Singularities when applying Stokes' theorem

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The discussion focuses on applying Stokes' theorem to a vector field with a singularity at the origin. The integral over the cone surface and its boundary leads to a calculated answer of A = -2π, which is confirmed as correct. The main concern raised is whether ignoring the singularity is justified, given that the line and surface integrals do not intersect the origin. It is noted that as long as the closed curve and the surface integral do not encompass the singularity, it can be disregarded in the calculations. The conversation emphasizes the importance of considering singularities in the context of Stokes' theorem and related theorems.
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Homework Statement
I am arguing with myself if the way i solved this problem is right
Relevant Equations
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It is more or less a generic problem of stokes theorem:
##\int_{\gamma} F dr##, where ##F = (-y/(x²+y²) + z,x/(x²+y²),ln(2+z^10))## and gamma is the intersection of ##z=y^2, x^2 + y^2 = 9## oriented in such way that its projection in xy is traveled clockwise.

So, i decided to apply stokes theorem to the cone surface, considering its boundary to be the one with z=0 and the other being the intersection said above. Call the answer A

##A + \int_{\alpha} (-y/(x^2+y^2), x/(x^2+y^2), ln(2)) * dC = \int \int grad F * ds##

Alpha is the circle at z=0, x²+y²=9, and the right side of the equation will give zero if you calculate it. Doing all the integrals it end as:

##A + 2\pi = 0 => A = -2\pi##

In fact the answer is right. But that is not the problem, i want to know about the singularities!
I mean i just ignored it, following the reasoning that
1: the grad will be calculate on the cylinder surface, which does not intersect the origin
2: the line integral of the circle at z=0 will also not intersect the origin

So, taking in account that the integral line and integral surface "does not know" about the singularity at the origin, i just ignored it. Is it right? (If it were necessary to calculate div F over the volume, in this case, i would need to pay attention at the singularity since the volume spread over there)
 
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You are right to think twice about singularities when talking about Stokes theorem. I can say that the divergence theorem has difficulties when the input vector field has singularity. The classic example is ##\frac{1}{|\vec{x}|}## at the origin. If the divergence theorem has these problems, then I'm inclined to say that Stokes theorem has them too, since they're both versions of the generalized Stokes theorem for the exterior derivative. I don't have a counterexample on hand, though.

Hopefully, someone who's studied their cohomology more recently me than me can expand on this.
 
You talk about the gradient of F when you actually mean the curl of F, right?
And yes as long as the closed curve in which you calculate the line integral, and the surface through which you calculate the curl surface integral do not contain the singularity, you don't need to worry about it.
 
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