Naked Science Forum
On the Lighter Side => That CAN'T be true! => Topic started by: alright1234 on 05/05/2019 23:45:55

Stokes' theorem is equating a line integral with a surface integral. The result of the line integral is a length which is not equivalent to the the surface area of the surface integral; units do not seem to match
m =/ m^2.............................................................equ 1

Three points.
Mathematical theorems are true.
You have forgotten to provide any reference, so it's impossible to know what you are talking about
It really would be better if you tried to learn a bit about a subject before getting all shouty about it.

This would be a good place to start: https://en.wikipedia.org/wiki/Stokes'_theorem
It does seem somewhat counterintuitive that the path integral of a closed loop would give any info about what is inside... but recall that it only applies in specific cases, ie. when looking at "smooth" and "orientable" surfaces (manifolds), which means that it is already limited to cases in which information about one specific point also has some bearing on its immediate neighbors (and their neighbors, and their neighbors, etc.)thus information is somewhat delocalized.
The theorem proves that this type of delocalization ultimately allows limited (no pun intended) information about an ndimensional manifold to be deduced by information gained by analyzing the n1 dimensional boundary of that manifold.
It's been quite a while since I studied or used multivariate calculus, but I recall that some of the derivations of the theorem were pretty straight forward (at least in specific casesI wouldn't even know where to start with the general form!)

Three points.
Mathematical theorems are true.
Until they are proven false.
The result of the Stoke's line integral is a length which is not equivalent to the the surface area of the surface integral.

Oh dear. Any integral has a higher dimensionality than the integrand. That was lesson 1 at the Open University access course for those who had not studied physical sciences. It's probably on the shelves at Harvard, covered in dust and unread by philosophers..

The result of the Stoke's line integral is a length which is not equivalent to the the surface area of the surface integral.
The integral of a function over a line is NOT the length of the line (HINT: it must relate to the function as well). Stokes' theorem does NOT equate a surface area with a lengthwhich is what you appear to be claiming.
Please seek a better understanding of this topic before knocking it.

Until they are proven false.
No.
Once again, you are showing that you don't know what you are talking about.
An idea in maths is only called a theorem after it has been proved to be true.
And that's irrevocable.
2+2 won't stop being 4

Is a length equal to an area?

Is a length equal to an area?
No, but, as has been pointed out, an integral along a path is not a length.

Is a length equal to an area?
No, but, as has been pointed out, an integral along a path is not a length.
Your statement is patently incorrect.

Is a length equal to an area?
No, but, as has been pointed out, an integral along a path is not a length.
Your statement is patently incorrect.
Among the things I can calculate as a path integral is the energy stored in a compressed gas.
Are you saying that the energy in a compressed gas is a line?
Again, it really would be better if you learned more before getting shouty

Mathematically, Is a length equal to an area?

Mathematically, Is a length equal to an area?
No.
Integrals are often areas.

Stokes' theorem is equating a line integral with a surface integral. The result of the line integral is a length which is not equivalent to the the surface area of the surface integral which proves Stokes' theorem is mathematically invalid.

Stokes' theorem is equating a line integral with a surface integral. The result of the line integral is a length which is not equivalent to the the surface area of the surface integral which proves Stokes' theorem is mathematically invalid.
Sorry. This only proves that you don't know what you're talking about. Multiple people have provided explanations or links to detailed explanations in an effort to educate you, while you have only repeated the same ridiculous straw man argument over and over. Please try to understand, or at least realize that you don't understand. This is the last post I will make in this thread.

Please indicate in your own words how Stokes' theorem justifies equating a line integral with a surface integral. The links you prove are incomprehensible to me. Please use simple words and examples that I can understand. A couple of example would be good since if it is possible It would be monumental proof.

Please indicate in your own words how Stokes' theorem justifies equating a line integral with a surface integral.
You need to show that it does equate them.

∫ F ⋅dr =∬ curl F ⋅ dS
http://tutorial.math.lamar.edu/Classes/CalcIII/StokesTheorem.aspx

Curl is a derivative operator.
So the units work fine.

Curl is a derivative operator.
So the units work fine.
The surface integral has precedence before the derivative of the curl.

It hardly matters what order you do it in
Take F
Differentiate it using curl
integrate it
That takes you back to something with the same units as F
Then integrate it again.
That gets you something in units of "integrated F"
On the other side of the equation, you integrate F
That gets you something in units of "integrated F"
I'm not qualified to say if they are the same thing (except to say of course they are it's a mathematical theorem).
But they clearly have the same units.
You should stop preaching about this.

The definition of madness is repeating the same thing over and over again and expecting a different result. No matter how many times you argue that everyone else is wrong it won't make you right. Just picking random topics and arguing against them doesn't make you smart. If you want to waste your time that way go right ahead. But you may well soon be arguing with yourself.

The area integral has precedence, The result of an integral is a length, area or volume. I my opinion any other application is false since Eudoxus (b. 390 BC) discovered that the area of a circle is proportional to the radius squared, and, Archimedes (b. 212 BC) derived the area equation for a circle using pi which was the most advance mathematics of the time. Determine the exact area of an equation was monumental achievement that modern physicists are attempting to destroy.

The result of an integral is a length, area or volume. I my opinion any other application is false since Eudoxus (b. 390 BC)
Sorry, but science and maths have moved on since then. Important you keep up to date and don’t post false information.

Determine the exact area of an equation was monumental achievement that modern physicists are attempting to destroy.
Can you clarify how scientists, who use calculus every day are undermining calculus?
I my opinion any other application is false
Get a better opinion.

The result of an integral is a length, area or volume.
Obviously not. The time integral of acceleration is velocity, as every driver knows. The time integral of replication rate is population. The triple spatial integral of obtuseness is a waste of space.