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I'd like to understand how to calculate the components
of Newtonian tidal accelaration tensor in polar coordinates.
Is any available Internet source which clearly explains the
technique with details?
Reading James B. Hartle "Gravity" textbook I stumbled on the following
Example from Chapter 21.
=======================================================
Example 21.1 Tidal Acceleration Oitside a Spherical Mass.
The Newtonian gravitational potential outside a spherically symmetric
distribution of mass is (G = 1 units)
[tex]
\Phi = \frac {- M }{r} (21.6)
[/tex]
where
[tex]
r = \sqrt {x^2+y^2+z^2 }
[/tex]
is the distance from the center od symmetry.
Evaluating the tidal gravitational acceleration tensor using
the rectangular coordinates gives:
[tex]
a_{ij} \equiv - \frac{\partial^2\Phi}{\partial x^i\partial x^j} =
-(\delta_{ij}-3n_{i}n_{j})\frac{M}{r^3} (21.7)
[/tex]
where
[tex]
n_{i} \equiv \frac {x^i}{r}
[/tex]
are the components of a unit vector in a radial direction.
In an orthonormal basis
[tex]
\vec{e_{\hat{r}}} , \vec{e_{\hat{\theta}}} , \vec{e_{\hat{\phi}}}
[/tex]
oriented along coordinate directions of a polar coordinates (r,\theta,\phi)
the nonvanishing components of the tidal acceleration tensor are
[tex]
a_{\hat{r}\hat{r}} = \frac{2M}{r^3},
a_{\hat{\theta}\hat{\theta}}=a_{\hat{\phi}\hat{\phi}}=-\frac{M}{r^3} (21.8)
[/tex]
=======================================================
Q1. From (21.6) I see that gravitational potential does depend only
on 'r' but not on theta and phi. Why in that case its theta, phi
partial derivatives described in (21.8) are non-zero?
Looks like (21.8) are calculated in some other polar coordinates,
but how those coordinates related to another ones desribed in (21.6)?
Q2. Let's assume that I'm wrong in Q1, and (21.6) does represent
gravitational potential only in Cartesian coordinates, but not
in a polar ones. Then I tried another way. I applied Chain Rule of
differentiation to (21.7) to calculate (21.8) based
on standard Cartesian --> Polar transformations:
[tex]
x=r\sin{\theta}\cos{\phi}, y=r \sin{\theta} \sin{\phi}, z=r\cos{\theta}
[/tex]
I calculated
[tex]
a_{{\hat{\theta}}{\hat{\theta}}}
[/tex]
from (21.8) applying Chain Rule to (21.7) but got
[tex]
- \frac {2 M} {r}
[/tex]
which differs from (21.8).
Did I make a mistake in my calculations, or Chain Rule
simply cannot be applied here?
of Newtonian tidal accelaration tensor in polar coordinates.
Is any available Internet source which clearly explains the
technique with details?
Reading James B. Hartle "Gravity" textbook I stumbled on the following
Example from Chapter 21.
=======================================================
Example 21.1 Tidal Acceleration Oitside a Spherical Mass.
The Newtonian gravitational potential outside a spherically symmetric
distribution of mass is (G = 1 units)
[tex]
\Phi = \frac {- M }{r} (21.6)
[/tex]
where
[tex]
r = \sqrt {x^2+y^2+z^2 }
[/tex]
is the distance from the center od symmetry.
Evaluating the tidal gravitational acceleration tensor using
the rectangular coordinates gives:
[tex]
a_{ij} \equiv - \frac{\partial^2\Phi}{\partial x^i\partial x^j} =
-(\delta_{ij}-3n_{i}n_{j})\frac{M}{r^3} (21.7)
[/tex]
where
[tex]
n_{i} \equiv \frac {x^i}{r}
[/tex]
are the components of a unit vector in a radial direction.
In an orthonormal basis
[tex]
\vec{e_{\hat{r}}} , \vec{e_{\hat{\theta}}} , \vec{e_{\hat{\phi}}}
[/tex]
oriented along coordinate directions of a polar coordinates (r,\theta,\phi)
the nonvanishing components of the tidal acceleration tensor are
[tex]
a_{\hat{r}\hat{r}} = \frac{2M}{r^3},
a_{\hat{\theta}\hat{\theta}}=a_{\hat{\phi}\hat{\phi}}=-\frac{M}{r^3} (21.8)
[/tex]
=======================================================
Q1. From (21.6) I see that gravitational potential does depend only
on 'r' but not on theta and phi. Why in that case its theta, phi
partial derivatives described in (21.8) are non-zero?
Looks like (21.8) are calculated in some other polar coordinates,
but how those coordinates related to another ones desribed in (21.6)?
Q2. Let's assume that I'm wrong in Q1, and (21.6) does represent
gravitational potential only in Cartesian coordinates, but not
in a polar ones. Then I tried another way. I applied Chain Rule of
differentiation to (21.7) to calculate (21.8) based
on standard Cartesian --> Polar transformations:
[tex]
x=r\sin{\theta}\cos{\phi}, y=r \sin{\theta} \sin{\phi}, z=r\cos{\theta}
[/tex]
I calculated
[tex]
a_{{\hat{\theta}}{\hat{\theta}}}
[/tex]
from (21.8) applying Chain Rule to (21.7) but got
[tex]
- \frac {2 M} {r}
[/tex]
which differs from (21.8).
Did I make a mistake in my calculations, or Chain Rule
simply cannot be applied here?
Last edited: