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Continuity equation derivation in Griffiths - why partial derivative?

  1. Sep 17, 2013 #1

    In Griffiths E&M, 3rd. Ed., on page 214, the following is part of the derivation of the continuity equation (the same derivation is shown on the Wikipedia article for the current density, under the continuity equation section: http://en.wikipedia.org/wiki/Current_density)

    [itex]\int[/itex][itex]\nabla[/itex][itex]\cdot[/itex]Jd[itex]\tau[/itex]=-[itex]\frac{d}{dt}[/itex][itex]\int[/itex][itex]\rho[/itex]d[itex]\tau[/itex]=-[itex]\int[/itex][itex]\frac{\partial}{\partial t}[/itex][itex]\rho[/itex]d[itex]\tau[/itex],

    where the integral is over the volume. Why isn't the relation


    Ie., in commutating the integral and the derivative, why is it switched to a partial derivative ∂/∂t rather than keeping the total derivative d/dt?

    Furthermore, how would the [itex]\frac{\partial x}{\partial t}[/itex], [itex]\frac{\partial y}{\partial t}[/itex], and [itex]\frac{\partial z}{\partial t}[/itex] terms in the total derivative [itex]\frac{d\rho}{dt}[/itex]=[itex]\frac{\partial \rho}{\partial t}[/itex]+[itex]\frac{\partial \rho}{\partial x}[/itex][itex]\frac{\partial x}{\partial t}[/itex]+[itex]\frac{\partial \rho}{\partial y}[/itex][itex]\frac{\partial y}{\partial t}[/itex]+[itex]\frac{\partial \rho}{\partial z}[/itex][itex]\frac{\partial z}{\partial t}[/itex] be interpreted?

    I think the answer to my first question might be that those terms are 0, but it's hard to think about whether that's true since I am having trouble interpreting those terms in this equation.

    Anyway, thanks for any help that you can give.

    -HJ Farnsworth
  2. jcsd
  3. Sep 17, 2013 #2


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    The total derivative was for the integral of the density; once you have integrated over the parameter, that parameter is absorbed into the integrated function.

    But when you pass the derivative through the integral, it becomes a partial derivative because you must not alter the functional relationship with the integration parameter.

    Also see http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign

    Note that the application of the generalized chain rule for total or partial derivatives is almost never clear from the notation: you have to know from context if all of the variables are allowed to vary, or if only one is allowed to vary. But in the case of differentiating an integral you already know that the integration parameters cannot be allowed to vary - otherwise you will mess up the integration.
  4. Sep 17, 2013 #3
    Thanks for the quick reply, UltrafastPED.

    I think I understand - basically, it can be thought of in terms of free and bound variables. Even though ρ is a function of x, y, z, and t, the definite integral of ρ over a given volume is not dependent on any of the position coordinates - at that point, it is just a constant with respect to x, y, and z, though it could still be a function of t. Therefore, in taking the derivative of the volume integral of ρ with respect to t, the expression [itex]\frac{d}{dt}[/itex]Q(x,y,z,t) is misleading; what we are really doing is taking [itex]\frac{d}{dt}[/itex]Q(t). In commutating the derivative and the integral, we must take this fact into account, so the total derivative becomes a partial derivative.

    Inidentally, the link you sent me quickly led to the article on the Leibniz integral rule, http://en.wikipedia.org/wiki/Leibniz_integral_rule, which had a very easy-to-follow proof for the 2-variable case in the Proofs section.

    Does this sound correct?

    Thanks very much again!
  5. Sep 17, 2013 #4


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    Yup .... you got it!
  6. Sep 17, 2013 #5
    The d/dt in this equation is not a material time derivative. It is just what we would regard as an ordinary time derivative. It is calculating the rate of change of mass within a stationary control volume. So you take the mass within the fixed control volume at time t + Δt, subtract the mass within the control volume at time t, and divide by Δt. The second equality is equivalent to this. However, within the control volume, the density is a function of spatial position, so you have to use the partial derivative with respect to time at each location.

  7. Sep 17, 2013 #6


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    There's only one caveat left. The continuity equation is the local equation
    [tex]\partial_t \rho+\vec{\nabla} \cdot \vec{j}=0.[/tex]
    Integrating this over a volume [itex]V[/itex] with boundary [itex]\partial V[/itex] (oriented in the usual way with the area normal vectors pointing outside of the volume)
    [tex]\int_V \mathrm{d^3} \vec{x} \partial_t \rho =-\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{j},[/tex]
    using Stokes integral theorem.

    This is equivalent to the time derivative of the charge inside the volume if and only if this volume is stationary, i.e., independent of time, because then and only then you can write
    [tex]\int_V \mathrm{d^3} \vec{x} \partial_t \rho=\frac{\mathrm{d}}{\mathrm{d} t} \int_V \mathrm{d^3} \vec{x} \rho.[/tex]

    If the volume is time-dependent there is an additional term cf. the Leibniz rule
    [tex]\frac{\mathrm{d}}{\mathrm{d} t} \int_V \mathrm{d}^3 \vec{x} \rho = \int_V \mathrm{d}^3 \vec{x} \partial_t \rho + \int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \rho \vec{v},[/tex]
    where [itex]\vec{v}=\vec{v}(t,\vec{x})[/itex] is the velocity of the boundary of the time-dependent volume.
  8. Sep 17, 2013 #7
    You can find out more details about what vanhees71 is talking about here by googling Reynolds' transport theorem.
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