Time dilation in an elliptical orbit

In summary: Can it be deduced that time dilation in an elliptical orbit at perihelion ist = \sqrt{1 - \left( GM...((2/r) - (1/a))) + 2GM/r / c^2}t = \sqrt{1 - \left( 2GM/r - GM/a + 2GM/r / c^2}t = \sqrt{1 - \left( 4GM/r - GM/a)/c^2}t = \sqrt{1 - GM((4/r) - (1/a))/c^2
  • #1
Keermalec
8
0
The formula for time dilation in a circular orbit is readily available but the literature seems to indicate it would not be so simple in the case of an elliptical orbit, and no simple formula seems to be available.

Given that time dilation in a circular orbit adds the velocity effect (GM/r) to the gravitational effect (2GM/r)

t = sqrt(1 - (GM/r + 2GM/r)/c²)
t = sqrt(1 - 3GM/rc²)

And given that velocity at perihelion in an elliptical orbit is

vp = sqrt(GM((2/r) - (1/a)))

Can it be deduced that time dilation in an elliptical orbit at perihelion is

t = sqrt(1 - (GM((2/r) - (1/a)) + 2GM/r)/c²)
t = sqrt(1 - (2GM/r - GM/a + 2GM/r)/c²)
t = sqrt(1 - (4GM/r - GM/a)/c²)
t = sqrt(1 - GM((4/r) - (1/a))/c²)
 
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  • #2
You're more likely to get replies if you mark up your math properly in LaTeX. We have a howto here: https://www.physicsforums.com/help/latexhelp/ . It's very difficult for people to read math that isn't written like math.

In general, the time elapsed on a path through spacetime is given by ##s=\int ds##, where ##ds^2=g_{ij}dx^idx^j##.

In your #1, you seem to be flipping back and forth between talking about two different things: (a) the time dilation as an object moves past a particular point in a static spacetime, and (b) the time dilation of an object that moves along a path (a periodic orbit) in spacetime. If a is meaningful, then it can't depend on the rest of the path. It's not completely obvious to me, however, that a is meaningful. To define what this really means, you'd have to relate it to at least some hypothetical, possible measurement process, which would involve synchronizing clocks and then later reuniting and comparing them. In the case of a static spacetime, the implicit understanding about what we mean by gravitational time dilation is that we synchronize clocks at P, transport one clock to some other spatial point of interest Q, leave it there for a very long time, and then transport it back for comparison. Since the spacetime is static, a spatial point like Q has a well-defined identity, and you can leave the clock at Q for as long as you like, so that any time dilation incurred during transport to and from Q can be made negligible.

Is this the kind of thing you have in mind, or are you thinking of something more like clock synchronization via radio signals? If the latter, then Doppler shifts are not going to be cleanly separable from time dilation.
 
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  • #3
bcrowell said:
You're more likely to get replies if you mark up your math properly in LaTeX. We have a howto here: https://www.physicsforums.com/help/latexhelp/ . It's very difficult for people to read math that isn't written like math.

In general, the time elapsed on a path through spacetime is given by ##s=\int ds##, where ##ds^2=g_{ij}dx^idx^j##.

In your #1, you seem to be flipping back and forth between talking about two different things: (a) the time dilation as an object moves past a particular point in a static spacetime, and (b) the time dilation of an object that moves along a path (a periodic orbit) in spacetime. If a is meaningful, then it can't depend on the rest of the path. It's not completely obvious to me, however, that a is meaningful. To define what this really means, you'd have to relate it to at least some hypothetical, possible measurement process, which would involve synchronizing clocks and then later reuniting and comparing them. In the case of a static spacetime, the implicit understanding about what we mean by gravitational time dilation is that we synchronize clocks at P, transport one clock to some other spatial point of interest Q, leave it there for a very long time, and then transport it back for comparison. Since the spacetime is static, a spatial point like Q has a well-defined identity, and you can leave the clock at Q for as long as you like, so that any time dilation incurred during transport to and from Q can be made negligible.

Is this the kind of thing you have in mind, or are you thinking of something more like clock synchronization via radio signals? If the latter, then Doppler shifts are not going to be cleanly separable from time dilation.
Thanks, here is the question with better wording and math:

The formula for time dilation in a circular orbit is readily available but the literature seems to indicate it would not be so simple in the case of an elliptical orbit, and no simple formula seems to be available.

Given that time dilation in a circular orbit adds the velocity effect (##v²##) to the gravitational effect (##2GM/r##)

[tex]t = \sqrt{1 - \left( v^2 + 2GM/r \right) / c^2}[/tex]
[tex]t = \sqrt{1 - \left( GM/r + 2GM/r \right) / c^2}[/tex]
[tex]t = \sqrt{1 - 3GM / rc^2}[/tex]

And given that velocity in an elliptical orbit is

[tex]v = \sqrt{GM \left( \frac{2}{r} - \frac{1}{a} \right) }[/tex]

Can it be deduced that time dilation in an elliptical orbit is

[tex]t = \sqrt{1 - \left(GM \left( \frac{2}{r} - \frac{1}{a} \right) + 2GM/r \right) / c^2}[/tex]
[tex]t = \sqrt{1 - GM \left( \frac{4}{r} - \frac{1}{a} \right) / c^2}[/tex]

I have no specific measurement process in mind. Just wanted to know if the algebraic logic holds true in this case and if time dilation at any point in an elliptical orbit can be predicted using the last equation.

NB, r is radius, a is semi-major axis, G is gravitational constant, M is mass of orbited body, c is the velocity of light
 
  • #4
Is this an orbit that's at velocities small compared to c? If not, then your expression for v will not be accurate.

Time dilation doesn't give a formula for the time on a clock. In fact, if you look at your formulas, you'll see that they don't have units of time.
 
  • #5
bcrowell said:
Is this an orbit that's at velocities small compared to c? If not, then your expression for v will not be accurate.

Time dilation doesn't give a formula for the time on a clock. In fact, if you look at your formulas, you'll see that they don't have units of time.
I realize that. I am not familiar with the correct way of writing "rate of change in time with respect to an immobile observer" but that is what I mean by "t".
I am looking for the formula that defines rate of change in time for small OR large velocities, ie orbitting a small OR large mass M. The question, if I may repeat it is, will the following formula allow the calculation of time dilation for an observer in an elliptical orbit?

[tex]t = \sqrt{1 - GM \left( \frac{4}{r} - \frac{1}{a} \right) / c^2}[/tex]
 
  • #6
Keermalec said:
I realize that. I am not familiar with the correct way of writing "rate of change in time with respect to an immobile observer" but that is what I mean by t.
You could write it as a ratio like ##t_1/t_2##.
Keermalec said:
I am looking for the formula that defines rate of change in time for small OR large velocities, ie orbitting a small OR large mass M.
This is a lot more complicated. You assumed Newtonian conservation of energy, which doesn't apply. See https://en.wikipedia.org/wiki/Schwarzschild_geodesics .
 
  • #7
  • #9
bcrowell said:
The answer is no.
So what is the equation for calculating time dilation in an elliptical orbit?
 
  • #10
Keermalec said:
So what is the equation for calculating time dilation in an elliptical orbit?

One thing you'll find out if you read the WP article I linked to is that there are no elliptical orbits.

You need to do some work in carefully defining and researching your question. Asking people for "the equation" isn't a substitute for that.
 
  • #11
bcrowell said:
One thing you'll find out if you read the WP article I linked to is that there are no elliptical orbits.

You need to do some work in carefully defining and researching your question. Asking people for "the equation" isn't a substitute for that.
Your position would be correct if I was a student; but I really am just looking for an answer.
According to the answer to the following post, the equation, simplified, would be at perihelion:

[tex]\delta t = \sqrt{1 - GM \left( 1 - \frac{a}{r} \right) \left( 1 - e^2 \right) /rc^2}[/tex]
https://www.quora.com/Is-the-subjective-relativistic-time-experienced-by-a-body-in-circular-orbit-more-less-or-the-same-as-a-body-moving-in-an-elliptical-orbit

Many thanks for your help.
 
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  • #12
Keermalec said:
I really am just looking for an answer.

A question that is not well-defined doesn't have a well-defined answer.

Keermalec said:
According to the answer to the following post

Which also isn't well-defined. What elliptical orbit (or trajectory that is approximately an elliptical orbit--as bcrowell pointed out, in GR there are no closed elliptical orbits) is being compared with a given circular orbit? In other words, how am I supposed to pick out the (trajectory that is approximately an) elliptical orbit that is "comparable" to a given circular orbit? You don't say, and neither does the quora post. (The answer you are referring to assumes that the "comparable" circular orbit is one whose radius is equal to the semi-major axis of the elliptical orbit; but the question itself doesn't say that, and it's not clear that the questioner even understands that that's a key issue.)

Keermalec said:
the equation, simplified, would be at perihelion:

$$
\delta t = \sqrt{1 - G M ( 1 - \frac{a}{r} ) ( 1 - e^2 ) / rc^2}
$$

This doesn't look right. The equation in the quora post (which is, btw, already a Newtonian approximation, so it won't be valid for large velocities or strong fields, and you said you were looking for an answer valid in those regimes) is

$$
d\tau = dt \sqrt{1 - \frac{2m}{r} - \frac{ma}{r^2} (1 - e^2) \left( 1 + \frac{e^2 \sin^2 \theta}{(1 - 2m/r)(1 - e \cos \theta)^2} \right)}
$$

They use ##m## to mean ##GM / c^2##, so it looks to me like this would simplify, for the case of an orbit in the equatorial plane for which ##\theta = \pi / 2##, to

$$
d\tau = dt \sqrt{1 - GM \left( 2 - \frac{a}{r - 2 GM / c^2} \left[ 1 - \frac{2GM}{c^2r} (1 - e^2) - e^4 \right] \right) / r c^2}
$$
 
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Related to Time dilation in an elliptical orbit

1. What is time dilation in an elliptical orbit?

Time dilation in an elliptical orbit refers to the phenomenon where time moves slower or faster depending on the speed and distance of an object in an elliptical orbit around another object, usually a planet or star.

2. How does time dilation occur in an elliptical orbit?

Time dilation occurs in an elliptical orbit due to the effects of gravity. The closer an object is to the center of mass, the stronger the gravitational pull and the slower time moves. This means that time will move slower when an object is closer to the planet in an elliptical orbit, and faster when it is farther away.

3. Does time dilation affect all objects in an elliptical orbit?

Yes, time dilation affects all objects in an elliptical orbit, regardless of their size or mass. However, the effects of time dilation may be more noticeable on larger objects due to their stronger gravitational pull.

4. How does time dilation in an elliptical orbit compare to that of a circular orbit?

The amount of time dilation in an elliptical orbit is greater than that of a circular orbit, as the distance and speed of the object in an elliptical orbit varies more compared to a circular orbit. This means that time will move slower in an elliptical orbit than in a circular orbit at the same distance from the center of mass.

5. Can time dilation in an elliptical orbit be observed?

Yes, time dilation in an elliptical orbit can be observed through experiments and calculations. Scientists have been able to measure and confirm time dilation in elliptical orbits through various methods, including atomic clocks and gravitational lensing.

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