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Charge distribution of point charges in spherical coordinates

  1. Aug 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi! This is not really a problem. I'm just confused on how to express the charge distribution of a set of point charges in spherical coordinates. From our discussion,
    [itex]ρ(\vec{r})=\sum\limits_{i=1}^N q_i δ(\vec{r}-\vec{r}')[/itex]
    where [itex]\vec{r}[/itex] is the position of the point where charge density is being evaluated and [itex]\vec{r}'[/itex] is the position of the point charge.

    Three-dimensional Dirac delta function in general is
    [itex]δ(\vec{r})= \frac{1}{h_1 h_2 h_3} δ(u_1 - u_1 ')δ(u_2 -u_2 ') δ(u_3 - u_3 ')[/itex]

    So in spherical coordinates, it is
    [itex]δ(\vec{r})= \frac{1}{r^2 sin(θ)} δ(r - r')δ(θ -θ') δ(\phi - \phi')[/itex]

    The continuous charge distribution of these point charges is therefore
    [itex]ρ(\vec{r})=\sum\limits_{i=1}^N \frac{1}{r^2 sin(θ)} δ(r - r'_i)δ(θ -θ'_i) δ(\phi - \phi'_i)[/itex]

    So here's the part where I'm confused. For the factor [itex]\frac{1}{r^2 sin(θ)}[/itex]. do I retain it as is (i.e. r and θ are coordinates of position vector [itex]\vec{r}[/itex] or do I subsitute the coordinates of my point charge (i.e. the coordinates of [itex]\vec{r}'[/itex])? Thank you very much.

    2. Relevant equations
    See above.

    3. The attempt at a solution
    I've seen a problem and an online solution and it seems that he substituted the coordinates of [itex]\vec{r}'[/itex] but I can't reconcile it from the definition of 3-D Dirac delta function I've read where the coordinates of [itex]\vec{r}[/itex] are substituted instead. Thank you very much.
     
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  3. Aug 11, 2014 #2

    vanhees71

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    But it doesn't make a difference, whether you put the primed or unprimed coordinates into the prefactor, because with a grain of salt, [itex]\delta(x)=0[/itex] for [itex]x \neq 0[/itex]. Thus you always have
    [tex]\frac{1}{r^2 \sin \theta} \delta(r-r') \delta(\theta-\theta') \delta(\phi-\phi')=\frac{1}{r'^2 \sin \theta'} \delta(r-r') \delta(\theta-\theta') \delta(\phi-\phi').[/tex]
    Note, however, that you never should hit the singularities of the coordinate system, because there the Jacobian becomes 0 in the denominator.
     
  4. Aug 12, 2014 #3
    Thanks vanhees71. :smile: By Jacobian, you mean [itex]\frac{1}{h_1 h_2 h_3}[/itex] right?

    Also, about your answer. Sorry for asking again, but will it really not make a difference? I'm asking this question because I'm supposed to get the multipole moment of four charges with spherical coordinates [itex](a, \frac{\pi}{2}, 0) [/itex], [itex](a, \frac{\pi}{2}, -\frac{\pi}{2}) [/itex], [itex](a, \frac{\pi}{2}, -\pi) [/itex] and [itex](a, \frac{\pi}{2}, \frac{3\pi}{2}) [/itex] respectively.

    The equation for multipole moment is [itex] q_{l m} = \int Y^*_{l m}(\theta', \phi')r^{'l}\rho(\vec{r}')d^3x'[/itex]. Therefore, if the factor [itex]\frac{1}{r^2 sin(\theta)}[/itex] is primed, it becomes part of the integration but if it is unprimed, I can take it out of the integral like a constant. So in this case, should the factor [itex]\frac{1}{h_1 h_2 h_3}[/itex] be primed or unprimed?

    Also, I am using this equation number 19 from this reference: http://web.mst.edu/~hale/courses/411/411_notes/Chapter1.Appendix.Dirac.Delta.pdf which as you see, has unprimed factors for the three-dimensional Dirac delta function in spherical coordinates. Thanks again.
     
  5. Aug 12, 2014 #4
    Thanks vanhees71. :smile: By Jacobian, you mean [itex]\frac{1}{h_1 h_2 h_3}[/itex] right?

    Also, about your answer. Sorry for asking again, but will it really not make a difference? I'm asking this question because I'm supposed to get the multipole moment of four charges with spherical coordinates [itex](a, \frac{\pi}{2}, 0) [/itex], [itex](a, \frac{\pi}{2}, -\frac{\pi}{2}) [/itex], [itex](a, \frac{\pi}{2}, -\pi) [/itex] and [itex](a, \frac{\pi}{2}, \frac{3\pi}{2}) [/itex] respectively.

    The equation for multipole moment is [itex] q_{l m} = \int Y^*_{l m}(\theta', \phi')r^{'l}\rho(\vec{r}')d^3x'[/itex]. Therefore, if the factor [itex]\frac{1}{r^2 sin(\theta)}[/itex] is primed, it becomes part of the integration but if it is unprimed, I can take it out of the integral like a constant. So in this case, should the factor [itex]\frac{1}{h_1 h_2 h_3}[/itex] be primed or unprimed?

    Also, I am using this equation number 19 from this reference: http://web.mst.edu/~hale/courses/411/411_notes/Chapter1.Appendix.Dirac.Delta.pdf which as you see, has unprimed factors for the three-dimensional Dirac delta function in spherical coordinates. Thanks again.
     
  6. Aug 12, 2014 #5

    Orodruin

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    As was said, it really does not matter. If you put the unprimed coordinates they will simply be replaced by the primed when you integrate.
     
  7. Aug 12, 2014 #6
    Okay, I see. Thank you very much Orodruin. :)
     
  8. Jun 14, 2017 #7

    Orodruin

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    This is a good example why you should never use a coordinate system that has your delta on the boundary. It is bound to lead to confusions like this one.

    The only reasonable definition of the delta is in terms of a distribution. In this case, the three-dimensional delta distribution should satisfy
    $$
    \int_V \delta^{(3)}(\vec x) f(\vec x) dV = f(0).
    $$
    With your assertion that the integral of the one-dimensional delta at the boundary is 1/2 and not 1, you would get this result wrong. You have to look at what distribution you are actually considering. You should generally also treat the delta as a distribution rather than a function. The bottom line is: If you want to treat it as a function with the integral property, don't use a coordinate system such that your delta is on its boundary.
     
  9. Jun 14, 2017 #8

    Orodruin

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    What makes you think you can't do this in cylinder or spherical coordinates? It is just a matter of knowing what the metric tensor is. There is absolutely no need to go to Cartesian coordinates.

    I believe the OP's question was settled satisfactorily three years ago.

    This is simply not true. The basis vectors (which is what you really mean, a unit vector can have a non-zero time derivative regardless of the coordinate system) depend on the position, there is no "time" intrinsic to Euclidean space. If the position depends on time, yes, the basis will change with the position. However, the concept of having a basis that is independent of the position (such as a Cartesian basis) is something that is very particular to Euclidean space. As soon as you try to generalise this to more general spaces you will run into problems as the tangent spaces of different points are different. I would say it is even detrimental to try to cling to the concept of having "the same basis" at different points.
     
  10. Jun 14, 2017 #9

    Orodruin

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    It is an orthonormal right-handed basis. It holds that
    $$
    \vec e_r \times \vec e_\theta = \vec e_\varphi
    $$
    and all cyclic permutations. So your case would be
    $$
    \vec a \times \vec b = (a_2 b_3 - a_3 b_2)\vec e_r + (a_3 b_1 - a_1 b_3)\vec e_\theta + (a_1 b_2 - a_2 b_1)\vec e_\varphi.
    $$

    If you have seen it or not is not really relevant. Tensors are central in many applications in physics. Anyway, it doesn't, so ...

    You never said anything about a rotating coordinate system, I think you are mixing things up. A rotating coordinate system will have time varying basis vectors - whether Cartesian or not.
     
  11. Jun 14, 2017 #10

    Orodruin

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    This seems seriously taken out of context. If presented like this, it is simply wrong. A curvilinear coordinate system in itself has absolutely nothing to do with time. There is the possibility I alluded to in #10: For an object moving in a curvilinear coordinate system, the basis vectors will change with time because the position changes with time and the basis vectors change with position. This is not a drawback of curvilinear coordinates - it is a virtue. It can be used to describe things such as fictitious forces easily. Still, I would say that it is not good presentation to say that the basis vectors change with time - it would be more appropriate to say that they change with position (because they are functions of position, not time).
     
  12. Jun 14, 2017 #11

    Orodruin

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    So I found the text in the Caltech e-library.

    It is exactly what I said. It is a book on dynamics and when they talk about cylinder coordinates they have implicitly assumed a motion of some object. I suggest reading a text that is actually about vector analysis and curvilinear coordinate systems.
     
  13. Jun 14, 2017 #12

    Orodruin

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    You are completely missing the point and your statement was a blanket one just generally stating:
    All of those statements are false. A priori, there is not even a "time" connected to a curvilinear coordinate system at all. The coordinate system remains the same in time. However, if there is motion, then there is variation in space, which the basis vectors do depend on.

    Furthermore, curvilinear coordinates are used for so much more than describing the motion of single point objects - what was being treated in this very thread being a prime example. I would also still call the statement in the H&H text misleading without further qualifiers and I believe that they assume that you have already had some introduction to curvilinear coordinate systems. You need to realise that you are not reading a book on the mathematics behind vector analysis. You are reading an excerpt from a book on dynamics, which is using vector analysis as a tool and within that context (and only within that context) can you trust what they say. Although the proper way of presenting it would be to state "the basis vectors change in space and so the basis at the point of the object changes with time due to the motion of the object".
     
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