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The continuity equation applies to more than just mass and charge. It applies to any conserved property. This is the form of the conservation law derived from Noether's theorem. IMO The article should be changed somewhat to reflect this. 63.205.41.128 03:23, 31 Jan 2004 (UTC)

${\displaystyle {\frac {\partial \varphi }{\partial t}}+\nabla \cdot f=s}$

should use the substantial derivative:${\displaystyle {\frac {D\varphi }{Dt}}+\nabla \cdot f=s}$

No! It should be the partial derivative, as on the Navier Stokes page and as in the derivation of the Maxwell case. I have changed this back. Paul Matthews (talk) 17:52, 1 February 2008 (UTC)

INTEGRAL'S FORM

May your describe the integral's form?*

Derivation

When the article talks about Ampère's law as one of Maxwell's equations, it should be noted that the formula given given includes Maxwell's correction. —Preceding unsigned comment added by Iain marcuson (talk • contribs) 19:31, 21 October 2007 (UTC)

Yeah I thought that Maxwell's correction was introduced into Ampere's Law specifically to satisfy the continuity equation. Which would mean that the continuity equation cannot be derived from Maxwell's equations at all, only it is consistent with them.

I changed "Ampere's law" to the more specific "Ampere's law with Maxwell correction". The term "Ampere's law" is vague, it can refer to either with or without the correction. Ampere's law with Maxwell correction is one of the Maxwell's equations, and the continuity equation for charge can be derived from Maxwell's equations, as proven in the article. :-) --Steve (talk) 21:23, 12 November 2010 (UTC)

q and A

The revisions to the main statement of the equation don't make sense, or at least I don't understand them.

• "q = quantity associated with φ" -- what does that mean? For example, if φ is mass density, then q is...what??
• "${\displaystyle A,\mathbf {\hat {n}} \,\!}$" -- A is an area and n is its unit normal. But area of what? I guess a surface? Why is there a surface? Who put a surface there? Is it a real surface or an imaginary surface? Is it a closed surface or an open surface? Is it flat or curved?
• ${\displaystyle \varphi ={\frac {\partial q}{\partial V}}\,\!}$ -- There seems to be a very basic misunderstanding of calculus here. I guess V is the volume of some imaginary blob in space. When you increase the size of the blob, q increases or decreases, I guess. You can calculate the differential rate of change of q and of V, and divide one by the other, a derivative. But there are many ways of increasing the size of the blob: You can pull out a lobe, you can expand it out from one center or another. All will generally lead to different relative rates of change of q and V. Therefore the derivative ${\displaystyle {\frac {\partial q}{\partial V}}\,\!}$ does not seem to be mathematically meaningful.
• ${\displaystyle \mathbf {f} =\mathbf {\hat {n}} {\frac {\partial ^{2}q}{\partial A\partial t}}\,\!}$ -- according to this equation, the flux points in the direction ${\displaystyle \mathbf {\hat {n}} }$. In other words, the surface is always exactly normal to the local flux. I was assuming until now that this was an arbitrary imaginary surface that I can define however I want, but I guess not. It is a very specific surface that everywhere points normal to the flux. Hold on though, this surface may not be closed, so it may not have a volume V. What happens in this case? Then there is the other problem of what it means to differentiate with respect to A. Again, I don't think it can be mathematically defined.

Anyway, my confusion piles up. My best guess is, this is an attempt to state the integral form of the continuity equation, and not just the differential form. Is that right? That's a fine and reasonable thing to do...provided, of course, that it is done correctly and clearly! :-) I'm happy to do so myself. But as is, I suggest restoring the definition to the previous version. Thanks in advance! Looking forward to understanding better what's going on here. :-) --Steve (talk) 18:12, 13 July 2011 (UTC)

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