# General relativity: time dilation and speed of light

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## Main Question or Discussion Point

Hello everyone,

I'll go straight to the question. The gravitational time dilation is equal to tearth = tspace*sqrt(1 - rs/r), with rs = 2GM/c2.
However, the formula for speed of light in gravitational field is equal to v = c(1 - rs/r).

My intuition tells me that these two formulas must be the same, since time dilation must compensate the speed of light change. Am I wrong or there are other parameters that must be taken into account (like length contraction etc). Or maybe the formulas don't describe the same situation (free fall in one case and g-force in the other case).

Thank you

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stevendaryl
Staff Emeritus
Hello everyone,

I'll go straight to the question. The gravitational time dilation is equal to tearth = tspace*sqrt(1 - rs/r), with rs = 2GM/c2.
However, the formula for speed of light in gravitational field is equal to v = c(1 - rs/r).

My intuition tells me that these two formulas must be the same, since time dilation must compensate the speed of light change. Am I wrong or there are other parameters that must be taken into account (like length contraction etc).

Thank you
If we just look at radial motion, then the metric for a spherical gravitational source is given in spherical Schwarzschild coordinates by:

$ds^2 = (1-\frac{r}{r_s}) c^2 dt^2 - \frac{1}{1-\frac{r}{r_s}} dr^2$

$ds^2$ can be used both to figure out the lengths of objects and the proper times between events.

If you have two events that take place at the same value of $r$, but at different times $t$, then the proper time between the events (the time that would be shown on a clock at that location) is:

$\delta \tau = \sqrt{1-\frac{r}{r_s}} \delta t$

If you have a rod that is at rest in the coordinate system, aligned vertically, then its proper length will be given by:

$\delta L = \frac{1}{\sqrt{1-\frac{r}{r_s}}} \delta r$

So a light signal traveling from one end of the rod to the other in time $\delta t$ will have:
1. Coordinate speed = $\frac{\delta r}{\delta t}$
2. Speed as measured by a local observer (using locally inertial coordinates) = $\frac{\delta L}{\delta \tau}$
The second one will be equal to $c$, because the local speed of light has the same value in every local inertial coordinate system. So we have the equation:

$c = \frac{\delta L}{\delta \tau} = \frac{\frac{1}{\sqrt{1-\frac{r}{r_s}}} \delta r}{\sqrt{1-\frac{r}{r_s}} \delta t} = \frac{1}{1-\frac{r}{r_s}} \frac{\delta r}{\delta t}$

Or inverting that, $\frac{\delta r}{\delta t} = (1-\frac{r}{r_s})c$. That's the speed of light in the coordinates $(r,t)$.

• sha1000
If we just look at radial motion, then the metric for a spherical gravitational source is given in spherical Schwarzschild coordinates by:

$ds^2 = (1-\frac{r}{r_s}) c^2 dt^2 - \frac{1}{1-\frac{r}{r_s}} dr^2$

$ds^2$ can be used both to figure out the lengths of objects and the proper times between events.

If you have two events that take place at the same value of $r$, but at different times $t$, then the proper time between the events (the time that would be shown on a clock at that location) is:

$\delta \tau = \sqrt{1-\frac{r}{r_s}} \delta t$

If you have a rod that is at rest in the coordinate system, aligned vertically, then its proper length will be given by:

$\delta L = \frac{1}{\sqrt{1-\frac{r}{r_s}}} \delta r$

So a light signal traveling from one end of the rod to the other in time $\delta t$ will have:
1. Coordinate speed = $\frac{\delta r}{\delta t}$
2. Speed as measured by a local observer (using locally inertial coordinates) = $\frac{\delta L}{\delta \tau}$
The second one will be equal to $c$, because the local speed of light has the same value in every local inertial coordinate system. So we have the equation:

$c = \frac{\delta L}{\delta \tau} = \frac{\frac{1}{\sqrt{1-\frac{r}{r_s}}} \delta r}{\sqrt{1-\frac{r}{r_s}} \delta t} = \frac{1}{1-\frac{r}{r_s}} \frac{\delta r}{\delta t}$

Or inverting that, $\frac{\delta r}{\delta t} = (1-\frac{r}{r_s})c$. That's the speed of light in the coordinates $(r,t)$.
Thank you very much for this response. So it's because of the length contraction.

I have a question.

Is it crucial for the rod to be aligned vertically to make the expressions above hold together? What would happen with these expressions if the rod is aligned horizontally?

stevendaryl
Staff Emeritus
Is it crucial for the rod to be aligned vertically to make the expressions above hold together? What would happen with these expressions if the rod is aligned horizontally?
If you define (as you would in Euclidean space)
• $z = r cos(\theta)$
• $x = r sin(\theta) cos(\phi)$
• $y = r sin(\theta) sin(\phi)$
If you have two nearby points that have the same value of $r$, then the proper distance between them will just be the Euclidean case:

$\delta L = \sqrt{(\delta x)^2 + (\delta y)^2 + (\delta z)^2}$

So the curvilinear coordinates don't affect distances in directions perpendicular to the radial direction.

• sha1000
pervect
Staff Emeritus
Does this mean that in direction perpendicular to the radial direction the speed of light is equal to v=c*sqrt(1-rs/r)? (Same expression as time dilation)
If we pick Schwarzschild coordinates (which you appear to be doing - sort of - but without actually stating your choice), $dr/dt$ or $d\theta/dt$ or $d\phi/dt$ are all numbers which we can compute, and have been computed by previous posters. We compute these numbers, as other posters have mentioned, by solving for ds=0, where ds is the line element of the metric in our chosen coordinates.

The conversion from $d\theta/dt$ to $v=c\,\sqrt{1-rs/r}$ is a bit problematic, you're suddenly using something other than Schwarzschild coordinates., because the Schwarzschild coordiantes are $r$, $\theta$#, $\phi$. So you appear to conflating the idea of a physical speed, which isn't dependent on one's choice of coordinates, with the notion of the rate of change of a coordinate.

To take another example, isotropic coordinates will have the line element https://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution#Alternative_form_in_isotropic_coordinates

$$ds^2= \left(1+\frac{Gm}{2c^2 r_1}\right)^{4}(dx^2+dy^2+dz^2) -c^2 dt^2 \left(1-\frac{Gm}{2c^2 r_1}\right)^{2}/\left(1+\frac{Gm}{2c^2 r_1}\right)^{2}$$

One will find that dx/dt, dy/dt, and dz/dt by solving for ds=0. The "speeds" dx/dt, dy/dt, and dz/dt will all be the same, so the speed will be independent of direction, but they'll be different than the numbers you've computed, as will the time dilation factor.

Not only are isotropic coordinates a valid coordinate choice, they're the usual choice in linearized gravity.

The best and most physical answer for the speed of light uses local rulers and local clocks, and is independent of the details of the choice of coordinates. This number for the speed of light is "c". 'Speed' numbers that are not c are not really physical speeds, but an artifact of a particular choice of coordinates one makes. If one specifies the coordinates one is choosing, one can compute these numbers, but you seem to want to NOT specify the associated coordinates. To me, this suggests you are actually thinking of speeds (which are independent of a coordiante choice). But the coordinate-independent "speed" is always "c". If and when the speed of light is anything other than "c", it's an indication that the coordinates one has chosen behave differently than local clocks and rulers would measure.

• sha1000
If you define (as you would in Euclidean space)
• $z = r cos(\theta)$
• $x = r sin(\theta) cos(\phi)$
• $y = r sin(\theta) sin(\phi)$
If you have two nearby points that have the same value of $r$, then the proper distance between them will just be the Euclidean case:

$\delta L = \sqrt{(\delta x)^2 + (\delta y)^2 + (\delta z)^2}$

So the curvilinear coordinates don't affect distances in directions perpendicular to the radial direction.
I'm still a little bit confused. The expression of the speed of light is as a function of r, v= c(1-rs/r). This expression is true for the speed of light in perpendicular direction (same r). So why we need to consider the movement of the light in vertically aligned rod and not horizontally aligned in order to make the demonstration from your first answer?

The best and most physical answer for the speed of light uses local rulers and local clocks, and is independent of the details of the choice of coordinates. This number for the speed of light is "c". 'Speed' numbers that are not c are not really physical speeds, but an artifact of a particular choice of coordinates one makes. If one specifies the coordinates one is choosing, one can compute these numbers, but you seem to want to NOT specify the associated coordinates. To me, this suggests you are actually thinking of speeds (which are independent of a coordiante choice). But the coordinate-independent "speed" is always "c". If and when the speed of light is anything other than "c", it's an indication that the coordinates one has chosen behave differently than local clocks and rulers would measure.
Thank you for this clarification.

stevendaryl
Staff Emeritus
I'm still a little bit confused. The expression of the speed of light is as a function of r, v= c(1-rs/r). This expression is true for the speed of light in perpendicular direction (same r). So why we need to consider the movement of the light in vertically aligned rod and not horizontally aligned in order to make the demonstration from your first answer?
I'm not sure I understand your point. The coordinate speed of light is not necessarily the same in every direction. I thought you were asking in the first post how to derive the result v = c(1 - rs/r). That is the coordinate speed of light in the radial direction, according to Schwarzschild coordinates.

• sha1000
I'm not sure I understand your point. The coordinate speed of light is not necessarily the same in every direction. I thought you were asking in the first post how to derive the result v = c(1 - rs/r). That is the coordinate speed of light in the radial direction, according to Schwarzschild coordinates.
And what is the speed of light in the perpendicular direction as a function of r ? Is it v = c*sqrt(1-rs/r) or am I talking non-sense?

stevendaryl
Staff Emeritus
And what is the speed of light in the perpendicular direction as a function of r ? Is it v = c*sqrt(1-rs/r) or am I talking non-sense?
The point made by @pervect is that it's not a particularly meaningful notion. But yes, you can certainly define a kind of velocity for light in the perpendicular direction:

$v =$ (distance traveled in the perpendicular direction)/(change in coordinate time), and that would turn out to be your expression.

pervect
Staff Emeritus
Distance in special relativity depends on the velocity of the observer. A moving observer won't measure the same distances that a static observer does. In this context, a static observer would be an observer hovering in place, at constant Schwarzschild r, theta, and phi coordinates.

For a static observer in the Schwarzschild metric, the perpendicular distance will be $r \, d\theta$, or $r \, \sin \theta \, d \phi$ where $\sin \theta$=1 for equatorial orbits ($\theta=90$).

The radial distance for a static observer will be $dr / \sqrt{1 - r_s/r}$. So r coordinate changes don't measure the radial distances measured by a static observer.

The time on a clock held by for a static observer will be $d\tau = \sqrt{1-r_s/r} \, dt$, where $d\tau$ is the proper time of the clock measured by the static observer, and dt is the Schwarzschild coordinate time. This difference in clock rates is usually called time dilation.

There is an unfortunate tendency for people to try to use the notion of Schwarzschild coordinate time as a drop-in replacmeent for the absolute time of Newtonian physics. This doesn't actually work, so I try to warn people about it, though it's unclear how many listen.

With this information, one should be able to compute static distances / static observer times, or static distances / coordinate times, or coordinate changes / coordinate times.

Of these quantities, the ratio of static distances / static observer times has the most physical significance when interpreted as a velocity, and it's equal to "c".

This post doesn't explain how non-static observers would measure distances, times, or velocities, however. There is one easy answer, of course. That is if one uses local definitions of distances and local measurements of time with local clocks and rulers, the velocity of light will be "c", as it always is. To understand why a moving observer in the Schwarzschild geometry measures the speed of light to be "c" isn't much harder - or much easier - than understanding why the velocity of light in special relativity is always "c". To change frames of reference (from a stationary frame to a moving one) one needs to take into account length contraction, time dilation, AND the relativity of simultaneity. The last factor is the one that is most likely to be omitted, but one will not get correct results without taking it into account.

• sha1000