# Equivalence of formulas for time reduction factors in GR

1. Dec 28, 2015

### Dema

Hello,

I’d have a question regarding two apparently different formulas for the time reduction factors for observers in orbit around a rotating black hole, as reported in this nice thread: Equation for time dilation of body in orbit around Kerr black hole?

The first one is:

$$A=\sqrt{g_{tt} + 2\Omega g_{\phi t}+\Omega^2 g_{\phi \phi}}$$

While the second one is:

$$A=\alpha\cdot\sqrt{1-v_{\pm}^2}$$

where $\alpha$ is the time reduction factor for a steady observer and $v_{\pm}$ the velocity with respect this observer.
I guess the first one is a direct consequence of the general relativistic formula for proper time $d \tau = \sqrt{g_{\mu \nu} \dot{x^{\mu}} \dot{x^{\nu}}}dt$, where the dot refers to derivative with respect the coordinate time, and is obtained for the special case of a circular orbit around a rotating black hole in the equatorial plane, where $\frac{d\phi}{dt} = \Omega$

On the other hand, I have no idea how the second one is derived.

A difference I noticed on the formulas, is that the velocity appearing on the first one is that observed at infinity, while the velocity appearing on the second one is that measured by a stationary observer.

So, I’m interested to know:
1. If my derivation of the first formula is correct
2. How the second formula is derived in the general case
3. How are they related
4. Are the general formulas $\frac{d \tau}{ dt}= \sqrt{g_{\mu \nu} \dot{x^{\mu}} \dot{x^{\nu}}}$ and $A=\alpha\cdot\sqrt{1-v_{\pm}^2}$ valid, with suitable substitutions, for any particular motion in a generic spacetime (I mean not only circular orbits in equatiorial planes for back holes)?
I would appreciate detailed calculations, if possible.

Thank you a lot!

D

2. Dec 28, 2015

### bcrowell

Staff Emeritus
This doesn't sound right to me. There is really no concept in GR of a velocity observed at infinity. There really can't be, because velocity is a vector, and transporting a vector to infinity will give different results depending on the path. Maybe what you mean here is what I would call a coordinate velocity, which is something that's not especially meaningful physically.

3. Dec 29, 2015

### Dema

Thank you for the clarification, I was a bit too quick in my explanation.
I understand there is no concept such velocity observed at infinity in GR and in the formula $\frac{d \tau}{ dt}= \sqrt{g_{\mu \nu} \dot{x^{\mu}} \dot{x^{\nu}}}$ the derivatives are those of coordinates with respect coordinate time, let's call them coordinate velocity.
However, for the specific case of particles in orbit around a black hole, I meant $\frac{d\phi}{dt} = \Omega$ to be the angular velocity as measured by a far away observer (a one sitting where the metric is almost flat) and satisfying Kepler's third law $M = \Omega^2 r^3$.
Provided these clarifications, what about the reported formulas for time reduction factors?

Thank you
D

4. Dec 29, 2015

### bcrowell

Staff Emeritus
I don't think $\frac{d\phi}{dt}$ has anything to do with a distant observer either. $\phi$ and $t$ are just coordinates.

5. Dec 29, 2015

### Staff: Mentor

For this to work, the faraway observer ("faraway" here means $r \rightarrow \infty$) also has to be at rest relative to the hole (i.e., with $\theta, \phi$ constant). (And you have to be using a particular coordinate chart, called "Boyer-Lindquist coordinates", on Kerr spacetime.)

Even with the specifications above, this will not be true if the hole is rotating; it only holds for the case of a non-rotating hole, i.e., Schwarzschild spacetime.

This is true as a general statement, but for the particular case of Kerr spacetime in Boyer-Lindquist coordinates, $d\phi / dt$ actually does have the physical interpretation stated (with the caveats I gave above).

6. Dec 29, 2015

### Staff: Mentor

The first thing to get clear is, what do you mean, physically, by "time reduction factors"? How would you actually measure them in an experiment?

7. Dec 30, 2015

### Dema

Thank you for the replies and clarifications.

I'll try to do my best to explain what I mean by "time reduction factor".
I would say that "time reduction factor" can be defined as the ratio of time durations of the same phenomenon as measured by a clock carried by an observer orbiting the BH ($\Delta T_{\text{orb}}$) and as measured by another identical clock carried by the faraway observer, at rest in a nearly flat spacetime region, let's say $\Delta T_{\text{inf}}$.
In Boyer-Lindquist coordinates $\Delta T_{\inf}$ would be the coordinate time interval $\Delta t$ (i.e. the proper time for the faraway guy). $\Delta T_{orb}$ on the other hand is the interval in proper time of the orbiting guy: $\Delta \tau$.
The time reduction factor would be therefore $A = \frac{\Delta T_{\text{orb}}}{\Delta T_{\inf}} = \frac{\Delta\tau}{\Delta t}$, which becomes $A = \frac{d\tau}{dt}$ when the measured durations goes to 0. By definition $A \leq 1$.
For the specific case of an observer orbiting a BH, $A < 1$. The phenomenon can be the orbital period: the faraway observer should measure a period $T_{\text{far}}$ longer than that measured by the orbiting observer $T_{\text{orb}}$, so that $T_{\text{far}} = T_{\text{orb}} / A > T_{\text{orb}}$.

Thanks
D.

8. Dec 30, 2015

### Staff: Mentor

How would the "time durations" of the phenomenon be measured? Let's make it easy by supposing that the phenomenon in question is taking place at the location of the observer orbiting the BH, so his clock can measure its duration $\Delta T_{\text{orb}}$ directly. How will the observer at infinity measure the time duration $\Delta T_{\text{inf}}$ of the same phenomenon?

9. Dec 31, 2015

### Dema

Hmmm....

the easiest way I can conceive is the following.

Let me assume for the moment that the metric is Schwartzschild (the BH does not rotate) and coordinates are the usual Schwartzschild ones ($t, r, \theta, \phi$)
Let's now assume that the orbiting observer emits a single light pulse toward the faraway one at the beginning and at the end of the phenomenon. For the orbiting observer, this elapsed proper time would be $$\Delta T_{\text{orb}} = \Delta \tau_{\text{orb}} = \sqrt {g_{tt} + g_{\phi\phi}\Omega^2} \Delta t = A \Delta t$$
The faraway observer will receive the light pulses with a time difference of $$\Delta T_{\text{inf}} = \Delta \tau_{\infty} \approx \Delta t$$
As a consequence, the ratio of measured durations for the same phenomenon would be:
$$\frac{\Delta T_{\text{inf}}}{\Delta T_{\text{orb}}} = \frac{1}{A}$$
This $A$ factor (and its inverse) is what I call "time reduction factor".

If my reasoning above is correct, this would explain the first formula in my original post, but not the second one.

Thanks
D.

10. Dec 31, 2015

### Staff: Mentor

This will only be true if the "orbiting" observer is not actually orbiting but "hovering" at a constant altitude and with constant angular coordinates; otherwise the distance between the orbiting observer and the faraway observer will change with time, so the light travel time for the two pulses will not be the same.

11. Jan 1, 2016

### Dema

True,
but if I already assume that the distance of the faraway observer is much greater than the orbit diameter, (approximation already needed to guarantee that $g_{tt} \approx 0$) than the difference in time travel of pulses due to the displacement on the orbit wouldn't be negligible as well?

If not, I could imagine that $\Delta T_{\text{inf}} = \Delta \tau_{\infty} \approx \Delta t +\delta t$ where the additional time delay $\delta t$ would be proportional to (the projection along the line of view of ) the displacement and the velocity, let's call them $\Delta S$ and $v_S$: $\delta t \varpropto \Delta S / v_S$. I'm not sure of what I've just written...

Thanks
D.

12. Jan 1, 2016

### Staff: Mentor

No, you can't assume this. (In fact, you can't even assume that the faraway observer can receive light signals from every point on the orbit; the central gravitating object could be in the way.) The difference in travel time of pulses from opposite sides of the orbit is comparable to the period of the orbit itself, and therefore is at least of the same order as (if not larger than) the duration of whatever phenomenon you are trying to measure. So if you ignore time differences of that order of magnitude, you can't even measure the duration of the phenomenon you're interested in.

Also, the limit for the faraway observer is not $g_{tt} \approx 0$, it's $g_{tt} \approx 1$. (Or $-1$, depending on which metric signature convention you are using.)

13. Jan 3, 2016

### Dema

Thank you for the clarification.
I tried to work out a way to include this in the definition of $A$.

On the simpler situation in which the emitting body is moving on a straight line (i.e. in a 1-dimensional problem) the duration observed by the far observer should be $\Delta T_{\text{inf}} \approx \Delta t (1+\frac{v}{c})$, where $v$ is the speed of the emitter, so that $v\Delta t$ is the additional path the emitter moved down during the (coordinate) time span $\Delta t$.
But when I come to consider the additional path ($\Delta s$) of light signals coming from an orbiting emitter, even when assuming a circular orbit, I come up with a cumbersome term which depends on the angle spanned on the orbit $\theta = \Omega t$, the orbit radius $R$ and on the distance $r_0$ between the observer and the emitter at the beginning of the transmission:
$$\Delta s = \sqrt{2R^2(1-\cos\theta) +2r_0 R (1-\cos\theta) + r_0^2} - r_0$$
I think there is a clever way to account for the emitter movement, but I can't find it now...

Ops... sorry: it was a typo.

Thanks
D.

14. Jan 3, 2016

### Staff: Mentor

I think you need to review the relativistic Doppler effect:

https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

Your formula doesn't look like the standard relativistic Doppler formula, which is what applies to this case (emitting body moving in a straight line). The formula assumes flat spacetime, but if the emitter is moving in a straight line you are basically assuming that anyway (if gravity were present its trajectory as seen by the stationary observer would be curved). I would recommend taking some time to understand this case before even trying to tackle the more complicated case of the emitter being in a circular orbit around a gravitating body.

15. Jan 9, 2016

### Dema

Thank you very much for your suggestion. I took some days to review the relativistic Doppler effect on some textbooks and I recognized the mistake I did in my previous evaluation, at least when the emitting body is moving on a straight line: I didn't take into account the special relativity time dilation factor $\gamma$.

On flat spacetime, the right formula for the ratio between observed wave lenght $\lambda_\text{o}$ and emitted wave lenght (as measured at the source $\lambda_\text{s}$) should be $$\frac{\lambda_\text{o}}{\lambda_\text{s}} \varpropto \frac{\Delta T_{\text{inf}}}{\Delta t / \gamma} = \gamma (1 + \frac{v}{c})$$ where $dt$ is the time, measured by the observer at rest, during which the source emits 2 subsequent wave crests, so the same time measured in the rest frame of the moving source is $dt/\gamma$

Unfortunately, as you already pointed out, this formula is valid on flat spacetime, so I'm not sure whether and how it could be applied on curved spacetime.

Here is my thinking, on the simple case where the observer is at rest at infinity and the source is falling on a spherical gravitational field (i.e. I suppose a Schwartzschild geometry) alond a radial path from Point 1 to Point 2 at radial coordinates $r_1$ and $r_2$.

The source is emitting 2 pulses, one at Point 1 and another at Point 2, in a proper time $\Delta\tau = \int_{r_1}^{r_2} d\tau = \int_{r_1}^{r_2}\sqrt{g_{tt} + g_{rr} dr/dt}\, dt = A \Delta t \varpropto \lambda_\text{s}$, where I supposed that P1 and P2 are close enough so that $A \approx \text{cost}$.
The faraway observer measures a time $\Delta T_{\text{inf}} = \Delta t \varpropto \lambda_\text{o}$ between the 2 pulses.
The source is moving, respective to the faraway observer, at velocity $v = dr/dt$, leading to a special relativity time dilation factor $\gamma = (1-\frac{v^2}{c^2})^{-1}$.
The source is therefore time delayed, with respect to infinity, by a gravitational $A$ factor and a special relativity $\gamma$ factor.
Putting all together, the ratio of wavelenght and time should be:
$$\frac{\lambda_\text{o}}{\lambda\text{s}} \varpropto \frac{\Delta T}{A \Delta T / \gamma} = \frac{\gamma}{A} \cdot (1+\frac{v}{c})$$
Could you please correct me if I'm wrong, pointing out the correct reasoning I have to do or a reference explaing this?

Thank you

D.

16. Jan 9, 2016

### Staff: Mentor

The formula you give is only true in the special case where the source and observer are moving directly away from each other. (It also gives the ratio of wavelengths; the ratio of frequencies, which is what is often talked about, is the inverse of this.) For the case of a source in a circular orbit, which is what you were talking about earlier, the formula is more complicated.

No, it isn't. $dr/dt$ is a coordinate velocity, not a special relativistic relative velocity. See below.

You have obtained the right result, but your reasoning to arrive at it is wrong. Here is the correct way of deriving the result.

First, imagine an observer at radius $r$ who sees the source free-falling past him. Since the observer and the source are in the same local inertial frame (because we're considering a short time period around the time when the source just passes the observer), we can assign a relative velocity $v$ to the source, with respect to this observer, and therefore a Doppler shift $\gamma \left( 1 + v/c \right) = \sqrt{(1 - v/c)/(1 + v/c)}$ in the wavelength of the light emitted by the source, relative to this observer.

But the observer we are really interested in is at infinity. So we now have to compute the additional redshift due to the light traveling from radius $r$ out to infinity along a purely radial path. This is given by $1 / A$, or $1 / \sqrt{1 - 2GM / c^2 r}$. So the final observed redshift is given by your formula; but not because the velocity $v$ is a "relative velocity" between the source and the observer at infinity. To see why, try working out the case of a source in a circular orbit, using the method I just used above (i.e., think of a family of "hovering" observers at radius $r$ who see the source going past them, emitting light towards the observer fixed at infinity--note that for all but one of these observers, the light won't be traveling radially outward). The source's coordinate velocity $v$ is constant (though it is now given by $r d\varphi / dt$ instead of $dr / dt$), and its radius $r$ is constant so the factor $A$ is constant, but the redshift of its light, seen by an observer at infinity, is not.

17. Jan 13, 2016

### Dema

Thank you again for having showed me the right way to do the calculation. I think now I got it.
As you suggested, I tried to apply it for a source on a circular orbit in the equatorial plane of a Schwarzschild BH, by considering a ring of static observers at radius R and an observer at rest at infinity, in the same equatorial plane.

For a generic hovering observer the source is measured with a velocity $v$ tangent to its local $e_{\hat \phi}$ axis. If I got the calculations right, the angle between the velocity and the direction of the faraway observer is $\pi / 2 + \phi$, where $\phi$ is the angle spanned by the source (i.e. the angle between the radius connecting the coordinate origin to the faraway observer and the radius connecting the coordinate origin to the hovering observer). So, for each static observer the time dilation factor is $\gamma (1+\frac{v \, \sin\phi}{c})$. Of course, only for the observer at $\phi = \frac{3}{2}\pi$ the velocity is parallel to the light.

The additional (gravitational) time dilation factor is then given by $\frac{1}{A} = \text{cost} = 1 / \sqrt{1-2GM / c^2 R}$, providing the total time dilation factor (or the ratio of wavelenghts) relative to the observer at infinity: $$\frac{\lambda_\text{o}}{\lambda\text{s}} \varpropto \frac{\gamma}{A} \cdot (1+\frac{v \, \sin\phi}{c})$$

If this result is correct, I can therefore prove that the "average" factor is $\frac{\gamma}{A}$.

Thanks
D.

18. Jan 13, 2016

### Staff: Mentor

On an initial skim it looks correct. Note, however, that what you are calling the "time dilation factor" is better termed a Doppler shift factor; it is the ordinary relativistic Doppler shift of the light from the source, as measured by the hovering observer.

Also note that we are assuming that the gravitational factor is the same even though all of the light rays except for one (the one emitted by the source just as it passes the radial line from the center of coordinates to the faraway observer) do not move on purely radial paths. This assumption turns out to be valid, but it might not be immediately obvious why it is.

19. Jan 14, 2016

### Dema

If the gravitational factor is given by $1 / \sqrt{1-2GM / c^2 R}$, why should it not appear trivially constant, as it depends only on the radius R of the circular orbit?

Thanks
D.

20. Jan 14, 2016

### Staff: Mentor

For a given circular orbit, i.e., a given $R$, it is constant; light emitted from anywhere on that orbit and received by an observer at rest at infinity has the same gravitational redshift factor--more precisely, light emitted by a static observer anywhere at the same $R$ has the same redshift factor when received at infinity, and this is what is meant by the "gravitational redshift factor". What varies from point to point along the orbit is the extra factor due to the motion of the emitter, relative to a static observer at the same $R$.