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WHere:

G is the gravitational constant

M is the solar mass

c is the speed of light in vacuum

a the Mercury's semi-major axis

e the eccentricity of Mercury's orbit

I have tried several times but from the dimensional analysis it turns out that the right-hand side of that equation is dimensionless. Am I wrong? I have taken this formula and the correspondent value from this web site of Rhode Island university https://phys.uri.edu/gerhard/PHY520/mln21.pdf and also at pag. 6 of this paper https://arxiv.org/pdf/0910.3800 then in formula (13) of this paper: https://pubs.aip.org/aapt/ajp/article/90/11/857/2820267/Simple-precession-calculation-for-Mercury-A

and most importantly in eq. (1) of this article https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://academic.oup.com/mnras/article-pdf/468/2/1405/11126982/stx548.pdf&ved=2ahUKEwjI9LbY75CJAxUE_rsIHXLWH2wQFnoECDwQAQ&usg=AOvVaw0TLWmAQ2Jk1NFVVWTPHOyg

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    $\begingroup$ Please define your symbols are at the least provide a link to the source that does. $\endgroup$
    – Hilmar
    Commented Oct 15 at 13:26
  • $\begingroup$ I have just provided the link you asked $\endgroup$
    – Roberto Napolitano
    Commented Oct 15 at 13:58
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    $\begingroup$ None of your references have $\dot\omega$, which is an angular acceleration, on the left side. They all have a dimensionless angle on the left. Why are you misrepresenting your own references? $\endgroup$
    – Ghoster
    Commented Oct 15 at 16:51
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    $\begingroup$ It is clearly wrtitten that the equation is in radians/revolution but is dimensionless instead. Radians are a dimensionless unit. The number of revolutions is dimensionless. Their ratio is therefore dimensionless. $\endgroup$
    – Ghoster
    Commented Oct 15 at 19:15
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    $\begingroup$ This is not an angular velocity. Suppose the perihelion precesses by 1 arc second every revolution. That is a meaningful angle regardless of whether the revolution took 1 year or 1 day. If you want the angular velocity of precession you have to divide by the period of the revolution. $\endgroup$
    – Ghoster
    Commented Oct 15 at 19:22

1 Answer 1

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I can rewrite you formula as:

$$ \dot\omega = 3\pi \frac{R_s} {a(1-\epsilon^2)} $$

The RHS is the ratio of two distances, so that is dimensionless.

The LHS, you have:

$$ \omega \rightarrow [T]^{-1}$$

$$ \frac d {dt} \rightarrow [T]^{-1}$$

so that has $[T]^{-2}$.

Something is wrong. $\omega$ must not be an angular frequency: what is it?

Per Zero's comment: the LHS is (from the link):

"The angle δθ of precession per cycle"

Here, and change in an angle is dimensionless (radians notwithstanding), while cycle is a bit confusing. It's not a period; rather it is in integer count, so, the LHS is:

$$ \frac{[{\rm rad}]}{[1]} \rightarrow [1] $$

which is dimensionless.

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  • $\begingroup$ They are close my question. Anyway if you remake the dimensional analisys you will see that the right hand side is dimensionless $\endgroup$
    – Roberto Napolitano
    Commented Oct 15 at 14:21
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    $\begingroup$ @RobertoNapolitano nobody challenges that the RHS is dimensionless: the point is the LHS is not dimensionless as written. It has units of $[T]^{-2}$ as a rate of change of an angular velocity. Also note that your first link does not give the lhs as $\dot\omega$ but as $\delta\vartheta$ which is dimensionless. $\endgroup$
    – ZeroTheHero
    Commented Oct 15 at 16:14
  • $\begingroup$ I have learmed at university that LHS and RHS of an equation must have the same dimensions, or must be both dimensionless, otherwise the equation is wrong $\endgroup$
    – Roberto Napolitano
    Commented Oct 15 at 16:40
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    $\begingroup$ OP's equation is wrong... $d\theta$ is an integral of $\omega$, not a derivative of it. $\endgroup$
    – Connor Behan
    Commented Oct 15 at 16:51
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    $\begingroup$ Radians are dimensionless; it can’t be dimensionless “instead” $\endgroup$
    – controlgroup
    Commented Oct 15 at 18:49

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