- #1
Rene Meyer
Hello,
I hope you don't mind this elementary question, but I got stuck with
it this evening:
GSW state in their equ. (2.1.44) that a combined reparametrisation and
a Weyl scaling obeying
d^aA^b + d^bA^a = Gamma n^ab
don't change the choosen gauge h^ab=n^ab. Now I tried to proof this.
The metric varies according to
delta n^ab = Gamma n^ab
and this should be zero. As n^ab is definately not zero, Gamma = 0.
Then d^aA^b = - d^bA^a, which automatically implies eq. (2.1.19) to
vanish too. But then, in the coordinates sigma^+ and sigma^-,
d/d^sigma^+ A^+ = 0 = d/d^sigma^- A^-
Thus the statement that A^+=A^+(sigma^+) and similarily for A^- is
wrong, though I know is it right from other sources. So what is wrong
with my line of reasoning?
Hope someone can find some time and help me.
René.
--
René Meyer
Student of Physics & Mathematics
Zhejiang University, Hangzhou, China
I hope you don't mind this elementary question, but I got stuck with
it this evening:
GSW state in their equ. (2.1.44) that a combined reparametrisation and
a Weyl scaling obeying
d^aA^b + d^bA^a = Gamma n^ab
don't change the choosen gauge h^ab=n^ab. Now I tried to proof this.
The metric varies according to
delta n^ab = Gamma n^ab
and this should be zero. As n^ab is definately not zero, Gamma = 0.
Then d^aA^b = - d^bA^a, which automatically implies eq. (2.1.19) to
vanish too. But then, in the coordinates sigma^+ and sigma^-,
d/d^sigma^+ A^+ = 0 = d/d^sigma^- A^-
Thus the statement that A^+=A^+(sigma^+) and similarily for A^- is
wrong, though I know is it right from other sources. So what is wrong
with my line of reasoning?
Hope someone can find some time and help me.
René.
--
René Meyer
Student of Physics & Mathematics
Zhejiang University, Hangzhou, China