Relativity Anomaly

[HydrogenBond]

If we look at a satellite orbiting the earth, the orbital motion is balancing the pull of gravity to create an orbit. But if we look at the slowing of its clock, the time dilation is not showing this canceling of two effects. Clocks should speed up with elevation, while the balancing velocity of the orbit will slow it down. Why do we get a net time effect (clock net slows) when the two opposing effects should cancel each other?

[Janus]

If we look at a satellite orbiting the earth, the orbital motion is balancing the pull of gravity to create an orbit. But if we look at the slowing of its clock, the time dilation is not showing this canceling of two effects. Clocks should speed up with elevation, while the balancing velocity of the orbit will slow it down. Why do we get a net time effect (clock net slows) when the two opposing effects should cancel each other?

 

They oppose each other, but except for one particular orbital altitude, they do not cancel each other out. This is easily shown by what you have already stated; that clocks speed up with elevation. But what happens to orbital velocity as you increase the elevation of the orbit? it goes down. Thus as the altitude of the orbit increases, speeding up the clock due to altitude increases, and the slowing down of the clock due to velocity decreases.

 

Below a certain altitude, the velocity factor dominates and the clocks run slow, above it the gravity effect dominates and clocks run fast.

[CraigD]

Clocks should speed up with elevation, while the balancing velocity of the orbit will slow it down. Why do we get a net time effect (clock net slows) when the two opposing effects should cancel each other?

I agree with what Janus wrote, but would further point out some suspicious assumptions and descriptions in both posts.

 

The speed of a satellite in a circular orbit is

[math]v= \sqrt{\frac{u}{r}}[/math]

where [math]u = M G[/math], the Earth’s standard gravitational parameter (its mass times the gravitational constant), and

[math]r[/math] is it’s orbital radius (measured from the center of the Earth).

 

So the orbital speed of a satellite, and thus its time dilation due to speed, decreases with altitude, as does its gravitational time dilation.

 

The only reason there is any domain where clocks in orbit tick slower than clocks on the ground is because clocks on the ground move much slower than the orbital speed of a ground-skimming satellite. If it were possible (ie: if the Earth didn’t have a prohibitively atmosphere), the orbit where clocks ticked the slowest would be a ground-skimming one.

 

One might think that clocks at the poles, which have zero speed relative to the Earth, would tick slightly faster than clocks at the equator, which have a speed of about 478 m/s, but because the Earth’s surface at the equator is over 21000 m further from the Earth’s center than at the poles, they actually tick slightly slower.

 

It’s a good and pretty easy exercise to take the equations for orbital speed, speed time dilation, and gravitational time dilation, and calculate the orbital radius (and altitude above ground) at which clocks in circular orbits tick at the same rate as clocks on the surface at the equator, or the same rate as clocks on the surface at a pole. Interested readers are invited to try the exercise, and post their result (a radius or altitude).

[Qfwfq]

If we look at a satellite orbiting the earth, the orbital motion is balancing the pull of gravity to create an orbit. But if we look at the slowing of its clock, the time dilation is not showing this canceling of two effects.

Arguing by analogy is always prone to non sequitur. The only conclusive way to work it out is by computing according to GR.

 

Clocks should speed up with elevation, while the balancing velocity of the orbit will slow it down. Why do we get a net time effect (clock net slows) when the two opposing effects should cancel each other?

If you look at it how it is for the distant observer, the effect of potential and that of velocity are not opposite at all. The speeding up you mention is a decrease in the effect of potential; you are reasoning according to an observer who's arse is sitting on the surface of this planet at sea level, so it depends on the planet's radius for the same M value. For an observer sitting on a high mountain peak it's already a bit different.

 

As you see, there's no reason for expecting the two effects to cancel out.

[CraigD]

If you look at it how it is for the distant observer, the effect of potential and that of velocity are not opposite at all.

...

As you see, there's no reason for expecting the two effects to cancel out.

I agree.

 

If you put the equations for speed and gravitational time dilation or a body in circular orbit in simple forms, they differ only in a single constant:

[math]{\frac{t_s}{t_f}}_V = \sqrt{1-\frac{u}{c^2r}}[/math]

for speed, and

[math]{\frac{t_s}{t_f}}_G = \sqrt{1-\frac{2u}{c^2r}}[/math]

for gravity, where [math]t_s[/math] is the proper time for the “slow” observer orbiting at radius [math]r[/math],

[math]t_f[/math] is the proper time of the infinitely distant observer at rest relative to the center of the Earth, and

[math]u[/math] is the Earth’s standard gravitational parameter.

 

You can get a situation like HBond describes if, instead of having a body in orbit, you have it on a tremendously high, rigid tower rising from the equator (ignoring the engineering impracticality of such a structure), because in this case, rather than its speed being

[math]v= \sqrt{\frac{u}{r}}[/math]

which decreases as [math]r[/math] increases, it’s

[math]v = \frac{2 \pi r}{1 \,\mbox{day}}[/math]

which increases.

 

(I’ve just given away all the equations needed to solve my calculating challenge in post #3, if anyone on all the internet’s motivated to answer it ;))

 

I don’t know how to exactly calculate the proper time of a body on a tower, however, because I don’t understand how to calculate the time dilation due to the centrifugal acceleration it experiences, or if this should be done. This pseudo-acceleration is small ([math]a = \left( \frac{\pi}{43200} \right)^2 r[/math], about 0.78 m/s/s for r=148125892 m, the radius at which the product of speed and gravitational time dilation for a body on a tower equal 1, “canceling each other”, so practically negligible, but needs to be accounted for. My grasp of GR fundamentals, alas, are inadequate to the task.

 

Help, anyone? :shrug:

[Qfwfq]

it’s

[math]v = \frac{2 \pi r}{1 \,\mbox{day}}[/math]

which increases.

but still wouldn't compensate the effect of potential [imath]\forall r[/imath].

 

In any case, for planet Earth, the effect of potential is very slight; even at sea level the radius is far greater than [imath]R_{\rm S}[/imath] and, though I haven't got numerical values, I suspect it is smaller than the effect of orbital velocity even for pretty large circular orbits.

[modest]

It’s a good and pretty easy exercise to take the equations for orbital speed, speed time dilation, and gravitational time dilation, and calculate the orbital radius (and altitude above ground) at which clocks in circular orbits tick at the same rate as clocks on the surface at the equator, or the same rate as clocks on the surface at a pole. Interested readers are invited to try the exercise, and post their result (a radius or altitude).

 

Velocity time dilation:

[math]\tau = t \left(1-\dfrac{\Delta v^2}{c^2} \right )^{-1/2}[/math]

Grav. time dilation:

[math]\tau= t \left(1+ \dfrac{2 \Delta \phi}{c^2} \right)[/math]

For proper time of both clocks to be equal vel. time dilation must equal grav. time dilation:

[math]t \left(1-\dfrac{\Delta v^2}{c^2} \right )^{-1/2} = t \left(1+ \dfrac{2 \Delta \phi}{c^2} \right)[/math]

 

I can't seem to simplify this into an expression r₂ = ...:shrug:

 

Oh well, subbing:

[math]\Delta \phi = \frac{GM}{r_1}-\frac{GM}{r_2}[/math]

and

[math]\Delta v^2 = \frac{GM}{r_2}+\left( \frac{2 \pi r_1}{6378100} \right)^2[/math]

 

Plugging this into excel I find r₂ = 9,567,148.2 m at which point gravitational potential adds and velocity subtracts 2.318 x 10^-10 seconds from a reference clock on Earth's equator. I have a feeling this is off a little. What do you have, Craig?

 

I don’t know how to exactly calculate the proper time of a body on a tower, however, because I don’t understand how to calculate the time dilation due to the centrifugal acceleration it experiences, or if this should be done. This pseudo-acceleration is small ([math]a = \left( \frac{\pi}{43200} \right)^2 r[/math], about 0.78 m/s/s for r=148125892 m, the radius at which the product of speed and gravitational time dilation for a body on a tower equal 1, “canceling each other”, so practically negligible, but needs to be accounted for. My grasp of GR fundamentals, alas, are inadequate to the task.

 

Help, anyone? :shrug:

 

Perhaps the Kerr metric.

 

~modest

[CraigD]

it’s

[math]v = \frac{2 \pi r}{1 \,\mbox{day}}[/math]

which increases.

but still wouldn't compensate the effect of potential [imath]\forall r[/imath].

There's no limit to the speed of a body on the top of a tower / end of a pole at the Earth's equator can have, so its net speed and gravitational time dilation

[math]{\frac{t_s}{t_f}} = \left( \sqrt{1- \frac{\pi^2}{43200^2 \,\mbox{s} \cdot c^2} r} \right) \left( \sqrt{1-\frac{2u}{c^2r}} \right)[/math]

can be the same for many [imath]r[/imath] pairs, including [imath]r \dot= 6578137 \,\mbox{m}[/imath] (Earth’s equatorial radius) and [imath]r \dot= 148125893 \,\mbox{m}[/imath] (about 1/3rd of the distance to the Moon).

 

A minimum net time dilation / fastest proper time (about .99999999984251019) occurs at [imath]r \dot= 42233136 \,\mbox{m}[/imath] (very close to a geosynchronous orbit, about 42364000 m ). The closeness of min time dilation to geostationary orbit doesn’t appear meaninglessly coincidental – it holds for many different day lengths, not just the standard 86400 seconds.

 

The top of a really tall (about 4.12e12 m, almost to Neptune) tower at Earth’s equator would travel at the speed of light, and thus be “infinitely” time dilated ([imath]{\frac{t_s}{t_f}} = 0[/imath]). As, even if some super strong, super light material capable of being made into such a structure could be made, spinning it up to speed would require not only more angular momentum than the Earth has, but unless the tower and body had zero invariant mass, infinite energy – but why worry about practical details? ;)

[UncleAl]

There is no relativistic effect of centripetal force, only the transverse Doppler effect,

 

Physics Today 58(3) 34 (2005)

Time passage, equator vs. poles

 

http://fourmilab.to/etexts/einstein/specrel/specrel.pdf

Comments on the paper: Clock Behavior and the Search for an Underlying Mechanism for Relativity Phenomena

Einstein

http://www.physics.gatech.edu/people/faculty/finkelstein/relativity.pdf

Longitudinal and transverse mass

 

Relativity in the Global Positioning System

Physics of GPS relativistic time delay « Unused Cycles

Relativistic effects on orbital clocks

[CraigD]

It’s a good and pretty easy exercise to take the equations for orbital speed, speed time dilation, and gravitational time dilation, and calculate the orbital radius (and altitude above ground) at which clocks in circular orbits tick at the same rate as clocks on the surface at the equator, or the same rate as clocks on the surface at a pole. Interested readers are invited to try the exercise, and post their result (a radius or altitude).

...

Plugging this into excel I find r₂ = 9,567,148.2 m at which point gravitational potential adds and velocity subtracts 2.318 x 10^-10 seconds from a reference clock on Earth's equator. I have a feeling this is off a little. What do you have, Craig?

Plugging the usual speed and gravitational time dilation and circular orbit equations into the following numeric approximation MUMPS program:

[noparse]USER>s R="XBS" x XLX w XBS
n (R) s D=$p(R,",",3)-$p(R,",",2),I=$p(R,",",4),P=$p(R,",",5),C=P<0'=(D<0),F=C!$p(R,",",6),I=$s('D:0,'I:1,F:$s(C:I/-2,1:I/2),$tr(D,"-")>$tr(P,"-"):I*-2,1:I*2),A=R+I,R=A_",,"_$p(R,",",3)_","_$s(R-A:I,1:"")_","_D_","_F ;XBS: binary search for $p(R,",",2)=$p(R,",",3) by varying $p(R,",")
USER>s c=299792458,u=398600441800000,r0=6578137,r0p=6356752,v0=r0*$zpi*2/86400,F10=1-(v0/c**2)**.5,F20=1-(2*u/c/c/r0)**.5,F20p=1-(2*u/c/c/r0p)**.5
USER>s R=",,1",r=r0 f  s v=u/r**.5,F1=1-(v*v/c/c)**.5,F2=1-(2*u/c/c/r)**.5,$p(R,",",1,2)=r_","_(F1*F2/F20p) x XBS s r=+R i '$p(R,",",4) w r q
9535127.9921875[/noparse]

We’re in the same rough (32 km) neighborhood. ;)

 

This wouldn’t be too difficult to solve algebraically - it’s a degree 3 (cubic) polynomial - but numeric approximations are quick and easy... yet another example of how computers can make mathematical sluggards of a once honest math student. :phones:

There is no relativistic effect of centripetal force, only the transverse Doppler effect, ...

Thanks, Al – I’ll read these links, and try to re-acquire my practical GR calculating skill. Unlike SR, I seem unable to maintain a brain-lock on it – if I haven’t used it in the past year or so, it becomes opaque to my mechanical intuition. :(

[UncleAl]

Phys. Rev. 129 (1963): Walter Kündig - Measurement of the Transverse...

http://www.physicsfoundations.org/PIRT_XI/papers/KHOLMETSKII%20PAPER%202008.doc

A Mössbauer experiment in a rotating system on the second-order Doppler shift: confirmation of the corrected result by Kündig

http://arxiv.org/pdf/0812.4507

Kündig's experiment on the transverse Doppler shift re-analyzed

 

Complicated, very very complicated.

[modest]

Complicated, very very complicated.

 

Amen to that.

 

Craig, I was looking at the Kerr metric trying to figure out how much effect the body's rotation has on the proper time of a clock in the field (gravitomagnetic effect). It's complicated and the best source I could find:

 

Let us then briefly analyze the behaviour of clocks around a spinning object....

 

For example, when a clock that co-reotates very slowly around the spinning Earth, at ~6000 km altitude, returns to its starting point, it finds itself advanced relative to a clock kept there at ‘rest’ (with respect to ‘distant stars’) by

 

[math]\oint \frac{g_{0i}}{g_{00}}dx^i \sim \frac{4 \pi J_{\oplus}}{r} \sim 5 \times 10^{-17} \ s[/math]

 

where [math]g_{0i} \sim 2J_{\oplus}/r^2[/math] is the Earth’s gravitomagnetic field and [math]J_{\oplus} \cong 145 \ cm^2[/math] is the Earth’s angular momentum. Similarly, a clock that counter-rotates very slowly around the spinning Earth finds itself retarded relative to a clock kept at ‘rest’ there by the same amount. However, a larger clock effect...

 

Gravitation: from the Hubble length ... - Google Books

 

I believe this refers to the gravitomagnetic effect only, so that gets smaller with larger r and goes to zero when [math]J_{\oplus}[/math] goes to zero. If this is the case, then you can safely ignore this effect as 10^-17 s / 6378100 s is not comparable to the first order Schwartzchild approximation.

 

In other words, I think you can safely ignore the rotating earth except for the velocity it gives the point on the tower as you have done.

 

~modest

[modest]

A minimum net time dilation / fastest proper time (about .99999999984251019) occurs at [imath]r \dot= 42233136 \,\mbox{m}[/imath] (very close to a geosynchronous orbit, about 42364000 m ). The closeness of min time dilation to geostationary orbit doesn’t appear meaninglessly coincidental – it holds for many different day lengths, not just the standard 86400 seconds.

 

I was talking about something similar the other day:

 

But, this is no coincidence because in gtr every particle wants to find a path where its proper time is greatest. If earth were to become too spherical while spinning then parts on the surface would want to fall from the poles to the equator to maximize their proper time.

 

A geodesic represents the greatest proper time between events:

 

A geodesic between two events could also be described as the curve joining those two events which has the maximum possible length in time

Geodesic (general relativity) - Wikipedia, the free encyclopedia

 

With your thought experiment of the lengthening tower, geosynchronous motion is the only point where the path is a geodesic—the only point where the clock at the top of the tower is not prevented from following the path that it would prefer to follow. I guess it shouldn't be too surprising that within those constraints it has the greatest proper time.

 

~modest

[freeztar]

Darn! :(

 

I thought I had geodesics nailed down, but this throws me for a loop:

 

A geodesic between two events could also be described as the curve joining those two events which has the maximum possible length in time — for a timelike curve — or the minimum possible length in space — for a spacelike curve.

 

I'm familiar with the latter, but I'm having a conceptual brain-fart trying to reconcile the two, seemingly opposite, definitions. What am I missing? Perhaps the equations could be broken down in a way that's more understandable (algebraically)?

[modest]

An event is a point in space and time. So, consider two events: one on earth today and another on the moon 5 days from now (5 days as measured from the earth). There are many different paths between these two events. It's only important that the person take off from earth exactly when and where the first event happens and lands on the moon exactly when and where the second event happens.

 

Under these constraints, the straight path directly from one event to the other, takes the greatest amount of time. Consider that two spaceships take off from earth at the first event. The first ship goes directly to the moon on a 5-day journey landing at event 2. The second ship accelerates very quickly and visits the nearest star to the sun then returns landing at the second event. Because of time dilation the ship which took the roundabout trip would have experienced less time between events. So, both ships took off and landed collocated and simultaneously, but the shortest and most-direct path through space between events takes the longest.

 

It's known in gtr as the The Principle of Extremal Aging. The inertial path between events represents the greatest proper time.

 

~modest

[Qfwfq]

What am I missing? Perhaps the equations could be broken down in a way that's more understandable (algebraically)?

The Minkowski metric and the twin that ages less. :(

[modest]

The Minkowski metric and the twin that ages less. ;)

 

I understand that well enough as time-like separation is given by:

 

[math]d \tau = \sqrt{d t^2 - d x^2}[/math]

 

where c=1. The largest tau is where dx = 0. But, the part of the quote that Freezy bolded...

 

...or the minimum possible length in space — for a spacelike curve

 

I'm having trouble reconciling this with Minkowski's metric where a space-like interval is given by:

 

[math]d \sigma = \sqrt{d x^2 - d t^2}[/math]

 

As length contraction would demand, sigma is smaller where t has nonzero value. In other words, a straight line between events with space-like separation seems to be the largest rather than the smallest distance. How does that reconcile with the quote above?

 

~modest