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

[jeff1evesque]

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

 

[Cyosis]

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?

 

[jeff1evesque]

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.

 

[Cyosis]

Then do you understand now why [itex]\hat{a} \cdot \vec{ds} = ds[/itex] when they are parallel?

 

[jeff1evesque]

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.

 

[Cyosis]

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?

 

[jeff1evesque]

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]

 

[Cyosis]

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.

 

[jeff1evesque]

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]

 

[jeff1evesque]

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.

 

[jeff1evesque]

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

 

[jeff1evesque]

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.