Why is \hat{a} \bullet \vec{ds} = ds? Explaining Stokes Theorem.

In summary, Stokes theorem is the generalized equation of Green's theorem and is used to calculate the flux of a substance through a 3-dimensional surface. To understand why \hat{a} \cdot \vec{ds} = ds, one must be familiar with inner products and the angle between parallel vectors. \hat{a} is the unit normal vector which has been normalized to a length of 1. If the angle between two vectors is 0, then \hat{a} \cdot \vec{ds} = ds is true. However, this condition is not always true. Stokes theorem is derived using this condition.
  • #1
jeff1evesque
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For stokes theorem, can someone tell me why [tex]\hat{a} \bullet \vec{ds} = ds[/tex]? My notes say it's because they are parallel, but I'm not sure what that means.

Also to get things clear, Stokes theorem is the generalized equation of Green's theorem. The purpose of Stokes theorem is to provide a means of calculating the "flux" [not the curl- which is the tendency for a point to rotate], or the amount of given substance moving through a surface [in our case a 3dimensional].THanks,
JL
 
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  • #2
Are you familiar with inner products. For vectors [itex]\vec{a},\vec{b}[/itex] the inner product, or dot product in this case, is given by [itex]\vec{a} \cdot \vec{b}=|\vec{a}|| \vec{b}|\cos\theta[/itex], with [itex]\theta[/itex] the angle between the two vectors. What is the angle between parallel vectors?
 
  • #3
Cyosis said:
Are you familiar with inner products. For vectors [itex]\vec{a},\vec{b}[/itex] the inner product, or dot product in this case, is given by [itex]\vec{a} \cdot \vec{b}=|\vec{a}|| \vec{b}|\cos\theta[/itex], with [itex]\theta[/itex] the angle between the two vectors. What is the angle between parallel vectors?

Yep, you can do a dot product between two vectors, or you can dot a vector to get the projection onto an axis.
 
  • #4
Then do you understand now why [itex]\hat{a} \cdot \vec{ds} = ds[/itex] when they are parallel?
 
  • #5
Cyosis said:
Then do you understand now why [itex]\hat{a} \cdot \vec{ds} = ds[/itex] when they are parallel?

No haha, that's my question, but I'll try to think about it more.
 
  • #6
It's a dot product. Write this dot product in the form I listed in post 2 then ask yourself what is the angle between two parallel vectors?
 
  • #7
Cyosis said:
Are you familiar with inner products. For vectors [itex]\vec{a},\vec{b}[/itex] the inner product, or dot product in this case, is given by [itex]\vec{a} \cdot \vec{b}=|\vec{a}|| \vec{b}|\cos\theta[/itex], with [itex]\theta[/itex] the angle between the two vectors. What is the angle between parallel vectors?

Ok, so it is:

[tex] \hat{a} _{n} \cdot \vec{ds}=| \hat{a} _n| |\vec{ds}|\cos \theta[/tex]
 
  • #8
Yep that is correct. Now for my second question, what's the value of [itex]\theta[/itex] if those vectors are parallel? Secondly what is the length of [itex]\hat{a}[/itex]? If you don't know try to explain to me what [itex]\hat{a}[/itex] is.
 
  • #9
Cyosis said:
Yep that is correct. Now for my second question, what's the value of [itex]\theta[/itex] if those vectors are parallel? Secondly what is the length of [itex]\hat{a}[/itex]? If you don't know try to explain to me what [itex]\hat{a}[/itex] is.

[itex]\hat{a}[/itex] is the unit normal vector, thus it has been "normalized" (to a length of 1). If we set [tex]\theta = 0[/tex] then we get [tex]\vec{ds} = |ds| = ds[/tex]
 
  • #10
But what if [tex]\theta[/tex] is not zero, then this is false. So it is not parallel in all conditions- this is part of the reason I am horrible with proofs.
 
  • #11
jeff1evesque said:
But what if [tex]\theta[/tex] is not zero, then this is false. So it is not parallel in all conditions- this is part of the reason I am horrible with proofs.

Ohh, never mind. What I needed was to show that particular condition (which you helped me discover) was true- not whether it was always true. By having that condition being true, I can use this in the derivation of Stokes theorem.Thanks,JL
 
  • #12
Cyosis said:
Yep that is correct. Now for my second question, what's the value of [itex]\theta[/itex] if those vectors are parallel? Secondly what is the length of [itex]\hat{a}[/itex]? If you don't know try to explain to me what [itex]\hat{a}[/itex] is.

This would have helped a while back ago, \vec{ds}, \hat{a} _n are by definition parallel.
 

FAQ: Why is \hat{a} \bullet \vec{ds} = ds? Explaining Stokes Theorem.

1. Why is the dot product of a unit normal vector and a small displacement vector equal to the magnitude of the displacement vector?

The dot product of two vectors represents the projection of one vector onto the other. In this case, the unit normal vector represents the direction perpendicular to the surface, while the small displacement vector represents a small change in position on the surface. Therefore, the dot product of these two vectors gives the magnitude of the displacement vector along the direction perpendicular to the surface.

2. How does the dot product relate to the surface area?

The dot product of the unit normal vector and small displacement vector is equal to the magnitude of the displacement vector multiplied by the cosine of the angle between the two vectors. This angle represents the slope of the surface at that point. Therefore, the dot product gives us the component of the displacement vector that is perpendicular to the surface, which is equivalent to the surface area.

3. What is the significance of using a unit normal vector?

A unit normal vector is a vector that has a magnitude of 1 and points in the direction perpendicular to the surface. Using this vector allows us to isolate the perpendicular component of the displacement vector, which is necessary for calculating the surface area. Additionally, using a unit vector ensures that the result is independent of the orientation of the surface.

4. How does Stokes Theorem relate to the fundamental theorem of calculus?

Stokes Theorem is a generalization of the fundamental theorem of calculus for higher dimensions. It relates the line integral of a vector field around a closed curve to the surface integral of the curl of the vector field over the enclosed surface. This allows us to solve line integrals by evaluating surface integrals, which can often be easier to calculate.

5. Can Stokes Theorem be applied to any surface and vector field?

Stokes Theorem can be applied to any smooth surface and vector field, as long as the surface is closed and the vector field is continuously differentiable. This means that the surface must have a well-defined tangent plane at every point and the vector field must have continuous partial derivatives. In practice, Stokes Theorem is most commonly used for planar surfaces and vector fields that are smooth and well-behaved.

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