1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Feynman & Tidal Forces

  1. Feb 25, 2016 #1
    I used to think I understood this - big mistake!

    I've just watched a great classic Feynman lecture posted by another PF user (a superb way to spend 50 mins of you can spare the time)
    https://www.physicsforums.com/threads/i-found-the-best-teacher-of-physics.855335/

    In that lecture, Feynman describes the tidal effect of the moon on the earth (26 mins in), and gives two components to it; the (obvious?) inverse cube tidal effect on the different sides of the earth, and the (lesser discussed?) centrifugal effect arising from the earth and moon rotating around a common point of focus which is inside the earth, but always on moon-side of the earth and therefore the water on the side furthest from the moon is "thrown off" more than the other side (Fenyman's words).

    I wanted to understand more about these two components and if they balance out exactly to produce an identical effect on both sides of the earth and ran into confusion. Articles fall into 4 camps - I've linked to examples, but they're not the only ones I looked at.

    i. The centrifugal force** should be ignored as it's mixing up a rotating frame of reference from POV of the earth and a static frame from a tidal force POV
    https://www.lhup.edu/~dsimanek/scenario/tides.htm
    (The article correctly differentiates between centripetal force and centrifugal force. I'll continue to use Centrifugal force in the way the articles do from here on)

    ii. It's all down to Centrifugal force
    http://www.moonconnection.com/tides.phtml

    iii. It's both
    http://www.livescience.com/29621-what-causes-the-tides.html
    https://tidesandcurrents.noaa.gov/restles3.html

    iv. And probably the least helpful... It's like this - No you're wrong, I'm right - NO! I'm right - No It's ME!
    https://en.wikipedia.org/wiki/Talk:Tidal_force
    This last one even says categorically that Feynman was wrong in his description. And then later someone else says he wasn't. So that wasn't exactly conclusive!

    So whats my question? Well if I'm allowed two of them:

    1. Are the "two components" actually just two ways of looking at the same force in rotating and a static reference frame, or are there really two separate components that make up the the overall force in play?

    2. Does whatever the answer is to 1, result in (theoretical) identical sized bulges on both sides of the earth?

    And I suppose there's a third... does anyone have a reference source they firmly believe to be "the right answer"?

    Thanks
    Matt
     
  2. jcsd
  3. Feb 26, 2016 #2

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    In the inertial frame the only force is gravity, so it's gradient must explain ALL of the tidal deformation. In all the non-inertial frames, the gravity gradient and the deformation are is still the same, so the gravity gradient still can explain ALL of the tidal deformation. The additional inertial forces (linear, centrifugal, Coriolis) that arise in those non-inertial frames are therefore not needed to explain the tidal deformation. Their deformative effects must cancel out.

    No, since the gravity field is non-linear.

    I like this article, which analyzes different frames of reference:
    http://www.vialattea.net/maree/eng/index.htm
     
    Last edited: Feb 26, 2016
  4. Feb 26, 2016 #3
    Thanks @A.T. The first part makes total sense - and I read it now and almost wonder why I was confused! It's amazing how a clear explanation can shake off the ambiguities you get when consulting multiple sources of unknown accuracy. Thanks for the link to the article as well - I'd not even seen reference to Coriolis force being treated alongside Centrifugal in the other articles.

    The answer to the second point is a real surprise to me as all of the simplified tidal diagrams show the bulges as the same height. I tried googling this specific question before I posted the original post and got more of the same articles that led to my confusion on point 1. Do you have a source that would help me understand this further?

    Thanks again
    Matt
     
  5. Feb 26, 2016 #4

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    Yes the one I already posted:

    http://www.vialattea.net/maree/eng/index.htm

    It computes acceleration differences between the near / far points and the center of mass, for every frame. The near and far acceleration differ from the COMs acceleration by a different amounts, meaning that the deformation will not be symmetrical.

    To get symmetrical deformation you would need a field that is proportional to 1/r (not 1/r2) and parallel (not radial), I guess.
     
  6. Feb 26, 2016 #5
    I think I follow that - the parallel /radial point at least seems really obvious now you point it out!!
    A much more thorough read of the article is clearly required! Thanks again for your help.

    A further search following your reply has pointed me to a Cambridge University article that other PF readers may enjoy. This article also links and makes favorable reference to a less academic article on the NOAA website. The NOAA site (to me) looks excellent.
    www.cambridge.org/gb/download_file/188298/
    https://tidesandcurrents.noaa.gov/restles1.html
     
  7. Feb 26, 2016 #6

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    To me it looks like the usual, flawed explanation based on misunderstanding centrifugal force (see Figure 1).
     
  8. Feb 26, 2016 #7
    The Cambridge paper also does this, but in both cases it's a means to an end. They summarize it as

    "we first find directly the magnitude and direction of the tidal forces for the two-body (earth and moon) problem for selected locations on the earth before presenting the general case....
    ... We then analyze the general case by writing the potentials of the moon and the centrifugal force in the rotating frame of reference of the moon-earth orbit and then expressing their sum in terms of earth centered coordinates. "
     
  9. Feb 26, 2016 #8

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    As far as I understand, what the Cambridge paper calls "centrifugal force" is not the radial inertial force due to the rotation of the frame (as usually understood), but the uniform inertial force due to the centripetal acceleration of the frame. This is indicated by this statement:

    "...The centrifugal force has the same magnitude and direction for observers anywhere on the earth..."

    The actual centrifugal force is radial, and therefore cannot have the same magnitude and direction for different points. The article by Paolo Sirtoli clarifies the difference between a rotating frame and a frame translating along a circle:

    http://www.vialattea.net/maree/eng/index.htm (see Case 4)

    As you can see in Fig 16 there is no centrifugal force (radial), just a uniform inertial force field with a time-dependent direction. See also Fig. 12 and the following paragraphs on why these inertial forces do not contribute to deformation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Feynman & Tidal Forces
  1. Tidal Forces (Replies: 4)

Loading...