<p>The Schwarzschild black hole is the simplest, being just an isolated, nonrotated, uncharged black hole. If $\tau$ is the proper time measured by a clock moving along an arbitrary path in this spacetime, then in Schwarzschild coordinates
$$-\mathrm{d}\tau^2 = -\left(1-\frac{R}{r}\right)\mathrm{d}t^2 + \left(1-\frac{R}{r}\right)^{-1}\mathrm{d}r^2 + r^2\left(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\varphi^2\right)\text{,}$$
where $R = 2GM/c^2$ is the Schwarzschild radius. If the observer is stationary, then neither the radial nor angular coordinates change ($\mathrm{d}r = \mathrm{d}\theta = \mathrm{d}\varphi = 0$). So, for example, relative the rate of time of stationary observers at some radial coordinates $r_1$ and $r_2$ is just
$$\frac{\mathrm{d}\tau_1}{\mathrm{d}\tau_2} = \frac{\mathrm{d}\tau_1/\mathrm{d}t}{\mathrm{d}\tau_2/\mathrm{d}t} = \sqrt{\frac{1-R/r_1}{1-R/r_2}}\text{.}$$</p>

<blockquote>
  <p>how might that affect our measurements?</p>
</blockquote>

<p>In the sense of the measurement process, it doesn't, since a measurement apparatus is not time-dilated relative to itself. In the sense of what results we get, then of course it does.</p>

<blockquote>
  <p>So it would logically follow that events happening near a black hole (a very large gravitational field) would seem to us to happen much more slowly than if they happened outside that field.</p>
</blockquote>

<p>Yes. If the first stationary observer is shifted closer to the horizon ($r_1\to R$), then $(1-R/r_1)\to 0$, so the above time dilation goes to $0$ as well. So, for example, if that observer shines a laser beam toward the other, it will be redshifted when it arrives to the other.</p>

<p>This is really the same thing as time dilation. The laser beam has a certain frequency, so it takes a certain amount of time for one oscillation. Thus a frequency shift is just another way of talking about time dilation, except that time dilation means that this happens to every process, not just light: e.g, if the first observer taps his foot at a certain frequency (say, a tap per second), the second observer will see him tapping at a lower frequency (say, a tap per two seconds). Slowed. We can can still call this "redshift" in the sense that the frequency was shifted in the same way, although the term is usually used specifically for electromagnetic radiation.</p>

<blockquote>
  <p>On a related note, does light being 'bent' around a black hole experience time dilation or red/blue shift from our viewing perspective?</p>
</blockquote>

<p>Light itself <em>experiences</em> only trivial time dilation in the sense that the proper time along any lightlike trajectory is exactly $0$. Above, we were talking about <em>stationary</em> observers, and light is not stationary. Rather, $\mathrm{d}\tau = 0$ for light beams, always. So strictly speaking, light doesn't experience anything at all.</p>

<p>However, if you mean what the stationary observers would get when measuring light bent around the black hole, then it is blueshifted when coming closer to the black hole and redshifted when moving away--in the sense that this would be what stationary observers closer or farther away would measure. The details of the path it took between the observers do not matter; the ratio of measured frequencies will just be the inverse of the time dilation factor between the observers.</p>

CC BY-SA 3.0
2014-12-05T05:02:39.823Z
So if we (here on Earth) viewed a star's light that was bent around a black hole is it blue shifted from what we would have seen if the light were not bent around the black hole?
Tracy Cramer
CC BY-SA 3.0
2014-12-05T05:10:05.993Z
@TracyCramer since both emission and detection happened far away from the black hole, no, it's the frequency is not shifted.
Stan Liou
CC BY-SA 3.0
2014-12-05T18:52:44.167Z
How about emissions near the black hole, like stars near the event horizon?  Wouldn't the photons be blue shifted (compressed) when leaving the gravity well of the black hole?  Would they 'stretch out' after leaving that high gravity field?
Tracy Cramer
CC BY-SA 3.0
2014-12-05T21:55:43.437Z
@TracyCramer: as explained in the answer, they would be redshifted.
Stan Liou
CC BY-SA 3.0
2014-12-05T04:09:59.020Z
2014-12-05T04:09:59.020Z
Stan Liou
<p>The first experiment that proved Einstein's theory of general relativity was the bending of light around our sun. So if the sun can effect our measurement, in this case the light from a star behind the sun, a black hole sure can.</p>

<p>Secondly, gravitational lensing is observed by galaxy clusters where multiple images of a background galaxy are formed around the foreground lens. Each of these images are red-shifted or blue-shifted by different amounts, the specific value which can be measured by studying the variability of the background source observed in the different images formed.</p>

<p>So, in order to really say whether the light will be red-shifted or blue-shifted, we will have to understand the gravitational potential of the black hole. The position of the background source with respect to the foreground lensing object (black hole in this case) also matters.</p>

CC BY-SA 3.0
2014-12-05T19:33:26.713Z
2014-12-05T19:33:26.713Z
Poruri Sai Rahul
<p><a href="https://astronomy.stackexchange.com/a/8142/1142">This</a> answer made me wonder if matter/energy/events near a black hole are viewed (by us) in a very different time frame, how might that affect our measurements?</p>

<pre><code>The people on the smaller planet would observe the people on the extremely large planet 
moving in "slow motion". (note: due to gravitational time dilation)
</code></pre>

<p>On a related note, does light being 'bent' around a black hole experience time dilation or red/blue shift from our viewing perspective? (as possibly suggested by <a href="https://astronomy.stackexchange.com/a/1331/1142">this</a> answer).</p>

CC BY-SA 3.0
2014-12-04T23:47:02.587Z
Events are observed over a different period of time than when they happen, and thus in a different time frame (in the usual English vernacular sense). But that has very little to do with GTR, so what do you mean by 'time frame' in this context?
Stan Liou
CC BY-SA 3.0
2014-12-05T00:11:14.850Z
I know next to nothing about GTR but based my question on the first answer I referenced.  In it he said observers in a low gravitational field would observe things in a larger field as though they happened more slowly.  Extrapolating from a 'large world' to the extreme yields a black hole.  So it would logically follow that events happening near a black hole (a very large gravitational field) would seem to us to happen much more slowly than if they happened outside that field.  So I wondered, if that were true how would that manifest itself in our measurements?
Tracy Cramer
<gravity><general-relativity><gravitational-lensing>
Are our 'observations' of or near a black hole affected by gravity induced time dilation?
Tracy Cramer
# Are our 'observations' of or near a black hole affected by gravity induced time dilation?

(Asked by: username_0 on 2014-12-04)

This answer made me wonder if matter/energy/events near a black hole are viewed (by us) in a very different time frame, how might that affect our measurements?
The people on the smaller planet would observe the people on the extremely large planet 
moving in "slow motion". (note: due to gravitational time dilation)

On a related note, does light being 'bent' around a black hole experience time dilation or red/blue shift from our viewing perspective? (as possibly suggested by this answer).

**Tags:** gravity, general-relativity, gravitational-lensing
> **Comments:**
> **username_1**: Events are observed over a different period of time than when they happen, and thus in a different time frame (in the usual English vernacular sense). But that has very little to do with GTR, so what do you mean by 'time frame' in this context?  
> **username_0**: I know next to nothing about GTR but based my question on the first answer I referenced.  In it he said observers in a low gravitational field would observe things in a larger field as though they happened more slowly.  Extrapolating from a 'large world' to the extreme yields a black hole.  So it would logically follow that events happening near a black hole (a very large gravitational field) would seem to us to happen much more slowly than if they happened outside that field.  So I wondered, if that were true how would that manifest itself in our measurements?  

---

## 2 answers

### Answer by username_1 on 2014-12-05. Score: 3 | Accepted answer

The Schwarzschild black hole is the simplest, being just an isolated, nonrotated, uncharged black hole. If $\tau$ is the proper time measured by a clock moving along an arbitrary path in this spacetime, then in Schwarzschild coordinates
$$-\mathrm{d}\tau^2 = -\left(1-\frac{R}{r}\right)\mathrm{d}t^2 + \left(1-\frac{R}{r}\right)^{-1}\mathrm{d}r^2 + r^2\left(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\varphi^2\right)\text{,}$$
where $R = 2GM/c^2$ is the Schwarzschild radius. If the observer is stationary, then neither the radial nor angular coordinates change ($\mathrm{d}r = \mathrm{d}\theta = \mathrm{d}\varphi = 0$). So, for example, relative the rate of time of stationary observers at some radial coordinates $r\_1$ and $r\_2$ is just
$$\frac{\mathrm{d}\tau\_1}{\mathrm{d}\tau\_2} = \frac{\mathrm{d}\tau\_1/\mathrm{d}t}{\mathrm{d}\tau\_2/\mathrm{d}t} = \sqrt{\frac{1-R/r\_1}{1-R/r\_2}}\text{.}$$

> how might that affect our measurements?

In the sense of the measurement process, it doesn't, since a measurement apparatus is not time-dilated relative to itself. In the sense of what results we get, then of course it does.

> So it would logically follow that events happening near a black hole (a very large gravitational field) would seem to us to happen much more slowly than if they happened outside that field.

Yes. If the first stationary observer is shifted closer to the horizon ($r\_1\to R$), then $(1-R/r\_1)\to 0$, so the above time dilation goes to $0$ as well. So, for example, if that observer shines a laser beam toward the other, it will be redshifted when it arrives to the other.

This is really the same thing as time dilation. The laser beam has a certain frequency, so it takes a certain amount of time for one oscillation. Thus a frequency shift is just another way of talking about time dilation, except that time dilation means that this happens to every process, not just light: e.g, if the first observer taps his foot at a certain frequency (say, a tap per second), the second observer will see him tapping at a lower frequency (say, a tap per two seconds). Slowed. We can can still call this "redshift" in the sense that the frequency was shifted in the same way, although the term is usually used specifically for electromagnetic radiation.

> On a related note, does light being 'bent' around a black hole experience time dilation or red/blue shift from our viewing perspective?

Light itself *experiences* only trivial time dilation in the sense that the proper time along any lightlike trajectory is exactly $0$. Above, we were talking about *stationary* observers, and light is not stationary. Rather, $\mathrm{d}\tau = 0$ for light beams, always. So strictly speaking, light doesn't experience anything at all.

However, if you mean what the stationary observers would get when measuring light bent around the black hole, then it is blueshifted when coming closer to the black hole and redshifted when moving away--in the sense that this would be what stationary observers closer or farther away would measure. The details of the path it took between the observers do not matter; the ratio of measured frequencies will just be the inverse of the time dilation factor between the observers.

> **Comments:**
> **username_0**: So if we (here on Earth) viewed a star's light that was bent around a black hole is it blue shifted from what we would have seen if the light were not bent around the black hole?  
> **username_1**: @TracyCramer since both emission and detection happened far away from the black hole, no, it's the frequency is not shifted.  
> **username_0**: How about emissions near the black hole, like stars near the event horizon?  Wouldn't the photons be blue shifted (compressed) when leaving the gravity well of the black hole?  Would they 'stretch out' after leaving that high gravity field?  
> **username_1**: @TracyCramer: as explained in the answer, they would be redshifted.  

---

### Answer by username_2 on 2014-12-05. Score: 1

The first experiment that proved Einstein's theory of general relativity was the bending of light around our sun. So if the sun can effect our measurement, in this case the light from a star behind the sun, a black hole sure can.

Secondly, gravitational lensing is observed by galaxy clusters where multiple images of a background galaxy are formed around the foreground lens. Each of these images are red-shifted or blue-shifted by different amounts, the specific value which can be measured by studying the variability of the background source observed in the different images formed.

So, in order to really say whether the light will be red-shifted or blue-shifted, we will have to understand the gravitational potential of the black hole. The position of the background source with respect to the foreground lensing object (black hole in this case) also matters.


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