# Asymmetry in Stokes' theorem & Gauss' theorem

• I
Search "double cone".

You confusion arises because a simple closed curve bounds at most one* single continuous area whereas a closed curve that is not simple (i.e. crosses itself) may bound 2 or more disconnected areas.

Exactly the same is true for surfaces - a simple closed surface bounds at most one volume but a surface that crosses (i.e. intersects) itself may* bound 2 or more.

* note degenerate cases e.g. line on a Möbius strip, Klein bottle. Excercise - must a crossing curve enclose at least one area? What about a simple surface?
Thank you. How would you explain 1 surface enclosing only 1 volume in applying Gauss' theorem? How will it apply to a double cone?

pbuk
Gold Member
Thank you. How would you explain 1 surface enclosing only 1 volume in applying Gauss' theorem?
I wouldn't. The divergence theorem (which is how we now refer to the vector calculus generalisation of Gauss's law in electrostatics), is not restricted to simple closed surfaces.
How will it apply to a double cone?
Well it would have to be a double truncated cone to be closed. Why do you think there is a difficulty in applying Gauss's law here?

I wouldn't. The divergence theorem (which is how we now refer to the vector calculus generalisation of Gauss's law in electrostatics), is not restricted to simple closed surfaces.

Well it would have to be a double truncated cone to be closed. Why do you think there is a difficulty in applying Gauss's law here?
My difficulty is, for Stokes' theorem 1 curve can enclose multiple surfaces but in Gauss' 1 surface encloses 1 volume

pbuk
Gold Member
My difficulty is, for Stokes' theorem 1 curve can enclose multiple surfaces but in Gauss' 1 surface encloses 1 volume
You keep stating this but you are not responding to a number of different people pointing out that a single "volume" in what you call Gauss's law is not necessarily continuous.

PeroK
Homework Helper
Gold Member
You keep stating this but you are not responding to a number of different people pointing out that a single "volume" in what you call Gauss's law is not necessarily continuous.
Also, this is the third post on the subject. It was already answered in the threads mentioned in post #12 above. Here, for example, is a good answer to the question:

https://www.physicsforums.com/threa...d-by-the-same-curve-in-stokes-theorem.989709/

fresh_42
fresh_42
Mentor
Also, this is the third post on the subject. It was already answered in the threads mentioned in post #12 above. Here, for example, is a good answer to the question:

https://www.physicsforums.com/threa...d-by-the-same-curve-in-stokes-theorem.989709/
@feynman1 You apparently got your answer. If you still have problems, then we may have communication problems, which is not unlikely since we are restricted to verbal only communication.

My suggestion is: Quote the original theorem you have a question to. Famous theorems are sometimes differently phrased. It looks as if you have trouble to understand the conditions of Gauß law. So the more precise you ask, the better the answers will be.

PeroK