Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Time derivative of tensor expression

  1. Sep 15, 2015 #1
    I was trying to compute the time derivative of the following expression:

    [tex] \mathbf{p_k} = \sum_i e_{ki}\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)!} \mathbf{r_{ki}}(\mathbf{r_{ki}\cdot \nabla})^n \delta(\mathbf{R_k}-\mathbf{R}) [/tex]

    I am following deGroot in his Foundations of Electrodynamics. He says, "Taking the derivative [itex] \partial_0 = \partial/\partial ct [/itex]" of the above equation, one finds

    [tex] \partial_0 \mathbf{p_k} + \frac{1}{c} \mathbf{\dot{R}_k \cdot \nabla p_k} - \frac{1}{c} \sum_i e_{ki} \sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)!} (\mathbf{\dot{r}_{ki} r_{ki} \cdot \nabla } + n\mathbf{r_{ki}\dot{r}_{ki} \cdot \nabla})(\mathbf{r_{ki}\cdot \nabla})^{n-1} \delta(\mathbf{R_k}-\mathbf{R}) = 0 [/tex]

    since "the time derivative of the delta function is equal to [itex] -\mathbf{\dot{R}_k \cdot \nabla} [/itex] acting on it." By this I assume he means that the total time derivative of the delta function is 0.

    Some context: deGroot is talking about a stable group of particles labeled by k. So [itex] \mathbf{R_{k}} [/itex] is the position of the center of mass of the group and the [itex] \mathbf{r_{ki}} [/itex] are the internal coordinates , which specify the positions of the constituent particles ki with respect to the stable group k. So we are dealing with the electric multipole moments of the group of particles. [itex] e_{ki} [/itex] is the charge of the ki particle.

    This is my attempt (it seems above that he is taking the total derivative of both sides and subtracting one from the other):
    Let's just focus on the [itex]\mathbf{r_{ki}}(\mathbf{r_{ki}\cdot \nabla})^n \delta(\mathbf{R_k}-\mathbf{R}) [/itex] term. I get
    [tex] \mathbf{\dot{r}_{ki}}(\mathbf{r_{ki}\cdot \nabla})^n \delta(\mathbf{R_k}-\mathbf{R}) + \mathbf{r_{ki}}\frac{d}{dt}[(\mathbf{r_{ki}\cdot \nabla})^n\delta(\mathbf{R_k}-\mathbf{R})][/tex]

    So the remaining part is to find [tex]\mathbf{r_{ki}}\frac{d}{dt}[(\mathbf{r_{ki}\cdot \nabla})^n\delta(\mathbf{R_k}-\mathbf{R})][/tex] which according to deGroot should be [tex] (n\mathbf{r_{ki}\dot{r}_{ki} \cdot \nabla})(\mathbf{r_{ki}\cdot \nabla})^{n-1} \delta(\mathbf{R_k}-\mathbf{R}) [/tex]

    This would usually make sense using the chain rule, but I don't immediately see how you can use the chain rule since the nabla operator is inside. The chain rule would give
    [tex] \frac{d}{dt}[(\mathbf{r_{ki}\cdot \nabla})^n] = n (\mathbf{r_{ki}\cdot \nabla})^{n-1} \frac{d}{dt}(\mathbf{r_{ki}\cdot \nabla}) [/tex]

    Is it really mathematically correct to say that [itex] \frac{d}{dt}(\mathbf{r_{ki}\cdot \nabla}) = (\mathbf{\dot{r}_{ki}\cdot \nabla}) [/itex]? Is it because you have to think of the operator being applied to the delta function first before taking the derivative and since the delta function is independent of time it makes no difference?
  2. jcsd
  3. Sep 17, 2015 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think your guess as to the reason may be correct.

    That is, perhaps it is not the case that ##
    \frac{d}{dt}(\mathbf{r_{ki}\cdot \nabla}) = (\mathbf{\dot{r}_{ki}\cdot \nabla})##

    but it is the case that
    \frac{d}{dt}\left[(\mathbf{r_{ki}\cdot \nabla})g(\mathbf{x})\right] = (\mathbf{\dot{r}_{ki}\cdot \nabla})g(\mathbf{x})$$
    where ##g## is a function only of the variable ##\textbf{x}##, which does not involve ##t##.

    Let's see:

    $$\frac{d}{dt}\left[(\mathbf{r_{ki}\cdot \nabla})g(\mathbf{x})\right] =
    \frac{d}{dt}\left[\mathbf{r_{ki}\cdot (\nabla}g(\mathbf{x}))\right] =
    \frac{d}{dt}[\mathbf{r_{ki}]\cdot (\nabla}g(\mathbf{x}))
    \mathbf{r_{ki}}\cdot \frac{d}{dt}[\nabla g(\mathbf{x})]

    \mathbf{\dot{r}_{ki}\cdot (\nabla}g(\mathbf{x}))
    \mathbf{r_{ki}}\cdot \mathbf{0}
    \mathbf{\dot{r}_{ki}\cdot (\nabla}g(\mathbf{x})) =
    (\mathbf{\dot{r}_{ki}\cdot \nabla})g(\mathbf{x})

    So it looks OK.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Time derivative of tensor expression