Doubt in the divergence

Discussion in 'Physics' started by logearav, Mar 14, 2012.

  1. logearav

    Thread Starter Member

    Aug 19, 2011
    248
    0
    For the position vector r = ix+jy+kz, show that div(r/r^3) = 0
    (i ∂/∂x + j ∂/∂y + k ∂/∂z) . ( ix + jy + kz/r^3)
    ∂/∂x(x/r^3) + ∂/∂y(y/r^3) + ∂/∂z(z/r^3)
    so 1/r^3 + 1/r^3 + 1/r^3.
    But my coursebook depicts the differentiation as 1/r^3 + x(-3r^-4∂r/∂x) for the first term. Like wise they have used the x^n = nx^n-1 formula for differentiating r^3 in ∂/∂y and ∂/∂z terms too.
    Revered Members,
    I am confused. It is partial differentiation with respect to x, then whey should differentiate r by using the power formula. Any help in this regard will be highly appreciated.
     
  2. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    It's just an equivalent form they are using. If you carefully write out what they have, you'll see they are equivalent, although a little confusing to look at.

    The whole thing gets a lot easiier if you just work in spherical coordinates, since there is only dependence on r and there is only one vector component, since the vector points in the radial direction.

    In spherical coordinates the vector is simply the radial component as follows (note that r-hat is the unit vector in the radial direction).

     \vec{A}={{1}\over{r^2}}\hat{r}

    Hence, the radial component is the only component as follows.

    {A_r}= {{1}\over{r^2}}

    Since there is no angular components to the vector, the divergence reduces to

     \nabla \cdot \vec{A}= {{1}\over{r^2}} {{\partial}\over{\partial r}}(r^2 A_r)= {{1}\over{r^2}} {{\partial}\over{\partial r}}(r^2 {{1}\over{r^2}}) = {{1}\over{r^2}} {{\partial}\over{\partial r}}(1) =0
     
    Last edited: Mar 14, 2012
  3. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    It's difficult to appreciate the film when you come in halfway through.

    This is a standard result about Laplace's theorem that your book has used to set a problem.

    I will start a little bit back to answer your question in stages by starting with a small lemma or standard result.

    If r the distance of a variable point P from a fixed point O. This distance function is a scalar function where r is equal to the length of the distance vector r.

    As Steve says it can also be thought of as a scalar multiplier times a unit vector in that direction r - hat

    {\bf{r}} = r\widehat{\bf{r}}

    Now let us consider a function of r

    V\left( r \right) = {r^m}

    where m is any real number.

    The level surfaces (surfaces of equal distance raised to the power m) are concentric spheres with centre O. (This is why Steve suggests spherical coords). The unit normal to these spheres is thus a radius parallel to the position vector r

    So considering the gradient of this function we can derive a useful formula.

    \nabla {r^m} = \frac{{\partial {r^m}}}{{\partial n}}{\bf{n}} = \frac{{\partial {r^m}}}{{\partial r}}\widehat{\bf{r}} = m{r^{m - 1}}\widehat{\bf{r}} = m{r^{m - 2}}{\bf{r}}

    Put m = -1 and presto

    \nabla \left( {\frac{1}{r}} \right) =  - \frac{1}{{^2}}\widehat{\bf{r}} =  - \frac{1}{{{r^3}}}{\bf{r}}

    Recognise it? It's the expression in your question.

    Now for your question about differentiation of a product.

    This again is a standard result. The product in question is the product of a scalar function u and a vector function v

    div\left( {u{\bf{v}}} \right) = \nabla u{\bf{.v}} + udiv\left( {\bf{v}} \right)

    Can you see the similarity to simple differentiation of a product for single variable calculus y= u(x) w(x) = uw?

    So using this

    {\nabla ^2}{r^m} = \nabla \left( {\nabla {r^m}} \right) = \nabla \left( {m{r^{m - 2}}{\bf{r}}} \right) = m\nabla \left( {{r^{m - 2}}{\bf{r}}} \right)

    So our scalar function is

    {{r^{m - 2}}}

    and our vector function is just r

    So

    m\nabla \left( {{r^{m - 2}}{\bf{r}}} \right) = m\left( {\nabla {r^{m - 2}}{\bf{.r}} + {r^{m - 2}}div({\bf{r}})} \right)

    now div(r) = 3 for 3 dimensions

    Now can you finish it?

    Hint set

    {\nabla ^2}{r^m}

    equal to zero and solve the algebra for m.

    It turns out that this is zero when m = -1.

    So that

    {\nabla ^2}\left( {\frac{1}{r}} \right) = 0

    is a solution of Laplace's equation.
     
    steveb likes this.
  4. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    By the way, this part of your calculation is not right. The variable r is a function of x, y and z (i.e. r(x,y,z)).

    Hence, ∂/∂x(x/r^3) is not equal to 1/r^3, because r=r(x). You can apply the division rule for differentiation to obtain
    ∂/∂x(x/r^3)=1/r^3-3x^2/r^5. From here you can see that ∂/∂x(x/r^3) + ∂/∂y(y/r^3) + ∂/∂z(z/r^3)=0
     
Loading...