- #1

JustinLevy

- 895

- 1

[tex]\vec{j}(\vec{r},t) = \sum_\alpha q_\alpha \dot{\vec{r}}_\alpha(t) \delta^3(\vec{r}_\alpha(t) - \vec{r})[/tex]

If I wanted to take the time derivative of this, what would be the best way to write it notationally?

I assume this is clear enough:

[tex]\dot{\vec{j}}(\vec{r},t) = \sum_\alpha [ q_\alpha \ddot{\vec{r}}_\alpha(t) \delta^3(\vec{r}_\alpha(t) - \vec{r}) + q_\alpha \dot{\vec{r}}_\alpha(t) \frac{d}{dt}\delta^3(\vec{r}_\alpha(t) - \vec{r}) ]

[/tex]

It is my understanding that the derivative of a delta "function" can be written:

[tex] \frac{d}{dt} \delta(t) = -\frac{1}{t}\delta(t)[/tex]

which I assume means I can write (where [itex]a[/itex] is some constant)

[tex] \frac{d}{dt} \delta(t-a) = -\frac{1}{t-a}\delta(t-a)[/tex]

correct?

So I could manipulate that last term like this?

[tex]\frac{d}{dt}\delta^3(\vec{s}(t) - \vec{r}) = \frac{d}{dt}\delta(s_x(t) - x)\delta(s_y(t) - y)\delta(s_z(t) - z) = \frac{d}{dt} \sum_{i} \frac{1}{|\dot{s}_x(t_i)|}\frac{1}{|\dot{s}_y(t_i)|}\frac{1}{|\dot{s}_z(t_i)|} \delta(s_x(t_i) - x)\delta(s_y(t_i) - y)\delta(s_z(t_i) - z)[/tex]

where [tex]t_i[/tex] is a time satisfying [tex]\vec{s}(t_i) = \vec{r}.[/itex]

Then I have:

[tex]\frac{d}{dt}\delta^3(\vec{s}(t) - \vec{r}) = \sum_{i} \left(\frac{1}{s_x(t_i) - x} + \frac{1}{s_y(t_i) - y} + \frac{1}{s_z(t_i) - z}\right)\frac{1}{|\dot{s}_x(t_i)|}\frac{1}{|\dot{s}_y(t_i)|}\frac{1}{|\dot{s}_z(t_i)|} \delta(s_x(t_i) - x)\delta(s_y(t_i) - y)\delta(s_z(t_i) - z)[/tex]

First of all, is that correct?

Second of all, is there a better notational way to write that?