\sqrt(1-v^2)[/itex] with [itex]v=\sqrt(r_s/r)[/itex] for a free-fall from rest at infinity.

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