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Legality and Generality of a Simplifying Method

  1. Jan 5, 2015 #1
    I have a question regarding the legality and generality of the following method for simplifying a vector expression. The specific expression that arose my question is ##\left[\left(\vec{P}\cdot\hat{r}\right)\left(\hat{r}\times\vec{M}\right) + \left(\vec{M}\cdot\hat{r}\right)\left(\vec{P}\times\hat{r}\right)\right]## where ##\hat{r}## is the spherical unit vector pointing away from the origin. This is part of a longer expression that needs to be integrated over a certain volume. I know that it can be simplified using a lengthy manipulation involving multiple usages of triple product formulas but even then, one still needs to assume that ##\vec{M}\times\vec{P}## lies along some known axis (like the z-axis) to do the integral and then bring it back into coordinate free form (if it is relevant or interesting to someone, this method can be found here). What I propose, however, is to suppose that, for instance, ##\vec{M} = M\hat{z}## and ##\vec{P} = P\hat{y}##. Then, plug these into the above expression, get the final answer (since the expression will be integrated over a volume, there will be no dependence on the spherical coordinates) and only then express the result in terms of ##\vec{M}\times\vec{P}##.
    Is there any loss of generality in this approach? (It does give the correct answer, I'm worried about its legality and generality).

    By the way, if there is any "test" I can perform to test for generality in such case, I would be glad to know about it.

    Any comments/corrections/suggestions will be greatly appreciated!
     
  2. jcsd
  3. Jan 5, 2015 #2

    mfb

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    Staff: Mentor

    You don't need assumptions about MxP, but you can freely choose the direction of MxP if the integration volume has a spherical symmetry.
    I don't think you can fix the direction of both M and P independently. That is beyond the freedom the spherical symmetry gives. Imagine P=M, then the integral is zero.

    You can split M into a part parallel to P and a part orthogonal to P, however, this is always possible. Both components are easier to evaluate (one is zero) and the final result is linear in M, so you can just add the two components. That should mean your result depends on MxP only, right.
     
  4. Jan 7, 2015 #3
    Sorry for the late reply and thank you very much. I have finally got around to redoing this problem with your method and succeeded.
     
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