I Orbital Precession Calculation: Unit Explained

[NODARman]

TL;DR Summary

.

Hi, I've just calculated the orbital precession for the earth using the sigma formula of general relativity.
$$
\sigma=\frac{24 \pi^{3} R^{2}}{T^{2} c^{2}\left(1-e^{2}\right)}=\frac{24\pi^3×1.5×10^{11}}{3×10^7×3×10^8(1-0.0034^2)}=0.012
$$
What is the unit of the result? Degrees per century or arcsec?

 

[PeterDonis]

What is the unit of the result?

What are the units of the quantities you plugged in to the formula? The units follow the same relationship as the quantities in the formula.

 

[PeterDonis]

I've just calculated the orbital precession for the earth

Not correctly since several quantities are squared in the formula but you didn't square them in your calculation.

 

[NODARman]

Not correctly since several quantities are squared in the formula but you didn't square them in your calculation.

My bad
$$
\frac{24 \pi^3 \times 1.5^2 \times\left(10^{11}\right)^2}{\left(3 \times 10^7 \times 3 \times 10^8\right)^2 \times\left(1-0.0034^2\right)}=2×10^{-7}
$$
But there are no more dimensions.
meter²×sec²/(meter²×sec²)

 

[PeterDonis]

there are no more dimensions

That's right: the number you get is a dimensionless number. Physically, it represents the fractional precession per orbit.

 

[DAH]

That's right: the number you get is a dimensionless number. Physically, it represents the fractional precession per orbit.

Is it the angle in radians per orbit? and could be converted into arcsec/century.

 

[PeterDonis]

Is it the angle in radians per orbit?

Read the second sentence of mine that you quoted in post #6 againl. What does it say?

 

[DAH]

Hi @PeterDonis
Its just that I recently did a calculation for the precession of Mercury and the formula gave me the angle in radians per orbit. I used: $$\Delta \phi=\frac{6\pi G(M+m)}{a(1-e^2)c^2}$$
Looking at the OP's formula and since $$T^2=\frac{4\pi^2 a^3}{G(M+m)}$$
Then its the same formula which I think gives the angle in radians.

Apologies if by saying fractional precession you mean radians.

 

[vanhees71]

Sure, in calculation you always use angles in "radians", i.e., dimensionless numbers. I've not seen this formula with $(M+m)$ in the numerator (where does this come from)? Instead of $(M+m)$ the standard treatment gives the mass of the Sun. Then your $\Delta \phi$ is the angle of perihelion precession for one period of the radial motion, i.e., from one perihelion to the next.

Usually the astronomers give the perihelion shift per century. For mercury one century means about 415 periods of its radial motion. So you have to multiply your result by 415 and then convert it to degrees (or rather arc minutes) using $1^\circ \hat{=} \pi/180 \text{rad}$ and $1^{\circ}=3600''$. The result is about 43'' perihelion shift per century.

 

[PeterDonis]

Then its the same formula which I think gives the angle in radians.

More precisely, it gives the extra angle in radians, over and above the $2 \pi$ radians that represent one complete circle, that the orbit covers between one perihelion and the next. That's what I meant by "fractional" precession. Radians are unitless (since they represent a ratio of lengths: arc length along a circular arc that covers a given angle, to radius of the circle), which is why the units in the formulas all cancel out.

 

[DAH]

More precisely, it gives the extra angle in radians, over and above the $2 \pi$ radians that represent one complete circle, that the orbit covers between one perihelion and the next. That's what I meant by "fractional" precession. Radians are unitless (since they represent a ratio of lengths: arc length along a circular arc that covers a given angle, to radius of the circle), which is why the units in the formulas all cancel out.

Thanks for clarifying. I was surprised you didn't say radians for the OP since it would be easy enough to convert to seconds of arc per century as @vanhees71 shows. The OP probably already knew this anyway.

 

[DAH]

I've not seen this formula with $(M+m)$ in the numerator (where does this come from)? Instead of $(M+m)$ the standard treatment gives the mass of the Sun.

In my book it gives $6\pi GM$ in the numerator but also states that $M$ is the total mass of the system. I suppose with the Sun totally dominating then just plugging in the Suns mass won't really alter the final calculation. Thanks.

 

[PeterDonis]

In my book

Which book?

but also states that $M$ is the total mass of the system.

You could use a model in which the total mass $M$ of the solar system, including the sun and planets and anything else considered significant enough, is taken to be at the barycenter of the system, and all of the objects in the system are treated as point masses orbiting the barycenter. Such a model would not be exact, but it would be a reasonable approximation for many purposes.

For a perihelion shift calculation, however, such a model wouldn't necessarily be good enough, because the actual observed perihelion shift of any planet contains contributions due to the perturbations induced by the other planets that are much larger than the GR contribution that we are calculating here. And those perturbations depend on the actual relative positions of the planets, so a simple barycentric model with the total mass at the center won't correctly capture them.

For even more accurate calculations, one could use a framework like the Einstein-Infeld-Hoffmann equations, which also would use a barycentric coordinate system, but which are multi-body equations in which each mass is treated separately. As I understand it, a framework like this is how the perturbations of the various planets on each other are calculated.

For calculating the GR contribution itself, you are correct that it is a good enough approximation to just use the Sun's mass, because it contains the vast majority of the mass of the entire solar system and any GR corrections to the Newtonian effects of the other planets are too small to matter.

I'm not aware of any context where it would make sense to use Newtonian formulas but plug in the sum of just the sun's mass and the mass of one planet.

 

[DAH]

Which book?You could use a model in which the total mass $M$ of the solar system, including the sun and planets and anything else considered significant enough, is taken to be at the barycenter of the system, and all of the objects in the system are treated as point masses orbiting the barycenter. Such a model would not be exact, but it would be a reasonable approximation for many purposes.

For a perihelion shift calculation, however, such a model wouldn't necessarily be good enough, because the actual observed perihelion shift of any planet contains contributions due to the perturbations induced by the other planets that are much larger than the GR contribution that we are calculating here. And those perturbations depend on the actual relative positions of the planets, so a simple barycentric model with the total mass at the center won't correctly capture them.

For even more accurate calculations, one could use a framework like the Einstein-Infeld-Hoffmann equations, which also would use a barycentric coordinate system, but which are multi-body equations in which each mass is treated separately. As I understand it, a framework like this is how the perturbations of the various planets on each other are calculated.

For calculating the GR contribution itself, you are correct that it is a good enough approximation to just use the Sun's mass, because it contains the vast majority of the mass of the entire solar system and any GR corrections to the Newtonian effects of the other planets are too small to matter.

I'm not aware of any context where it would make sense to use Newtonian formulas but plug in the sum of just the sun's mass and the mass of one planet.

Its a level 3 text book as I study part time with the OU in the UK, but its only an introduction to relativity so not advanced GR. I just re read the section on precession and it also states that you can ignore the mass of the planet so there was no need for me to include $M+m$. The prediction using the formula above for the perihelion advance of Mercury actually matches observation which I think is due to the semi major axis being small and a high eccentricity, but for other planets the formula may not be so precise.

Thanks for the detailed explanation on perturbations from other bodies in the solar system, I will probably look into this sometime in the future once I get to grips on the basics of GR.

 

[PeterDonis]

Its a level 3 text book as I study part time with the OU in the UK, but its only an introduction to relativity so not advanced GR.

All of this is interesting, but it doesn't answer my question. When you're asked for a reference here, the expectation is that you will give the specific reference: the title and author(s) of the book, or paper, or whatever. A web link is also appreciated if applicable.

 

[PeterDonis]

The prediction using the formula above for the perihelion advance of Mercury actually matches observation

Yes, to within the accuracy of our observations.

which I think is due to the semi major axis being small and a high eccentricity

No, those things make the formula, if anything, less accurate. It's just that any theoretical corrections are too small to matter even for the case of Mercury, and are even more so for the other planets.

but for other planets the formula may not be so precise.

No, it's even more precise, because the corrections are even smaller.

 

[DAH]

All of this is interesting, but it doesn't answer my question. When you're asked for a reference here, the expectation is that you will give the specific reference: the title and author(s) of the book, or paper, or whatever. A web link is also appreciated if applicable.

The book is Relativity, Gravitation and Cosmology.
Author: Robert J.A. Lambourne
Consultant author for the chapter is: Aiden Droogan
You can find it here:https://is.muni.cz/el/ped/podzim2015/FY2BP_KOSM/Lambourne_Relativity__gravitation_and_cosmology.pdf

Chapter 7 page 204

 

[DAH]

Yes, to within the accuracy of our observations.No, those things make the formula, if anything, less accurate. It's just that any theoretical corrections are too small to matter even for the case of Mercury, and are even more so for the other planets.No, it's even more precise, because the corrections are even smaller.

I think you're right, its the accuracy of the observations that are better for the precession of Mercury.

 

[PeterDonis]

its the accuracy of the observations that are better for the precession of Mercury.

Not the accuracy of observations, but the size of the effect as compared with the accuracy of our observations (i.e., the minimum effect size that we can reliably detect).

IIRC our observations now are accurate enough to detect the GR precession for all of the planets out to Mars.

 

[vanhees71]

For calculating the GR contribution itself, you are correct that it is a good enough approximation to just use the Sun's mass, because it contains the vast majority of the mass of the entire solar system and any GR corrections to the Newtonian effects of the other planets are too small to matter.

I'm not aware of any context where it would make sense to use Newtonian formulas but plug in the sum of just the sun's mass and the mass of one planet.

Well, arguing with the Newtonian approximation it makes some sense, because the two-body Kepler problem (inclusing the motion of the Sun) is equivalent to the uniform motion of the center of mass and the motion of a "quasi particle" with the reduced mass $\mu=m_1 m_2/M$ moving in the gravitational field of a body with mass $M=m_1+m_2$.

I'm only not so sure, whether it makes sense to use the Schwarzschild solution for a body of mass $M$ instead of $m_1$ (mass of the Sun) to get a more accurate approximation than just using the Schwarzschild metric with the Sun as the source mass. It's of course a good approximation in our solar system, because the Sun is so much heavier than all planets.

It's different if you consider binary-star systems like pulsars. There you need to use systemat post-Newtonian approximations to the GR two-body problem, which afaik is not exactly solved yet. A thorough discussion of this method can be found in Weinberg, Gravitation and Cosmology (1971).

 

[PeterDonis]

I'm only not so sure, whether it makes sense to use the Schwarzschild solution for a body of mass $M$ instead of $m_1$ (mass of the Sun) to get a more accurate approximation than just using the Schwarzschild metric with the Sun as the source mass.

If you mean plugging in the total mass of the solar system, no, that doesn't make sense, because the mass is not all contained in a single body at the center.

Astronomers doing solar system calculations don't even use the bare Schwarzschild metric with the sun's mass; our measurements are accurate enough that we can see the errors in that approximation (although it's fine for quick and dirty calculations). They use one of several frameworks that allows them to include perturbations from the planets.