- #1
nonequilibrium
- 1,439
- 2
Although I am an aspiring physicist, I cannot cope with the physicist's love for vagueness when it comes to yielding math. Exactness is simply not a luxury that can be ignored, certainly not in theoretical physics.
But okay, I realize the dirac delta function can be made exact by the use of, for example, distributions, although the measure-theoretic concept seems prettier to me, so maybe I should delve into that. Anyway, something I cannot make sense of, even with distributions:
[tex]\vec \nabla \cdot \left( \frac{\vec{r}}{r^3} \right) = 4 \pi \delta^3(\vec r)[/tex]
(source: p70 from Griffiths' Electrodynamics)
I suppose it can't be anything else than the definition of the gradient of the function between parentheses in such a way that the good ol' divergence theorem
[tex]\int\int\int_V \vec \nabla \cdot \left( \frac{\vec{r}}{r^3} \right) \mathrm dV = \int \int_{\partial V} \frac{\vec{r}}{r^3} \cdot \mathrm d \vec S[/tex]
is once again true, because of course normally the theorem is simply not defined for the function (and domain) at hand, so we can view the definition as a generalization of the divergence theorem by use of distributions.
But one thing is bugging me: is it clear that the definition is well-defined? In the sense that: how do we know that there is not another situation where [tex]\vec \nabla \cdot \left( \frac{\vec{r}}{r^3} \right) [/tex] can turn up as the limiting, normally non-defined case of a certain equality (just like the divergence theorem), in which case we no longer have a choice of definition and in which case we must substitute the dirac delta function as defined? Do we know that the same definition will extend also that equality (again: by use of distributions)?
Thank you! I'm learning to cope... (although I don't know if that's good or not)
But okay, I realize the dirac delta function can be made exact by the use of, for example, distributions, although the measure-theoretic concept seems prettier to me, so maybe I should delve into that. Anyway, something I cannot make sense of, even with distributions:
[tex]\vec \nabla \cdot \left( \frac{\vec{r}}{r^3} \right) = 4 \pi \delta^3(\vec r)[/tex]
(source: p70 from Griffiths' Electrodynamics)
I suppose it can't be anything else than the definition of the gradient of the function between parentheses in such a way that the good ol' divergence theorem
[tex]\int\int\int_V \vec \nabla \cdot \left( \frac{\vec{r}}{r^3} \right) \mathrm dV = \int \int_{\partial V} \frac{\vec{r}}{r^3} \cdot \mathrm d \vec S[/tex]
is once again true, because of course normally the theorem is simply not defined for the function (and domain) at hand, so we can view the definition as a generalization of the divergence theorem by use of distributions.
But one thing is bugging me: is it clear that the definition is well-defined? In the sense that: how do we know that there is not another situation where [tex]\vec \nabla \cdot \left( \frac{\vec{r}}{r^3} \right) [/tex] can turn up as the limiting, normally non-defined case of a certain equality (just like the divergence theorem), in which case we no longer have a choice of definition and in which case we must substitute the dirac delta function as defined? Do we know that the same definition will extend also that equality (again: by use of distributions)?
Thank you! I'm learning to cope... (although I don't know if that's good or not)