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Question about Newtonian Tensor from Zee's EGR book

  1. Feb 16, 2014 #1
    I'm working through A. Zee's new EGR book, and I came to a step on tidal forces I couldn't follow. He presents the gravitational potential

    [itex]V(\vec{x})=-GM/r[/itex]

    and asks us to verify that the tensor [itex]R^{ij}(\vec{x})\equiv\partial^{i}\partial^{j}V(\vec{x})[/itex] is, in this case,

    [itex]R^{ij}=GM(\delta^{ij}r^{2}-3x^{i}x^{j})/r^{5}[/itex].

    (From all context I can find within the chapter, his notation [itex]\partial^{i}[/itex] refers to [itex]\frac{\partial^{i}}{\partial x^{i}}[/itex]). Can anyone tell me how to derive this expression explicitly? I've experimented with a lot of differentiation but I can't figure out how to get the term with the 3 in it. All I can do is see that the formula does provide the result he wants, namely that

    [itex]
    R=\frac{GM}{r^3} \left( \begin{array}{ccc}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & -2 \end{array} \right)\
    [/itex]

    Many thanks!
     
  2. jcsd
  3. Feb 16, 2014 #2

    WannabeNewton

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    Science Advisor

    Write ##r## as ##r = (x_k x^k)^{1/2}## so that ##V = -GM(x_k x^k)^{-1/2}## hence ##\partial^j V = GM (x_k x^k)^{-3/2}x_l \partial^j x^l = GM r^{-3} x^j##.

    Go ahead and compute ##\partial^i \partial^j V## in the same manner and you'll get the desired result.

    By the way, does the result remind you of anything from mechanics?
     
  4. Feb 16, 2014 #3
    Worked like a charm! Thanks very much.

    I knew to represent ##r## that way, but I didn't think to bring in the new index ##l## when I did the chain rule. So I had these new mis-indexed ##x^k## terms that messed me up.

    In retrospect, I can sort of see why we're summing over the index ##l## by thinking of my ##x^k## as ranging over the conventional coordinates ##x##, ##y##, ##z##; but I could still use some extra clarification here. Is there a rule I should know about bringing in a new index during the chain rule?

    As for the result from mechanics, I see that multiplying by a displacement vector ##\vec{r}## gets us our usual force due to gravity. Is there another thing I should be recognizing? Thanks again.
     
  5. Feb 20, 2014 #4
    I have problems with the math editor on here, never used it before. For the partial with the superscript on the right, the denominator has a subscript on the partial on the left side. The notation is just the opposite of what you would think. That is why you wind up with all superscripts in the final answer. You are taking the partial with a subscript in the denominator. I have problems with the editor and have not used it. On page 72 equation (15) Zee defines the subscript partial. So the index on the left side denominator is a superscript. Also you do not put the i in the numerator like you have it, because two of the same indexes implies Einstein summation, there should be only one index like on page 72 equation (15). Sorry about the math editor. If you see two of the same indexes then summation is assumed. When you did the problem that is how you got the delta function. You had the derivative of x with some index on top and on the bottom you had the derivative of x with respect to some index. If the indexes are the same you get dx/dx which is one.
     
  6. Feb 23, 2014 #5
    For the steps left out.

    Wanabenewton had it but,all the steps were not shown , but it is right and that was the hard part. With his answer the rest would be easy. You need to write r = sqrt [itex]\delta[/itex][itex]^{kl}[/itex]x[itex]^{k}[/itex]x[itex]^{l}[/itex] Then take der. of each x and change that index to what ever index you are taking the dev of. Since two of the same indexes are dummy variables you can change them to whatever you want(That was the point of the problem). When you hit the x twice you will get 2x after you contract the Kronecker delta's. Than you will get a 1/2 from taking the der of the square root and it will cancel the 2. I hate this editor. I will leave out the x's for the second time you take the der. you will get partials of x on top and bottom with indexes, i/j without the x's. This will be equal to the Kronecker delta, again without the delta you get (delta)ij. If it were dx/dx then you get one or zero if top and bottom not same variable. The answer above is correct but some steps were left out. Finally as long as you have two indexes you can call them whatever you want as long as you change both to the same symbol. It would take me a day using this editor. Also you should not be using the chain rule. The only time that comes into play is when you are going from one coordinate system or frame to another, which in this case you are not.

    I use MathType and word. Then convert it to a PDF file after I write the paper.
     
    Last edited: Feb 23, 2014
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