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davidge
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How would one prove the Stokes' theorem for general cases? Namely that $$ \int_{\partial M} W = \int_M \partial W$$ where ##M## is the manifold.
Ok. I have tried a derivation. Can this be considered valid?Orodruin said:The exterior derivative of the differential form ##\omega## is normally written ##d \omega##.
So are you saying that ##w## is not equal to an infinite sum of its differentials ##dw##Orodruin said:To be perfectly honest, it seems that you are not at all familiar with these concepts. Essentially nothing that you proposed above is a valid identity
davidge said:##M = \int dM## and (maybe) ##\int_{a}^{b}dM \approx dM## if ##|b-a| <<1##
davidge said:So, ##M## could be written as $$M = \sum_{n=1}^{\infty} n \times \lim_{(b\ -\ a) \longrightarrow 0} \int_{a}^{b}dM = \sum_{n=1}^{\infty} n \ dM$$
davidge said:So, $$\int_M dw = \int_{\sum_{n=1}^{\infty} n \ dM} dw \ \text{,}$$ which we can split in $$\int_{dM} dw + \int_{dM} dw + \int_{dM} dw \ + \ ...$$
davidge said:Now, deriving both sides of ##\int_{\partial M} w = \int_M dw## with respect to ##M## (and using the above), we get ##w = \sum_{n=1}^{\infty}n \ dw##, which I guess is correct.
davidge said:I know that this is not a derivation of the theorem, anyways maybe it is a proof that the theorem is correct.
Oh, I thought it was a small part of ##M##.Orodruin said:##\partial M## that appears in Stokes' theorem is the boundary of ##M##, not a small part of ##M##.
Orodruin said:This makes no sense whatsoever. You need to make up your mind whether ##M## is the manifold or whether it is a differential form. Regardless there is no integral over an interval from ##a## to ##b## and even in the simple case of a one-dimensional integral over a single variable ##M## that does not make sense. Also, an integral of a differential form is not a manifold nor a differential form, it is a number.
Yes, I know it. So a way of doing things right would be to map ##M## into an interval in ##\mathbb{R}^m## (##m## is the dimension of ##M##), through a function ##\phi##, so that we would change all the ##M## in the integrals by ##\phi (M)##? If so, would my indentities become valid?Orodruin said:You cannot differentiate with respect to a manifold. What do you think the result would be?
Stokes' theorem is a fundamental theorem in multivariable calculus that relates the surface integral of a vector field over a closed surface to the line integral of the same vector field over the boundary of the surface. It is named after Irish mathematician George Gabriel Stokes.
Stokes' theorem is significant because it provides a powerful tool for evaluating surface integrals, which can be difficult to compute directly. It also helps to establish a connection between differential forms and vector calculus, making it useful in various areas of physics and engineering.
Stokes' theorem can be derived from the divergence theorem, which relates the triple integral of a vector field over a solid region to the surface integral of the same vector field over the boundary of the region. By applying the divergence theorem to a small region around a point on the surface, we can derive the general form of Stokes' theorem.
Stokes' theorem holds for any smooth surface in three-dimensional space, as long as the surface is closed and oriented consistently with the orientation of the boundary curve. Additionally, the vector field must be continuously differentiable in the region bounded by the surface and its boundary curve.
Stokes' theorem has many practical applications in physics and engineering, such as in fluid mechanics, electromagnetism, and computer graphics. It is used to calculate fluid flow rates, electric and magnetic fields, and surface areas in 3D models, among other things. It is also used in the study of differential forms and their applications in geometry and topology.