Continuity equation derivation in Griffiths - why partial derivative?

In summary: Basically, if you are considering a time-dependent volume, you need to take into account the change in the volume with respect to time and the flow of the quantity through the changing volume boundaries. The term \vec{v} represents the velocity of the boundaries, which will appear in the continuity equation if the volume is time-dependent.So in summary, the reason why the derivative in the continuity equation is a partial derivative is because we are considering a time-dependent volume, where the density is a function of both time and space. To properly account for the change in the volume and the flow of the quantity through the changing boundaries, we must use a partial derivative.
  • #1
HJ Farnsworth
128
1
Greetings,

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

[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{d}{dt}[/itex][itex]\rho[/itex]d[itex]\tau[/itex]?

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
 
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  • #2
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.
 
  • #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!
 
  • #4
Yup ... you got it!
 
  • #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.

Chet
 
  • #6
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.
 
  • #7
You can find out more details about what vanhees71 is talking about here by googling Reynolds' transport theorem.
 

1. What is the purpose of using a partial derivative in the continuity equation derivation in Griffiths?

The partial derivative is used to account for changes in a specific variable while holding other variables constant. In the continuity equation, it is necessary to consider the change in charge density with respect to time while keeping the other variables, such as position, constant.

2. Why is the continuity equation important in physics?

The continuity equation is important because it represents the fundamental principle of conservation of charge. It states that the amount of charge entering a given region must equal the amount of charge leaving that region, which is essential in understanding the behavior of electric and magnetic fields.

3. Can the continuity equation be used in other areas of science?

Yes, the continuity equation can be applied in various fields of science, including fluid dynamics, thermodynamics, and quantum mechanics. In these contexts, it represents the conservation of mass, energy, and probability, respectively.

4. How is the continuity equation derived in Griffiths?

In Griffiths, the continuity equation is derived by considering a small volume element and examining the change in charge density within that volume due to the flow of charge. This leads to the general equation of continuity, which can then be applied to specific scenarios, such as the flow of current in a wire.

5. Is the continuity equation always valid?

The continuity equation is a fundamental principle in physics and is always valid in any system where charge is conserved. However, it may not always be applicable in certain scenarios, such as when dealing with non-conservative fields or in extreme conditions, such as near black holes.

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