Relative motion and time dilation

[TurtleMeister]

I've been reading the threads in this forum about relative motion and time dilation and I would like to know if I am understanding things correctly.

The atomic clocks aboard the GPS satellites are programmed with a "factory offset" (according to Wikipedia) of +7us/day for relative motion and -45us/day for gravitational time dilation for a total of -38us/day. So, if I were in the same Earth orbit as the GPS satellite with two atomic clocks, and I want to send one of them to the Earth's surface, how would I need to program it's clock so that it would be in sync with the one that remains with me on the satellite?

My answer is +52us/day. Is this correct?

 

[yuiop]

My initial guess is +38us/day.

 

[TurtleMeister]

My initial guess is +38us/day.

That was my initial guess also. But then I changed my mind after reading some threads here. So I guess I'm still confused. :)

 

[TurtleMeister]

Now I'm thinking that +38us/day is the correct answer. My reason: The clock hypothesis, which has been experimentally tested to be correct to a high degree of precision, states that the tick rate of a clock when measured in an inertial frame depends only upon its velocity relative to that frame, and is independent of its acceleration or higher derivatives. The key words here being "inertial frame" and "relative to that frame". So if I consider the Earth and the satellite as being in the same frame, then in order for the frame to be inertial I must consider the frame to be referenced to the center of mass. Viewing it this way it becomes obvious that the velocity of the clock on the satellite is higher than the velocity of the clock on the Earth's surface. Therefore, time adjustments to a clock sent from the satellite to the Earth's surface must be adjusted -7us/day (for relative motion) because of the decrease in velocity.

Is this correct, or am I still confused?

 

[TurtleMeister]

Does anyone here know if +38us/day is the correct answer? Regardless of whether my reason is correct or not.

 

[JesseM]

Now I'm thinking that +38us/day is the correct answer. My reason: The clock hypothesis, which has been experimentally tested to be correct to a high degree of precision, states that the tick rate of a clock when measured in an inertial frame depends only upon its velocity relative to that frame, and is independent of its acceleration or higher derivatives. The key words here being "inertial frame" and "relative to that frame". So if I consider the Earth and the satellite as being in the same frame, then in order for the frame to be inertial I must consider the frame to be referenced to the center of mass. Viewing it this way it becomes obvious that the velocity of the clock on the satellite is higher than the velocity of the clock on the Earth's surface. Therefore, time adjustments to a clock sent from the satellite to the Earth's surface must be adjusted -7us/day (for relative motion) because of the decrease in velocity.

No frame covering a non-infinitesimal region of curved spacetime can qualify as "inertial", inertial frames only exist in flat spacetime, or "locally inertial frames" in infinitesimal regions of curved spacetime (see the http://www.aei.mpg.de/einsteinOnline/en/spotlights/equivalence_principle/index.html ).

Aside from that, can you explain you're getting these numbers? Pervect's post #35 here suggests you can't actually break down the time dilation experienced by satellite clocks in an Earth-centered coordinate system into a simple sum of velocity-based time dilation and gravitational time dilation, as you seem to be suggesting.

 

[TurtleMeister]

Thanks for the reply JesseM. So my reason is incorrect but not my answer?

No frame covering a non-infinitesimal region of curved spacetime can qualify as "inertial", inertial frames only exist in flat spacetime, or "locally inertial frames" in infinitesimal regions of curved spacetime (see the equivalence principle).

I guess I'm not understanding this at all because it seems contradictory to me. Isn't all "real" space time curved?

According to Wikipedia a COM frame can be an inertial frame.

A special case of the center of momentum frame is the center of mass frame: an inertial frame in which the center of mass (which is a physical point) is at the origin at all times. In all COM frames, the center of mass is at rest, but it may not necessarily be at rest at the origin of the coordinate system.

Aside from that, can you explain you're getting these numbers? Pervect's post #35 here suggests you can't actually break down the time dilation experienced by satellite clocks in an Earth-centered coordinate system into a simple sum of velocity-based time dilation and gravitational time dilation, as you seem to be suggesting.

I got the numbers from Wikipedia. And it appears that the programming of the GPS clocks for time dilation are a simple sum of relative motion based time dilation and gravitational time dilation. However, I admit that I may be misinterpreting this.

 

[yuiop]

Aside from that, can you explain you're getting these numbers? Pervect's post #35 here suggests you can't actually break down the time dilation experienced by satellite clocks in an Earth-centered coordinate system into a simple sum of velocity-based time dilation and gravitational time dilation, as you seem to be suggesting.

Starting with this equation given by pervect here


$$c^2 \left( \frac{d\tau}{dt} \right)^2 = c^2 \, \left(1-\frac{r_s}{r}\right)^2\ - r^2 \left(\frac{d\phi}{dt}\right)^2$$

[EDIT] I think pervect made a small typo here and it should have been:

$$c^2 \left( \frac{d\tau}{dt} \right)^2 = c^2 \, \left(1-\frac{r_s}{r}\right)\ - r^2 \left(\frac{d\phi}{dt}\right)^2$$

(I have edited the rest of the text to allow for this)

and given that instantaneous tangential coordinate velocity u is related to the angular velocity by:

$$u = r \left(\frac{d\phi}{dt}\right)$$

then pervect's equation can be rewritten as:

$$c^2 \left( \frac{d\tau}{dt} \right)^2 = c^2 \, \left(1-\frac{r_s}{r}\right)\ - u^2$$

so:

$$\left( \frac{d\tau}{dt} \right)^2 = \, \left(1-\frac{r_s}{r}\right)\ - \frac{u^2}{c^2}$$

Now u in the above equation is the instantaneous tangential velocity as measured by the Schwarzschild observer at infinity.
The local velocity v is related to the coordinate velocity by:

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

[EDIT] I made an error here. It should have been:

$$u = v \sqrt{1-\frac{r_s}{r}}$$

(Further editing has been done below, to correct the propogation of my error in the text.)

Substituting:

$$\left( \frac{d\tau}{dt} \right)^2 = \, \left(1-\frac{r_s}{r}\right)\ - \frac{v^2}{c^2} \left(1-\frac{r_s}{r}\right)$$

Factoring:

$$\left( \frac{d\tau}{dt} \right)^2 = \, \left(1-\frac{r_s}{r}\right)\ \left(1 - \frac{v^2}{c^2} \right)$$

Taking the square root of both sides:

$$\frac{d\tau}{dt} = \, \sqrt{1-\frac{r_s}{r}}\ \sqrt{1 - \frac{v^2}{c^2} }$$

so there appears (if I have done the calculations right) to be a sense in which the total time dilation is the product of the gravitational time dilation and the velocity based time dilation, when the local instantaneous velocity is used.

 

[JesseM]

Taking the square root of both sides:

$$\frac{d\tau}{dt} = \, \left(1-\frac{r_s}{r}\right)\ \sqrt{1 - \frac{v^2}{c^2} }$$

so there appears (if I have done the calculations right) to be a sense in which the total time dilation is the product of the gravitational time dilation and the velocity based time dilation, when the local instantaneous velocity is used.

Didn't pervect show that it was approximately but not exactly the product of the two at the end of his post? He derived this equation:

This can be rewritten as

$$\frac{d\tau}{dt}= \sqrt{ g_{tt} } \sqrt{1 - \frac{1}{g_{tt}} \left( \frac{r \frac{d\theta}{dt} }{c} \right)^2$$


This is almost in the form of the product of the GR and SR time dilation but not quite exactly.

Since the object is assumed to be orbiting at constant angular velocity, its tangential velocity should be equal to $$r \frac{d\theta}{dt}$$, so if $$g_{tt}$$ is close to one the right side looks almost like the SR equation $$\frac{d\tau}{dt}= \sqrt{1 - \frac{v^2}{c^2}}$$, but it's not exact because of that extra factor of $$\frac{1}{g_{tt}}$$.

 

[yuiop]

Didn't pervect show that it was approximately but not exactly the product of the two at the end of his post? He derived this equation:

Since the object is assumed to be orbiting at constant angular velocity, its tangential velocity should be equal to $$r \frac{d\theta}{dt}$$, so if $$g_{tt}$$ is close to one the right side looks almost like the SR equation $$\frac{d\tau}{dt}= \sqrt{1 - \frac{v^2}{c^2}}$$, but it's not exact because of that extra factor of $$\frac{1}{g_{tt}}$$.

We both started with the same equation. The reason pervect did not get exactly the product of the two is that he was using coordinate (angular) velocity and I am using local velocity.

 

[JesseM]

We both started with the same equation. The reason pervect did not get exactly the product of the two is that he was using coordinate (angular) velocity and I am using local velocity.

Ah, I see now the steps where you noted that the coordinate velocity is $$u = r \left(\frac{d\phi}{dt}\right)$$, and then later said "The local velocity v is related to the coordinate velocity by: $$u = v \left(1-\frac{r_s}{r}\right)$$". Does "local velocity" in this case mean velocity in the locally inertial frame of a freefalling observer passing next to the orbiting body? If so what is being assumed about the freefalling observer's own path through spacetime--presumably freefalling observers passing the object in different directions would get different values for the object's velocity in their own local inertial frames?

 

[yuiop]

Ah, I see now the steps where you noted that the coordinate velocity is $$u = r \left(\frac{d\phi}{dt}\right)$$, and then later said "The local velocity v is related to the coordinate velocity by: $$u = v \left(1-\frac{r_s}{r}\right)$$". Does "local velocity" in this case mean velocity in the locally inertial frame of a freefalling observer passing next to the orbiting body? If so what is being assumed about the freefalling observer's own path through spacetime--presumably freefalling observers passing the object in different directions would get different values for the object's velocity in their own local inertial frames?

By local velocity (in Schwarzschild coordinates) I had in mind the velocity as measured by an (accelerating) observer located at radius r with zero angular coordinate velocity and zero radial coordinate velocity. For example an observer located on a tower of height r on a non rotating planet.

 

[JesseM]

By local velocity (in Schwarzschild coordinates) I had in mind the velocity as measured by an (accelerating) observer located at radius r with zero angular coordinate velocity and zero radial coordinate velocity. For example an observer located on a tower of height r on a non rotating planet.

But how is this observer defining "velocity", if not in terms of change in Schwarzschild coordinate position over change in Schwarzschild coordinate time (which would just give us the velocity u)? Are you defining v in terms of change in Schwarzschild coordinate position over change in clock time for a set of clocks hovering at that radius (and synchronized in Schwarzschild coordinates)?

 

[yuiop]

But how is this observer defining "velocity", if not in terms of change in Schwarzschild coordinate position over change in Schwarzschild coordinate time (which would just be u)? Are you defining v in terms of change in Schwarzschild coordinate position over change in clock time for a set of clocks hovering at that radius (and synchronized in Schwarzschild coordinates)?

The local observer could use a horizontal ruler. The ruler is calibrated by timing light signals. Let's say it takes one second for a light signal to travel the length of this ruler, then if this horizontal ruler is transported to any radial height it will always take a light signal one second to traverse the length of the ruler. The clocks at a given radius always run at the same rate so there is no problem with using more than one clock and of course a mirror could always be placed at one end of the ruler and an average of the two way light travel time taken using a single clock. So the local velocity is measured using a local ruler and local clocks. How to tell if the rulers and clocks have zero angular velocity? Simply place two objects in orbit going in opposite directions. If they both have equal velocitities by your local measurement and maintain an orbital radius of r, then you have zero angular velocity.

 

[TurtleMeister]

Here are a couple of references for my previous posts:

http://www.phys.lsu.edu/mog/mog9/node9.html

The system is based on the principle of the constancy of c in a local inertial frame: the http://en.wikipedia.org/wiki/Earth-centered_inertial" . Time dilation of moving clocks is significant for clocks in the satellites as well as clocks at rest on earth.

http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html

Because an observer on the ground sees the satellites in motion relative to them, Special Relativity predicts that we should see their clocks ticking more slowly (see the Special Relativity lecture). Special Relativity predicts that the on-board atomic clocks on the satellites should fall behind clocks on the ground by about 7 microseconds per day because of the slower ticking rate due to the time dilation effect of their relative motion.

Further, the satellites are in orbits high above the Earth, where the curvature of spacetime due to the Earth's mass is less than it is at the Earth's surface. A prediction of General Relativity is that clocks closer to a massive object will seem to tick more slowly than those located further away (see the Black Holes lecture). As such, when viewed from the surface of the Earth, the clocks on the satellites appear to be ticking faster than identical clocks on the ground. A calculation using General Relativity predicts that the clocks in each GPS satellite should get ahead of ground-based clocks by 45 microseconds per day.

The combination of these two relativitic effects means that the clocks on-board each satellite should tick faster than identical clocks on the ground by about 38 microseconds per day (45-7=38)!

 

[JesseM]

I guess I'm not understanding this at all because it seems contradictory to me. Isn't all "real" space time curved?

Yes, SR is an idealization. But the practical upshot of the equivalence principle is that if you pick a region of spacetime that's small enough that tidal forces (which measure spacetime curvature) are negligible according to your equipment, then you can construct a coordinate system in this region in which the laws of physics are close enough to those of ideal inertial frames in SR that your equipment won't notice the difference (though with more sensitive equipment you'd be able to).

According to Wikipedia a COM frame can be an inertial frame.

COM = center of mass? They might have been talking about classical mechanics where there's no spacetime curvature, or they might have been talking in the sort of approximate sense I discussed above, where the curvature due to the object's mass is too small to make any measurable difference.

I got the numbers from Wikipedia. And it appears that the programming of the GPS clocks for time dilation are a simple sum of relative motion based time dilation and gravitational time dilation. However, I admit that I may be misinterpreting this.

Which wikipedia page? From pervect and kev's calculations it seems like it'd be more accurate to treat total time dilation as a product of gravitational and SR time dilation.

edit: I see you posted a link to a site in your last post, unfortunately the author doesn't give any details as to how the numbers were arrived at...

 

[TurtleMeister]

According to Wikipedia a COM frame can be an inertial frame.

COM = center of mass? They might have been talking about classical mechanics where there's no spacetime curvature, or they might have been talking in the sort of approximate sense I discussed above, where the curvature due to the object's mass is too small to make any measurable difference.

Yes COM is center of mass. You must not have followed the link I provided for "Earth-Centered Inertial Frame" (ECI).

ECI coordinate frames are not truly inertial since the Earth itself is accelerating as it travels in its orbit about the Sun. In many cases, it may be assumed that the ECI frame is inertial without adverse effect.

The ECI coordinate system is the one used for GPS.

edit: I see you posted a link to a site in your last post, unfortunately the author doesn't give any details as to how the numbers were arrived at...

What other details are needed to show that the author's method is simple addition of SR time dilation and gravitational time dilation?

 

[JesseM]

Yes COM is center of mass. You must not have followed the link I provided for "Earth-Centered Inertial Frame" (ECI).

Which link? I didn't see a link to a wikipedia article in your posts.

What other details are needed to show that the author's method is simple addition of SR time dilation and gravitational time dilation?

None, but I was saying it was unfortunate that the author doesn't provide details so we could see how this squares with the derivation of total time dilation given by pervect and kev.

 

[pervect]

Yes, it should be a product. However, the product should be able to be approximated by a sum.

(1+alpha)(1+beta) = 1 + alpha + beta + alpha*beta. But the last term is small, of second order when alpha << 1 and beta << 1.

BTW, as kev points out, with the right definition of velocity, the multiplicative relationship is exact.

 

[TurtleMeister]

Which link? I didn't see a link to a wikipedia article in your posts.

The link is in the underlined "Earth-Centered Inertial or ECI frame". Just click on it. Sorry, I should have made it a separate link for clarity.

 

[yuiop]

I made some errors in my calculations in https://www.physicsforums.com/showpost.php?p=2446850&postcount=8" and I have taken the liberty of editing that post to avoid confusion to anyone referencing the calculations in the future. I hope I have made it clear where I have made the changes.

 

[yuiop]

Because an observer on the ground sees the satellites in motion relative to them, Special Relativity predicts that we should see their clocks ticking more slowly (see the Special Relativity lecture). Special Relativity predicts that the on-board atomic clocks on the satellites should fall behind clocks on the ground by about 7 microseconds per day because of the slower ticking rate due to the time dilation effect of their relative motion.

Further, the satellites are in orbits high above the Earth, where the curvature of spacetime due to the Earth's mass is less than it is at the Earth's surface. A prediction of General Relativity is that clocks closer to a massive object will seem to tick more slowly than those located further away (see the Black Holes lecture). As such, when viewed from the surface of the Earth, the clocks on the satellites appear to be ticking faster than identical clocks on the ground. A calculation using General Relativity predicts that the clocks in each GPS satellite should get ahead of ground-based clocks by 45 microseconds per day.

The combination of these two relativistic effects means that the clocks on-board each satellite should tick faster than identical clocks on the ground by about 38 microseconds per day (45-7=38)!

Assuming:

Gravitational constant (G) = 6.67488E-11 m³/kg/s²
Mass of Earth (M) = 5.98E24 kgs
Orbital radius of GPS satellite ($R_S$) = 26578000 m
Equatorial radius of Earth ($R_E$) = 6378000 m
Sidereal day on the surface of the Earth (D) = 86164 s
Speed of light (c) = 299792458 m/s

The orbital period of the satellite is given by:

$$P = 2\pi \sqrt{\frac{R_S^3}{GM}} = 43092.7 s$$

The tangential velocity of the satellite is:

$$V_S = 2\pi R_S /P = 3875.2 m/s$$

The tangential velocity on the surface of the Earth is:

$$V_E = 2\pi R_E /D = 465.1 m/s$$

The ratio of the velocity time dilation factor for the Earth surface and the satellite is:

$$\frac{\sqrt{1-V_E^2/c^2}}{\sqrt{1-V_S^2/c^2}} = 1.000000000082342569791133268$$

During a sidereal day on the surface of the Earth, the orbiting clock falls behind by:

D - D*1.000000000082342569791133268 = -0.000007094965183483206886590666862 s

due to velocity time dilation, which is about 7 microseconds.


The ratio of the gravity time dilation factor for the Earth surface and the satellite is:

$$\frac{\sqrt{1-2GM/(R_E\ c^2)}}{\sqrt{1-2GM/(R_S\ c^2)}} = 0.9999999994707970061739772883$$

During a sidereal day on the surface of the Earth, the orbiting clock gets ahead by:

D - D*0.9999999994707970061739772883 = 0.00004559824676002542093384086709 s

due to gravitational time dilation, which is about 45 microseconds.


The total additive time offset of the satellite per sidereal day is then:

0.00003850328157654221404725020023 s or about 38 microseconds.


Multiplying the velocity time dilation ratio by the gravitational time dilation ratio gives a total multiplicative time dilation ratio of:

$$0.9999999995531395759215346216$$

During a sidereal day on the surface of the Earth, the orbiting clock gets ahead by:

D - D*0.9999999995531395759215346216 = 0.00003850328158029689086344090986 s

due to the total multiplicative time dilation, which is also about 38 microseconds and only differs from the additive time dilation after 14 decimal places! However in the strongly curved spacetime near a black hole the additive time dilation approximation will differ greatly from the multiplicative time dilation formula.

The multiplicative ratio given above is the preset factory offset applied to the gps satellite clock frequency before it is launched. There are many other effects that may have to be taken into account (e.g. for military gps receivers) such as Sagnac effect, ionosphere delays, Shapiro delay, orbit eccentricity, etc but these effects are generally much smaller than the main two relativistic effects.

 

[TurtleMeister]

Thank you for the informative post Kev. And thanks for taking the time to present it in a way that is easy to understand (for a layman). It is greatly appreciated. I understand completely. :)

 

[yuiop]

Thank you for the informative post Kev. And thanks for taking the time to present it in a way that is easy to understand (for a layman). It is greatly appreciated. I understand completely. :)

Thanks It is nice to have some positive feedback now and then. Cheers

 

[starthaus]

Starting with this equation given by pervect here $$c^2 \left( \frac{d\tau}{dt} \right)^2 = c^2 \, \left(1-\frac{r_s}{r}\right)^2\ - r^2 \left(\frac{d\phi}{dt}\right)^2$$

You start with the wrong metric, therefore you get the wrong result.

The correct metric is:

$$(ct\tau)^2=(1-\frac{r_s}{r})(cdt)^2-(rd\phi)^2$$

There is no $$(1-\frac{r_s}{r})^2$$

This was a rare goof by pevect , perpetrated by you.

 

[yuiop]

You start with the wrong metric, therefore you get the wrong result.

The correct metric is:

$$(ct\tau)^2=(1-\frac{r_s}{r})(cdt)^2-(rd\phi)^2$$

There is no $$(1-\frac{r_s}{r})^2$$

This was a rare goof by pevect , perpetrated by you.

If you bothered to read the very next line in #8 you would see that I have stated the formula given by pervect should be :

$$c^2 \left( \frac{d\tau}{dt} \right)^2 = c^2 \, \left(1-\frac{r_s}{r}\right)\ - r^2 \left(\frac{d\phi}{dt}\right)^2$$

and have allowed for it in subsequent calculations. It is not as if you are immune to errors and typos.

... The correct metric is:

$$(ct\tau)^2=(1-\frac{r_s}{r})(cdt)^2-(rd\phi)^2$$

Actually the correct metric is:

$$(c d\tau)^2=(1-\frac{r_s}{r})(cdt)^2-(r d\phi)^2$$

 

[starthaus]

If you bothered to read the very next line in #8 you would see that I have stated the formula given by pervect should be :

$$c^2 \left( \frac{d\tau}{dt} \right)^2 = c^2 \, \left(1-\frac{r_s}{r}\right)\ - r^2 \left(\frac{d\phi}{dt}\right)^2$$

and have allowed for it in subsequent calculations. It is not as if you are immune to errors and typos.
Actually the correct metric is:

$$(c d\tau)^2=(1-\frac{r_s}{r})(cdt)^2-(r d\phi)^2$$

Actually, I stopped at the second error:

$$u=v(1-\frac{r_s}{r})$$

Looks like you corrected it as well. Good for you.

 

[starthaus]

If you bothered to read the very next line in #8 you would see that I have stated the formula given by pervect should be :

$$c^2 \left( \frac{d\tau}{dt} \right)^2 = c^2 \, \left(1-\frac{r_s}{r}\right)\ - r^2 \left(\frac{d\phi}{dt}\right)^2$$

and have allowed for it in subsequent calculations. It is not as if you are immune to errors and typos.



Actually the correct metric is:

$$(c d\tau)^2=(1-\frac{r_s}{r})(cdt)^2-(r d\phi)^2$$

Sorry, this is still wrong on several planes:

-The correct metric is

$$(cd\tau)^2=(1-r_s/r)(cdt)^2-(1-r_s/r)^{-1}(dr)^2-(rd\theta)^2-(rd\phi sin\theta)^2$$

-You need to make

$$d\theta=dr=0$$

-v is equal to:

$$r\frac{d\phi}{dt}$$ and not $$r\frac{d\theta}{dt}$$

You need to look up the definition of Schwarzschild metric and the Schwarzschild coordinates and you'll understand the mistakes.