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Griffiths, introduction to E.M, Pr. 10.8 (Lorentz gauge)

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Confirm that the retarded potentials satisfy the Lorentz gauge condition.


    2. Relevant equations

    [itex]\vec{A}(\vec{r}, t) = \frac{\mu_{0}}{4\pi}\int\frac{\vec{J}(\vec{r'},t_{r})}{R}d\tau'[/itex]

    [itex]V(\vec{r}, t) = \frac{1}{4\pi\epsilon_{0}}\int\frac{\rho(\vec{r'},t_{r})}{R}d\tau'[/itex]

    Where:

    [itex] \vec{r} [/itex]: position of measurement

    [itex] \vec{r'} [/itex]: integration variable (running on charge and current densities)

    [itex] R = |\vec{R}| = |\vec{r} - \vec{r'}| [/itex]

    [itex] t_{r} [/itex] is the retarded time: [itex] t_{r} = t - \frac{R}{c} [/itex]

    [itex] \vec{J}(\vec{r'},t_{r}) [/itex] is the current density (estimated at the point [itex] \vec{r'}[/itex] at the retarted time), and [itex] \rho(\vec{r'},t_{r}) [/itex] is the charge density.


    Lorentz gauge means:

    [itex] \nabla \cdot \vec{A} = -\epsilon_{0}\mu_{0} \frac{\partial V}{\partial t} [/itex]

    3. The attempt at a solution

    Now, Griffiths gives a rather lengthy hint containing different formulas that one could apply in order to prove this. But I found his method cumbersome and could not understand why it's necessary (he uses relations between primed and unprimed operators and eventually gets there). For me, my "straight-forward" method is simpler, but for some reason it doesn't work. I'd really appreciate your insights!


    [itex] \nabla \cdot \vec{A} = \frac{\mu_{0}}{4\pi}\int\nabla \cdot (\frac{\vec{J}(\vec{r'},t_{r})}{R}) d\tau' = \frac{\mu_{0}}{4\pi}\int [\frac{1}{R}\nabla \cdot \vec{J} - \frac{1}{R^{2}} \hat{R} \cdot \vec{J}]d\tau'[/itex]

    Meanwhile:

    [itex] -\epsilon_{0}\mu_{0} \frac{\partial V}{\partial t} = - \frac{\mu_{0}}{4\pi}\int \frac{\partial}{\partial t} (\frac{\rho(\vec{r'},t_{r})}{R})d\tau' = - \frac{\mu_{0}}{4\pi}\int \frac{1}{R}\frac{\partial \rho}{\partial t} d\tau' = \frac{\mu_{0}}{4\pi}\int \frac{1}{R}\nabla \cdot \vec{J} d\tau' [/itex]

    Where in the last inequality I've used the continuity equation: [itex]\frac{\partial \rho}{\partial t} = -\nabla \cdot \vec{J}[/itex]

    So somehow the next equation should hold:

    [itex] \frac{\mu_{0}}{4\pi}\int \frac{1}{R}\nabla \cdot \vec{J} d\tau' = \frac{\mu_{0}}{4\pi}\int [\frac{1}{R}\nabla \cdot \vec{J} - \frac{1}{R^{2}} \hat{R} \cdot \vec{J}]d\tau'[/itex]

    However, unless the integral containing [itex]\frac{1}{R^{2}} \hat{R} \cdot \vec{J} [/itex] vanishes, this is obviously wrong. And I can't seem to prove it vanishes.
    Am I doing something wrong? Is any step here false? THANK YOU!

    Tomer
     
  2. jcsd
  3. Oct 21, 2012 #2

    gabbagabbahey

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    The continuity equation tells you [itex]\frac{\partial \rho ( \mathbf{r}, t)}{\partial t} = -\mathbf{\nabla} \cdot \mathbf{J} ( \mathbf{r}, t)[/itex] (note the arguments of the charge and current densities!). You seem to be claiming that this means [itex]\frac{\partial \rho ( \mathbf{r}', t_r)}{\partial t} = -\mathbf{\nabla} \cdot \mathbf{J} ( \mathbf{r}', t_r)[/itex]....is that really the case?

    There's a reason that Dr. Griffiths' hint makes you pay special attention to the primed and unprimed coordinates :wink:.
     
    Last edited: Oct 21, 2012
  4. Oct 21, 2012 #3
    Thanks very much for the reply!

    Isn't it true that [itex] \frac{\partial}{\partial t} \rho (\vec{r'}, t_{r}) = -\nabla \cdot \vec{J}(\vec{r'}, t_{r}) [/itex] ? I mean, what is the difference between [itex] \vec{r} [/itex] and [itex] t [/itex] to [itex] \vec{r'} [/itex] and [itex] t_{r} [/itex] ? These are just another coordinate set, why should the continuity equation not hold there? *confusing*

    Is there a way to handle this? or do I really have to follow Griffith's method?
     
    Last edited: Oct 21, 2012
  5. Oct 21, 2012 #4

    gabbagabbahey

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    No. If you pay attention to the functional dependence of [itex]\rho[/itex] and [itex]\mathbf{J}[/itex], you should see that the continuity equation [itex]\frac{\partial \rho ( \mathbf{r}, t)}{\partial t} = -\mathbf{\nabla} \cdot \mathbf{J} ( \mathbf{r}, t)[/itex] only tells you that [itex]\frac{\partial \rho ( \mathbf{r}', t_r)}{\partial t_r} = -\mathbf{\nabla}' \cdot \mathbf{J} ( \mathbf{r}', t_r)[/itex]. That is, if you are going to make the substitutions [itex]\mathbf{r} \to \mathbf{r}'[/itex] & [itex]t \to t_r[/itex] in the continuity equation, it will affect the derivative operators as well as the charge & current densities.

    I would say Griffiths' method is the quickest, but yours should work too if you can relate [itex]\frac{\partial \rho ( \mathbf{r}', t_r)}{\partial t}[/itex] to [itex] \frac{\partial \rho ( \mathbf{r}', t_r)}{\partial t_r}[/itex] (should be fairly easy!) and [itex]\mathbf{\nabla} \cdot \mathbf{J} ( \mathbf{r'}, t_r)[/itex] to [itex]\mathbf{\nabla}' \cdot \mathbf{J} ( \mathbf{r'}, t_r)[/itex] (more difficult).
     
  6. Oct 21, 2012 #5
    Of course! Gotcha!
    Thanks a lot for the help!
     
  7. Oct 22, 2012 #6
    Hey again gab,
    sorry, but apparently there's still something I'm missing.
    I'm not sure how to get:

    [itex] \nabla' \cdot \vec{J}(\vec{r'},t_{r}) = -\frac{\partial \rho(\vec{r'},t_{r})}{\partial t_{r}} -\frac{1}{c}\dot{\vec{J}}(\vec{r'}, t_{r}) \cdot \nabla'(R) [/itex]

    This is what Griffiths suggests we should prove.

    You mentioned that the next equation holds:

    [itex] \nabla' \cdot \vec{J}(\vec{r'},t_{r}) = -\frac{\partial \rho(\vec{r'},t_{r})}{\partial t_{r}} [/itex]

    which seems to contradict what Griffiths wanted us to show.

    the first equality (the one Griffiths mentions) seems to make sense, although writing the "proof" looks ridiculous to me. When deriving the current density with respect to the primed coordinated, I understand there are both an explicit part and implicit, hidden in the retarded time. But it looks really weird when writing it. For example:

    [itex] \frac{\partial J_{x}(\vec{r'},t_{r})}{\partial x'} = \frac{\partial J_{x}(\vec{r'},t_{r})}{\partial x'} + \frac{\partial J_{x}(\vec{r'},t_{r})}{\partial t_{r}} \frac{\partial t_{r}}{\partial x'} = ... [/itex]

    The first part (after doing so for all components) will result, after use of the continuity equation in the form you mentioned (with the primes) in the derivative of rho, whereas the rest results in the second term Griffith mentioned.

    However:
    1. It looks really weird that the partial derivative with respect to x' appears twice in the equation above, but I don't know how else to express the fact we have to perform these two derivatives...
    2. What Griffiths wants to prove seems to contradict the "primed" continuity equation, although he uses it!

    :-\
     
    Last edited: Oct 22, 2012
  8. Oct 22, 2012 #7

    gabbagabbahey

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    Dr Griffiths is correct, and my earlier post was not.

    In the continuity equation, [itex]\frac{\partial \rho ( \mathbf{r}, t)}{\partial t} = -\mathbf{\nabla} \cdot \mathbf{J} ( \mathbf{r}, t)[/itex], it is clear that the divergence should only affect the first argument of [itex]\mathbf{J}[/itex]. When you make the substitution [itex]t \to t_r[/itex], there is an implicit coordinate dependence in [itex]t_r[/itex], so in order for this to be a valid substiution, you must expressly hold [itex]t_r[/itex] constant when taking the divergence. Thus the continuity equation becomes:

    [tex]\frac{\partial \rho ( \mathbf{r}', t_r)}{\partial t_r} = -\left. \mathbf{\nabla}' \cdot \mathbf{J} ( \mathbf{r}', t_r) \right|_{\text{constant } t_r}[/tex]

    Whereas the total (primed) divergence of [itex]\mathbf{J} ( \mathbf{r}', t_r)[/itex] is

    [tex]\mathbf{\nabla}' \cdot \mathbf{J} ( \mathbf{r}', t_r) = \left. \partial_i' J_i \right|_{\text{constant } t_r} + \frac{\partial J_i}{\partial t_r} \partial_i' t_r = \left. \mathbf{\nabla}' \cdot \mathbf{J} ( \mathbf{r}', t_r) \right|_{\text{constant } t_r} + \frac{\partial J_i}{\partial t_r} \partial_i' t_r [/tex]

    If this bothers you, remember that [itex]\frac{d}{dt} f(q_1(t), q_2(t), \ldots q_n(t)) = \frac{\partial f}{\partial q_1}\frac{\partial q_1}{\partial t} +\frac{\partial f}{\partial q_2}\frac{\partial q_2}{\partial t} + \ldots + \frac{\partial f}{\partial q_n}\frac{\partial q_n}{\partial t} = \frac{\partial f}{\partial q_i}\frac{\partial q_i}{\partial t}[/itex], where it is understood that when taking the partial derivative of [itex]f[/itex] with respect to [itex]q_1[/itex], for example, all the other [itex]q_i[/itex] are held fixed.
     
  9. Oct 22, 2012 #8

    TSny

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    It does look a little odd because the term on the left side looks like the first term on the right side even thought they represent different quantities.

    The term on the left is the "full" rate of change of J with respect to x' including the explicit dependence of J on x' as well as the implicit dependence on x' through the retarded time ##t_r##.

    The first term on the right represents the rate of change of J' due to varying x' in the explicit dependence but not varying x' in the retarded time. This is the quantity that appears in the continuity equation and therefore is related to the time rate of change of the charge density.

    I don't see any inconsistency with what Griffiths does, but the notation is confusing. I generally find that working with expressions that depend on retarded time gives me a headache.
     
  10. Oct 23, 2012 #9
    Thanks a lot, both of you, for the time you give me.

    Usually the "full" rate of change would be written as: [itex] \frac{d}{dx'}(Something)[/itex], Like gab mentioned in his last equation.
    The weird thing is that we do the same thing with the partial derivative!
    You know the feeling that you lose confidence in things you were highly confident in? I think that's why this question drives me crazy!
    On the one hand, [itex] \frac{\partial f(x,y(x),z)}{\partial x}[/itex] should mean the derivative of f(x,y(x),z) according to x (first variable!) alone, while the others are regarded as constants. So "y" stays fixed, despite its dependence on x. On the other hand, if I have the function: [itex] f(x,y) = x * y + z [/itex], and I use [itex] y(x) = x^{2}[/itex], I obviously have to calculate:

    [itex] \frac{\partial f(x,y(x),z)}{\partial x} = \frac{\partial f(x,y(x),z)}{\partial x} + \frac{\partial f(x,y(x),z)}{\partial y(x)}\frac{\partial y(x)}{\partial x} = y(x) + x * y'(x) = x^{2} + x* 2x = 3 x^{2}[/itex], where at the first derivative I treated y(x) as constant.
    I mean, when setting y(x) inside the function, you don't *have* y anymore to keep constant! Then you do really only derive with respect to x! So the difference is weather we actually set y to be y(x) or not.

    So what's up with the same symbol standing for two different things? I could write the partial derivative with this vertical line implying y is constant, like you did, gab (I hate this symbol ever since taking Thermodynamics!), but I really don't understand why this isn't partial derivative per se! From wiki:

    "In mathematics, a partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant."

    According to Wiki, what I calculated above isn't the partial derivative at all, but the full derivative. However, it *is* the partial derivative, cause if I wanna perform the partial derivative of [itex] f(x,y(x), z) = x * x^{2} + z = x^{3} + z [/itex], I'll get the answer above, where I had to derive with respect to y(x) as well!

    Ahh.
     
    Last edited: Oct 23, 2012
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