# Time dilation effects of relative velocity vs gravitational force?

1. Dec 1, 2012

### Fredster1765

I've been wondering.. If a person (A) is, say, on top of Mt. Everest, he would be moving faster than a guy (B) at the foot of the mountain since A is further from the center of the earth, but he would also be experiencing a weaker gravitational force than B.
So, for which of the invidividuals would time appear to be travelling slower?

In other words, can it be said that gravitation has a greater or lesser impact on time dilation than relative motion?

Hope the question isn't completely nonsensical..

2. Dec 1, 2012

### pervect

Staff Emeritus
All clocks on the geoid (roughly speaking, at seal level) tick at the same rate. Since the clock on the mountain is presumably above the geoid, it will tick faster.

3. Dec 1, 2012

### Staff: Mentor

The gravitational effect is greater than the velocity effect for objects near the surface of the earth at rest in the earth centered frame.

4. Dec 1, 2012

### tom.stoer

The following calculation may help. We start with the proper time τ for a worldline C

$$\tau = \int_C d\tau$$

Now let's use the Schwarzschild metric for the gravitational field of the earth; we use proper time τ and coordinates (t,r,Ω):

$$d\tau^2 = f\,dt^2 - f^{-1}\,dr^2 - r^2\,d\Omega^2$$

with

$$f(r) = \left(1-\frac{r_s}{r}\right)$$

rs is the so-called Schwarzschild radius.

Let's assume vanishing radial velocity, i.e. dr=0, and constant angular velocity ω at fixed height r. The proper time becomes

$$d\tau^2 = dt^2\left(f(r) - r^2\,\omega^2\right)$$

b/c all terms are time-independent the integral is simply

$$\tau = \int_0^t dt \sqrt{\left(1-\frac{r_s}{r}\right) - r^2\,\omega^2} = t \; \sqrt{\left(1-\frac{r_s}{r}\right) - r^2\,\omega^2}$$

One finds that the gravitational effect in the square root is ~1/r whereas the effect due to the velocity is ~v² = ω² r²

(note: as usual I set c=1)

Now one can compare proper times for different observers following different world lines; each world line is specified by a fixed radius r, a fixed angular velocity ω and a "duration" t. This coordinate time t is the proper time of a stationary observer with ω=0 at r=∞:

$$\tau_{\omega=0,\,r\to\infty} = \int_0^t dt = t$$

Last edited: Dec 1, 2012
5. Dec 1, 2012

### Fredster1765

Perhaps I should have clarified that the answer I was looking for was more conceptual, like the two first responses :P
I don't study physics (or math for that matter), so those calculations are entirely meaningless to me - I feel bad now, seeing as you went through all the trouble of laying it out mathematically! :)

6. Dec 1, 2012

### andrien

If someone has certain speed moving in gravitational field,then time will slow down as speed increases and if he goes higher than,gravitation effects will become lesser and his clock runs fast.It will be the combination of these two effects which will decide whether his clock is running fast or getting slow down.

7. Dec 1, 2012

### Fredster1765

Yes, hence my question of which of the two influences would be more substantial. Although I see now that it's naive to ask that question and expect a concrete non-mathematical answer.. Guess I should've taken physics/math A levels instead! Oh well.

8. Dec 1, 2012

### Staff: Mentor

Dalespam gave you the concrete non-math answer in post #3.

9. Dec 1, 2012

### Fredster1765

Of course, but my question was quite specific. If I want a more general sense of which force is greater given different velocities and distances from the gravitational center of an object, which I assume the math says something about (if not, then we're back to my lacking qualifications), I'd say it becomes a prerequisite to understand the equations. That probably wasn't clear from my earlier post.
But yes, I know the answer to my initial question was provided by posts #2 and 3.

10. Dec 1, 2012

### tom.stoer

we are very close to a quantitative comparison of velocity-dependent and gravitational effect; and the formulas will become much simpler; interested?

11. Dec 1, 2012

### phinds

Just FYI, I can't give you a link or the math to support it, but I remember reading on this forum that in order to keep you from driving through cornfields, the GPS system has to account for both effects and the result is roughly this: special relativity (speed) has an effect of 7 microseconds per day and general relativity (gravity) has an effect of 45 microseconds per day.

1) Those maybe should be nanoseconds, not microseconds ... I don't remember --- I focused on the ratio of 7 to 45
2) These specific figures depend on the speed and distance from the center of the Earth of the GPS satellites.

12. Dec 1, 2012

### Staff: Mentor

It's microseconds. But a key point is that the two effects are in opposite directions: the SR effect makes GPS clocks run slow relative to clocks on the Earth's surface, while the GR effect makes them run faster. The combined effect is that GPS clocks, if not compensated, run 45 - 7 = 38 microseconds a day fast compared to clocks on the Earth's surface. (I say "if not compensated" because there is an extra oscillator on board each satellite that corrects its clock rate to match the rate of ground clocks, by applying a frequency offset to the base clock signal.)

The best source of info I know on relativistic effects in the GPS system is this paper by Neil Ashby:

http://relativity.livingreviews.org/Articles/lrr-2003-1/ [Broken]

There are actually a number of other, smaller effects in addition to the two we've discussed. You're correct that all of these effects depend on the exact orbital parameters, which are slightly different for each satellite (and which also change over time as adjustments are made to the orbits).

Last edited by a moderator: May 6, 2017
13. Dec 1, 2012

### phinds

Ah. I DO remember that now, and should have thought it through, but just posted without thinking. Thanks for that clarification.

14. Dec 1, 2012

### pervect

Staff Emeritus
Near the Earth's surface, raising a clock makes it tick faster. But as you get further away from the Earth's surface, things get more complicated, depending on the exact velocity / height relationship. If you consider objects that appear not to move relative to the Earth's surface, for instance, at some height the object's velocity r*omega will approach the speed of light, and the time dilation will approach infinity

So the answer that is correct and sufficeint in the context of comparing "surface to mountaintop" might not apply to the situation of "satellite v satellite" or "my mountain is really a space-elevator".

15. Dec 1, 2012

### Staff: Mentor

The actual equations that I've seen for GPS (for example, in the Ashby paper I linked to) explicitly separate velocity from height, even for an observer that's at rest on the Earth's surface. These calculations use the Earth-Centered Inertial (ECI) frame, which is not a rotating frame (i.e., it does not rotate with the Earth). So an observer at rest on the Earth's surface has a non-zero velocity relative to this frame, and that appears in the formulas along with his height. (The time coordinate in ECI is scaled, IIRC, so that coordinate time happens to be the same as proper time for an observer at rest on the geoid, rotating with the Earth. But that doesn't change the equations, just the scaling of the time coordinate.)

16. Dec 2, 2012

### jartsa

A photon climbing to the top of Mt. Everest loses energy, because of the gravity.

A photon climbing to the top of Mt. Everest gains some energy, because of the centrifugal force. This gain is about 3/1000 of the loss.

The photon's frequency is changed, 0.3 % of the change is caused by the motion, 99,7 % is caused by the gravity.

So the answer is that 99.7 % of the clock rate change is by the gravity.

I assumed that g is 10 m/s, and that Mt. Everest is at the equator, where centrifugal acceleration is 0.03 m/s

17. Dec 2, 2012

### tom.stoer

I still propose to do calculations instead of speculations

18. Dec 2, 2012

### arindamsinha

Yes, that is really what resolves this easily.

Assuming that the OP's question is based on both observers having the same angular velocity around the center of Earth, there will be a specific altitude where the gravitational and velocity effects cancel out (giving zero time dilation w.r.t. Earth surface). For simplicity's sake, the observer on Earth surface can be assumed to have zero velocity without affecting the result by much.

Below the zero time dilation altitude, higher clocks will tick faster than on Earth surface. Above, higher clocks will tick slower.

'Dominate' (i.e. greater of lesser impact) is rather a subjective word in this context, but at the height of Mt. Everest, gravitational effects will 'dominate' (i.e. velocity time dilation will be negligible in comparison).

I haven't done the calc, but it should be pretty easy. If you don't mind, tom.stoer, please post the zero time dilation altitude answer.

Last edited: Dec 2, 2012
19. Dec 2, 2012

### tom.stoer

OK, let's continue with

$$\tau = t \; \sqrt{\left(1-\frac{r_s}{r}\right) - r^2\,\omega^2}$$

First we make a Taylor expansion for the square root; that means we assume

$$\frac{r_s}{r} \ll 1$$
$$r^2\,\omega^2 \ll 1$$

(again: c=1)

$$\tau = t \; \left(1-\frac{1}{2}\left(\frac{r_s}{r} + r^2\,\omega^2\right)\right)$$

Then we introduce the radius of the earth rE and the altitude h:

$$r = r_E + h$$

and make another Taylor expansion in h/rE

$$\tau = t \; \left(1-\frac{1}{2}\left(\frac{r_s}{r_E(1+h/r_E)} + r^2\,\omega^2\right)\right) = t \; \left(1-\frac{1}{2} \left( \frac{r_s}{r_E} \left( 1 - \frac{h}{r_E} \right) + r^2\,\omega^2\right)\right)$$

Remember that the coordinate t is the proper time of a stationary = non-corotating observer at r → ∞. For a non-corotating observer at h=0 (she will sit at h=0 and will see a fast moving surface of the earth!!) the result is

$$\tau(h=0,\, \omega=0) = t \; \left(1-\frac{r_s}{2r_E} \right)$$

So in general we have

$$\tau(h,\, \omega) = \tau(h=0,\, \omega=0) + \Delta\tau(h,\, \omega)$$

with

$$\Delta\tau(h,\, \omega) = \frac{t}{2} \; \left(\frac{r_sh}{r_E^2} - \frac{r^2\,\omega^2}{c^2}\right)$$

where I re-intruduced c to allow for explicit calculations

The missing piece is the Schwarzschild radius

$$r_s = \frac{2GM_E}{c^2}$$

In the next step we get rid of the unobservable coordinate t and express t in terms of the proper time of the non-corotating observer at h=0; in addition we use (again) the approximation

$$r \simeq r_E$$

$$\frac{\Delta\tau(h,\, \omega)}{\tau(0,\,0)} = \frac{1}{2} \frac{ \frac{r_sh}{r_E^2} - \frac{r^2\,\omega^2}{c^2} }{ 1-\frac{r_s}{2r_E} } \simeq \frac{1}{2} \frac{ \frac{r_sh}{r_E^2} - \frac{r_E^2\,\omega^2}{c^2} }{ 1-\frac{r_s}{2r_E} }$$

This is the final result for small velocities v = rω and small altitude h.

As you can see there is increasing (Delta of) proper time τ with increasing h (weaker gravitational field) and decreasing (Delta of) proper time with increasing v = rω; the second term v²/2c² is well-known from SR. The denominator is a correction due to the fact that already at h=0 we have some gravitational field which slows down time.

This formula should be applicable for an airport at h=0 and an airplane at e.g. h=10km which is small compared to rE. Please remember that ω=0 does apply to a non-corotating observer, so in order to use the formula for airports and airplanes one would introduce the velocities

$$v_\text{airport} = r_E\,\omega_\text{airport}$$

and

$$v_\text{airplane} = v_\text{airport} + \Delta v_\text{airplane}$$

where Δv is the velocity of the airplain w.r.t. the airport, e.g. ±800km/h for westward / eastward direction; this velocity is of course small compared to c=300000km/h. So using this formula you could immediately analyse the Hafele–Keating experiment

http://en.wikipedia.org/wiki/Hafele–Keating_experiment

20. Dec 2, 2012

### tom.stoer

remark: of course in principle there is no need for an approximation; one can use

$$\frac{\tau(r_E+h,\,\omega_E + \omega)}{\tau(r_E,\,\omega_E) } = \sqrt{\frac{ 1-\frac{r_s}{r_E+h} - \frac{(r_E+h)^2\,(\omega_E+\omega)^2}{c^2} }{ 1-\frac{r_s}{r_E} - \frac{r_E^2\,\omega_E^2}{c^2} }}$$

directly