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Potential of a tidal force

  1. Apr 27, 2013 #1
    Given the tidal force
    $$
    -\alpha\left(\frac{\hat{\mathbf{d}}}{d^2}-\frac{\hat{\mathbf{d}}_0}{d_0^2}\right)
    $$
    How is the potential of the tidal force
    $$
    -\alpha\left(\frac{1}{d}+\frac{x}{d_0^2}\right)
    $$
    where ##-\nabla U_{tide} = \mathbf{F}_{tide}##.

    Idon't see how we get the force by taking the negative of the gradient of U.
     
  2. jcsd
  3. Apr 27, 2013 #2

    jambaugh

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    Gold Member

    The getting of a force by taking the negative gradient of a potential is the definition of the potential, we can't express any force this way but when we can we can by assumption and by definition. If you can take the gradient of some potential and get a specific force then that is a fact of the calculus and an observation about the specific force (that it is a conservative force). If you cannot (testable by seeing if the force has non-zero curl) then that is another possibility.

    In general you take a given expression for a force, test for whether it is conservative (= curl free = can be expressed as a gradient) and if so you integrate the gradient relationship to find the potential the same as you integrate any derivative/differential equations. I'm not clear what you question actually is. Are you unclear as to the calculus steps? or unclear as to the physics of why its possible? i think I addressed the latter. If your assignment was to "do the math" then follow those instructions. If you are asking for math help then follow the guidelines (see link in my signature) and show what you've done so far and where you're getting stuck.
     
  4. Apr 27, 2013 #3
    I don't see how they arrived to that result mathematically.
     
  5. Apr 27, 2013 #4
    How do we even show the curl is zero?

    ##\hat{\mathbf{d}} =\langle \hat{\mathbf{x}},\hat{\mathbf{y}}, \hat{\mathbf{z}}\rangle## and ##\hat{\mathbf{d}}_0 =\langle \hat{\mathbf{x}}_0, \hat{\mathbf{y}}_0, \hat{\mathbf{z}}_0\rangle##

    so ##d^2 = x^2 + y^2 + z^2## and ##d_0^2 = x_0^2 + y_0^2 + z_0^2##
     
  6. Apr 27, 2013 #5

    jambaugh

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    I'm not clear on which "result". How they took the gradient to confirm the potential functions gradient is the force in question? Or how to start with the force and find the potential to which it is a gradient?

    I can explain either or both but not if you aren't up to speed on the basic calculus. The gradient is the gradient. You should know how to calculate it. The reverse process is a matter of starting with one component and integrating with respect to one variable treating others as constants. When you get a solution you have a "constant" of integration which may depend on the remaining variables. you repeat this for each variable in turn until you have the full potential. (That's an overview and I'd be happy to work through an example.)
     
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