alejandrito20 Posted September 22, 2011 Share Posted September 22, 2011 (edited) i need to evaluate the follows integral: [math]\oint dy (A' e^{A})' [/math] where ' is derivation respect [math]y[/math] and [math]A(y)=A(y+2\pi)[/math] and too [math]A(y)'=A(y+2\pi)'[/math] where [math]y \in (-\pi,\pi)[/math] ia a angular coordinate , and [math]A' [/math] is discontinuous in [math]-\pi,0,\pi[/math] the result is zero, but i think that is NOT CORRECT to say: [math]\oint dy (A' e^{A})' = A' (\pi)e^{A(\pi)}- A' (-\pi)e^{A(-\pi)} =0 [/math], since [math]A' [/math] is discontinuous in -pi and pi Edited September 22, 2011 by alejandrito20 Link to comment Share on other sites More sharing options...
timo Posted September 22, 2011 Share Posted September 22, 2011 (edited) I think you meant that A' is discontinuous in -pi and pi, not A. Why not simply integrate from 1 to (1+2pi), and not have your problem? EDIT: Oh, I overlooked something: If A' is discontinuous: How can (A' exp(A))' exist? Edited September 22, 2011 by timo Link to comment Share on other sites More sharing options...
alejandrito20 Posted September 22, 2011 Author Share Posted September 22, 2011 (edited) I think you meant that A' is discontinuous in -pi and pi, not A. Why not simply integrate from 1 to (1+2pi), and not have your problem? yes , is discontinuous in A' wy not simply integrate from 1 to (1+2pi), and not have your problem? your says [math]\int_1^{1+2\pi} dy (A' exp(A))' [/math]????? EDIT: Oh, I overlooked something: If A' is discontinuous: How can (A' exp(A))' exist? (A' exp(A))' there is Not exist in -pi,0,pi Edited September 22, 2011 by alejandrito20 Link to comment Share on other sites More sharing options...
DrRocket Posted September 23, 2011 Share Posted September 23, 2011 i need to evaluate the follows integral: [math]\oint dy (A' e^{A})' [/math] where ' is derivation respect [math]y[/math] and [math]A(y)=A(y+2\pi)[/math] and too [math]A(y)'=A(y+2\pi)'[/math] where [math]y \in (-\pi,\pi)[/math] ia a angular coordinate , and [math]A' [/math] is discontinuous in [math]-\pi,0,\pi[/math] the result is zero, but i think that is NOT CORRECT to say: [math]\oint dy (A' e^{A})' = A' (\pi)e^{A(\pi)}- A' (-\pi)e^{A(-\pi)} =0 [/math], since [math]A' [/math] is discontinuous in -pi and pi The integral of the derivative of a periodic function over a period will always be zero. But as Timo observes, you have imposed additional conditions that make the function non-differentiable, in contrtadiction to your problem statement. You have something messed up in the problem statement. Link to comment Share on other sites More sharing options...
alejandrito20 Posted September 23, 2011 Author Share Posted September 23, 2011 The integral of the derivative of a periodic function over a period will always be zero. But as Timo observes, you have imposed additional conditions that make the function non-differentiable, in contrtadiction to your problem statement. You have something messed up in the problem statement. in the text (http://arxiv.org/PS_cache/hep-th/pdf/0011/0011225v2.pdf eq2.13) says: "in a compact internal space without boundary, the integral vanishes" the inicial space is [math] z \in -\infty, \infty[/math], but [math] z = r \phi [/math] , whit r radio konstant Link to comment Share on other sites More sharing options...
DrRocket Posted September 23, 2011 Share Posted September 23, 2011 in the text (http://arxiv.org/PS_...1/0011225v2.pdf eq2.13) says: "in a compact internal space without boundary, the integral vanishes" the inicial space is [math] z \in -\infty, \infty[/math], but [math] z = r \phi [/math] , whit r radio konstant What you are faced with is an application of (the general differential goemoetry version of) Stokes theorem. This is sraightforward, but you have missed the point that your problem statement is self-contradictory. It is also clear that you are in way over yoir head. You need to master a great deal of more basic mathematics and physics before you attempt advanced M theory. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now