Resultant time dilation from both gravity and motion

[espen180]

When a frame is moving in relation to an observer in his rest frame at infinity, and the frame is in a gravitational well, is the resultant time dilation simply the sum of the motional and gravitational dilation, e.g.

$$t=\tau\left(\gamma^{-1}+\gamma_g^{-1}\right)=\tau\left(\sqrt{1-\frac{v^2}{c^2}}+\sqrt{1-\frac{GM}{c^2r}}\right)$$

Where $$\tau$$ is proper time and $$t$$ is measured by the observer?

If, not what is the correct expression?

 

[starthaus]

When a frame is moving in relation to an observer in his rest frame at infinity, and the frame is in a gravitational well, is the resultant time dilation simply the sum of the motional and gravitational dilation, e.g.

$$t=\tau\left(\gamma^{-1}+\gamma_g^{-1}\right)=\tau\left(\sqrt{1-\frac{v^2}{c^2}}+\sqrt{1-\frac{GM}{c^2r}}\right)$$

Where $$\tau$$ is proper time and $$t$$ is measured by the observer?

If, not what is the correct expression?

There is no reason why it would be the sum , you can calculate the expression easily from the Schwarzschild metric:

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

Make $$d\theta=dr=0$$

 

[JesseM]

For an object in a circular orbit, the total time dilation is a product of gravitational and velocity-based time dilation--see kev's post #8 on this thread and post #10 here. But cases other than a circular orbit would probably be more complicated.

 

[Jonathan Scott]

In non-relativistic situations, you can simply fall back on Newtonian theory:

The fractional time dilation (that is, the difference in time rate divided by the original time rate) due to velocity is equal to the ratio of kinetic energy to rest energy.

The fractional time dilation due to gravity is equal to the ratio of potential energy to rest energy.

The combined effect simply adds the fractions together to give the overall fraction (which is equivalent to multiplying the time dilation factors for each of the two effects).

For free fall (including any shape of orbit around a static mass), the sum of kinetic energy and potential energy is constant, so the time dilation is constant (and so is the total energy, as in Newtonian theory).

The relative time rates for different orbits can be compared using Newtonian potential theory.

 

[espen180]

Thank you very much. All replies were very useful.

 

[starthaus]

For an object in a circular orbit, the total time dilation is a product of gravitational and velocity-based time dilation--see kev's post #8 on this thread and post #10 here. But cases other than a circular orbit would probably be more complicated.

Hi Jesse,

I don't think the expressions put down by kev in that post are correct. The correct result is derived from the Schwarzschild metric, the periods of two clocks situated at radiuses $$r_1$$ and $$r_2$$ respectively is expressed by the ratio:

$$\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-r_s/r_1}{1-r_s/r_2}}\sqrt{\frac{1-(r_1sin\theta_1\omega/\sqrt{1-r_s/r_1})^2}{1-(r_2sin\theta_2\omega/\sqrt{1-r_s/r_2})^2}}$$

where $$r_s$$ is the Schwarzschild radius.The above is valid for a uniform density sphere.
Start with the Schwarzschild metric:

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

and make $$d\theta=dr=0$$ for an object orbiting at $$r=constant$$.

If $$d\theta=d\phi=0$$ we get the expression for an object moving radially, which is still different from kev's expressions. In kev's notation:

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

where $$v=\frac{dr}{dt}$$

 

[JesseM]

If $$d\theta=d\phi=0$$ we get the expression for an object moving radially, which is still different from kev's expressions.

kev wasn't talking about an object moving radially, as I said before he was dealing with the scenario of an object in circular orbit. pervect also found that for this case, the total time dilation was "almost" a product of SR and GR time dilations here...I think the difference was just because pervect was using coordinate velocity in Schwarzschild coordinates in the part of the equation that looked "almost" like SR time dilation, whereas kev was using the local velocity as seen in a freefalling frame for an observer whose coordinate velocity in Schwarzschild coordinates is zero at the moment the orbiting object passes it.

 

[DrGreg]

I believe that the equation

$$\frac{dt}{d\tau} = \frac{1}{\sqrt{1-v^2/c^2}\sqrt{1 - 2GM/rc^2}}$$​

always applies (for radial, tangential or any other motion) where v is speed relative to a local hovering observer using local proper distance and local proper time.

I derived this in posts #9 and #7 of the thread "Speed in general relativity"[/color] (and repeated in post #46).

 

[starthaus]

kev wasn't talking about an object moving radially, as I said before he was dealing with the scenario of an object in circular orbit. pervect also found that for this case, the total time dilation was "almost" a product of SR and GR time dilations here...I think the difference was just because pervect was using coordinate velocity in Schwarzschild coordinates in the part of the equation that looked "almost" like SR time dilation, whereas kev was using the local velocity as seen in a freefalling frame for an observer whose coordinate velocity in Schwarzschild coordinates is zero at the moment the orbiting object passes it.

kev's expression for radial motion is not correct (see post #6 above). It is very easy to obtain the correct expressions.

 

[starthaus]

I believe that the equation

$$\frac{dt}{d\tau} = \frac{1}{\sqrt{1-v^2/c^2}\sqrt{1 - 2GM/rc^2}}$$​

Yes, this is correct, provided "v" in your case is defined as:

$$\frac{dr/dt}{1-r_s/r}$$

or as:

$$\frac{r*sin\theta* d\phi/dt}{\sqrt{1-r_s/r}}=\frac{\omega rsin\theta}{\sqrt{1-r_s/r}}$$

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

(see post 6)

 

[espen180]

So the correct expression is

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

, right?

Do you then define $$r\frac{\text{d}\phi}{\text{d}t}$$ as coordinate velocity?

 

[JesseM]

kev's expression for radial motion is not correct (see post #6 above). It is very easy to obtain the correct expressions.

Your post #6 seems to be addressing a different question than pervect and kev, since you are finding the ratio of ticking rates of two clocks orbiting at finite radius, while pervect and kev were deriving time dilation of an orbiting clock relative to a stationary clock at infinity (as in the commonly-used equation for gravitational time dilation). I suppose your expression would probably have a well-defined limit as r2 approaches infinity though. Anyway, it might be easier to deal with pervect's derivation rather than kev's, since pervect's equation is expressed entirely in Schwarzschild coordinates rather than including a non-Schwarzschild notion of "velocity". Do you disagree with pervect's conclusions here? If so, where's the first line you would dispute?

 

[starthaus]

Your post #6 seems to be addressing a different question than pervect and kev,
since you are finding the ratio of ticking rates of two clocks orbiting at finite radius

Precisely. It addresses the question in the OP. (post 1). That is, what is the difference in rates for atomic clocks on the geoid.

while pervect and kev were deriving time dilation of an orbiting clock relative to a stationary clock at infinity (as in the commonly-used equation for gravitational time dilation). I suppose your expression would probably have a well-defined limit as r2 approaches infinity though.

No, the first formula in post 6 is derived from :

$$\frac{d\tau_1}{dt}=...$$

and

$$\frac{d\tau_2}{dt}=...$$

where $$\frac{d\tau}{dt}$$ is derived straight from the metric:

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

Make $$d\theta=d\phi=0$$:

$$\frac{d\tau}{dt}=\sqrt{1-r_s/r}\sqrt{...}$$

Anyway, it might be easier to deal with pervect's derivation rather than kev's, since pervect's equation is expressed entirely in Schwarzschild coordinates rather than including a non-Schwarzschild notion of "velocity". Do you disagree with pervect's conclusions here?

Pervect's formula in the post you linked is identical to mine. So, no dispute.

 

[starthaus]

So the correct expression is

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

, right?

Yes.

Do you then define $$r\frac{\text{d}\phi}{\text{d}t}$$ as coordinate velocity?

I don't define anything.

 

[espen180]

I don't define anything.

How do I interpret it then?

 

[espen180]

So the correct expression is

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

But it looks from the Schwartzschild metric that it would be

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

?

 

[espen180]

The Schwartzschild metric for constant r and $$\theta=\frac{\pi}{2}$$ gives us

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

If we divide both sides with c² we get

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

"Factoring out" $1-\frac{r_s}{r}$ on the right side gives

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

then, taking the square root gives the result in #16;

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

I don't see where the mistake is. Would you please point it out to me?

 

[starthaus]

The Schwartzschild metric for constant r and $$\theta=\frac{\pi}{2}$$ gives us

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

If we divide both sides with c² we get

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

"Factoring out" $1-\frac{r_s}{r}$ on the right side gives

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

then, taking the square root gives the result in #16;

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

I don't see where the mistake is. Would you please point it out to me?

yes, fine

 

[yuiop]

Hi Jesse,

I don't think the expressions put down by kev in that post are correct. The correct result is derived from the Schwarzschild metric, the periods of two clocks situated at radiuses $$r_1$$ and $$r_2$$ respectively is expressed by the ratio:

$$\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-r_s/r_1}{1-r_s/r_2}}\sqrt{\frac{1-(r_1sin\theta_1\omega/\sqrt{1-r_s/r_1})^2}{1-(r_2sin\theta_2\omega/\sqrt{1-r_s/r_2})^2}}$$

where $$r_s$$ is the Schwarzschild radius.The above is valid for a uniform density sphere.
Start with the Schwarzschild metric:

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

and make $$d\theta=dr=0$$ for an object orbiting at $$r=constant$$.

If $$d\theta=d\phi=0$$ we get the expression for an object moving radially, which is still different from kev's expressions. In kev's notation:

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

where $$v=\frac{dr}{dt}$$

I was using a notion of local velocity (v' = dr'/dt') as measured by a stationary observer at r.

Since $$v' = dr'/dt' = (dr/dt)/(1-r_s/r)$$

the value of v' can be directly substituted into your expression to obtain:

$$\frac{d\tau}{dt}=\sqrt{1-\frac{r_s}{r}}\sqrt{1-(\frac{1}{c^2}\frac{dr/dt}{1-r_s/r})^2}=\sqrt{1-\frac{r_s}{r}}\sqrt{1-(\frac{v'}{c})^2}$$

The two forms are numerically the same and in agreement with #8 by DrGReg here:

I believe that the equation

$$\frac{dt}{d\tau} = \frac{1}{\sqrt{1-v^2/c^2}\sqrt{1 - 2GM/rc^2}}$$​

always applies (for radial, tangential or any other motion) where v is speed relative to a local hovering observer using local proper distance and local proper time.

I derived this in posts #9 and #7 of the thread "Speed in general relativity"[/color] (and repeated in post #46).

 

[starthaus]

I was using a notion of local velocity (v' = dr'/dt') as measured by a stationary observer at r.

Since $$v' = dr'/dt' = (dr/dt)/(1-r_s/r)$$

There is no mention of any such convention in this post. Actually, there is no derivation, the expression is put in by hand, you simply multiplied the kinematic factor by the gravitational factor.

 

[yuiop]

There is no mention of any such convention in this post. Actually, there is no derivation, the expression is put in by hand, you simply multiplied the kinematic factor by the gravitational factor.

It was not meant to be a derivation, just a statement of facts from various references, put into context and interelated to each other. If you want a derivation, Dr Greg has done some perfectly good ones that come to the same conclusion. In the post you linked to, I made it clear in the surrounding text that I was talking about the the local velocity.

But it looks from the Schwartzschild metric that it would be

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

?

Yes, that equation is correct.

There are two motion/gravity time dilation equations if purely Schwarzschild coordinate measurements are used.

The time dilation ratio for orbital motion is:

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

The time dilation ratio for radial motion is:

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

Now if I define v' = dx'/dt' as the local velocity of the moving test particle as measured by a stationary observer at r using local clocks and rulers (where dx' can be a vertical or horizontal distance), then a single equation is obtained:

$$\frac{\text{d}\tau}{\text{d}t}=\sqrt{1-\frac{r_s}{r}}\sqrt{1-\frac{v'^2}{c^2}}$$

which is equally valid for horizontal or vertical motion of the particle.

To try and make the concept of "local velocity" even clearer, this is the velocity calculated by a local stationary observer orientating a ruler of proper length (dx') parallel to the motion of the test particle and timing the interval (dt') it takes for the test particle to traverse the ruler according to the stationary observers local clock.

 

[espen180]

Thank you very much kev! Everything fits now. :)

 

[yuiop]

Thank you very much kev! Everything fits now. :)

You are very welcome.

The equation I gave

$$\frac{\text{d}\tau}{\text{d}t}=\sqrt{1-\frac{r_s}{r}}\sqrt{1-\frac{v '^2}{c^2}}$$

uses an odd mix (something DrGreg alluded to) of velocity measured locally (v') and Schwarzschild coordinate gravitational time dilation.

A more general equation is:

$$\frac{\text{d}\tau}{\text{d}t}= \sqrt{\frac{1-r_s/r}{1-r_s/r_o}} \sqrt{1- \left (\frac{1-r_s/r_o}{1-r_s/r} \right )^2 \left (\frac{\text{d}r}{c\text{d}t} \right )^2 - \left (\frac{1-r_s/r_o}{1-r_s/r} \right ) \left(\frac{r \text{d}\theta}{c\text{d}t}\right)^2 - \left (\frac{1-r_s/r_o}{1-r_s/r} \right) \left(\frac{r \sin \theta \text{d}\phi}{c\text{d}t}\right)^2 }$$

where $r_o$ is the Schwarzschild radial coordinate of the stationary observer and r is the Schwarzschild radial coordinate of the test particle and dr and dt are understood to be measurements made by the stationary observer at $r_o$ in this particular equation.

For $r_o = r$ the time dilation ratio is:

$$\frac{\text{d}\tau}{\text{d}t} = \sqrt{1-\frac{v'^2}{c^2}}$$

in agreement with the generally accepted fact that local measurements made in a gravitational field are Minkowskian.

 

[JesseM]

Pervect's formula in the post you linked is identical to mine. So, no dispute.

Then why did you dispute kev's equations? He explicitly stated in post #8 here (which I linked to earlier) that he was just working from pervect's derivation, but with the substitution of a "local velocity" v for the Schwarzschild coordinate velocity u, related by $$u = v \sqrt{1-\frac{r_s}{r}}$$.

 

[starthaus]

Then why did you dispute kev's equations? He explicitly stated in post #8 here (which I linked to earlier) that he was just working from pervect's derivation, but with the substitution of a "local velocity" v for the Schwarzschild coordinate velocity u, related by $$u = v \sqrt{1-\frac{r_s}{r}}$$.

Because kev's equations did not apply to the OP. Since then, the threads have been split.

 

[JesseM]

Because kev's equations did not apply to the OP.

Why do you say that? The original post asked "When a frame is moving in relation to an observer in his rest frame at infinity, and the frame is in a gravitational well, is the resultant time dilation simply the sum of the motional and gravitational dilation", it wasn't asking about the ratio between ticks of clocks at different finite radii. I brought up the result kev derived because it gives the time dilation in one special case--for a clock in a perfect circular orbit in a Schwarzschild spacetime--relative to an observer at infinity, which seemed to me to be relevant to the OP.

 

[JesseM]

To try and make the concept of "local velocity" even clearer, this is the velocity calculated by a local stationary observer orientating a ruler of proper length (dx') parallel to the motion of the test particle and timing the interval (dt') it takes for the test particle to traverse the ruler according to the stationary observers local clock.

Would this be the same as the velocity measured in the locally inertial frame of a free-falling observer who happens to be instantaneously at rest (in Schwarzschild coordinates) at the moment the orbiting clock passes him? I assumed this was what was meant by "local" velocity but from your description above I'm not sure if it's the same...

 

[starthaus]

Why do you say that? The original post asked "When a frame is moving in relation to an observer in his rest frame at infinity, and the frame is in a gravitational well, is the resultant time dilation simply the sum of the motional and gravitational dilation", it wasn't asking about the ratio between ticks of clocks at different finite radii.

kev didn't derive any result, kev puts in results by hand.

I brought up the result kev derived because it gives the time dilation in one special case--for a clock in a perfect circular orbit in a Schwarzschild spacetime--relative to an observer at infinity, which seemed to me to be relevant to the OP.

The question came up in the different thread, the one about "Why do all clocks tick at the same rate on the geoid" by Dmitry7. I pointed out repeatedly to you why kev's formulas were not appropiate for answering that thtrad.

 

[starthaus]

Then why did you dispute kev's equations? He explicitly stated in post #8 here (which I linked to earlier) that he was just working from pervect's derivation, but with the substitution of a "local velocity" v for the Schwarzschild coordinate velocity u, related by $$u = v \sqrt{1-\frac{r_s}{r}}$$.

Because post #8(and all subsequent posts based on it) by kev contains a glaring mistake. I have corrected it in post 25.
Citing kev's posts does nothing but perpretrate mistakes.

 

[yuiop]

Would this be the same as the velocity measured in the locally inertial frame of a free-falling observer who happens to be instantaneously at rest (in Schwarzschild coordinates) at the moment the orbiting clock passes him? I assumed this was what was meant by "local" velocity but from your description above I'm not sure if it's the same...

My understanding or interpretation (up to now) is that the local velocity is measured by a non inertial accelerating observer that is stationary at r with respect to the Schwarzschild coordinates. i.e. the velocity of this non inertial observer is $$dr/dt = r d\theta/dt = r d\phi =0.$$ It might be better to think of it terms of inertial observer that happens to be apogee at r when the orbiting particle (with orbital radius r) passes close by. As I understand it, the clock rates and ruler measurements of a non-inertial accelerating observer that is stationary at r are the same as the measurements made by an inertial observer that momentarily happens to be at apogee at r at the same time. I might have to think about it some more. There might be a limitation to how far this "equivalence" (aplication of the clock hypothosis) can be taken when it concerns measurements of acceleration. That is something I am working on.