1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Two problems while reading Feynman lectures (vector field))

  1. Jul 14, 2011 #1
    Question 1:

    solved!



    Question 2:

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




    Many thanks.
     

    Attached Files:

    • 1.jpg
      1.jpg
      File size:
      24.8 KB
      Views:
      78
    • 2.jpg
      2.jpg
      File size:
      28.9 KB
      Views:
      80
    Last edited: Jul 14, 2011
  2. jcsd
  3. Jul 14, 2011 #2

    EWH

    User Avatar

    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.
     
  4. Jul 15, 2011 #3
    In fact I just can't understand why we have

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

    Why the same form ...
     

    Attached Files:

    • 3.jpg
      3.jpg
      File size:
      34.3 KB
      Views:
      63
  5. Jul 15, 2011 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

  6. Jul 15, 2011 #5
  7. Jul 15, 2011 #6
    if i am getting your question right ... you are asking why A X (AT) = (A X A)T = 0
    is that right?
     
  8. Jul 15, 2011 #7

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  9. Jul 16, 2011 #8
    Yes! And I'm still confused now!!
     
  10. Jul 16, 2011 #9

    BruceW

    User Avatar
    Homework Helper

    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.
     
  11. Jul 16, 2011 #10
    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 [itex]\vec{A} X \vec{B} \ = \ AB \ sin\theta \ \hat{n}[/itex]

    where [itex]\theta[/itex] is and b/w [itex]\vec{A} \ \ and \ \ \vec{B}[/itex]

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

    thus (A X A)T = 0
     
  12. Jul 16, 2011 #11
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Two problems while reading Feynman lectures (vector field))
  1. Feynman lectures Problem (Replies: 13)

Loading...