Two problems while reading Feynman lectures (vector field))

[Proust]

Question 1:

solved!



Question 2:

Why it's zero? I think we cannot get zero unless it's an exact differential form?




Many thanks.

 

[EWH]

Well, I don't know how to explain it better than Feynman, but the curl of the gradient of a scalar function is always zero.

 

[Proust]

Well, I don't know how to explain it better than Feynman, but the curl of the gradient of a scalar function is always zero.

In fact I just can't understand why we have

A X (AT) = (A X A) T

Why the same form ...

 

[Dick]

Your second thumbnail in the first post explains it. Mixed partial derivatives are equal. See for example http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives

 

[Proust]

Your second thumbnail in the first post explains it. Mixed partial derivatives are equal. See for example http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives

From which we can conclude A X (AT) = (A X A) T ?

 

[cupid.callin]

if i am getting your question right ... you are asking why A X (AT) = (A X A)T = 0
is that right?

 

[SammyS]

I think he's asking why ᐁ X (ᐁT) = (ᐁ X ᐁ)T is of the same form as A X (AT) = (A X A)T.

 

[Proust]

if i am getting your question right ... you are asking why A X (AT) = (A X A)T = 0
is that right?

Yes! And I'm still confused now!

 

[BruceW]

These kinds of relations are easy when you use the levi-civita tensor instead of vector form.
Unfortunately, I guess tensors shouldn't be included in introductory physics?

Instead, you could write out all the components explicitly, and see that the equality holds.

 

[cupid.callin]

These kinds of relations are easy when you use the levi-civita tensor instead of vector form.
Unfortunately, I guess tensors shouldn't be included in introductory physics?

Instead, you could write out all the components explicitly, and see that the equality holds.

All i can understand and tell you is that, if A is a vector and T is some scalar constant then its kind of a basic rule of vectors that A X (AT) = (A X A)T because no matter f you multiply the scalar before of after solving cross product ... answer comes same.

and also $\vec{A} X \vec{B} \ = \ AB \ sin\theta \ \hat{n}$

where $\theta$ is and b/w $\vec{A} \ \ and \ \ \vec{B}$

so angle b/w $\vec{A} \ \ and \ \ \vec{A}$ is 0 and sin(0) = 0

thus (A X A)T = 0

 

[tharinduuuu]

i think post#10 explains it very well. just want to add that Feynman was probably trying to make you think of the del operator as just another vector and the scalar field(T) as just a scalar by showing the similarity between the two expressions. The fact that one of the expressions evaluate to zero should then help you guess that maybe the other one is zero too which in turn will help you in proving that it indeed is.And it will also help you remember and have an intuitive understanding of identities like this without having to memorize everything.