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Homework Help: Difficult equation to understand in Goldstein's Mechanics

  1. Sep 19, 2012 #1
    1. The problem statement, all variables and given/known data

    If anyone is familiar with (or has nearby) Goldstien's Mechanics text, I am having some difficulty seeing why equation 1-34 is true.

    Can someone explain or perhaps hint as to why, in the given context, the gradient of a potential function of distance (in terms of a difference of position vectors) equals the difference in position vectors times an arbitrary scalar function?

    Perhaps I should also lay out some other inquiries I had regarding this discussion:

    Part of Goldstein's development of the internal potential energy for a system of a system of particles is motivated by "sasifying the strong law of action and reaction." Is this satisfaction a necessary feature of conservative fields, or he is simply doing this because many typical fields satisfy this criteria? (page 10 of edition 2)

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Sep 19, 2012
  2. jcsd
  3. Sep 19, 2012 #2
    You just need to differentiate the 1.32 to understand that. Observe that the magnitude of a vector is the square root of its scalar product with itself. The gradient of that is the vector multiplied by a scalar.
  4. Sep 19, 2012 #3
    Can you explain where I went wrong (or how to continue) the explicit derivation I have attached?

    Attached Files:

  5. Sep 19, 2012 #4
    \nabla |\mathbf{n}|
    = \nabla \sqrt{X^2 + Y^2 + Z^2}
    [/tex]Because ## X = x + a, \ Y = y + b, \ Z = z + c ## (since ## \mathbf{n} = \mathbf{r} - \mathbf{r}_0 ##)[tex]
    \nabla |\mathbf{n}|
    = \nabla \sqrt{X^2 + Y^2 + Z^2}
    = \frac 1 {|\mathbf{n}|} (X, Y, Z)
    = \frac {\mathbf{n}} {|\mathbf{n}|}
    = f(|\mathbf{n}|)|\mathbf{n}|
    [/tex] The same is true when ## \mathbf{r} ## and ## \mathbf{r}_0 ## are exchanged.
  6. Sep 19, 2012 #5
    Ah, thank you voko. I was thinking a bit too abstractly- for some reason I was considering the components of the vector (which you call 'n') as multivariate functions, when clearly this is not the case, as n is simply a difference of position vectors.

    With that being said, as a minor note, shouldn't we have X = xr - xr0? where X is the xr component of r and xr0 is the x component of r0, i.e. since n = r - r0?

    Also, besides |n|-1, is V'(|n|) also a factor in the final scalar function (by the chain rule)? I'd just like to verify this.

    Lastly, Goldstein goes on to state that if the scalar potential were in adiition a function of, say, the difference in velocity vectors of two internally interacting particles, then their mutual forces would still be equal and opposite, only not directed along the line connecting the particles. I am trying to explicity derive this as well. My only trouble seems to be taking the gradient of a function of form: V(|ri - rj|,|vi - vj|). Any pointers?
    Last edited: Sep 19, 2012
  7. Sep 19, 2012 #6
    Yes to both questions.

    Regarding the generalized potential, let ## V = V(a, b) = V(|\mathbf{r}_i - \mathbf{r}_j|, |\mathbf{v}_i - \mathbf{v}_j|) ##, then ## \nabla_i V = \frac {\partial V} {\partial a} \nabla_i |\mathbf{r}_i - \mathbf{r}_j| + \frac {\partial V} {\partial b} \nabla_i |\mathbf{v}_i - \mathbf{v}_j| ##. This is just using the chain rule and linearity of ## \nabla ##.
  8. Sep 20, 2012 #7
    voko, please check you private message inbox (incase you haven't already since yesterday).
  9. Sep 20, 2012 #8
    Also, it seems a scalar constant- the same from (1-34)- is missing from the r.h.s. of Goldstein's equation (1-35). Is this in fact the case?
    Last edited: Sep 20, 2012
  10. Sep 20, 2012 #9
    I do not have the text at hand now, but I could have a look tomorrow.
  11. Sep 21, 2012 #10
    No, I don't think so. 1-34 is a formula for ## \nabla V_{ij} ##, while 1-35 is a formula for something else, where ## \nabla V_{ij} ## is not present. The latter formula is derived without regard to 1-34, the derivation simply uses the fact that ## \int_{\mathbf{r}}^{\mathbf{R}} \nabla f(\mathbf{r})\cdot d\mathbf{r} = f(\mathbf{R}) - f(\mathbf{r}) ## for any (differentiable) ## f(\mathbf{r}) ##.
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