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Homework Help: Newtonian tidal acceleration tensor in polar coordinates

  1. Dec 23, 2006 #1
    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)

    \Phi = \frac {- M }{r} (21.6)


    r = \sqrt {x^2+y^2+z^2 }

    is the distance from the center od symmetry.
    Evaluating the tidal gravitational acceleration tensor using
    the rectangular coordinates gives:

    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)


    n_{i} \equiv \frac {x^i}{r}

    are the components of a unit vector in a radial direction.
    In an orthonormal basis

    \vec{e_{\hat{r}}} , \vec{e_{\hat{\theta}}} , \vec{e_{\hat{\phi}}}

    oriented along coordinate directions of a polar coordinates (r,\theta,\phi)
    the nonvanishing components of the tidal acceleration tensor are

    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)


    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:

    x=r\sin{\theta}\cos{\phi}, y=r \sin{\theta} \sin{\phi}, z=r\cos{\theta}

    I calculated


    from (21.8) applying Chain Rule to (21.7) but got

    - \frac {2 M} {r}

    which differs from (21.8).

    Did I make a mistake in my calculations, or Chain Rule
    simply cannot be applied here?
    Last edited: Dec 23, 2006
  2. jcsd
  3. Dec 23, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    If you want to work the problem in polar coordinates, you need to be familiar with covariant derivatives, a formalism from differential geometry that allows one to specify gradients, etc. in arbitrary coordinate systems.

    While there is some information on them on the internet, for example at the Wikipedia, I'm not sure if you'll be able to get everything you need without a textbook.

    Basically, you need information about the Christoffel symbols / the connection in order to take the covariant derivative of a vector field. And the gradient of your scalar potential function is a vector field. So you need to understand how to take the derivative of a vector field in arbitrary coordinates to answer your question.

    In the notation of covariant derivatives, what you compute is
    F_j = \nabla_j \Phi

    F being the force, a vector (and Phi is a scalar).

    a_{ij} = \nabla_i F_j

    I'm sure there is someway to write this all in one line, rather than two separate equations, but I'm not quite sure how to do it correctly ...
    Last edited: Dec 23, 2006
  4. Dec 23, 2006 #3
    Oh, I see the way to go now.

    Thanks pervect!
  5. Dec 24, 2006 #4

    Chris Hillman

    User Avatar
    Science Advisor

    Form of tidal tensor in spherically symmetric gravitational field

    Hi, ss29593,

    I am not sure whether you are asking for a more elementary derivation than what you found in the book by Hartle, but in a 8 Mar 1998 post to sci.physics.relativity, titled "Tidal Forces in Newtonian Theory and GTR", I offered an elementary derivation of the Coulomb form [tex]m/r^3 \operatorname{diag} (-2,1,1)[/tex], common to both Newtonian gravitation and gtr in the case of a spherically symmetric gravitational field, using only Newton's force law and the small angle approximation; see http://www.math.ucr.edu/home/baez/RelWWW/group.html [Broken] for this post.

    Note that I seem to be using the opposite sign for the potential from Hartle, which ensures that the result I obtained agrees with the tidal tensor for the Schwarzschild vacuum solution in general relativity. In the Coloumb form which I noted above, the negative radial component indicates radial tidal tension while the positive components orthogonal to the radial direction indicate tidal compression. This happens because as a small ball of dust falls in a spherically symmetric vacuum field, it elongates radially and is compressed orthogonally. In Newtonian terms, the radial elongation happens because the closer part of the dust ball is subjected to a slightly stronger gravitational force than the further part, so the dust ball elongates. The orthogonal compression happens because the force vectors pointing toward the center of mass squeeze the ball slightly orthogonally to the direction of infall.

    If you want a reference to another textbook discussion of the Newtonian tidal tensor, the clearest and most extensive discussion I have seen is in sections 1.8-9 of Ohanian and Ruffini, Gravitation and Spacetime, 2nd Ed., Norton, 1994.

    Oh wow, I am very glad to see that Hartle uses frame fields!

    (See http://en.wikipedia.org/w/index.php?title=Frame_fields_in_general_relativity&oldid=42117350 for an overview of frame fields, especially on applications of them in gtr.)

    Phi depends only on r because we are assuming that the potential is spherically symmetric. His eq. (21.8) gives the components in the frame field (hatted indices are often used to emphasize that components refer to a frame or anholonomic or orthonormal basis, rather than to the coordinate basis). In the case of a polar spherical chart on [tex]E^3[/tex], the coordinate covectors are [tex] dr, \; d\theta, \; d\phi[/tex], while the coordinate vector fields are [tex] \frac{\partial}{\partial r}, \; \frac{\partial}{\partial \theta}, \; \frac{\partial}{\partial \phi}[/tex]. Rescaling, the covectors in the ON coframe are
    [itex] \sigma^1 = dr, \; \sigma^2 = r \, d\theta, \; \sigma^3 = r \sin(\theta) \, d\phi[/itex]
    (these can be read right off the line element written in the polar spherical chart). The dual frame field (three vector orthonormal fields) is
    [itex] \vec{e}_1 = \frac{\partial}{\partial r}, \; \vec{e}_2 = \frac{1}{r} \, \frac{\partial}{\partial \theta}, \; \vec{e}_2 = \frac{1}{r \, \sin(\theta)} \, \frac{\partial}{\partial \phi} [/itex]

    You were getting warm, as the above shows; the ON frame corresponds to the idea of a "local Cartesian frame" which is valid very close to a given point. Similarly, in gtr, frame fields correspond to the idea of a "local Lorentz frame" valid very near a given event.

    The result given by Hartle is of course correct; you might try the elementary approach first as a sanity check. Next, to learn about frames, you might try Harley Flanders, Differental Forms with Applications to the Physical Sciences, Dover, 1989 reprint of 1963 original, an engaging and readable book.
    Last edited by a moderator: May 2, 2017
  6. Dec 24, 2006 #5
    Thanks Chris!

    Links you provided are very helpful.

    I see where I was wrong now.

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