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 (http://www.physicsforums.com/showthread.php?t=355378) and post #10 here (http://www.physicsforums.com/showthread.php?t=398557). 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 (http://www.physicsforums.com/showthread.php?t=355378) and post #10 here (http://www.physicsforums.com/showthread.php?t=398557). 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 (http://www.physicsforums.com/showthread.php?p=2445122)...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" (http://www.physicsforums.com/showthread.php?t=383884) (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 (http://www.physicsforums.com/showthread.php?p=2445122)...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\lef t(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 (http://www.physicsforums.com/showthread.php?p=2445122)? 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 (http://www.physicsforums.com/showthread.php?p=2445122)?

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\lef t(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\lef t(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}{\te xt{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=\lef t(1-\frac{r_s}{r}\right) - \left(\frac{r\frac{\text{d}\phi}{\text{d}t}}{c}\ri ght)^2

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

\left(\frac{\text{d}\tau}{\text{d}t}\right)^2=\lef t(1-\frac{r_s}{r}\right)\left(1 - \frac{1}{1-\frac{r_s}{r}}\left(\frac{r\frac{\text{d}\phi}{\te xt{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}{\te xt{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=\lef t(1-\frac{r_s}{r}\right) - \left(\frac{r\frac{\text{d}\phi}{\text{d}t}}{c}\ri ght)^2

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

\left(\frac{\text{d}\tau}{\text{d}t}\right)^2=\lef t(1-\frac{r_s}{r}\right)\left(1 - \frac{1}{1-\frac{r_s}{r}}\left(\frac{r\frac{\text{d}\phi}{\te xt{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}{\te xt{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" (http://www.physicsforums.com/showthread.php?t=383884) (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 (http://www.physicsforums.com/showpost.php?p=2690159&postcount=10) 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 (http://www.physicsforums.com/showpost.php?p=2690159&postcount=10) 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}{\te xt{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\sqr t{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. :smile:

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 (http://www.physicsforums.com/showthread.php?t=355378) (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 (http://www.physicsforums.com/showthread.php?t=355378) (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 (http://www.physicsforums.com/showthread.php?t=407425), 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 (http://www.physicsforums.com/showthread.php?t=355378) (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 (http://www.physicsforums.com/showpost.php?p=2446850&postcount=8)(and all subsequent posts based on it) by kev contains a glaring mistake. I have corrected it in post 25 (http://www.physicsforums.com/showpost.php?p=2747110&postcount=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.

[JesseM]

kev didn't derive any result, kev puts in results by hand.
True, kev didn't derive the relation between local velocity and coordinate velocity, but unless you had definite reason to think the relation he used was incorrect (as opposed to possibly correct but not sufficiently justified in his post), I don't see why you would say "I don't think the expressions put down by kev in that post are correct."
The question came up in the different thread (http://www.physicsforums.com/showthread.php?t=407425), 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.
What do you mean by "the question"? My post bringing up kev's result was in direct response to espen180's OP on this thread, so it doesn't seem to make any sense to cite some completely different thread in order to back up your claim that "kev's equations did not apply to the OP."

[starthaus]

True, kev didn't derive the relation between local velocity and coordinate velocity, but unless you had definite reason to think the relation he used was incorrect (as opposed to possibly correct but not sufficiently justified in his post), I don't see why you would say "I don't think the expressions put down by kev in that post are correct."

What do you mean by "the question"? My post bringing up kev's result was in direct response to espen180's OP on this thread, so it doesn't seem to make any sense to cite some completely different thread in order to back up your claim that "kev's equations did not apply to the OP."

Why don't you read post 29? kev's formulas that you keep citing are wrong, ok?

[JesseM]

Because post #8 (http://www.physicsforums.com/showpost.php?p=2446850&postcount=8)(and all subsequent posts based on it) by kev contains a glaring mistake. I have corrected it in post 25 (http://www.physicsforums.com/showpost.php?p=2747110&postcount=25).
Citing kev's posts does nothing but perpretrate mistakes.
In post 25 you say that the mistake was originally pervect's which was just perpetuated by kev, but then earlier in post 13 (http://www.physicsforums.com/showthread.php?p=2745606#post2745606) you said you didn't dispute pervect's results, I guess you changed your mind? It does look like pervect used the wrong formula there.

[yuiop]

Because post #8 (http://www.physicsforums.com/showpost.php?p=2446850&postcount=8)(and all subsequent posts based on it) by kev contains a glaring mistake. I have corrected it in post 25 (http://www.physicsforums.com/showpost.php?p=2747110&postcount=25).
Citing kev's posts does nothing but perpretrate mistakes.

The very next line in that post http://www.physicsforums.com/showpost.php?p=2446850&postcount=8 states that I have found and corrected for pervect's rare mistake. I made the edit over a year ago and cleary state that the remaining calculations have been edited to correct for the typo by pervect.

[starthaus]

The very next line in that post http://www.physicsforums.com/showpost.php?p=2446850&postcount=8 states that I have found and corrected for pervect's rare mistake. I made the edit over a year ago and cleary state that the remaining calculations have been edited to correct for the typo by pervect.

You are right, you also corrected another error that you made further down in your post.

[JesseM]

You are right, you also corrected another error that you made further down in your post. But the derivation in post 8 applies to orbital motion, the equation in this (http://www.physicsforums.com/showpost.php?p=2690159&postcount=10) post cited by JesseM). is also for orbital motion and not applicable to this thread.
Why do you think an equation for the time dilation experienced by an orbiting object (an equation which you now agree is correct, I take it?) is "not applicable to this thread"? The OP didn't say anything about the precise state of motion of the object, just that it was in a gravity well and was moving (which would certainly be true for an orbiting object!):
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

[starthaus]

Why do you think an equation for the time dilation experienced by an orbiting object (an equation which you now agree is correct, I take it?) is "not applicable to this thread"? The OP didn't say anything about the precise state of motion of the object, just that it was in a gravity well and was moving (which would certainly be true for an orbiting object!):

You are going around in circles. Let's put a stop to this, I gave you the correct expressions , including the derivations for both orbital and radial motion at post 6. My post 6 really belongs in the Dmitry7 thread, whoever split the threads made a mistake.
The reason for all the confusion is that espen180 thread was split from the Dmitry7 thread. The two threads (espen180 and Dmitry7) deal with different situations. The answer I gave you at post 6, stands, the correct answer to Dmitry7 question is not the kev posts you cite but the answer I derived.
Spcifically:

-The correct answer to Dmitry7's question is:

\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/c\sqrt{1-r_s/r_1})^2}{1-(r_2sin\theta_2\omega/c\sqrt{1-r_s/r_2})^2}}

-The answer to espen180 question is :


\frac{d\tau}{dt}=\sqrt{1-\frac{r_s}{r}}\sqrt{1-(\frac{r\omega sin(\theta)/c}{\sqrt{1-r_s/r}})^2}

[JesseM]

You are going around in circles. Let's put a stop to this, I gave you the correct expressions , including the derivations for both orbital and radial motion at post 6.
I don't dispute your expressions, but it seems you also do not dispute that kev's expressions are correct in the case of an orbiting object or that they are relevant to the OP by espen180, correct? So will you acknowledge that all your previous argumentative statements saying that kev was wrong and that I was wrong to cite him were made too hastily?
My post 6 really belongs in the Dmitry7 thread, whoever split the threads made a mistake.
The reason for all the confusion is that espen180 thread was split from the Dmitry7 thread. The two threads (espen180 and Dmitry7) deal with different situations. The answer I gave you at post 6, stands, the correct answer to Dmitry7 question is not the kev posts you cite but the answer I derived.
But I was the one who originally brought up kev's derivation, and I brought it up in response to espen180's post, I never claimed that it was supposed to be relevant to Dmitry7's post. Did you misunderstand who I was responding to? My original post (http://www.physicsforums.com/showpost.php?p=2744027&postcount=3) on this thread was made 8 minutes before Dmitry7's first post (http://www.physicsforums.com/showthread.php?t=407425) according to the timestamps on the upper left, so even before the thread split my post should have appeared before his and it should have been clear that I was not responding to his question.

[starthaus]

I don't dispute your expressions, but it seems you also do not dispute that kev's expressions are correct in the case of an orbiting object or that they are relevant to the OP by espen180, correct?

Yes, but NOT in the context of the original thread as started by Dmitry7. This is where my objections started. With post 6. Do you now understand what my objection is to your citing the inappropriate material for answering Dmitry7's OP?




So will you acknowledge that all your previous argumentative statements saying that kev was wrong and that I was wrong to cite him were made too hastily?
No. See above.



But I was the one who originally brought up kev's derivation, and I brought it up in response to espen180's post, I never claimed that it was supposed to be relevant to Dmitry7's post.

The thread started as one thread, the Dmitry7 thread. Your citation was inappropriate in the context. It is quite possible that when the split was made, the timestamps were messed up as well. Anyways, I have posted clearly what formula goes with what thread.

-The correct answer to Dmitry7's question is:

\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/c\sqrt{1-r_s/r_1})^2}{1-(r_2sin\theta_2\omega/c\sqrt{1-r_s/r_2})^2}}

-The correct answer to espen180's question is :


\frac{d\tau}{dt}=\sqrt{1-\frac{r_s}{r}}\sqrt{1-(\frac{r\omega sin(\theta)}{c \sqrt{1-r_s/r}})^2}

I hope that this clarifies things once and for all.

[JesseM]

The thread started as one thread, the Dmitry7 thread.
No it didn't, this seems to be your basic misunderstanding. As I already said, you can look at the timestamps in the upper left of each post to see that my post responding to espen180's post was posted 8 minutes before Dmitry7's very first post. The actual time displayed on your browser may depend on your time zone, but on my browser espen180's OP was from Jun2-10, 02:43 PM, my post #3 responding to him (and citing kev's posts) was from Jun2-10, 03:03 PM, while Dmitry7's first post on the split thread (http://www.physicsforums.com/showthread.php?t=407425) was from Jun2-10, 03:11 PM.
It is quite possible that when the split was made, the timestamps were messed up as well.
Isn't it a little more likely that your memory is playing tricks on you? For myself, I remember pretty clearly that espen180's post was in fact the original post when I responded to it.

[starthaus]

No it didn't, this seems to be your basic misunderstanding. As I already said, you can look at the timestamps in the upper left of each post to see that my post responding to espen180's post was posted 8 minutes before Dmitry7's very first post. The actual time displayed on your browser may depend on your time zone, but on my browser espen180's OP was from Jun2-10, 02:43 PM, my post #3 responding to him (and citing kev's posts) was from Jun2-10, 03:03 PM, while Dmitry7's first post on the split thread (http://www.physicsforums.com/showthread.php?t=407425) was from Jun2-10, 03:11 PM.

Isn't it a little more likely that your memory is playing tricks on you? For myself, I remember pretty clearly that espen180's post was in fact the original post when I responded to it.

espen180 thread was split from Dmitry7 thread. Besides, if you paid attention to the correct formulas, they both need to contain sin(\theta) and \omega is \frac{d\phi}{dt}, not \frac{d\theta}{dt}. The reason for the error is that kev picked up not only a wrong formula from pervect but also a truncated one. It is the \phi coordinate that describes the complete circle, not \theta. See here (http://en.wikipedia.org/wiki/Schwarzschild_coordinates#Definition). So, kev's post 8 is still wrong becuse he started with the wrong metric and used the wrong definitions all along.

[espen180]

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.

Let's first combine \text{d}\theta^2+\sin^2\theta \text{d}\phi^2=\text{d}\Omega^2 and so simplify the equation to

\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}\Omega}{c\text{d}t}\right)^2 }

Working backwards to get back to the metric gives me

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

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

I was hoping that doing this would lead me to an explanation as to where the \frac{1-\frac{r_s}{r}}{1-\frac{r_s}{r_0}} came from, but it seems it did not.

I do observe that in modeling this metric the metric coefficients are found by taking the ratio of the coefficients of the particle wrt an observer at infinity to the coefficients of the observer at r_0 to the same observer at infinity, but could I have an explanation of why that works?

[JesseM]

espen180, can you settle this? When you originally wrote the OP, were you starting a new thread at the time or were you just responding to a prior thread that had been started by Dmitry67?

[espen180]

Isn't it a little more likely that your memory is playing tricks on you? For myself, I remember pretty clearly that espen180's post was in fact the original post when I responded to it.

espen180 thread was split from Dmitry7 thread.

This thread was not split from Dmitry7's thread. I started a new thread with the OP. I hope this settles that dispute.



Besides, if you paid attention to the correct formulas, they both need to contain sin(\theta) and \omega is \frac{d\phi}{dt}, not \frac{d\theta}{dt}. The reason for the error is that kev picked up not only a wrong formula from pervect but also a truncated one. It is the \phi coordinate that describes the complete circle, not \theta. See here (http://en.wikipedia.org/wiki/Schwarzschild_coordinates#Definition). So, kev's post 8 is still wrong becuse he started with the wrong metric and used the wrong definitions all along.

Why not just contract the angle differentials into \text{d}\theta^2+\sin^2\theta\text{d}\phi^2=\text{ d}\Omega^2 and avoid the problem alltogether?

Kev's post #8 is in agreement with all the references I can find on the Schwartzschild metric, and the algebra checks out. What, in your opinion, is the right metric and definitions?

[starthaus]

Why not just contract the angle differentials into
\text{d}\theta^2+\sin^2\theta\text{d}\phi^2=\text{ d}\Omega^2 and avoid the problem alltogether?

Because \phi and \theta are independent coordinates. So your hack is illegal.


Kev's post #8 is in agreement with all the references I can find on the Schwartzschild metric, and the algebra checks out. What, in your opinion, is the right metric and definitions?

Nope, it doesn't. Look it up.

[starthaus]

espen180, can you settle this? When you originally wrote the OP, were you starting a new thread at the time or were you just responding to a prior thread that had been started by Dmitry67?

Not relevant. What is relevant is that post 8 by kev is wrong. For a list of errors see here (http://www.physicsforums.com/showpost.php?p=2747325&postcount=28).

[espen180]

Because \theta and \phi are independent coordinates.

But you have spherical symmetry, and since the choice of the \theta axis is arbitrary, you can always define a new single coordinate which represents the total angular distance traversed, right?

Nope, it doesn't. Look it up.

I don't have a book handy to look it up in. I can only observe that other PF members like JesseM seem to have given him their support.

[starthaus]

But you have spherical symmetry, and since the choice of the \theta axis is arbitrary, you can always define a new single coordinate which represents the total angular distance traversed, right?

Nope. Like I said, you need to read about Schwarzschild metric and Schwarzschild coordinates.
Contrary to your beliefs, \thetaand \phi are not interchangeable.



I don't have a book handy to look it up in.

Google is your friend. Try "Schwarzschild metric", "Schwarzschild coordinates"

I can only observe that other PF members like JesseM seem to have given him their support.

This is not a scientific criterion.I know that you are a big fan of kev's from other encounters but this is not a scientific criterion either. I posted kev's errors in the thread where he did his derivation.

[espen180]

Nope. Like I said, you need to read about Schwarzschild metric and Schwarzschild coordinates.
Contrary to your beliefs, thetaand \phi are not interchangeable.




Google is your friend. Try "Schwarzschild metric"



This is not a scientific criterion. I posted kev's errors in the thread where he did his derivation.


I realize that the two independent angle coordinates hav different definitions, but you must also realize that there is no preferred coordinate systems.

Let me elaborate on the contraction. Define new angle coordinates \text{d}\theta^2 + \sin^2\theta\text{d}\phi^2=\text{d}\Sigma^2 + \sin^2\Sigma\text{d}\Omega^2 and define the orientation of this new coordinate system such that \Sigma = \frac{\pi}{2}. Since keeping the same angular orientation of the coordinates is not an issue due to spherical symmetry, there is no problem defining \text{d}\theta^2 + \sin^2\theta\text{d}\phi^2=\text{d}\Omega^2

As for the Schartzschild metric, I know it by heart. Kev is using the correct metric

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

If you read pervects original derivation you would know that kev was working with a particle in orbit around the equator, where \theta=\frac{\pi}{2}. But he would not need to. He could just have used the nagle contraction explained above to shift the coordinates such that motion around the equator was realized.

[starthaus]

As for the Schartzschild metric, I know it by heart. Kev is using the correct metric

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


Good for you.


If you read pervects original derivation you would know that kev was working with a particle in orbit around the equator, where \theta=\frac{\pi}{2}. But he would not need to. He could just have used the nagle contraction explained above to shift the coordinates such that motion around the equator was realized.

So what? his derivation is wrong just the same.

[espen180]

So what? his derivation is wrong just the same.

You are missing the point. Your original claim that kev is using the wrong metric is false. Now that we have established that there is nothing wrong with the definitions, please point to the spesific place the error occurs, and preferrably propose the correct result is its place.

[JesseM]

Not relevant.
You seemed to think it was relevant before when you said "Do you now understand what my objection is to your citing the inappropriate material for answering Dmitry7's OP?" The problem is that rather than sticking to a single criticism, you keep changing your line of attack, never really admitting that you made any mistakes in your previous attacks, as if you somehow believe that as long as you can show kev was wrong in some way, you have "won", even if the way you finally decide he is wrong had not even occurred to you at the moment you started attacking his post. Your latest criticism in post #41 about angles isn't any better than your previous attacks. The Schwarzschild metric is spherically symmetric, so although the fact that \theta only ranges from 0 to \pi means you can't have a full orbit with constant r and \phi, the time dilation equation is only talking about the instantaneous rate a clock is ticking relative to a clock at infinity over an infinitesimally short section of its orbit. It is certainly possible to have a circular orbit which for one half of the orbit has \theta varying from 0 to \pi while r has a constant value of R and \phi has a constant value of \pi/2 (so for any infinitesimal section of an orbiting object's worldline whose endpoints lie on this half of the orbit, dr and d\phi would be 0), while the other half of the orbit also has \theta varying from 0 to \pi and r having a constant value of R, but now with \phi having a constant value of -\pi/2 (so for any infinitesimal section of an orbiting object's worldline whose endpoints lie on this half of the orbit, dr and d\phi would still be 0). kev's derivation would work just fine in this case.

[starthaus]

You seemed to think it was relevant before when you said "Do you now understand what my objection is to your citing the inappropriate material for answering Dmitry7's OP?" The problem is that rather than sticking to a single criticism, you keep changing your line of attack, never really admitting that you made any mistakes in your previous attacks, as if you somehow believe that as long as you can show kev was wrong in some way, you have "won", even if the way you finally decide he is wrong had not even occurred to you at the moment you started attacking his post. Your latest criticism in post #41 about angles isn't any better than your previous attacks.

His derivation is a hack and you've been doing your darnest to defend it. Why is it so difficult for you to admit that it is wrong?


The Schwarzschild metric is spherically symmetric, so although the fact that \theta only ranges from 0 to \pi means you can't have a full orbit with constant r and \phi, the time dilation equation is only talking about the instantaneous rate a clock is ticking relative to a clock at infinity over an infinitesimally short section of its orbit. It is certainly possible to have a circular orbit which for one half of the orbit has \theta varying from 0 to \pi while r has a constant value of R and \phi has a constant value of \pi/2 (so for any infinitesimal section of an orbiting object's worldline whose endpoints lie on this half of the orbit, dr and d\phi would be 0), while the other half of the orbit also has \theta varying from 0 to \pi and r having a constant value of R, but now with \phi having a constant value of -\pi/2 (so for any infinitesimal section of an orbiting object's worldline whose endpoints lie on this half of the orbit, dr and d\phi would still be 0). kev's derivation would work just fine in this case.

Please read here (http://www.physicsforums.com/showpost.php?p=2747325&postcount=28). Sorry, but no matter how hard you may try, \omega is not \frac{d\theta}{dt}
We are talking about rigotous derivations, not about hacks, right?

[espen180]

Please read here (http://www.physicsforums.com/showpost.php?p=2747325&postcount=28).
We are talking about rigotous derivations, not about hacks, right?

Everything in that post has been adressed above.

Regarding "hacks", I would like to hear your definition of one, and why you think using algebra is "against the rules" if they don't conform to your rules (also state those rules, please).

[starthaus]

Everything in that post has been adressed above.

Regarding "hacks", I would like to hear your definition of one, and why you think using algebra is "against the rules" if they don't conform to your rules (also state those rules, please).

You mean using algebra badly? Like in truncating the metric by missing non-null terms? You claimed that you knew the metric by heart.
Like in using the wrong definition of angular speed?

[JesseM]

His derivation is a hack and you've been doing your darnest to defend it. Why is it so difficult for you to admit that it is wrong?
But currently your only basis for saying it's wrong is the argument in post #41. Regardless of whether that argument is valid, can you not admit that all your previous unrelated arguments which had nothing to do with \phi vs. \theta were on the wrong track?
Please read here (http://www.physicsforums.com/showpost.php?p=2747325&postcount=28).
We are talking about rigotous derivations, not about hacks, right?
Yes, I already did read that post, it's the same argument as the one I was responding to when I referred to "Your latest criticism in post #41 about angles". It is perfectly "rigorous" to consider a circular orbit which has constant r=R and constant \phi=\pi/2 for one half, and constant r=R and constant \phi=-\pi/2 for the other half, so that for any infinitesimal section of the object's worldline on either half, dr = d\phi = 0; do you deny that such an orbit should be physically possible in the Schwarzschild spacetime? It may be true that for the purposes of a derivation, it might be a bit more "elegant" to consider a different orbit where r and \theta remain constant for the whole orbit, but there's nothing physically wrong or non-rigorous about the way kev did it.

[starthaus]

Yes, I already did read that post, that's exactly what I was responding to above. It is perfectly "rigorous" to consider a circular orbit which has constant r=R and constant \phi=\pi/2 for one half, and constant r=R and constant \phi=-\pi/2 for the other half, so that for any infinitesimal section of the object's worldline on either half, dr = d\phi = 0; do you deny that such an orbit should be physically possible in the Schwarzschild spacetime? It may be true that for the purposes of a derivation,

What about the missing terms in \phi? What about the v=r\frac{d\theta}{dt}. Wouldn't it be easier for you to admit that you are backing the wrong formulas rather than patching in all kinds of special pleads? I gave you the correct general formula, it does not agree with kev's formula. I gave you the general derivation, it does not agree with the pervect/kev derivation. Can you at least decide which is right and which is wrong?

it might be a bit more "elegant" to consider a different orbit where r and \theta remain constant for the whole orbit, but there's nothing physically wrong or non-rigorous about the way kev did it.

Isn't this the problem that needs to be solved? Isn't this the problem I solved at post 2?

[espen180]

In #2 you gave the metric, which of course is at the heart of all the results suggested in the thread. Personally I am unsure where the problem is. For my part, I choose dr/dt=0 and theta=pi/2 at the beginning of the derivation, but the argument here was that keeping these zero is not neccesary. The calculation is just as valid, for example at the apogee of the particle's trajectory, is what I think JesseM meant.

The other disputes have been focused on individual pieces of the derivation, like how to treat the metric or how to define certain variables.

[JesseM]

What about the missing terms in \phi?
It's true that kev did not write out the full metric, but given that he was assuming an orbit where for any infinitesimal segment you'd have d\phi = 0, those extra terms would disappear anyway so this wouldn't affect his final results. And kev never claimed he was starting from the full metric, he said in post #8 (http://www.physicsforums.com/showpost.php?p=2446850&postcount=8) that he was "Starting with this equation given by pervect", and pervect had already eliminated terms that went to zero.
What about the v=r\frac{d\theta}{dt}.
What about it? That would appear to be an equation for Schwarzschild coordinate velocity (as opposed to kev's 'local velocity') for an object in circular orbit with varying \theta coordinate, as with the type of orbit I described--again, do you agree that the type of orbit I described is a physically valid one? If you agree there would be a valid physical orbit with that type of coordinate description (with \phi having one constant value for half the orbit and a different constant value for the other half), do you disagree that the above equation would be the correct coordinate velocity for an object in this orbit?
Wouldn't it be easier for you to admit that you are backing the wrong formulas rather than patching in all kinds of special pleads?
I think you don't understand what special pleading (http://www.nizkor.org/features/fallacies/special-pleading.html) is, the fact that I and others respond to each of your various arguments with counterarguments, resulting in you continually abandoning your previous arguments in favor of new arguments you have invented on the spot, does not qualify as "special pleading". Yes or no, do you acknowledge that the arguments you made against kev's derivation prior to the new argument you've made in the posts here (http://www.physicsforums.com/showpost.php?p=2747325&postcount=28) and here (http://www.physicsforums.com/showpost.php?p=2747305&postcount=41) were flawed?
I gave you the correct general formula, it does not agree with kev's formula.
kev's formula is not intended to be a "general" one for arbitrary motion, it deals specifically with the case of an object in circular orbit. And since the OP was asking about whether total time dilation was a sum of gravitational and velocity-based time dilation, I thought it would be interesting to point out that for this specific case, total time dilation was actually a product of the two (whereas your more general formula does not relate in any obvious way to the formulas for gravitational and velocity-based time dilation)
I gave you the general derivation, it does not agree with the pervect/kev derivation.
Do you deny that the general formula would reduce to the specific formulas found by pervect/kev in the specific case they were considering, namely an infinitesimal section of a circular orbit where the radial coordinate and one of the two angular coordinates are constant?
Can you at least decide which is right and which is wrong?
If a general formula reduces to a more specific formula under the specific conditions assumed in the derivation of the specific formula, I'd say that both are right.

[starthaus]

whereas your more general formula does not relate in any obvious way to the formulas for gravitational and velocity-based time dilation

Are you even reading what I am writing? Can you re-read posts 2,6,39,47?

[starthaus]

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}}

No, you are missing a lot of terms.

[JesseM]

Are you even reading what I am writing? Can you re-read posts 2,6,39,47?
Post 47 was by espen180, and as to the others, yes you derived equations for special cases that were closer to a product of gravitational time dilation and something else, but since you didn't use the concept of local velocity the "something else" (i.e. the second part of the product that didn't look like the gravitational time dilation equation) didn't look very much like the SR velocity-based time dilation equation. Since espen180's original post was asking about the total time dilation in relation to the gravitational time dilation and velocity-based time dilation formulas, I thought kev and pervect's equations were relevant to the OP. Again, I'm not saying your equations are wrong in any way, but you haven't made a convincing case that kev's are wrong either--if you still maintain that, please answer the questions in my previous post.

[starthaus]

Post 47 was by espen180, and as to the others, yes you derived equations for special cases that were closer to a product of gravitational time dilation and something else, but since you didn't use the concept of local velocity the "something else" (i.e. the second part of the product that didn't look like the gravitational time dilation equation) didn't look very much like the SR velocity-based time dilation equation.

Because it isn't. Both effects are GR effects, in fact, there is only one effect. There is no such thing as an SR effect. The effect falls out the Schwarzschild metric.I have already explained this here (http://www.physicsforums.com/showpost.php?p=2744657&postcount=6)


Since espen180's original post was asking about the total time dilation in relation to the gravitational time dilation and velocity-based time dilation formulas, I thought kev and pervect's equations were relevant to the OP. Again, I'm not saying your equations are wrong in any way, but you haven't made a convincing case that kev's are wrong either--if you still maintain that, please answer the questions in my previous post.

I have answered your questions, I would like you to answer mine.

[JesseM]

Because it isn't. Both effects are GR effects. There is no such thing as an SR effect.
This is a strawman, I didn't say anything about "SR effect", I just said that the formula for total time dilation of an orbiting object calculated using GR (not SR) broke down into the product of two formulas that look like the formulas for gravitational time dilation for a stationary object and velocity-based time dilation for a moving object in SR.
I have answered your questions, I would like you to answer mine.
What questions of yours have I not answered? You didn't even ask any questions in this post! Anyway, there are several questions in that previous post #59 that I have asked variations on in the past and you have not answered, such as "again, do you agree that the type of orbit I described is a physically valid one?" (referring to the type of orbit I mentioned earlier in post 56) and "Yes or no, do you acknowledge that the arguments you made against kev's derivation prior to the new argument you've made in the posts here (http://www.physicsforums.com/showpost.php?p=2747325&postcount=28) and here (http://www.physicsforums.com/showpost.php?p=2747305&postcount=41) were flawed?" and "Do you deny that the general formula would reduce to the specific formulas found by pervect/kev in the specific case they were considering, namely an infinitesimal section of a circular orbit where the radial coordinate and one of the two angular coordinates are constant?" If you are interested in good-faith debate here, please answer the questions I ask you (and I will do likewise of course) rather than just picking one part of my post to criticize and ignoring everything else.

[Passionflower]

Because it isn't. Both effects are GR effects, in fact, there is only one effect. There is no such thing as an SR effect. The effect falls out the Schwarzschild metric.
That does not exclude the possibility to identify them as separate things.

For instance in the Gullstrand–Painlevé chart you can readily identify the Lorentz factor for relative motions.

But in non-stationary spacetimes this will obviously be a very daunting task.

[starthaus]

but since you didn't use the concept of local velocity

Because, contrary to kev's claims, the term \frac{dr/dt}{\sqrt{1-r_s/r}} does not represent "local velocity". It represents nothing. We had a very lengthy discussion in another thread, if you want the correct formula for local velocity you can find it in my blog, in the file "General Euler-Lagrange derivation for proper and coordinate acceleration".




the "something else" (i.e. the second part of the product that didn't look like the gravitational time dilation equation) didn't look very much like the SR velocity-based time dilation equation.

Because it doesn't. Because it has absolutely nothing to do with any "SR velocity-based time dilation".

What questions of yours have I not answered?

One very basic one: that the derivation and the result I gave are the rigorous, complete ansers. Yes or no?


You didn't even ask any questions in this post!

Sure I did, you need to look at the sentences ending with question marks.


Anyway, there are several questions in that previous post #59 that I have asked variations on in the past and you have not answered, such as "again, do you agree that the type of orbit I described is a physically valid one?" (referring to the type of orbit I mentioned earlier in post 56) and "Yes or no, do you acknowledge that the arguments you made against kev's derivation prior to the new argument you've made in the posts here (http://www.physicsforums.com/showpost.php?p=2747325&postcount=28) and here (http://www.physicsforums.com/showpost.php?p=2747305&postcount=41) were flawed?" and "Do you deny that the general formula would reduce to the specific formulas found by pervect/kev in the specific case they were considering, namely an infinitesimal section of a circular orbit where the radial coordinate and one of the two angular coordinates are constant?" If you are interested in good-faith debate here, please answer the questions I ask you (and I will do likewise of course) rather than just picking one part of my post to criticize and ignoring everything else.

If you are happy with hacks and with formulas that are correct for restrictive conditions , like only in the equatorial plane, the answer is yes. Happy?

[JesseM]

Because, contrary to kev's claims, the term \frac{dr/dt}{\sqrt{1-r_s/r}} does not represent "local velocity". It represents nothing. We had a very lengthy discussion in another thread, if you want the correct formula for local velocity you can find it in my blog.
What thread did you discuss it? Do you also dispute DrGreg's derivations which he linked to in post #8?
Because it doesn't. Because it has absolutely nothing to do with any "SR velocity-based time dilation".
"Has to do with" is a rather ill-defined phrase. I'd say that if it uses the same equation as "SR velocity-based time dilation", then it "has to do with" it in at least some limited sense.
One very basic one: that the derivation and the result I gave are the rigorous, complete ansers. Yes or no?
Your derivation seems rigorous but you'll have to define what you mean by "complete". Do you just mean that it's the most general case, and that all more specific answers would be derivable from it? If so then I agree. But if you're saying that from a pedagogical point of view it's "complete" in the sense that there's no point in discussing any more specific cases, I disagree, the more specific cases may be more helpful for gaining physical intuitions than the more general equation.
What questions of yours have I not answered? You didn't even ask any questions in this post!
Sure I did, you need to look at the sentences ending with question marks.
When I said "in this post" I meant the post I was responding to (the one where you said 'I have answered your questions, I would like you to answer mine'). There were no sentences ending with question marks in that post. If there were questions in other posts that you think I didn't address and you'd still like answers to, just point them out.
If you are happy with hacks and with formulas that are correct for restrictive conditions , like only in the equatorial plane, the answer is yes. Happy?
Not completely, because your comments about "hacks" and "restrictive conditions" still seem to imply you think there is something better about your own suggestion here (http://www.physicsforums.com/showpost.php?p=2747325&postcount=28) that we should set d\theta to 0 rather than d\phi, when in fact this would be every bit as restrictive in terms of the set of circular orbits that would meet this condition--do you disagree that both are equally restrictive? Anyway, as I think espen180 pointed out earlier, because of the spherical symmetry of the Schwarzschild spacetime, for any circular orbit you can always do a simple coordinate transformation into a coordinate system that still has the same metric but where the circular orbit now meets this condition, so in fact kev's final equation should apply to arbitrary circular orbits. If you're familiar with the phrase without loss of generality (http://en.wikipedia.org/wiki/Without_loss_of_generality) in proofs, kev could have said "without loss of generality, assume we're dealing with a circular orbit where d\phi = 0" and a physicist would understand the implied argument about why the final results should apply to all circular orbits.

[starthaus]

What thread did you discuss it?

Here (http://www.physicsforums.com/showthread.php?t=402135)


Do you also dispute DrGreg's derivations which he linked to in post #8?

DrGreg's time dilation formula is a subset of mine, so "no". The point is that the quantity in discussion (v) is not what you and kev claim it is. The correct formula can be found in my blog.



"Has to do with" is a rather ill-defined phrase. I'd say that if it uses the same equation as "SR velocity-based time dilation", then it "has to do with" it in at least some limited sense.

That's what you said. And I explained to you that it has nothing to do with any "SR-based time dilation. ".


Your derivation seems rigorous

It either is or it isn't. Can you answer with yes or no, please?


but you'll have to define what you mean by "complete". Do you just mean that it's the most general case, and that all more specific answers would be derivable from it? If so then I agree.

Good. This is what I meant.



your comments about "hacks" and "restrictive conditions" still seem to imply you think there is something better about your own suggestion here (http://www.physicsforums.com/showpost.php?p=2747325&postcount=28) that we should set d\theta to 0 rather than d\phi,

Absolutely.d \theta=0 means no motion along the meridian whereas d \phi=0 means no rotation, contradicting the problem statement. Why are we even discussing this?


when in fact this would be every bit as restrictive in terms of the set of circular orbits that would meet this condition--do you disagree that both are equally restrictive?

Of course I do, how many posts do we need to waste on this obvious issue?


Anyway, as I think espen180 pointed out earlier, because of the spherical symmetry of the Schwarzschild spacetime, for any circular orbit you can always do a simple coordinate transformation into a coordinate system that still has the same metric but where the circular orbit now meets this condition,

That's not the point. \phi and \theta are not intechangeable. They have different meanings , both mathematically and physically. They have different domains of definition, \omega is \frac{d\phi}{dt} (and not \frac{d\theta}{dt}). If you insist on interchanging them, you would need to exchange their domains of definition (\theta would need to be in the interval [0,2\pi]), you would also need to rewrite the Schwarzschild metric. This is not what kev did in his hack.




so in fact kev's final equation should apply to arbitrary circular orbits. If you're familiar with the phrase without loss of generality (http://en.wikipedia.org/wiki/Without_loss_of_generality) in proofs, kev could have said "without loss of generality, assume we're dealing with a circular orbit where d\phi = 0" and a physicist would understand the implied argument about why the final results should apply to all circular orbits.

This is incorrect, since \phi and \theta are not intechangeable.

[yuiop]

-The correct answer to Dmitry7's question is:

\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/c\sqrt{1-r_s/r_1})^2}{1-(r_2sin\theta_2\omega/c\sqrt{1-r_s/r_2})^2}}

-The correct answer to espen180's question is :


\frac{d\tau}{dt}=\sqrt{1-\frac{r_s}{r}}\sqrt{1-(\frac{r\omega sin(\theta)}{c \sqrt{1-r_s/r}})^2}

I hope that this clarifies things once and for all.

This is not the correct answer to the question posed by espen180 in the OP of this thread. Here it the original question again to remind you:

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?

Espen is asking about the resultant time dilation due to motional and gravitational dilation. He asks about the contribution due to motion but does not specify that the motion should be orbital. You supposedly "correct answer" is for the limiting case of orbital motion only. This might seem a bit picky, but you have set the standard here:

If you are happy with hacks and with formulas that are correct for restrictive conditions , like only in the equatorial plane, the answer is yes. Happy?
You have put a restrictive condition of only considering orbital (horizontal) motion. The complete general and correct answer to the OP was given by DrGreg in #8 and I quote him again 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" (http://www.physicsforums.com/showthread.php?t=383884) (and repeated in post #46).

DrGreg's conclusions agree with the conclusions of pervect and myself.


You seem to think that you equation given in #7:

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

and my equation:

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

are in disagreement, with yours right and mine wrong and fail to understand that they are numerically identical.

[EDIT] Well they would be numerically identical, when you correct the error in your equation. Your equation in #7 should read:

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

[starthaus]

This is not the correct answer to the question posed by espen180 in the OP of this thread.

:lol:
DrGreg's conclusions agree with the conclusions of pervect and myself.

How do you get the "general solution" from the truncated metric you've been using? Run this by us again, please.


Espen is asking about the resultant time dilation due to motional and gravitational dilation. He asks about the contribution due to motion but does not specify that the motion should be orbital. You supposedly "correct answer" is for the limiting case of orbital motion only. This might seem a bit picky, but you have set the standard here:

If you don't understand the use of Schwarzschild solution in deriving the answer, then ask and I'll try to help you.

[starthaus]

[EDIT] Well they would be numerically identical, when you correct the error in your equation. Your equation in #7 should read:

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

Congratulations, you found a missing closed/open parens after \frac{1}{c^2}. You are good at finding typos. :-)
The complete equation is obviously correct:

\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}

[yuiop]

This is incorrect, since \phi and \theta are not intechangeable.

Let us say an exam question asks what is the time dilation ratio d\tau/dt of a particle moving in Schwarzschild coordinates with velocity components dr/dt = 0, r d\theta/dt = 0 and r d\phi/dt = 0.5 and numerical values for G, M, c and r are given.

Student A states that is not possible to give a numerical solution because the value of \theta has not been given.

Student B realises the spherical symmetry of Schwarzschild coordinates allows him to re-orientate the axes of the coordinates and interchange d\phi and d\theta so that he can obtain a correct numerical solution, even though the value of \theta is not known.

Who deserves the most points for their answer? Student A who is plugging numbers into a formula or student B who is using understanding of the physical situation and using a bit of ingenuity to actually obtain the required numerical answer?

[starthaus]

Let us say an exam question asks what is the time dilation ratio d\tau/dt of a particle moving in Schwarzschild coordinates with velocity components dr/dt = 0, d\theta/dt = 0 and d\phi/dt = 0.5 and numerical values for G, M, c and r are given.

Student A states that is not possible to give a numerical solution because the value of \theta has not been given.

And he's right.



Student B realises the spherical symmetry of Schwarzschild coordinates allows him to re-orientate the axes of the coordinates and interchange d\phi and d\theta so that he can obtain a correct numerical solution, even though the value of \theta is not known.

Try doing that and I'll show you where the mistake is.



Who deserves the most points for their answer? Student A who is plugging numbers into a formula or student B who is using understanding of the physical situation and using a bit of ingenuity to actually obtain the required numerical answer?

Neither, student C who knows how to use the general (not the truncated) Schwarzschild solution such that he/she produces the correct general solution that has \theta as a parameter. How would you do this , kev, without resorting to silly hacks?

[yuiop]

Why not just contract the angle differentials into \text{d}\theta^2+\sin^2\theta\text{d}\phi^2=\text{ d}\Omega^2 and avoid the problem alltogether?Because \phi and \theta are independent coordinates. So your hack is illegal.

LOL. In which countries is this illegal?

I have seen this equation given by espen used in enough references to be fairly certain it is legitimate and pretty much standard procedure.

[starthaus]

LOL. In which countries is this illegal?

I have seen this equation given by espen used in enough references to be fairly certain it is legitimate and pretty much standard procedure.

calculus is not your strong suit.

[JesseM]

DrGreg's time dilation formula is a subset of mine, so "no". The point is that the quantity in discussion (v) is not what you and kev claim it is.
In post #8 DrGreg said that his v stood for "speed relative to a local hovering observer using local proper distance and local proper time". Do you disagree with this definition? If not, what do you think kev "claims" that is different? (I haven't claimed anything about it myself, so the phrase 'what you and kev claim' is pure imagination on your part)
That's what you said. And I explained to you that it has nothing to do with any "SR-based time dilation. ".
I don't think you understood my point. I was saying that "has to do with" is a semantically ambiguous phrase, and that under one reasonable definition of "has to do with", the fact that the term in the total GR time dilation equation for circular orbits looks just like the SR time dilation equation would by definition mean it "has to do with" the SR time dilation equation, since "has to do with" can be defined in a broad way that does not imply any deeper connection besides a superficial similarity in equations. You certainly never "explained" why it doesn't have to do with SR time dilation, since you never gave any meaningful definition of the vague phrase "has to do with".
It either is or it isn't. Can you answer with yes or no, please?
The reason I said "seems" is that I looked at your answer in post #13 and got the gist of how you derived it, the approach seemed fine but I didn't check the details of the math, trusting you probably got it right. If you insist that I double-check your work in detail, fine:


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


d\tau /dt = (1/c)\sqrt{(1-r_s/r)c^2-(1-r_s/r)^{-1}(dr/dt)^2-(rd\theta/dt)^2-(rd\phi/dt sin\theta)^2}

Strictly speaking this equation is itself the most general answer for the time dilation for an object moving along an arbitrary worldline where dr/dt, d\theta/dt, and d\phi/dt might all be nonzero. But then if we introduce the condition that we are looking at a portion of an orbit where dr = d\theta = 0 as you did in post #6, and set \omega = d\phi/dt, then this reduces to:

d\tau /dt = (1/c)\sqrt{(1-r_s/r)c^2 - (r sin\theta \omega)^2}

Factoring out \sqrt{(1-r_s/r)c^2} gives:

d\tau /dt = \sqrt{1 - r_s/r}\sqrt{1 - (r sin\theta \omega)^2 / (1-r_s/r)c^2}

or, to make it closer to the form you chose:

d\tau /dt = \sqrt{1 - r_s/r}\sqrt{1 - (r sin\theta \omega/ c\sqrt{1-r_s/r})^2}

Then for the ratio of times for two clocks in circular paths with r1 and r2 and \theta_1 and \theta_2 and d\phi_1/dt = \omega_1 and d\phi_2/dt = \omega_2 we'd have:


\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_1/c\sqrt{1-r_s/r_1})^2}{1-(r_2sin\theta_2\omega_2/c\sqrt{1-r_s/r_2})^2}}


This is almost like the equation you got, but you do seem to have made the minor error of leaving out the "c" that appears in my equation (yours is not dimensionally correct, since \omega has units of 1/time), and also you just wrote \omega in both parts of the fraction, neglecting to account for the possibility that \omega_1 differs from \omega_2. An additional thing to note is that if you want to have a circular orbit where d\theta = 0 as opposed to an arbitrary circular path, you must pick \theta = 0, so that sin\theta = 0 and the whole thing reduces to \frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-r_s/r_1}{1-r_s/r_2}}; a circular path with d\theta = 0 but \theta not equal to zero would be one where the center of the path did not coincide with the center of the Schwarzschild coordinate system at r=0, like how a line of latitude on the Earth forms a circle whose center does not coincide with the center of the Earth (unless it's latitude 0, in which case the line of latitude would just be the equator), so this would not actually be a physically valid "orbit".
when in fact this would be every bit as restrictive in terms of the set of circular orbits that would meet this condition--do you disagree that both are equally restrictive?
Of course I do, how many posts do we need to waste on this obvious issue?
In what sense do you imagine that your suggestion of setting d\theta = 0 is less "restrictive" than setting d\phi = 0? Setting d\theta=0 is equivalent to picking a circle of constant latitude on a globe (latitude varies 180 degrees from the South Pole to the North Pole, just like the \theta coordinate varies over half a circle from 0 to \pi), like the first image on this page (http://literacynet.org/sciencelincs/showcase/drifters/activity1b.html):

http://literacynet.org/sciencelincs/showcase/drifters/images/latitude.jpg

Whereas setting d\phi = 0 is equivalent to picking a circle of constant longitude on a globe (longitude varies 360 degrees from 180 west of the prime meridian to 180 east of it, just like \phi varies in a full circle from -\pi to \pi), like the second image:

http://literacynet.org/sciencelincs/showcase/drifters/images/longitude.jpg

It should be clear that all the infinite possible circles with d\phi = 0 have centers that coincide with the center of the coordinate system at r=0, whereas only a single circle with d\theta = 0 has a center that coincides with the center of the coordinate system (the one at \theta = 0). So if you're interested in circular orbits, d\theta = 0 seems more restrictive, not less restrictive! But either way you could also find an infinite number of circular orbits at different angles relative the coordinate system such that neither d\theta = 0 nor d\phi = 0 would apply.

Do you disagree with any of the above? If so, which part? If not, can you explain your reasoning behind calling d\theta = 0 the less "restrictive" condition?
That's not the point. \phi and \theta are not intechangeable. They have different meanings , both mathematically and physically. They have different domains of definition, \omega is \frac{d\phi}{dt} (and not \frac{d\theta}{dt}). If you insist on interchanging them, you would need to exchange their domains of definition (\theta would need to be in the interval [0,2\pi]), you would also need to rewrite the Schwarzschild metric.
No, you would not need to rewrite the Schwarzschild metric, that's exactly what I meant when I said "Anyway, as I think espen180 pointed out earlier, because of the spherical symmetry of the Schwarzschild spacetime, for any circular orbit you can always do a simple coordinate transformation into a coordinate system that still has the same metric but where the circular orbit now meets this condition". Just as the metric has exactly the same form in the different inertial coordinate systems used in flat spacetime, so it is also true that because of the spherical symmetry of the Schwarzschild metric, one can find a family of different spherical coordinate systems with the \theta = 0 and \phi = 0 axes oriented in different directions (all of them having the same r=0 point), but with the same Schwarzschild metric applying to all these coordinate systems. And for any circular orbit whose center is at r=0, there will be one member of this family of spherical coordinate systems where the orbit has a constant \theta coordinate, and another where the orbit can be divided into two halves that each have a constant \phi coordinate. Do you disagree?

[starthaus]

In post #8 DrGreg said that his v stood for "speed relative to a local hovering observer using local proper distance and local proper time". Do you disagree with this definition? If not, what do you think kev "claims" that is different? (I haven't claimed anything about it myself, so the phrase 'what you and kev claim' is pure imagination on your part)

I don't think you understood my point. I was saying that "has to do with" is a semantically ambiguous phrase, and that under one reasonable definition of "has to do with", the fact that the term in the total GR time dilation equation for circular orbits looks just like the SR time dilation equation would by definition mean it "has to do with" the SR time dilation equation, since "has to do with" can be defined in a broad way that does not imply any deeper connection besides a superficial similarity in equations. You certainly never "explained" why it doesn't have to do with SR time dilation, since you never gave any meaningful definition of the vague phrase "has to do with".

The reason I said "seems" is that I looked at your answer in post #13 and got the gist of how you derived it, the approach seemed fine but I didn't check the details of the math, trusting you probably got it right. If you insist that I double-check your work in detail, fine:


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


d\tau /dt = (1/c)\sqrt{(1-r_s/r)c^2-(1-r_s/r)^{-1}(dr/dt)^2-(rd\theta/dt)^2-(rd\phi/dt sin\theta)^2}

Strictly speaking this equation is itself the most general answer for the time dilation for an object moving along an arbitrary worldline where dr/dt, d\theta/dt, and d\phi/dt might all be nonzero. But then if we introduce the condition that we are looking at a portion of an orbit where dr = d\theta = 0 as you did in post #6, and set \omega = d\phi/dt, then this reduces to:

d\tau /dt = (1/c)\sqrt{(1-r_s/r)c^2 - (r sin\theta \omega)^2}

Factoring out \sqrt{(1-r_s/r)c^2} gives:

d\tau /dt = \sqrt{1 - r_s/r}\sqrt{1 - (r sin\theta \omega)^2 / (1-r_s/r)c^2}

or, to make it closer to the form you chose:

d\tau /dt = \sqrt{1 - r_s/r}\sqrt{1 - (r sin\theta \omega/ c\sqrt{1-r_s/r})^2}

Then for the ratio of times for two clocks in circular paths with r1 and r2 and \theta_1 and \theta_2 and d\phi_1/dt = \omega_1 and d\phi_2/dt = \omega_2 we'd have:


\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_1/c\sqrt{1-r_s/r_1})^2}{1-(r_2sin\theta_2\omega_2/c\sqrt{1-r_s/r_2})^2}}


This is almost like the equation you got, but you do seem to have made the minor error of leaving out the "c"

It's a typo, corrected in many subsequent posts. See 37 for example.




that appears in my equation (yours is not dimensionally correct, since \omega has units of 1/time), and also you just wrote \omega in both parts of the fraction, neglecting to account for the possibility that \omega_1 differs from \omega_2.

If you insist to have the objects rotating a different angular speeds, yes. But the post is about different clocks at different altitudes and different latitudes. Remember that I was answering Dmitry67's question. So, the clocks share the same \omega

[yuiop]

And he's right.

OK, I will concede this one. They are only interchangeable if \theta = \pi/2 which is what espen and pervect specified. It is often a practical convenience to orientate the axes so that this condition is met, a bit like orientating the x axes of two inertial frames with each other and parallel with the relative motion of the two frames in SR even though this is not the most general situation.

[JesseM]

It's a typo, corrected in many subsequent posts. See 37 for example.

If you insist to have the objects rotating a different angular speeds, yes. But the post is about different clocks at different altitudes and different latitudes. Remember that I was answering Dmitry67's question. So, the clocks share the same \omega
Fair enough, now can you please answer the questions I asked in that post? Specifically the sentences ending in question marks.

[starthaus]

No, you would not need to rewrite the Schwarzschild metric, that's exactly what I meant when I said "Anyway, as I think espen180 pointed out earlier, because of the spherical symmetry of the Schwarzschild spacetime, for any circular orbit you can always do a simple coordinate transformation into a coordinate system that still has the same metric but where the circular orbit now meets this condition". Just as the metric has exactly the same form in the different inertial coordinate systems used in flat spacetime, so it is also true that because of the spherical symmetry of the Schwarzschild metric, one can find a family of different spherical coordinate systems with the \theta = 0 and \phi = 0 axes oriented in different directions (all of them having the same r=0 point), but with the same Schwarzschild metric applying to all these coordinate systems.

I think you explained it to yourself. \theta gives you the latitude of the plane of the circle defined by 0<\phi<2\pi, \theta=constant. So, the angular speed is \frac{d\phi}{dt}, not \frac{d\theta}{dt} that you keep trying to justify. You wrote it yourself , remember? \omega_1=\frac{d\phi_1}{dt} and \omega_2=\frac{d\phi_2}{dt} .


And for any circular orbit whose center is at r=0, there will be one member of this family of spherical coordinate systems where the orbit has a constant \theta coordinate, and another where the orbit can be divided into two halves that each have a constant \phi coordinate. Do you disagree?

You are trying to justify the inadvertent replacement of \phi (the correct coordinate) with \theta (the incorrect coordinate).
If you insist on doing that, at least do it correctly, the new metric should look like this:

ds^2=(1-r_s/r)(cdt)^2-\frac{dr^2}{1-r_s/r}-(rd\phi)^2-(rsin(\phi)d\theta)^2

So, you now have to make dr=d\phi=0 in order to get the time dilation and you are now obviously stuck with the term in sin(\phi)

[starthaus]

OK, I will concede this one. They are only interchangeable if \theta = \pi/2 which is what espen and pervect specified.

They aren't interchangeable unless you are in the business of producing hacks.There is a reason for the presence of sin(\theta) in the metric.

[JesseM]

I think you explained it to yourself. \theta gives you the latitude of the plane of the circle defined by 0<\phi<2\pi, \theta=constant. So, the angular speed is \frac{d\phi}{dt}, not \frac{d\theta}{dt} that you keep trying to justify.
That would only be the angular speed of a circle that happened to coincide with the equator (which would actually be \theta = \pi/2 rather than \theta = 0 as I incorrectly stated earlier). Do you disagree that any great circle (http://en.wikipedia.org/wiki/Great_circle) on a sphere would correspond to a valid circular orbit, including a circle which could be divided into two halves of constant longitude (i.e. constant \phi), or plenty of circles where neither longitude nor latitude were constant?
If you insist on doing that, at least do it correctly, the new metric should look like this
No, you're just not getting it. I specifically said that despite the fact that the coordinate systems are different, the metric would have exactly the same form in each one:
Just as the metric has exactly the same form in the different inertial coordinate systems used in flat spacetime, so it is also true that because of the spherical symmetry of the Schwarzschild metric, one can find a family of different spherical coordinate systems with the \theta = 0 and \phi = 0 axes oriented in different directions (all of them having the same r=0 point), but with the same Schwarzschild metric applying to all these coordinate systems.
Do you understand the analogy with "different inertial coordinate systems used in flat spacetime"? If we have two coordinate systems in flat spacetime related by the Lorentz transformation, you'd agree that even though their coordinate axes point in different directions, they would both still have the same metric d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2), right? If you can understand that, you should be able to understand how two spherical coordinate systems with their axes pointed in different directions can nevertheless have the same metric too. I believe that's essentially what the "spherical symmetry" of the Schwarzschild metric means, just like the Lorentz symmetry of Minkowski spacetime can be taken to mean that the metric is unchanged in the different inertial coordinate systems related by the Lorentz transformation.

[JesseM]

They aren't interchangeable unless you are in the business of producing hacks.There is a reason for the presence of sin(\theta) in the metric.
And the only valid circular orbit (i.e. a circular path whose center coincides with the center of the Schwarzschild coordinate system at r=0) where d\theta = 0 is one in the "equatorial" plane where \theta = \pi /2, in which case sin(\theta) = 1. Do you disagree?

[starthaus]

In post #8 DrGreg said that his v stood for "speed relative to a local hovering observer using local proper distance and local proper time". Do you disagree with this definition?

This is a nit but if you want the coordinate speed, both at arbitrary r and at the "hovering point" r_0, you can get it from the file I pointed out to you.


If not, what do you think kev "claims" that is different?

You are banging on a nit, I call \frac{dr/dt}{\sqrt{1-r_s/r}} a nothing, you insist on calling it "speed relative to a local hovering observer using local proper distance and local proper time". How you name it does not affect the final result and that result is unique, it falls out the Schwarzschild metric. Do you dipute that?

I think I have answered all your sentences that end with "?". :-)

[starthaus]

That would only be the angular speed of a circle that happened to coincide with the equator (which would actually be \theta = \pi/2 rather than \theta = 0 as I incorrectly stated earlier).

Yes, so what?


Do you disagree that any great circle (http://en.wikipedia.org/wiki/Great_circle) on a sphere would correspond to a valid circular orbit, including a circle which could be divided into two halves of constant longitude (i.e. constant \phi), or plenty of circles where neither longitude nor latitude were constant?

The point is that it doesn't. The domain for \theta is [0,\pi]. Do you dispute that?

[starthaus]

And the only valid circular orbit (i.e. a circular path whose center coincides with the center of the Schwarzschild coordinate system at r=0) where d\theta = 0 is one in the "equatorial" plane where \theta = \pi /2, in which case sin(\theta) = 1. Do you disagree?

So what? There is an infinity of other circles that do not share \theta = \pi /2,.
None of these are captured by the solution that uses the truncated metric. Why is this so difficult for you to understand?

[starthaus]

Do you understand the analogy with "different inertial coordinate systems used in flat spacetime"? If we have two coordinate systems in flat spacetime related by the Lorentz transformation, you'd agree that even though their coordinate axes point in different directions, they would both still have the same metric d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2), right? If you can understand that, you should be able to understand how two spherical coordinate systems with their axes pointed in different directions can nevertheless have the same metric too. I believe that's essentially what the "spherical symmetry" of the Schwarzschild metric means, just like the Lorentz symmetry of Minkowski spacetime can be taken to mean that the metric is unchanged in the different inertial coordinate systems related by the Lorentz transformation.

Yes, I understand it very well, this is why I gave you the counter-example that shows what happens to the metric when you exchange the roles of \theta and \phi.



The point is that kev used a truncated metric. Do you dispute that?
Using a truncated metric, he got a particular solution, that isn't valid in the general case. Do you dispute that?
I posted the general solutions for both orbital and radial motion using the full metric. Do you dispute that?
Both solutions are correct. Do you dispute that?
Heck, I even posted a superset of the solution , using the Kerr metric. Do you have any complains about it?

[JesseM]

Yes, I understand it very well, this is why I gave you the counter-example that shows what happens to the metric when you exchange the roles of \theta and \phi.
You obviously don't understand at all, since doing a coordinate transformation which changes the direction the \theta and \phi axes point in, and then finding the new metric in this coordinate system, is NOT equivalent to switching the places of the \theta and \phi coordinates in the metric equation. My whole point is that the "spherical symmetry" of the Schwarzschild metric means you can reorient the \theta and \phi axes in arbitrary directions and the metric will always remain unchanged, just like the metric remains unchanged under a Lorentz transformation with an arbitrary choice of velocity. The wikipedia article on spherically symmetric spacetimes (http://en.wikipedia.org/wiki/Spherically_symmetric_spacetime) supports this by saying a spherically symmetric spacetime is often described as one whose metric is "invariant under rotations".
The point is that kev used a truncated metric.
I don't know what you mean by "truncated metric". The metric gives the proper time along an arbitrary path, and kev was considering a circular orbit, so he could set terms like dr/dt and d\phi/dt to 0, dropping some terms. You did exactly the same thing in your derivation, only with dr = d\theta = 0. Is that all you mean by "truncated"?

[starthaus]

I don't know what you mean by "truncated metric".

Missing terms right off the bat.

I understand very well, please stop talking down to me. I asked you a set of questions, would you please answer them as a set (all of them in one post)? Thank you

[JesseM]

So what? There is an infinity of other circles that do not share \theta = \pi /2,.
None of these are captured by the solution that uses the truncated metric. Why is this so difficult for you to understand?
There are no circular orbits where \theta is constant (so d\theta = 0) and has a value other than \theta = \pi / 2. There are other circular paths where the value of \theta is some other constant, but they are like circles of constant latitude on a globe (aside from the equator), the center of the path does not coincide with the center of the coordinate system and thus they are not valid free-fall orbits. Do you disagree?

[starthaus]

There are no circular orbits where \theta is constant (so d\theta = 0) and has a value other than \theta = \pi / 2. There are other circular paths where the value of \theta is some other constant, but they are like circles of constant latitude on a globe (aside from the equator),

Correct, these are precisely the circles covered by the solution I gave in post 6. You covered them just the same in the reconstruction of my sollution (see the rsin(\theta)?)



the center of the path does not coincide with the center of the coordinate system and thus they are not valid free-fall orbits. Do you disagree?

Who's talking about free-fall orbits? How many times do I need to tell you that the solution evolved from answering Dmitry67's question about the rate of ticking clocks a different latitudes? What do you think the different \theta's in the formula represent?

Could you please answer all my questions, in one post and without turning every point into your question?

[JesseM]

Missing terms right off the bat.
But kev explicitly said (http://www.physicsforums.com/showpost.php?p=2446850&postcount=8) that he was starting from an equation pervect derived, where certain terms had already been eliminated. Do you think pervect's derivation was "missing terms right off the bat"?
I understand very well
A person who doesn't understand something will sometimes also fail to understand that they don't understand it. It's pretty clear that you didn't understand what I was saying if you thought that changing the metric equation by switching the roles of \theta and \phi had anything to do with what I was talking about, since I said very clearly that the metric should be invariant under rotations.
please stop talking down to me
When you keep repeating the same mistaken ideas about what I'm saying even though my words clearly show otherwise, I'm going to highlight the fact that you're not understanding me, if that seems like "talking down", well, better that than being over-polite and allowing you to persist in your mistaken understanding.
I asked you a set of questions, would you please answer them? Thank you
You asked a bunch of questions over a series of posts less than half an hour old and it's obvious I'm in the process of answering them, so hold your horses please.

[JesseM]

You are banging on a nit, I call \frac{dr/dt}{\sqrt{1-r_s/r}} a nothing, you insist on calling it "speed relative to a local hovering observer using local proper distance and local proper time".
I don't care what you choose to call it, but you seemed to imply that kev was actually incorrect in his description of what it meant when you said "The point is that the quantity in discussion (v) is not what you and kev claim it is". Are you actually saying there was any error in what kev "claims" about this quantity, or is it just that you prefer not to describe it at all?
How you name it does not affect the final result and that result is unique, it falls out the Schwarzschild metric.
Final result for what? An object in circular orbit, or some more general case?

Going to bed now, will continue tomorrow...

[starthaus]

Do you think pervect's derivation was "missing terms right off the bat"?

Yes, obviously. pervect not only truncated the metric, he also got the g_{tt} wrong. Do you disagree?



A person who doesn't understand something will sometimes also fail to understand that they don't understand it. It's pretty clear that you didn't understand what I was saying if you thought that changing the metric equation by switching the roles of \theta and \phi had anything to do with what I was talking about, since I said very clearly

You may not realize but what you are talking about is not a rotation of coordinates. What you are talking about is exchanging the rotational motion in the plane \theta=constant with a pseudo-rotation in the plane \phi=constant while all along refusing to admit that you can't complete such a motion since \theta<\pi.
(Basically you are trying to convey the idea that a half circle is a full circle. )



When you keep repeating the same mistaken ideas about what I'm saying even though my words clearly show otherwise, I'm going to highlight the fact that you're not understanding me, if that seems like "talking down", well, better that than being over-polite and allowing you to persist in your mistaken understanding.

Goes both ways, I think that you refuse to understand something very basic and that you put up this strwaman in order not to admit that the solution you have been defending is incorrect and incomplete. We will not get any resolution on this item so I propose that we table this subject. OK?



You asked a bunch of questions over a series of posts less than half an hour old and it's obvious I'm in the process of answering them, so hold your horses please.

I'll wait. Please do not ask any more questions before answering all my questions. I would really appreciate that.

[starthaus]

v is speed relative to a local hovering observer using local proper distance and local proper time.
.

v, in your time dilation formula is a scalar. Isn't the above in contradiction with your defining v as four-speed here (http://www.physicsforums.com/showpost.php?p=2613007&postcount=9)?

[espen180]

You may not realize but what you are talking about is not a rotation of coordinates. What you are talking about is exchanging the rotational motion in the plane \theta=constant with a pseudo-rotation in the plane \phi=constant while all along refusing to admit that you can't complete such a motion since \theta<\pi.
(Basically you are trying to convey the idea that a half circle is a full circle. )

You keep nagging on this point. There is nothing wrong with moving along a circular orbit around \theta rather than \phi. Since the metic is spherically symmetric, ALL circular paths coinciding with the center of the coordinate system is a valid circular orbit. What you are saying seems similar to "You can have a circular orbit about the equator of the Earth, but not perpendicular to the equator (concidering Earth as a perfect nonrotating sphere)". You make it seem like there is a preferred coordinate system where validity of circular orbits is decided.

It's as if you are just looking at the maths, but completely ignoring the physics.

[yuiop]

Let's first combine \text{d}\theta^2+\sin^2\theta \text{d}\phi^2=\text{d}\Omega^2 and so simplify the equation to

\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}\Omega}{c\text{d}t}\right)^2 }

Working backwards to get back to the metric gives me

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

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

I was hoping that doing this would lead me to an explanation as to where the \frac{1-\frac{r_s}{r}}{1-\frac{r_s}{r_0}} came from, but it seems it did not.

I do observe that in modeling this metric the metric coefficients are found by taking the ratio of the coefficients of the particle wrt an observer at infinity to the coefficients of the observer at r_0 to the same observer at infinity, but could I have an explanation of why that works?

Hi espen,

You are right to question this and I apologise for any confusion caused. When trying to answer your question I realised that basically I got this bit completely wrong. This is how it should be done:

The proper time of a moving clock at r relative to the reference clock at infinity in Schwarzschild coordinates is (using your notation):

\text{d}\tau = \text{d}t \sqrt{1-r_s/r} \sqrt{1- (1-r_s/r)^{-2}(\text{d}r/c\text{d}t)^2 - (1-r_s/r)^{-1}(r \text{d}\Omega/c\text{d}t)^2 }

The proper time of an observers clock d\tau_0 relative to the reference clock at infinity with motion dr_o/dt_o and d\Omega_o/dt_o at radius r_o, is given by:

\text{d}\tau_o = \text{d}t \sqrt{1-r_s/r_o} \sqrt{1- (1-r_s/r_o)^{-2}(\text{d}r_o/c\text{d}t_0)^2 - (1-r_s/r_o)^{-1}(r_o \text{d}\Omega_o/c\text{d}t_o)^2 }

The ratio of the proper time of the two clocks is then:

\frac{\text{d}\tau}{\text{d}\tau_o} = \frac{\sqrt{1-r_s/r\;}}{\sqrt{1-r_s/r_o}} \frac{\sqrt{1- (1-r_s/r\;)^{-2}(\text{d}r\;/c\text{d}t\;)^2 - (1-r_s/r\;)^{-1}(r\: \text{d}\Omega\;/c\text{d}t\;)^2 }}{ \sqrt{1- (1-r_s/r_o)^{-2}(\text{d}r_o/c\text{d}t_0)^2 - (1-r_s/r_o)^{-1}(r_o \text{d}\Omega_o/c\text{d}t_o)^2 }}

This is the completely general case of the ratio of two moving clocks in the Schwarzschild metric.

If the observer is stationary at r_o the equation reduces to:

\frac{\text{d}\tau}{\text{d}\tau_o}= \sqrt{\frac{1-r_s/r}{1-r_s/r_o}} \sqrt{1- \frac{(\text{d}r/\text{d}t)^2}{c^2(1-r_s/r)^2} - \frac{(r \text{d}\Omega/\text{d}t)^2}{c^2(1-r_s/r)} }

<EDIT> THe above has been edited to correct a typo.

[DrGreg]

v, in your time dilation formula is a scalar. Isn't the above in contradiction with your defining v as four-speed here (http://www.physicsforums.com/showpost.php?p=2613007&postcount=9)?With hindsight, it was a poor choice of notation, as the same symbol v was being used in different, incompatible ways. That's why I rewrote the argument in different notation in post #46 (http://www.physicsforums.com/showpost.php?p=2636290&postcount=46) of that same thread.

[espen180]

Thanks kev. So

\frac{\text{d}\tau}{\text{d}\tau_o}

here is the ratio of the observer's clock and the observed clock(for lack of a better term) as seen by a second observer at infinity, correct?

It is probably just my intuitive understanding that is failing me, but is this ratio the same ratio as observed by the first observer at r_0?

I think an SR example can explain my confusion:

Define the frame S, in which there is an observer at rest. In S, there are frames S' and S'' with observers at rest going at speeds v_1=\beta_1 c and v_2=\beta_2 c respectively in S. Now the rest observer can measure

t=\gamma_1\tau_1=\gamma_2\tau_2

The rest observer in S measures the ratio of the proper times of the rest observers in S' and S'' to be
\frac{\text{d}\tau_1}{\text{d}\tau_2}=\frac{\gamma _2}{\gamma_1}=\sqrt{\frac{1-\beta_1^2}{1-\beta_2^2}}

In S', the observer in S'' is travelling at the speed

v_2^{\prime}=\frac{\beta_2-\beta_1}{1-\beta_1\beta_2}c

and since \tau_1=\gamma_2^{\prime}\tau_2 the observer in S' measures

\left(\frac{\text{d}\tau_2}{\text{d}\tau_1}\right) ^{\prime}=\sqrt{1-\left(\frac{\beta_2-\beta_1}{1-\beta_1\beta_2}\right)^2}=\frac{\sqrt{1+\beta_1^2\ beta_2^2-\beta_2^2-\beta_1^2}}{1-\beta_1\beta_2}

Unless I made an error underway (which is very possible, this was a messy calculation), obervers S and S' dont seem to agree on the value of \frac{\text{d}\tau_2}{\text{d}\tau_1}.

[starthaus]

With hindsight, it was a poor choice of notation, as the same symbol v was being used in different, incompatible ways. That's why I rewrote the argument in different notation in post #46 (http://www.physicsforums.com/showpost.php?p=2636290&postcount=46) of that same thread.

Thank you for the honest answer. This brings me to a follow-up question. Can you please show how you calculate the value for w?

[starthaus]

You keep nagging on this point. There is nothing wrong with moving along a circular orbit around \theta rather than \phi. Since the metic is spherically symmetric, ALL circular paths coinciding with the center of the coordinate system is a valid circular orbit. What you are saying seems similar to "You can have a circular orbit about the equator of the Earth, but not perpendicular to the equator (concidering Earth as a perfect nonrotating sphere)".

No, this is not what I'm saying. What I am saying is something very basic and totally different. Yet, you seem clearly unable to grasp it.




You make it seem like there is a preferred coordinate system where validity of circular orbits is decided.

Not at all. It is very basic, really but you are so insistent, I'll explain it (maybe JesseM) will also get this one). The Earth rotates about the NS axis. \theta represents the angle from the N pole, therefore \frac{d\theta}{dt} is a very bad choice to represent the Earth rotation. By contrast, \frac{d\phi}{dt} is the correct choice.


It's as if you are just looking at the maths, but completely ignoring the physics.

LOL

[starthaus]

Hi espen,

You are right to question this and I apologise for any confusion caused. When trying to answer your question I realised that basically I got this bit completely wrong. This is how it should be done:

The proper time of a moving clock at r relative to the reference clock at infinity in Schwarzschild coordinates is (using your notation):

\text{d}\tau = \text{d}t \sqrt{1-r_s/r} \sqrt{1- (1-r_s/r)^{-2}(\text{d}r/c\text{d}t)^2 - (1-r_s/r)^{-1}(r \text{d}\Omega/c\text{d}t)^2 }

OK.



The proper time of an observers clock d\tau_0 relative to the reference clock at infinity with motion dr_o/dt_o and d\Omega_o/dt_o at radius r_o, is given by:

\text{d}\tau_o = \text{d}t \sqrt{1-r_s/r_o} \sqrt{1- (1-r_s/r_o)^{-2}(\text{d}r_o/c\text{d}t_0)^2 - (1-r_s/r_o)^{-1}(r \text{d}\Omega_o/c\text{d}t_o)^2 }

This is wrong. Since you are putting in results by hand again, try deriving it from the basics and you'll find out why.


The ratio of the proper time of the two clocks is then:

\frac{\text{d}\tau}{\text{d}\tau_o} = \frac{\sqrt{1-r_s/r\;}}{\sqrt{1-r_s/r_o}} \frac{\sqrt{1- (1-r_s/r\;)^{-2}(\text{d}r\;/c\text{d}t\;)^2 - (1-r_s/r\;)^{-1}(r \text{d}\Omega\;/c\text{d}t\;)^2 }}{ \sqrt{1- (1-r_s/r_o)^{-2}(\text{d}r_o/c\text{d}t_0)^2 - (1-r_s/r_o)^{-1}(r \text{d}\Omega_o/c\text{d}t_o)^2 }}

No.

[starthaus]

Unless I made an error underway (which is very possible, this was a messy calculation), obervers S and S' dont seem to agree on the value of \frac{\text{d}\tau_2}{\text{d}\tau_1}.

You sure did, they must agree. You just botched the SR Doppler effect. Now, where did you make the blunder?You and kev are piling up them errors.

[espen180]

Not at all. It is very basic, really but you are so insistent, I'll explain it (maybe JesseM) will also get this one). The Earth rotates about the NS axis. \theta represents the angle from the N pole, therefore \frac{d\theta}{dt} is a very bad choice to represent the Earth rotation. By contrast, \frac{d\phi}{dt} is the correct choice.

This is exactly why I opted to contract the angles into a single one which is in the direction on tangential motion. Then there is no need to worry about which angle to rotate around.

[starthaus]

This is exactly why I opted to contract the angles into a single one which is in the direction on tangential motion. Then there is no need to worry about which angle to rotate around.

Oh, but you do. Do you think that a mathematical sleigh of hand fixes your misunderstanding of basics physics? What axis does the Earth rotate about?

[espen180]

Oh, but you do. Do you think that a mathematical sleigh of hand fixes your misunderstanding of basics physics? What axis does the Earth rotate about?

I thought we were working with the Schwartzschild geometry, a non-rotating body.

Still, what difference does Earth's rotation axis make for orbits around the Earth? It's not like the Earth exhibits a non-negligible amount of rotational frame-dragging.

[starthaus]

I thought we were working with the Schwartzschild geometry, a non-rotating body.

So, what is \frac{d\phi}{dt} again?


Still, what difference does Earth's rotation axis make for orbits around the Earth?

You are making the same mistakes as JesseM, we are talking about the delay experienced by clocks on the Erath surface due to Earth rotation. What do you think I have been trying to explain to you starting with post 6?


It's not like the Earth exhibits a non-negligible amount of rotational frame-dragging.

This is not what we are talking about.

[DrGreg]

Thank you for the honest answer. This brings me to a follow-up question. Can you please show how you calculate the value for w?I'm not sure the context you have in mind. In my original post, the whole point was to calculate w from everything else that was in the same equation. That's probably not what you meant. So what did you mean? (In other words, if you want w in terms of something else, what is the "something else"?)

[starthaus]

I'm not sure the context you have in mind. In my original post, the whole point was to calculate w from everything else that was in the same equation. That's probably not what you meant. So what did you mean? (In other words, if you want w in terms of something else, what is the "something else"?)

In this thread I am calculating \frac{d\tau}{dt} as a function of coordinate speed \frac{dr}{dt} from the Schwarzschild metric:

ds^2=(1-r_s/r)(cdt)^2-.....

So, it would appear that your w is equal to \frac{dr/dt}{1-r_s/r}. I asked this question before in the thread, in post 10, probably it got missed in the tremendous noise. Is this correct? How would you arrive to w's value in your derivation? You do not appear to use the same approach I am using, this is why I am interested.

[DrGreg]

In this thread I am calculating \frac{d\tau}{dt} as a function of coordinate speed \frac{dr}{dt} from the Schwarzschild metric:

ds^2=(1-r_s/r)(cdt)^2-.....

So, it would appear that your w is equal to \frac{dr/dt}{1-r_s/r}. I asked this question before in the thread, in post 10, probably it got missed in the tremendous noise. Is this correct? How would you arrive to w's value in your derivation? You do not appear to use the same approach I am using, this is why I am interested.First of all I had better come clean about a detail that I glossed over. In my derivation in the other thread I referred to w as speed measured in... the locally-Minkowski coordinates of a free-falling and momentarily-at-rest local observer at that event...whereas in post #8 of this thread I referred to what I am now calling w as...speed relative to a local hovering observer using local proper distance and local proper timeIn case somebody complains, I should point out that the two speeds must be the same. As far as relative velocity is concerned, it doesn't matter whether the observer is accelerating or not, the relative velocity (in this sense) will be the same. (Of course you cannot use that argument for other quantities such as acceleration.)

If you want to express w in terms of Schwarzschild coordinates, you could construct "locally-rescaled Schwarzchild coordinates" at the event of interest (that is multiply each coordinate by a constant such that the metric equals the Minkowski metric at that event only) and then w will be the coordinate velocity in those coordinates, which you can then rescale back into Schwarzschild coordinate velocity.

So, for radial motion only (where \theta and \phi are constant and can be ignored), change coordinates to

T = t \sqrt{1-r_s/r_0}
R = r / \sqrt{1-r_s/r_0}

where r0 is the value of r where you want to make the measurement, so that the metric becomes

ds^2=c^2\,dT^2-dR^2

at that point only. Then, along the worldline being measured,

w = \frac{dR}{dT} = \frac{dR/dr}{dT/dt}\cdot \frac{dr}{dt} = \frac{dr/dt}{1-r_s/r_0}

So, yes, you are correct about w in this case.

Note that if you simply want to calculate dt/d\tau in terms of dr/dt you don't really need to involve w at all, you just plug dr = (dr/dt)\,dt into the metric and it all falls out.

[starthaus]

First of all I had better come clean about a detail that I glossed over. In my derivation in the other thread I referred to w as speed measured in... ...whereas in post #8 of this thread I referred to what I am now calling w as...In case somebody complains, I should point out that the two speeds must be the same. As far as relative velocity is concerned, it doesn't matter whether the observer is accelerating or not, the relative velocity (in this sense) will be the same. (Of course you cannot use that argument for other quantities such as acceleration.)

If you want to express w in terms of Schwarzschild coordinates, you could construct "locally-rescaled Schwarzchild coordinates" at the event of interest (that is multiply each coordinate by a constant such that the metric equals the Minkowski metric at that event only) and then w will be the coordinate velocity in those coordinates, which you can then rescale back into Schwarzschild coordinate velocity.

So, for radial motion only (where \theta and \phi are constant and can be ignored), change coordinates to

T = t \sqrt{1-r_s/r_0}
R = r / \sqrt{1-r_s/r_0}

So, R and T are simply r and t rescaled to make 1-r_s/r "go away" from the Schwrazschild metric. I am having trouble assigning any physical properties to R and T and, consequently to w. To me, they are just rescaled versions of r,t,dr/dt.

where r0 is the value of r where you want to make the measurement, so that the metric becomes

ds^2=c^2\,dT^2-dR^2

at that point only. Then, along the worldline being measured,

w = \frac{dR}{dT} = \frac{dR/dr}{dT/dt}\cdot \frac{dr}{dt} = \frac{dr/dt}{1-r_s/r_0}

So, yes, you are correct about w in this case.



Thank you

My "w" simply falls out the metric (see post 6).

Note that if you simply want to calculate dt/d\tau in terms of dr/dt you don't really need to involve w at all, you just plug dr = (dr/dt)\,dt into the metric and it all falls out.

Yes, our methods are identical, I am just skipping the coordinate rescaling step.

[yuiop]

...
The proper time of a moving clock at r relative to the reference clock at infinity in Schwarzschild coordinates is (using your notation):

\text{d}\tau = \text{d}t \sqrt{1-r_s/r} \sqrt{1- (1-r_s/r)^{-2}(\text{d}r/c\text{d}t)^2 - (1-r_s/r)^{-1}(r \text{d}\Omega/c\text{d}t)^2 }
OK.
...
The proper time of an observers clock d\tau_0 relative to the reference clock at infinity with motion dr_o/dt_o and d\Omega_o/dt_o at radius r_o, is given by:

\text{d}\tau_o = \text{d}t \sqrt{1-r_s/r_o} \sqrt{1- (1-r_s/r_o)^{-2}(\text{d}r_o/c\text{d}t_0)^2 - (1-r_s/r_o)^{-1}(r \text{d}\Omega_o/c\text{d}t_o)^2 }

This is wrong. Since you are putting in results by hand again, try deriving it from the basics and you'll find out why.

OK, there is a typo in the second equation where I missed the "o" subscript for r in the Omega term. I think it is obvious what was intended from the method.

The second equation should be:

\text{d}\tau_o = \text{d}t \sqrt{1-r_s/r_o} \sqrt{1- (1-r_s/r_o)^{-2}(\text{d}r_o/c\text{d}t_0)^2 - (1-r_s/r_o)^{-1}(r_o \text{d}\Omega_o/c\text{d}t_o)^2 }

... Unless I made an error underway (which is very possible, this was a messy calculation), obervers S and S' dont seem to agree on the value of \frac{\text{d}\tau_2}{\text{d}\tau_1}. I will check it out. :wink:

[starthaus]

OK, there is a typo in the second equation where I missed the "o" subscript for r in the Omega term. I think it is obvious what was intended from the method.

The second equation should be:

\text{d}\tau_o = \text{d}t \sqrt{1-r_s/r_o} \sqrt{1- (1-r_s/r_o)^{-2}(\text{d}r_o/c\text{d}t_0)^2 - (1-r_s/r_o)^{-1}(r_o \text{d}\Omega_o/c\text{d}t_o)^2 }


Still wrong, still put in by hand.

[yuiop]

Still wrong, still put in by hand.

If the first equation is right, how can the second equation be wrong?

All I have done is change the names of the variables.

[starthaus]

If the first equation is right, how can the second equation be wrong?

All I have done is change the names of the variables.

You did more than that. Look it over carefully.

[yuiop]

You did more than that. Look it over carefully.

Ah, guessing games again. It was not my intention to do more than changing the names of the variables, so if that is not the case you must be talking about a typo I can not spot.

[JesseM]

That would only be the angular speed of a circle that happened to coincide with the equator (which would actually be \theta = \pi/2 rather than \theta = 0 as I incorrectly stated earlier).
Yes, so what?
"So what" is that since \theta = \pi/2 is the only case of the equation below (from my post 76, and basically the same as your own equation) that actually corresponds to a circular orbit rather than a non-orbiting circular path:

d\tau /dt = \sqrt{1 - r_s/r}\sqrt{1 - (r sin(\theta) \omega/ c\sqrt{1-r_s/r})^2}

...then for a circular orbit with d\theta = 0 it must be true that sin(\theta) = 1 and therefore the equation reduces to:

d\tau /dt = \sqrt{1 - r_s/r}\sqrt{1 - (r \omega/ c\sqrt{1-r_s/r})^2}

If you then make the substitution v = r \omega / \sqrt{1 - r_s/r}, which was exactly the substitution kev made in post #8 (http://www.physicsforums.com/showpost.php?p=2446850&postcount=8) (he defined u = r \left(\frac{d\phi}{dt}\right), equivalent to u = r\omega, and then he defined u = v \sqrt{1-\frac{r_s}{r}}, equivalent to v = r\omega / \sqrt{1 - r_s/r}), then this equation becomes:

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

So the equation is correct for the special case of a circular orbit where d\theta = 0. Do you disagree? If not, then the symmetry argument I already mentioned shows why this would hold for any circular orbit, even one where d\theta was not equal to 0 in our original coordinate system (since you could always rotate into a new coordinate system where d\theta was equal to 0 on the orbit, and the metric would be exactly the same in this new coordinate system since the Schwarzschild metric is invariant under rotations)

Actually now that I've looked back at kev's post #8 more carefully I have no idea why in post 28 (http://www.physicsforums.com/showpost.php?p=2747325&postcount=28) you criticized him by saying:
You need to make

d\theta=dr=0

-v is equal to:

r\frac{d\phi}{dt} and not r\frac{d\theta}{dt}
...since it appears to me he did make d\theta = 0, and he did define the velocity in terms of \frac{d\phi}{dt} rather than \frac{d\theta}{dt}! I guess I shouldn't have taken your word for it that he did it differently there (even though you could still get exactly the same final result by assuming a circular orbit where d\phi = 0 along a short segment, do you disagree? If you do disagree, I can demonstrate)
Do you disagree that any great circle (http://en.wikipedia.org/wiki/Great_circle) on a sphere would correspond to a valid circular orbit, including a circle which could be divided into two halves of constant longitude (i.e. constant \phi), or plenty of circles where neither longitude nor latitude were constant?
The point is that it doesn't.
The point is that what doesn't? You didn't answer my question about whether you disagree that there are valid circular orbits in Schwarzschild spacetime which, in a given coordinate system, would have a description like the one above. If you do disagree then I think you need to do some thinking about how spherical coordinates work, in particular what the coordinate description would look like for an "upright" circle whose plane was at a right angle to the "horizontal" \theta = \pi/2 plane.
The domain for \theta is [0,\pi]. Do you dispute that?
No, of course not, why do you imagine I would? In post #56 I gave the example of a complete circle where the coordinate description would be such that one half of the circle would have a constant r=R and \phi = \pi/2 while the other half would have constant r=R and \phi = -\pi/2, I thought it was fairly obvious that the points covered by each half would then be defined by varying \theta from 0 to \pi. Again, do you disagree that this would be a valid coordinate description for the set of points on a single continuous circle, one whose center is at r=0 and whose plane is at a right angle to the \theta = \pi/2 plane, and where d\phi = 0 along any infinitesimal segment of this circle? If not you should see why, despite the fact that kev actually made d\theta = 0 rather than d\phi = 0, it would have been perfectly valid for him to do the reverse, either way there'd be a valid circular orbit meeting this condition, there'd be nothing non-rigorous or "hack"-y about such a starting assumption.

[JesseM]

Who's talking about free-fall orbits? How many times do I need to tell you that the solution evolved from answering Dmitry67's question about the rate of ticking clocks a different latitudes?
Yes, I understand that's the question you were addressing in post #6, and (minor algebra errors aside) I don't dispute that your answer there is a good one to Dmitry67's post. But this whole debate got started when you disputed my answer to espen108's OP in post #3, in which I cited kev's result; you seem insistent that this is wrong somehow, although you seem to constantly change your mind about why it is wrong. kev's result is valid for all circular orbits, and it's relevant to espen108's OP because in that case the answer is a product of two equations that look just like GR time dilation and SR time dilation.
What do you think the different \theta's in the formula represent?
In your formula, you are assuming two circular paths which each have a different constant \theta coordinate. This makes sense as a response to Dmitry67's question about clocks at different latitudes.
Could you please answer all my questions, in one post and without turning every point into your question?
I reply to posts individually, so if you write one big post I'll respond with one big post, if you write a lot of little posts I'll have an equal number of responses. Up until recently we were going back and forth with big posts, but then for some reason you decided to start breaking up your responses (your posts 77, 80, 84 all respond to my post 76, while your posts 85 and 87 respond to my post 82) which resulted in our back-and-forth being spread out over many more posts.

[JesseM]

Yes, obviously. pervect not only truncated the metric, he also got the g_{tt} wrong. Do you disagree?
He made a minor error with g_{tt} which kev corrected in his response, but other than that I still don't know what you mean by "truncated", in his post (http://www.physicsforums.com/showpost.php?p=2445122&postcount=35) he started from the abstract form of the metric which included all the terms:

c^2 d\tau^2 = g_{tt} dt^2 + g_{rr} dr^2 + g_{\theta\theta} d\theta^2 + g_{\phi\phi} d\phi^2

Are you calling this "truncated" just because he didn't actually write out the equations for each "g"? There's nothing non-rigorous about this, anyone can look up what they'd all be in the Schwarzschild metric. Or do you think there was something wrong with pervect's next step of specifying that he was talking about a circular orbit where dr/dt and d\theta/dt would be zero, and eliminating the appropriate terms given these conditions?
A person who doesn't understand something will sometimes also fail to understand that they don't understand it. It's pretty clear that you didn't understand what I was saying if you thought that changing the metric equation by switching the roles of and had anything to do with what I was talking about, since I said very clearly
You may not realize but what you are talking about is not a rotation of coordinates. What you are talking about is exchanging the rotational motion in the plane \theta=constant with a pseudo-rotation in the plane \phi=constant while all along refusing to admit that you can't complete such a motion since \theta<\pi.
(Basically you are trying to convey the idea that a half circle is a full circle. )
Since nothing I said faintly resembles what you are saying, you need to actually explain where you got the idea that this is "what I am talking about", rather than just asserting it. I take it you think something I said implies this somehow? If so, what specific quote? Do you deny that in spherical coordinates, it is possible to find a valid continuous circle whose center is at r=0 and whose coordinates match the description I gave at the end of post 117? If you don't deny that, are you denying that it would be possible to rotate the original coordinate system into a new coordinate system where the same complete circle would now lie entirely in the \theta = \pi / 2 plane? If you deny either of these I suggest the problem lies with your understanding of how spherical coordinates work.
When you keep repeating the same mistaken ideas about what I'm saying even though my words clearly show otherwise, I'm going to highlight the fact that you're not understanding me, if that seems like "talking down", well, better that than being over-polite and allowing you to persist in your mistaken understanding.
Goes both ways, I think that you refuse to understand something very basic and that you put up this strwaman in order not to admit that the solution you have been defending is incorrect and incomplete. We will not get any resolution on this item so I propose that we table this subject. OK?
Sure, I was only responding to your complaint about "talking down".
You asked a bunch of questions over a series of posts less than half an hour old and it's obvious I'm in the process of answering them, so hold your horses please.
I'll wait. Please do not ask any more questions before answering all my questions. I would really appreciate that.
I believe I've now responded to all your replies to me, but I can't agree to the request not to ask any further questions, as that would make it impossible for me to pin you down on a lot of your ambiguous arguments (like the one above where you tell me 'what I am saying' when your summary bears no resemblance to what I actually said and you don't explain how you think it was implied by what I said).

[yuiop]

Let's start with this equation for the time dilation ratio:

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

The above equation is obtained directly from the Schwarzschild metric and I think we are all in agreement about its validity.

Now define local velocities u_x, u_y, u_z as measured by a stationary observer at r using his proper length (dr') and proper time (dt'):

u_x = dr'/dt' = \frac{dr/dt}{(1-r_s/r)}

u_y = r d\theta/dt' = \frac{r d\theta/dt}{\sqrt{1-r_s/r} }

u_z = r \sin\theta d\phi/dt' = \frac{r \sin \theta d\phi/dt}{\sqrt{1-r_s/r} }

Substitute these local velocities into the time dilation equation:

d\tau/dt = \sqrt{1-r_s/r} \sqrt{1- u_x^2/c^2 - u_y^2/c^2 - u_y^2/c^2 }

Now define the local 3 velocity as:

w = \sqrt{ u_x^2 + u_y^2 + u_y^2 }

and substitute this value in:

d\tau/dt = \sqrt{1-r_s/r} \sqrt{1- w^2/c^2 }

This is the result obtained in more detail and more rigorously by DrGreg in #8 and valid for all vertical/horizontal or radial/orbital motion of a test particle at r, just as DrGreg claimed.

I do not think there is anything Starthaus can dispute there.

[starthaus]

Let's start with this equation for the time dilation ratio:

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

Much better, how did you manage to get the winning combination after all the false starts?
All you needed to do is to start from the correct Schwarzschild metric and to factor out 1-r_s/r

[JesseM]

You are making the same mistakes as JesseM, we are talking about the delay experienced by clocks on the Erath surface due to Earth rotation. What do you think I have been trying to explain to you starting with post 6?
No, we are not talking about clocks on Earth's surface. You were talking about that in response to Dmitry67's post, not the OP by espen180 (you claim Dmitry67's post was originally on this thread and that it was later split, but I'm not even sure you're correct about that--given that you seemed to think Dmitry67's post was always the first one, it's quite possible the two threads were always separate and that you simply got confused and posted on this thread thinking you were still looking at the other thread, I don't remember Dmitry67's post ever being on this thread). The whole debate between you vs. everyone else got started because you claimed there was something dreadfully wrong with kev's derivation, which was meant to deal with the question of time dilation experienced by clocks in arbitrary circular orbits. The set of all valid circular orbits in Schwarzschild spacetime would include orbits with many different orbital planes, just like Pluto and Earth have different orbital planes despite orbiting the same Sun--some of these orbits would be ones where d\theta is equal to 0 (orbits in the \theta = \pi/2 plane), but others would be ones in a different plane where d\theta is not equal to 0. Do you disagree?

[starthaus]

No, we are not talking about clocks on Earth's surface. You were talking about that in response to Dmitry67's post, not the OP by espen180 (you claim Dmitry67's post was originally on this thread and that it was later split, but I'm not even sure you're correct about that--given that you seemed to think Dmitry67's post was always the first one, it's quite possible the two threads were always separate and that you simply got confused and posted on this thread thinking you were still looking at the other thread, I don't remember Dmitry67's post ever being on this thread). The whole debate between you vs. everyone else got started because you claimed there was something dreadfully wrong with kev's derivation,

Yes, look at the post above , kev finally got the right formula after a lot of false starts. Feel free to peruse all his false starts throughout this thread (posts 97, 112, etc). I am done.

[JesseM]

Yes, look at the post above , kev finally got the right formula after a lot of false starts. Feel free to peruse all his false starts throughout this thread. I am done.
I haven't been following the more recent posts between you and kev on this thread so I don't know if there were any "false starts" in his last few equations, but the original derivation I linked to was fine, and I pointed out in post 117 you could get the exactly the same formula from the equation you were using in post 6. If you want to use some minor error in a recent post by him as an excuse to avoid addressing all the flaws and confusions in your own arguments (in particular your own confused criticisms of my posts, not kev's), be my guest.

[Al68]

This is wrong. Since you are putting in results by hand again, try deriving it from the basics and you'll find out why.How about you tell us why instead? Or is this thread destined to exceed 300 posts, too?

Many readers miss out on opportunities to learn due to their understandable unwillingness to sort through 300+ posts of confrontational nonsense that could be avoided by you doing everyone a huge favor by just specifying what you object to and explaining why.

Just try it for once. You might even enjoy being helpful.

[starthaus]

How about you tell us why instead? Or is this thread destined to exceed 300 posts, too?

These are basic math errors, if you can't see them all by yourself, you should find a different hobby. This does not include your current one :trolling.
Enough said that kev understood his errors.

[JesseM]

These are basic math errors
If kev made any "basic math errors" in his recent posts to you they are apparently minor ones which don't affect the final equation he derived for circular orbits, as I showed you in post 117 using your own equation to derive it. And of course you have made plenty of minor math errors yourself, like not including the factor of c in post #6.

But never mind, you found a trivial error in someone else's argument, therefore you win the thread! Hooray! (claps very slowly)

[Al68]

These are basic math errors, if you can't see them all by yourself, you should find a different hobby. This does not include your current one :trolling.You keep accusing me of trolling, but you are the one who keeps destroying threads, making them completely worthless for most readers, with your shenanigans.

Back to the point, I'll take that as a NO, you either can't or won't explain (in any rational way) your claims.

[starthaus]

You keep accusing me of trolling, but you are the one who keeps destroying threads, making them completely worthless for most readers, with your shenanigans.

Back to the point, I'll take that as a NO, you either can't or won't explain (in any rational way) your claims.
I help people that are sincere, I don't help trolls. You contributted noothing to this thread.
If you can't spot the errors, you have no business (other than trolling) in this thread. But, I'll give you a hint, the errors have to do with the wrong variables in the expression.Compare against the correct final expression.

[Al68]

If you can't spot the errors, you have no business (other than trolling) in this thread.LOL. Yep, that's why you can't substantiate your claims: because if I can't "spot the errors" I have "no business in this thread".

Is that also why you're so rude and condescending?

Seriously, dude, it might feel good to say something constructive and useful in your posts.

[starthaus]

LOL. Yep, that's why you can't substantiate your claims: because if I can't "spot the errors" I have "no business in this thread".

Finally. You understand.



Is that also why you're so rude and condescending?

Standard response to your trolling <shrug>



Seriously, dude, it might feel good to say something constructive and useful in your posts.
So, you are unable to see the wrong variables. ....
I did, I put up quite a few solutions but you can only see the stuff that makes you tick, err troll.

[Al68]

LOL. Yep, that's why you can't substantiate your claims: because if I can't "spot the errors" I have "no business in this thread".

Is that also why you're so rude and condescending?

Seriously, dude, it might feel good to say something constructive and useful in your posts.I did, I put up quite a few solutions but you can only see the stuff that makes you tick, err troll.LOL. I was responding to your last post which contained nothing useful or constructive.

[starthaus]

LOL. I was responding to your last post which contained nothing useful or constructive.

Try reading the other posts, the ones that contain formulas. Can you read formulas?

[Al68]

Try reading the other posts, the ones that contain formulas. Can you read formulas?I read them. They weren't the ones I was referring to. Was that not obvious? Can you read English?

[ZapperZ]

Since we are at the point in this thread where people are no longer discussing the topic, but rather who is a troll and who isn't, I take it that the topic is no longer interesting. So this thread is done.

Note that if you think someone is trolling, DO NOT FEED THE TROLL!. Please use the REPORT button and report the post/thread to the Mentors.

Zz.