1. Not finding help here? Sign up for a free 30min 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!

Zwiebach, page 154

  1. Sep 23, 2007 #1
    1. The problem statement, all variables and given/known data
    Page 154, equation (9.19) is
    [tex]n\cdot\mathcal{P}^{\sigma} = 0[/tex]



    2. Relevant equations
    The text says:
    Indeed, [itex]n\cdot\mathcal{P}^{\sigma}[/itex] is required to vanish at the string endpoints to guarantee the conservation of
    [tex]n\cdot p = \int d\sigma n\cdot \mathcal{P}^{\sigma}[/tex]
    (consider equation (8.38) dotted with n.)
    Here is equation (8.38) on page 138
    [tex]\frac{dp_{\mu}}{d\tau} = \mathcal{P}_{\mu}^{\sigma}(\sigma = 0) - \mathcal{P}_{\mu}^{\sigma}(\sigma = \sigma_1)[/tex]

    3. The attempt at a solution
    So conservation of [itex]n\cdot p[/itex] gives (using (8.38))
    [tex]0 = \frac{d(n\cdot p)}{d\tau} = n\cdot \mathcal{P}^{\sigma}(\sigma = 0) - n\cdot\mathcal{P}^{\sigma}(\sigma = \sigma_1)[/tex]
    But we already knew that from equation (9.18):
    [tex]\frac{\partial}{\partial \sigma}n\cdot\mathcal{P}^{\sigma} = 0[/tex]

    It seems that in fact, Zwiebach means to apply equation (6.56) on page 103 which is the free endpoint condition and says (slightly edited)
    [tex]\mathcal{P}_{\mu}^{\sigma}(0) = \mathcal{P}_{\mu}^{\sigma}(\sigma_1) = 0[/tex]
    Am I correct? Actually, I doubt it because I think he means to imply that it is true regardless of the boundary conditions. Besides, if that was what he meant, then he could have applied it directly to eqn (9.19) without reference to (8.38).
     
    Last edited: Sep 24, 2007
  2. jcsd
  3. Oct 6, 2007 #2
    Maybe you have a different edition, but that sentence is not in my book.
     
  4. Oct 6, 2007 #3
    My copy has no information on which edition or printing, but I knew it was coming out before it did and bought it as soon as it became available, so I figure it must be first edition, first printing.

    That line "(consider equation (8.38) dotted with n.)", is just above equation (9.19). If you have a later printing, or later edition, then perhaps it was removed. If so, it would be an improvement and perhaps someone else already pointed it out to Professor Zwiebach. But even without that hint, my question still remains. Since by equation (6.56) we already know that [itex]\mathcal{P}[/itex] is 0 at the endpoints of all open strings, why say that [itex]n\cdot\mathcal{P}[/itex] needs to be 0 at the endpoints to guarantee conservation of np?
     
    Last edited: Oct 6, 2007
  5. Oct 6, 2007 #4

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    My edition has the sentence you mentioned. I must have a first edition.

    You are right that condition 9.18 is stronger than 8.38 so there is no new information from 8.38! If the sentence was removed it would confirm that this was a mistake in the first place.

    My *guess* is that Zwieback looked at 8.38 which says (in my edition):
    [itex] 0 = - {\cal P}_\mu^{\sigma} |_0^{\sigma 1} [/itex]
    and that he did not pay attention to the fact that it was evaluated between the two boundaries! So he read it as saying that the momentum vanishes at the boundaries instead of saying that the momentum is the same at the extremities of the string.

    So I agree with you that he probably meant 6.56 instead.

    No, the point is not to show that np is conserved (which was already shown). The point is to prove 9.19, that [itex] {\cal P}^{\sigma} \cdot n =0 [/itex] at all points along the string. Eq. 9.18 shows that n dot P is the same everywhere along the string and 6.56 shows that it is zero at the endpoints. Putting the two together implies that is is zero everywhere along the string. That's the way I interpret it.

    patrick
     
    Last edited: Oct 6, 2007
  6. Oct 6, 2007 #5
    Thank nrqed and Ehrenfest for taking a look at this. I included my own quote because I think you misinterpreted my question to be a statement. Zwiebach stated that [itex]n\cdot\mathcal{P}[/itex] needs to be 0 at the endpoints to guarantee conservation of np, not me. I am asking why he said it. If, as you agree, he was applying (6.56) then in the subsequent printing he should have deleted more than he apparently did. Ehrenfest, can you please quote the paragraph above equation (9.19) from your copy of the book
     
  7. Oct 6, 2007 #6
    I found the sentence "(consider equation (8.38) dotted with n^mu)" on page 151 in the paragraph below equation 9.4

    It seemed like it was around 9.19 in your book. It makes sense to me there and I think his reference to it is correct in the paragraph above 9.19 . See the attachment.
     

    Attached Files:

  8. Oct 7, 2007 #7

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member


    I stand for what I wrote earlier.

    Here is what Zwiebach's logic was (and where he made a mistake).

    His Logic in our printing of the book:


    a) I have shown from 9.18 that the value of (n P^sigma) is sigma independent

    b) therefore, if I can show that (n P^sigma) is zero anywhere along the string, it will be zero everyhwere

    c) the conservation of (b p) implied that (n P^sigma) is zero at the endpoints (MISTAKE IN HIS LOGIC HERE!)

    d) therefore, since (n P^sigma) is zero at the endpoints, by 9.18 it's zero everywhere



    The mistake is this: conservation of (np) required that (n P^sigma) is the same at the two endpoints, NOT that it is zero at both endpoints.

    Basically, Zwieback was looking for an argument showing that at least somewhere (n P^sigma) is zero.

    Does that make sense?

    Thanks to erhenfest for the scanned page, it looks as if there was a major rewriting of this paragraph.
     
  9. Oct 7, 2007 #8
    Thanks Ehrenest. Can you show us the relevant text at equation (9.4) too.
    Is there an edition number and/or printing number on the copyright page? A printing number would be a sequence of numbers with possibly the first few missing. Did he mention my name in the acknowledgments?
     
  10. Oct 8, 2007 #9
    Sorry it is a little blurred near the middle. There is only one edition. I am not sure what the printing number is:

    2003068835 is on some random number on the copyright page maybe that is it

    I did not see any mention of jimmy snyder in the acknowledgements.
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Zwiebach, page 154
  1. Zwiebach page 234 (Replies: 1)

  2. Zwiebach page 258 (Replies: 0)

  3. Zwiebach page 292 (Replies: 3)

  4. Zwiebach page 299 (Replies: 1)

  5. Zwiebach page 287 (Replies: 1)

Loading...