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Partial Differentiation with Einstein Notation

  1. Sep 18, 2011 #1
    Hi there,
    I'm writing a research paper and have hit a roadblock, (wikipedia did not help) and one of my collaborators sent me an e-mail that I do not understand.

    I am attempting to find when the following functional is stationary:

    T = \int\limits_{\lambda_{1}}^{\lambda_{2}}
    \sqrt{\sum_{I=1}^{n} \sum_{i=1}^{d}
    \left(\frac{d}{d\lambda}\left(\sum_{j=1}^{d} s(\lambda)R_{j}^{i}(\lambda)(q_{I}^{j}( \lambda) + a^{j}(\lambda))\right)\right)^2} d\lambda

    This is not necessarily important for anyone who can help answer this question to understand but it helps gives context:

    T is the length of all possible curves in configuration space between two particular systems of particles. I represents the number of particles, while i and j are two indices defining the dimension. s is a function describing scale, R is a function describing rotation, q describes the system of particles, and a describes translation. So, the similarity group has 3 elements: the subgroups T^{d} and SO(d) plus the element k, defined by a, R, and s, respectively.

    Now, math wise I'm attempting to use the euler-lagrange equation to solve for constraints on s, R, q, and a. Lambda is just a parameter defining where on the trial curve the system of particles is. So, given the euler-lagrange:

    \frac{\partial{f}}{\partial{x}} - \frac{d}{d \lambda}
    \left(\frac{\partial{f}}{\partial{\dot{x}}}\right) = 0

    Where f is the integrand of T and x is the variable defining g(\lambda) such that if g(\lambda) is taken to be s(\lambda) then x would be the scaling constant k. In order to avoid a long and drawn out derivation of the conditions on T, the following substitution must be observed, where [tex]\dot{g}(\lambda) = \frac{d}{d \lambda}(g(\lambda))[/tex]:
    \varepsilon_{I}^{i} = \frac{d}{d \lambda} \left( \sum\limits_{j=1}^{d} s(\lambda) R_{j}^{i}(\lambda)(q_{I}^{j}(\lambda) + a^{j}(\lambda)) \right) [/tex]

    Now, I essentially just need the rules for using partial differentiation with respect to Einstein notation.

    I got to the point where using regular derivative rules I need to compute:
    \frac{\partial}{\partial{a^{j}}} \left(\sum\limits_{j} (\dot{s} R_{j}^{i} q_{I}^{j} + s \dot{R}_{j}^{i} q_{I}^{j} + s {R}_{j}^{i} \dot{q}_{I}^{j} + \dot{s} R_{j}^{i} a^{j} + s \dot{R}_{j}^{i} a^{j} + s R_{j}^{i} \dot{a}^{j}) \right)
    Now, I am unsure what to do. My collaborator said this:

    "You are summing over $j$ in the bracketed expression but, in (6), you want to take derivatives with respect to $a_j$. These two $j$ are not the same!! Call one of them $k$ (say the one that is summed over). Then you get,
    \frac {\partial a_j} {\partial a_k} = \delta^j_k
    where [tex]\delta^j_k[/tex] is 1 if $j=k$ and zero otherwise. This will give a vector with index $j$ when you perform the sum over $k$, which will now be trivial. You must have a vector with these components because you are differentiating with respect to a vector.

    In the case of [tex]q_I^i[/tex], you will get two delta functions [tex]\delta^I_K[/tex] and [tex]\delta^i_k[/tex]. The sums over $I$ and $k$ can then be trivially taken. You will see that this leads to a very different result."

    Can anyone help me interpret this or help me out? If you have time I would love to just send you the paper on what I've done so far, it's about 17 pages and 4 of them are incorrect because they lead off of a mistake I made right at this point. If you need any clarification whatsoever I can e-mail you the paper in a .pdf or .tex if you post your e-mail here or send it to me in a pm.

    If anyone can help me that would be so unbelievably helpful, I've asked a professor at my university and he was not familiar with Einstein notation so he could not help.

    -Sam Reid
  2. jcsd
  3. Sep 18, 2011 #2


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    Write out one or two terms in the sum explicitly so you can see what is going on.

    For example
    [tex]\sum_j \dot s R_j^i a^j[/tex]
    is just
    [tex]\dot s R_1^i a^1 + \dot s R_2^i a^2 + \dot s R_3^i a^3[/tex]
    When you differentiate that wrt [itex]a^1[/itex] for example, the terms involving [itex]a^2[/itex] and [itex]a^3[/itex] don't contribute anything.
  4. Sep 18, 2011 #3
    Okay, I understand that. But I'm unsure of how to differentiate with respect to another index altogether. Would you be able to explain how I could take:

    \frac{\partial{a^k}}{\partial{a^{j}}} \left(\sum\limits_{j} (\dot{s} R_{j}^{i} q_{I}^{j} + s \dot{R}_{j}^{i} q_{I}^{j} + s {R}_{j}^{i} \dot{q}_{I}^{j} + \dot{s} R_{j}^{i} a^{j} + s \dot{R}_{j}^{i} a^{j} + s R_{j}^{i} \dot{a}^{j}) \right)
    The part that is confusing to me is about multiplying to get a delta function afterwards or something like that? It's my collaborators e-mail that is confusing to me.
  5. Sep 18, 2011 #4


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    One key point in the email is:

    If you keep on trying to use "j" for two different things in the same expression (as in your last post) that won't help to sort out your confusion.

    What you are writing is a bit like trying to make sense of

    [tex] \frac{d}{dt} \int_0^t f(t) dt[/tex]

    That sort of notation leads to confusion about whether you want to differentiate w.r.t the upper bound of the integral, or the function [itex]f[/itex] across the whole range of the integral, or whatever.

    Don't try to use the same letter to represent both a "summed" and an "un-summed" index, and then it shoud be clearer what you are doing - and if not, writing out the "sum" in full should make it absolutely clear.
  6. Sep 18, 2011 #5
    Okay, thank you very much for the response. The part that I am confused about is why the two "j"'s are not the same! I'm looking for clarification on what a delta function is and how it comes into play here, I appreciate the notation advice but it does not really answer my questions (I hope I am not being rude).

    So, I am essentially trying to understand how I can take a functional derivative in the case where the index I want to differentiate with respect to is not the one that is being summed over in the expression.
  7. Sep 18, 2011 #6


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    The two uses of j represent different ideas.

    [tex]\sum_j \cdots_j\cdots_j[/tex]
    the j is a "dummy variable". If you write out all the terms of the sum, the expanded expression does not contain any j's at all.

    [tex]\frac {\partial}{\partial a^j} = \cdots[/tex]
    the j stands for one particular value (1, 2, or 3), and the whole expression is equivalent to 3 separate expressions
    [tex] \begin{array}
    {} \frac {\partial}{\partial a^1} = \cdots \cr
    \frac {\partial}{\partial a^2} = \cdots \cr
    \frac {\partial}{\partial a^3} = \cdots
    where any j's in the right hand side of the equations are also replaced by the corresponding 1, 2, and 3.

    So, if you try to interpret
    [tex]\frac {\partial}{\partial a^j} \sum_j \cdots_j\cdots_j [/tex]
    either it could mean 3 separate expressions like
    [tex]\frac {\partial}{\partial a^1} \sum_j \cdots_1\cdots_1 [/tex]
    But then, there is nothing in each expression for the "j" to sum over.

    Or you would get something nonsensical like
    [tex]\frac {\partial}{\partial a^1} \sum_1 \cdots_j\cdots_j[/tex]
    but "summing over 1" doesn't mean anything.
  8. Sep 18, 2011 #7
    Okay, thank you that really helps! But, I am still unclear about what a delta function is and how it comes into play with this at all.
  9. Sep 20, 2011 #8
    I think I understand the Kronecker delta function and what is going on with the indices, but I've tried multiple times to solve the Euler-Lagrange and I've only gotten ridiculously messy expressions that would not simplify to something neat. I think I may be assuming something incorrectly when dealing with the delta function in the Euler-Lagrange for the term: [tex]\frac{\partial{f}}{\partial{\dot{x}}}[/tex], where x is either [tex]s, R_{j}^{i}, q_{I}^{j}[/tex], or [tex]a^{j}[/tex]

    Any ideas for how I distinguish the difference between just get a kronecker delta and how would I represent that if I'm differentiating with respect to "derivative of a", rather than just "a"?
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