Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lanczos "Variational Principles of Mechanics" Problem

  1. May 8, 2015 #1
    I have started going through Lanczos' book "The Variational Principles of Mechanics", and have got stuck on the first problem he gives at 16.9 to do with holonomic constraints

    Investigate the integrability of the following differential relation:
    x dz + (y2 - x2 - z) dx + (z - y2 - xy) dy = 0 (16.9)

    He then states that the condition is indeed holonomic and can be replaced by the finite relation:
    z = x2 - xy + y2 (16.10)

    I just can't get anywhere with this problem using the holonomic condition in (16.7). I can't even see how to go backwards from 16.10 to 16.9 using 16.2. Has anyone worked through it? Given 16.10 is symmetric in x and y whereas 16.9 isn't, I wonder if it's a misprint?
     
  2. jcsd
  3. May 8, 2015 #2
    Hi,
    z=x2-xy+y2[ (0)

    You'll get

    x dz + (xy-2x2) dx + (x2-2xy)dy = 0 (1)

    From (0):
    xy=x2+y2-z
    as well as:
    x2=z+xy-y2

    Substitute xy in the first term of (1), substitute x2 in the second term, q.e.d.

    Lanczos is pretty reliable :wink:

    Best wishes,
    Jens
     
  4. May 8, 2015 #3

    Philip Wood

    User Avatar
    Gold Member

    Start by getting 16.9 into the same form as 16.4, by dividing through by x and taking terms in dx and dy over to the other side. Take q1, q2 and q3 as x, y and z. It should then be clear what the coefficients B1 and B2 are as functions of q1 and q2. Do the partial differentiations required for 16.7. It does work, but there are lots of opportunities for slips!
     
    Last edited: May 8, 2015
  5. May 11, 2015 #4
    Thanks guys.

    I had used Philip Wood's procedure to get the result that 16.7 becomes 1=(-xy+y2-z)/x2. The next step seems to be to solve for z, whereas I was expecting just identity at this stage. Validity of that seems questionable to me...

    Anyway it seems that Lanczos is not pedagogical and I'm not sure I will get anything from doing the exercises!
     
  6. May 11, 2015 #5

    Philip Wood

    User Avatar
    Gold Member

    My solution is on the bottom of the attached thumbnail. You don't have to look at it if you don't want to!
    He is a great enthusiast, though, and really savours the beauty of classical mechanics. It's hard to resist succumbing. He's semi-pedagogical in that he does include exercises. Not enough to make for an easy learning experience, though.
     

    Attached Files:

  7. May 11, 2015 #6
    Thanks for that solution and confirms what I did above as well. As I say though I expected to come up with the LHS and RHS of 16.7 being an identity, as opposed to finding a solution for z that makes it so. I guess Lanczos is demonstrating his point of the procedure where he says "However it may happen that q3 does not drop out of the resulting equation...". It's all a bit hairy, but I think I have enough now to be satisfied and continue with the book:)
     
  8. May 12, 2015 #7

    Philip Wood

    User Avatar
    Gold Member

    I don't quite get why you say that the relationship between x, y and z is an equation rather than an identity. I thought the distinction was that an equation held only for particular values of the variables. This is not the case here, surely. I expect I'm missing something.

    I
     
    Last edited: May 12, 2015
  9. May 12, 2015 #8
    Lanczos procedure is that if our differential is integrable then 16.7 must be an identity

    Both you and I arrive at the equivalent of (using your working just before you rearrange):
    -xy-z+y2=-x2

    This isn't an identity unless we assume the "answer", the equation that Lanczos has given as the finite relation. But this is his answer to the problem and I don't think we should assume this, I thought it must be first shown that 16.7 is an identity for the differential to show that it is integrable, and not solve for z to make it an identity.
     
  10. May 13, 2015 #9

    Philip Wood

    User Avatar
    Gold Member

    I now think that we need to take z = –xy + y2 + x2 (the result of applying the exactness criterion (16.7)) as a trial relationship. In sloppy notation this differentiates to dz = (2x – y)dx + (2yx) dy. At first sight, the coefficients of dx and dy don't look much like the original P and Q. But if we use the trial relationship again, for example substituting y = (y2 + x2 - z)/x into (2xy) we do get back to the relationship (16.9) given by Lanczos for our examination. This finally nails it: 16.9 is equivalent to z = –xy + y2 + x2. I still wonder whether I've missed something; this procedure seems very cumbersome. Is there an easy way of seeing, as soon as z = –xy + y2 + x2 emerged by applying 16.7, that it had to be equivalent to 16.9 ?
     
    Last edited: May 13, 2015
  11. May 13, 2015 #10
    Yes I think that's what Lanczos might be demonstrating the need to do, following his discussion after 16.7. It is cumbersome, and I don't know of the importance of the need to know holonomic systems to this degree to try further to understand it
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Lanczos "Variational Principles of Mechanics" Problem
  1. The Pendulum Problem (Replies: 1)

  2. Optics Problem (Replies: 2)

  3. Kepler Problem (Replies: 5)

  4. Imaging problem (Replies: 1)

Loading...