<p>Gravity doesn't affect the <em>speed</em> of light. It affects the space-time geometry and hence the <em>paths</em> of light. However, this can have a similar effect.</p>

<p>Light emitted at source $S$ to pass a massive object $M$ that is very close on the otherwise (if <em>M</em> weren't there) straight path to an observer $O$ has to "go around" $M$, which takes longer than following the straight path in the absence of $M$. The light that reaches $O$ from $S$ is not emitted from $S$ in the "straight" (in the absence of $M$) direction to $O$, but slightly off that direction, such that the "bending" of its path by the gravity of $M$ "deflects" it onto $O$.</p>

<p>Of course, light is never bent, but always follows a straight path. What is bent is the space-time as compared to Euclidean space-time in the absence of distorting masses (see: <a href="http://en.wikipedia.org/wiki/Geodesics_in_general_relativity">geodesic</a>). This distortion in the fabric of space time is called <a href="http://en.wikipedia.org/wiki/Gravitational_lens">gravitational lensing</a>.</p>

CC BY-SA 3.0
2014-01-10T21:53:27.693Z
2014-01-12T19:49:01.547Z
2014-01-12T19:49:01.547Z
Walter
<p>This is a tough one, especially as I'm not used to giving explanations in non technical terms.</p>

<p>Starting at the top:</p>

<p>Conditionally yes. In the emptiest space possible - not that between stars, not that between galaxies, not that between families of galaxies and so on.....  in the very emptiest space between the super-clusters of galaxies, there it's at its fastest, where gravity is at its weakest.</p>

<p>If you had the time to do so, and a nice clear target black hole and shot a blue laser just at the event horizon on one side (say it's transmitting the whole of the works of Shakespeare followed by the rest of project Gutenberg) - in such a way that it skimmed all the way round and then exited back in your direction, like a slingshot of the moon the first orbit of the moon did, what would happen? Would the light colour change?</p>

<p>The closer the beam got to the event horizon the more stretched space is - think of it that way, then the light has to travel further, and the same all the way round the black hole - the closer to the event horizon the deeper the well, the more the space is stretched and the longer the light takes to get around. From your point of view, the black hole is X distance away, the path the light took is Y in apparent length. Using your handy slide rule you calculate the Taming of the shrew should arrive at time Z.</p>

<p>It doesn't turn up on time. Why? Remember the light had to take a very long path because of the gravity field density making the journey longer. When it does turn up finally What colour is it? Still blue - this does not depend on if the black hole is moving away or closer - there is no red or blue shift.
(I am being slightly disingenuous here as the wavelength would have shifted a minute amount to the red - it does this as it travels, the further it does the more it shifts, partly from collisions with free floating atoms that absorb then re emit at a lower frequency eg. the big bang (Very hot) - the light from this is very long wavelength indeed, (red shifted to the extreme) but space is expanding remember. To put it in a nutshell, entropy cannot be reversed.</p>

<p>The odd thing is the distance the light travels from the point of view of the observer who shot the laser, he would extrapolate that the light waves containing The Shrew, since they arrived so late must have not only slowed down but gotten closer together (blue shifted) - but when it comes back to the observer it's just the same colour as before. (Space lengthened apparently, that would explain this wouldn't it?)</p>

<p>To say that gravity slows light is the same as saying a watched kettle never boils, it has a kind of truth to it from a particular point of view - a perceptual point of view.</p>

<p>Looking at the whole universe, there are visible hot spots and cold spots, places with more and less matter - this can be observed. The trouble we're having at the moment is with dark matter, and dark energy.</p>

<p>We started with observations in our own solar system. Distant objects are all measured relative to each other. A large number of observations are made of many objects, their luminosity, their aggregate luminosity, their red or blue shift - and interestingly their change in Doppler shift. Various different type of star, pulsating ones, stars that emit hard radiation, co-orbiting stars of all sorts, the accretion disks at the center of galaxies and their temperatures, This accumulation of data since Copernicus, or at least since the Renaissance has all been put together, adjusting along the way taking into account  world changing paradigm shifts such as relativity, and huge advances in the resolution of our observations of the universe, from land and space based platforms has (we think!) made our estimations of the error margin of our calculations of the age of the universe smaller and smaller. </p>

CC BY-SA 3.0
2014-01-10T02:34:44.563Z
2014-01-10T23:46:18.077Z
2014-01-10T23:46:18.077Z
Jiminy Cricket.
<p>As Walter says, gravity doesn't bend light. Light travels along null geodesics, a particular type of straight path. Since (affine) geodesics don't change direction by definition, geometrically light trajectories are straight. Moreover, the speed of light in vacuum is $c$ in every inertial frame, regardless of whether or not spacetime is curved, although a curved spacetime inertial frames can only ever be local.</p>

<p>What can change, however, is the <em>coordinate speed</em> of light. Since coordinates are just labels for spacetime events, this is true even in completely flat spacetime. For example, in the Rindler coordinate chart, the Minkowski metric of flat spacetime takes the form
$$\mathrm{d}s^2 = -\frac{g^2x^2}{c^2}\,\mathrm{d}t^2 + \underbrace{\mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2}_{\mathrm{d}S^2_\text{Euclid}}\text{,}$$
where $g$ has units of acceleration. Since light travels along null ($\mathrm{d}s^2 = 0$) wordlines, the coordinate speed of light is
$$\frac{\mathrm{d}S}{\mathrm{d}t} = \frac{|gx|}{c}\text{,}$$
which is position-dependent and can even be $0$, as there is an apparent event horizon. An observer that's stationary in Rindler coordinates actually has proper acceleration $g$, so the Rindler chart of flat spacetime is a natural analogue of a "uniform gravitational field".</p>

<blockquote>
  <p>if gravity bends the course of light, does this imply that gravity the retards light so it is moving at a slower speed?</p>
</blockquote>

<p>No, but what we can say is this. For weak, slowly-changing gravitational fields, the following  metric is appropriate to describe spacetime in terms of the Newtonian gravitational potential $\Phi$:
$$\mathrm{d}s^2 = -\left(1+2\frac{\Phi}{c^2}\right)c^2\,\mathrm{d}t^2 + \left(1-2\frac{\Phi}{c^2}\right)\,\mathrm{d}S^2\text{,}$$
as we can readily calculate the coordinate speed of light (again $\mathrm{d}s^2 = 0$):
$$\frac{\mathrm{d}S}{\mathrm{d}t} = c\sqrt{\frac{1+2\Phi/c^2}{1-2\Phi/c^2}}\text{,}$$
and so by expanding its reciprocal in a Taylor-MacLaurin series, we find that light travels "<em>as if</em>" we had an <a href="http://en.wikipedia.org/wiki/Refractive_index">index of refraction</a>
$$n = c \frac{\mathrm{d}t}{\mathrm{d}S} \approx 1 - 2\frac{\Phi}{c^2} + \mathcal{O}\left(\frac{\Phi^2}{c^4}\right)\text{.}$$</p>

<p>If we keep in mind that we are dealing with just the coordinate speed of light, then yes, we could say that gravity (rather, gravitational potential) retards light. Another way of thinking of this is like so: if we pretend that we're dealing with ordinary flat Minwkoski spacetime in the usual inertial coordinates, then we need a medium with the above refractive index to reproduce the trajectories of light. But of course taking this literally is not legitimate, since (1) the metric affects more than the propagation of light, and (2) such an interpretation would fail to explain gravitational redshift.</p>

<p>The latter approach is morally similar to what's described in Walter's answer, since it depends on a hypothetical comparison with flat spacetime. The difference is that by limiting ourselves to talk about what happens far from the gravitating bodies, Walter can sidestep the gravitational redshift issue, but then cannot ascribe any local refractive index (on the plus side, his approach is not limited to weak, slowly-changing gravity).</p>

<blockquote>
  <p>And if gravity does affect the speed of light, what does that say about our measurements of the distance to the farthest observable object? Could we assume that all the gravity effects across 15 billion light-years even themselves out?</p>
</blockquote>

<p>Our cosmological models <a href="http://en.wikipedia.org/wiki/Cosmological_principle">assume</a> that the universe is on the large scale homogeneous and isotropic, an assumption which is backed by <a href="http://sci.esa.int/planck/51551-simple-but-challenging-the-universe-according-to-planck/">observations</a> of the parts of it we can see. In a homogeneous and isotropic universe, it's fairly easy to account for how light behaves when traversing it. So no, we don't need to assume that the effects gravity even themselves out--on the contrary, we use such gravitational effects on light to fit the parameters of our models.</p>

CC BY-SA 3.0
2014-01-13T16:50:37.883Z
Now there's an *answer*.  I only barely understood the English prose, not to say that I understood all the implications, and those equations are simply marvelous.  Thankyou!
Cyberherbalist
CC BY-SA 3.0
2014-01-11T02:40:57.950Z
2014-01-14T01:14:13.793Z
2014-01-14T01:14:13.793Z
Stan Liou
<p>The speed of light in a vacuum is presumably the fastest speed possible.  </p>

<p>if gravity bends the course of light, does this imply that gravity the retards light so it is moving at a slower speed?  If it affects its course, why cannot gravity affect its speed -- or does it?</p>

<p>And if gravity does affect the speed of light, what does that say about our measurements of the distance to the farthest observable object?  Could we assume that all the gravity effects across 15 billion light-years even themselves out?  Or is the actual distance across the observable universe subject to unknowable variations due to gravity effects?</p>

<gravity><light><general-relativity><speed><special-relativity>
Does gravity slow or speed light up?
Cyberherbalist
# Does gravity slow or speed light up?

(Asked by: username_0 on 2014-01-10)

The speed of light in a vacuum is presumably the fastest speed possible.

if gravity bends the course of light, does this imply that gravity the retards light so it is moving at a slower speed? If it affects its course, why cannot gravity affect its speed -- or does it?

And if gravity does affect the speed of light, what does that say about our measurements of the distance to the farthest observable object? Could we assume that all the gravity effects across 15 billion light-years even themselves out? Or is the actual distance across the observable universe subject to unknowable variations due to gravity effects?

**Tags:** gravity, light, general-relativity, speed, special-relativity

---

## 3 answers

### Answer by username_3 on 2014-01-11. Score: 13 | Accepted answer

As Walter says, gravity doesn't bend light. Light travels along null geodesics, a particular type of straight path. Since (affine) geodesics don't change direction by definition, geometrically light trajectories are straight. Moreover, the speed of light in vacuum is $c$ in every inertial frame, regardless of whether or not spacetime is curved, although a curved spacetime inertial frames can only ever be local.

What can change, however, is the *coordinate speed* of light. Since coordinates are just labels for spacetime events, this is true even in completely flat spacetime. For example, in the Rindler coordinate chart, the Minkowski metric of flat spacetime takes the form
$$\mathrm{d}s^2 = -\frac{g^2x^2}{c^2}\,\mathrm{d}t^2 + \underbrace{\mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2}\_{\mathrm{d}S^2\_\text{Euclid}}\text{,}$$
where $g$ has units of acceleration. Since light travels along null ($\mathrm{d}s^2 = 0$) wordlines, the coordinate speed of light is
$$\frac{\mathrm{d}S}{\mathrm{d}t} = \frac{|gx|}{c}\text{,}$$
which is position-dependent and can even be $0$, as there is an apparent event horizon. An observer that's stationary in Rindler coordinates actually has proper acceleration $g$, so the Rindler chart of flat spacetime is a natural analogue of a "uniform gravitational field".

> if gravity bends the course of light, does this imply that gravity the retards light so it is moving at a slower speed?

No, but what we can say is this. For weak, slowly-changing gravitational fields, the following metric is appropriate to describe spacetime in terms of the Newtonian gravitational potential $\Phi$:
$$\mathrm{d}s^2 = -\left(1+2\frac{\Phi}{c^2}\right)c^2\,\mathrm{d}t^2 + \left(1-2\frac{\Phi}{c^2}\right)\,\mathrm{d}S^2\text{,}$$
as we can readily calculate the coordinate speed of light (again $\mathrm{d}s^2 = 0$):
$$\frac{\mathrm{d}S}{\mathrm{d}t} = c\sqrt{\frac{1+2\Phi/c^2}{1-2\Phi/c^2}}\text{,}$$
and so by expanding its reciprocal in a Taylor-MacLaurin series, we find that light travels "*as if*" we had an [index of refraction](http://en.wikipedia.org/wiki/Refractive_index)
$$n = c \frac{\mathrm{d}t}{\mathrm{d}S} \approx 1 - 2\frac{\Phi}{c^2} + \mathcal{O}\left(\frac{\Phi^2}{c^4}\right)\text{.}$$

If we keep in mind that we are dealing with just the coordinate speed of light, then yes, we could say that gravity (rather, gravitational potential) retards light. Another way of thinking of this is like so: if we pretend that we're dealing with ordinary flat Minwkoski spacetime in the usual inertial coordinates, then we need a medium with the above refractive index to reproduce the trajectories of light. But of course taking this literally is not legitimate, since (1) the metric affects more than the propagation of light, and (2) such an interpretation would fail to explain gravitational redshift.

The latter approach is morally similar to what's described in Walter's answer, since it depends on a hypothetical comparison with flat spacetime. The difference is that by limiting ourselves to talk about what happens far from the gravitating bodies, Walter can sidestep the gravitational redshift issue, but then cannot ascribe any local refractive index (on the plus side, his approach is not limited to weak, slowly-changing gravity).

> And if gravity does affect the speed of light, what does that say about our measurements of the distance to the farthest observable object? Could we assume that all the gravity effects across 15 billion light-years even themselves out?

Our cosmological models [assume](http://en.wikipedia.org/wiki/Cosmological_principle) that the universe is on the large scale homogeneous and isotropic, an assumption which is backed by [observations](http://sci.esa.int/planck/51551-simple-but-challenging-the-universe-according-to-planck/) of the parts of it we can see. In a homogeneous and isotropic universe, it's fairly easy to account for how light behaves when traversing it. So no, we don't need to assume that the effects gravity even themselves out--on the contrary, we use such gravitational effects on light to fit the parameters of our models.

> **Comments:**
> **username_0**: Now there's an *answer*.  I only barely understood the English prose, not to say that I understood all the implications, and those equations are simply marvelous.  Thankyou!  

---

### Answer by username_1 on 2014-01-10. Score: 14

Gravity doesn't affect the *speed* of light. It affects the space-time geometry and hence the *paths* of light. However, this can have a similar effect.

Light emitted at source $S$ to pass a massive object $M$ that is very close on the otherwise (if *M* weren't there) straight path to an observer $O$ has to "go around" $M$, which takes longer than following the straight path in the absence of $M$. The light that reaches $O$ from $S$ is not emitted from $S$ in the "straight" (in the absence of $M$) direction to $O$, but slightly off that direction, such that the "bending" of its path by the gravity of $M$ "deflects" it onto $O$.

Of course, light is never bent, but always follows a straight path. What is bent is the space-time as compared to Euclidean space-time in the absence of distorting masses (see: [geodesic](http://en.wikipedia.org/wiki/Geodesics_in_general_relativity)). This distortion in the fabric of space time is called [gravitational lensing](http://en.wikipedia.org/wiki/Gravitational_lens).


---

### Answer by username_2 on 2014-01-10. Score: 3

This is a tough one, especially as I'm not used to giving explanations in non technical terms.

Starting at the top:

Conditionally yes. In the emptiest space possible - not that between stars, not that between galaxies, not that between families of galaxies and so on..... in the very emptiest space between the super-clusters of galaxies, there it's at its fastest, where gravity is at its weakest.

If you had the time to do so, and a nice clear target black hole and shot a blue laser just at the event horizon on one side (say it's transmitting the whole of the works of Shakespeare followed by the rest of project Gutenberg) - in such a way that it skimmed all the way round and then exited back in your direction, like a slingshot of the moon the first orbit of the moon did, what would happen? Would the light colour change?

The closer the beam got to the event horizon the more stretched space is - think of it that way, then the light has to travel further, and the same all the way round the black hole - the closer to the event horizon the deeper the well, the more the space is stretched and the longer the light takes to get around. From your point of view, the black hole is X distance away, the path the light took is Y in apparent length. Using your handy slide rule you calculate the Taming of the shrew should arrive at time Z.

It doesn't turn up on time. Why? Remember the light had to take a very long path because of the gravity field density making the journey longer. When it does turn up finally What colour is it? Still blue - this does not depend on if the black hole is moving away or closer - there is no red or blue shift.
(I am being slightly disingenuous here as the wavelength would have shifted a minute amount to the red - it does this as it travels, the further it does the more it shifts, partly from collisions with free floating atoms that absorb then re emit at a lower frequency eg. the big bang (Very hot) - the light from this is very long wavelength indeed, (red shifted to the extreme) but space is expanding remember. To put it in a nutshell, entropy cannot be reversed.

The odd thing is the distance the light travels from the point of view of the observer who shot the laser, he would extrapolate that the light waves containing The Shrew, since they arrived so late must have not only slowed down but gotten closer together (blue shifted) - but when it comes back to the observer it's just the same colour as before. (Space lengthened apparently, that would explain this wouldn't it?)

To say that gravity slows light is the same as saying a watched kettle never boils, it has a kind of truth to it from a particular point of view - a perceptual point of view.

Looking at the whole universe, there are visible hot spots and cold spots, places with more and less matter - this can be observed. The trouble we're having at the moment is with dark matter, and dark energy.

We started with observations in our own solar system. Distant objects are all measured relative to each other. A large number of observations are made of many objects, their luminosity, their aggregate luminosity, their red or blue shift - and interestingly their change in Doppler shift. Various different type of star, pulsating ones, stars that emit hard radiation, co-orbiting stars of all sorts, the accretion disks at the center of galaxies and their temperatures, This accumulation of data since Copernicus, or at least since the Renaissance has all been put together, adjusting along the way taking into account world changing paradigm shifts such as relativity, and huge advances in the resolution of our observations of the universe, from land and space based platforms has (we think!) made our estimations of the error margin of our calculations of the age of the universe smaller and smaller.


---

