Undergrad What is the role of the divergence theorem in deriving local laws in physics?

[dRic2]

As far as I can tell the divergence theorem might be one of the most used theorems in physics. I have found it in electrodynamics, fluid mechanics, reactor theory, just to name a few fields... it's literally everywhere. Usually the divergence theorem is used to change a law from integral form to differential (local) form.

Take for example Gauss's law in integral form:
$$\int_{\partial \Omega} \mathbf E \cdot d \mathbf a = \frac {Q_{int}} {\epsilon}$$
and in local form:
$$\nabla \cdot \mathbf E = \frac {\rho} {\epsilon}$$
If I remember correctly the argument goes like this:
1) use divergence theorem
2) write the RHS like a volume integral of some density function (here it is used the charge density but you can do it in different areas of physics and the same argument is used)
3) say something like this: "since this must be true for any control volume the integral can be dropped" (here may lie my doubt)

Now let's stick with Gauss' law to make the example easier. It happens that Gauss' law in integral form is not valid for changing electric fields while the differential form is still valid (locally) even for changing electric fields. This troubles me. Similar arguments can be presented for every other formula I ran into that was derived with this logic.

Since the differential form was derived starting from the integral form there must be something wrong in the proof OR there should be another way to derive the differential form which do not use divergence theorem and doesn't start from an integral law.

 

[kuruman]

$$\int_S \vec E \cdot \hat n~dA=\int_V \vec \nabla \cdot \vec E~dV$$I prefer to think that the divergence theorem is a relation between a volume integral and a surface integral. It basically says that if you walk around a closed surface and measure the flux through that surface, you don't need to look inside to figure out the sources (or sinks) that generate this flux. Conversely if you are inside a closed surface and thus you know the flux sources, you can predict the flux that will come out of the surface.

 

[dRic2]

I know that. Sorry but I don't see the connection with my question

 

[kuruman]

I probably misunderstood your question. Can you explain why you say that Gauss's Law in integral form is invalid when the electric field is changing?

 

[Dale]

It happens that Gauss's law in integral form is not valid for changing electric fields while the differential form is still valid (locally) even for changing electric fields. This troubles me.

Let’s make sure we are talking about the same thing. This law is what I know as Gauss’ law in integral form:

$$\epsilon_0 \int \int_S \mathbf E \cdot d\mathbf S = \int \int \int_V \rho \: dV$$

It is valid for changing E fields.

Edit: never mind I see that you posted Gauss’ law in integral form, which is equivalent to what I wrote. You are incorrect in your statement. It is, in fact, valid for changing E fields.

 

[dRic2]

Probably I am wrong, but consider the following example.

Imagine you have some charges and consider a very big surface that contains them . Now at $t=t_0$ let a new charge show up (imagine you bring it from very very far with a very very high speed that you could almost say it popped out of nowhere). Now the charge density inside the surface is different, but the electric field on the surface is the same as before because it takes a while to the information to arrive at the "boundary of the surface", i.e. where the flux is evaluated. So at the instant $t = t_0$ we have $\rho(t_0) \neq \rho_{old}$ but on the other hand $\int_{\partial \Omega} \mathbf E(t_0) \cdot d \mathbf a = \int_{\partial \Omega} \mathbf E_{old} \cdot d \mathbf a$

 

[Dale]

imagine you bring it from very very far with a very very high speed

As the charge approaches the boundary the net flux from this new charge is zero. At the moment the charge crosses the boundary the net flux increases. It doesn’t matter how fast the charge is moving.

Now the charge density inside the surface is different, but the electric field on the surface is the as before because it takes a while to the information to arrive at the "boundary of the sorface",

When the charge crosses the boundary then it is at the boundary and it does not take any time for the information to arrive since it is happening at the boundary.

 

[dRic2]

I don't see the solution to the problem. The moment the charge enters the surface the density inside the surface increases. So at $t=t_0$ you have a new values for the RHS of the equation. But the left hand side is the same because it takes a while to the information to reach the whole surface. Am I missing something obvious ?

 

[Dale]

But the left hand side is the same because it takes a while to the information to reach the whole surface. Am I missing something obvious ?

Here is your mistake. Why would it take a while to reach the surface? It crosses the surface at the surface. The distance is zero.

Edit: Maybe the problem is that you think the information has to reach the “whole surface”. It just has to reach any part of the surface.

 

[dRic2]

Ok I understand. In my mind I imagined that the charge popped out of nowhere like it was created, but that is unphysical.

To be honest I read that the integral form of Gauss' law is not right on the internet, but this argument convinced me so much that I though it had to be true. Thanks God I asked immediately rather than stick with this false interpretation.

Thank you very much

 

[Dale]

In my mind I imagined that the charge popped out of nowhere like it was created, but that is unphysical

Sounds like you got it now! This is exactly right, and because of this fact you can actually derive a continuity equation from Maxwell’s equations. Gauss’ law and the others require charge continuity.

 

[dRic2]

Yes, because in the continuity equation there is no "source" therm: that means that charge can't be created from nothing. It is the simples case of balance equation.