feynman1 said:Stokes theorem relates a closed line integral to surface integrals on any arbitrary surface bounded by the same curve. Gauss theorem relates a closed surface integral to the volume integral within a unique volume bounded by the same surface. What causes this asymmetry in these 2 theorems, in the language of calculus?
Right. Why is there such a difference? Mathematically or philosphically.PeroK said:A closed surface defines a unique volume. A closed curve defines a unique area in 2D, but not a unique 3D volume.
There's no difference except the dimension. Why is a cube different from a square?feynman1 said:Right. Why is there such a difference? Mathematically or philosphically.
Great. Could you please give an example of bridges for a spherical surface or anything simple?fresh_42 said:There is no asymmetry. A closed 2D surface can also enclose multiple 3D volumes if you allow "bridges", what you obviously did in the 2D case. The intersection point of a figure 8 can likewise be used to connect surfaces in 3D.
Then in this example of yours, the 2 balls form 1 surface, then what is/are the enclosed 3d volume?fresh_42 said:Just take two balls and connect them by a straight line. Then you have a closed surface. But even without, the two spheres alone are one closed surface. You confuse closed with connected.
I agree with you , this asymmetry exists indeed, and it must be something related to the topology of ##\mathbb{R^3}## and ##\mathbb{R^n}## in general. But as I was pretty weak in topology during my undergraduate studies, I can't seem to give a deeper explanation on why this asymmetry is present.feynman1 said:A 2d closed surface can only enclose a unique 3d volume while a 1d close curve can enclose multiple 2d surfaces. Why’s that asymmetry?
The reason I ask this is due to applications of Guass' and Stokes theorems.Delta2 said:I agree with you , this asymmetry exists indeed, and it must be something related to the topology of ##\mathbb{R^3}## and ##\mathbb{R^n}## in general. But as I was pretty weak in topology during my undergraduate studies, I can't seem to give a deeper explanation on why this asymmetry is present.
feynman1 said:A 2d closed surface can only enclose a unique 3d volume while a 1d close curve can enclose multiple 2d surfaces. Why’s that asymmetry?
Yes in ##R^4## there isn't such asymmetry but i believe there we would have another asymmetry present (something like a 3D volume can enclose only one 4D volume but a 3D surface can enclose many 4D volumes).PeroK said:There is no asymmetry:
2D surface - unique 3D volume (in 3D space) - multiple 3D volumes (in 4D space)
1D curve - unique 2D surface (in 2D) - multiple 2D surfaces (in 3D)
It's the same thing one dimension higher.
I doubt about this. What if the 1D curve isn't planar, then there's no 2D surface in 2D in your language.feynman1 said:1D curve - unique 2D surface (in 2D) - multiple 2D surfaces (in 3D)
The original question is here:Delta2 said:Yes in ##R^4## there isn't such asymmetry but i believe there we would have another asymmetry present (something like a 3D volume can enclose only one 4D volume but a 3D surface can enclose many 4D volumes).
I doubt about this. What if the 1D curve isn't planar, then there's no 2D surface in 2D in your language.feynman1 said:1D curve - unique 2D surface (in 2D) - multiple 2D surfaces (in 3D)
The spheres are the 2D surfaces, the balls the 3D volume.feynman1 said:Then in this example of yours, the 2 balls form 1 surface, then what is/are the enclosed 3d volume?
Yes, so there's no more than 1 volume enclosed?fresh_42 said:The spheres are the 2D surfaces, the balls the 3D volume.
Why that? There are two spheres, hence they enclose two volumes / balls.feynman1 said:Yes, so there's no more than 1 volume enclosed?
But when applying Gauss' theorem, one has to enclose both spheres. It's not a choice to enclose which of the 2.fresh_42 said:Why that? There are two spheres, hence they enclose two volumes / balls.
I think you still confuse closed and connected. But maybe I don't understand your point. A picture would be helpful.feynman1 said:But when applying Gauss' theorem, one has to enclose both spheres. It's not a choice to enclose which of the 2.
Could you post a pic of your idea of 1 2D surface enclosing more than 1 choice of 3D volume?fresh_42 said:I think you still confuse closed and connected. But maybe I don't understand your point. A picture would be helpful.
No, since I don't know what choice means in this context.feynman1 said:Could you post a pic of your idea of 1 2D surface enclosing more than 1 choice of 3D volume?
Search "double cone".feynman1 said:Could you post a pic of your idea of 1 2D surface enclosing more than 1 choice of 3D volume?
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 said: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?
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.feynman1 said:Thank you. How would you explain 1 surface enclosing only 1 volume in applying Gauss' theorem?
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?feynman1 said:How will it apply to a double cone?
My difficulty is, for Stokes' theorem 1 curve can enclose multiple surfaces but in Gauss' 1 surface encloses 1 volumepbuk said: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?
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.feynman1 said:My difficulty is, for Stokes' theorem 1 curve can enclose multiple surfaces but in Gauss' 1 surface encloses 1 volume
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:pbuk said: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.
@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.PeroK said: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/
