Question: Does Epsilon = i A^+/p^+ for Zwiebach Page 178?

  • Thread starter Thread starter ehrenfest
  • Start date Start date
ehrenfest
Messages
2,001
Reaction score
1

Homework Statement


Below equation 10.77, Zwiebach claims that the gauge epsilon = i A^+/p^+ makes the vector potential component A^+ vanish.

However, I thought epsilon was supposed to be infinitesimal and a constant, both of which are violated by setting epsilon equal to the expression?


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
I thought epsilon was supposed to be infinitesimal and a constant, both of which are violated by setting epsilon equal to the expression?
Actually, just above equation (10.68) on page 177, he refers you back to Section 3.3 where on page 43, just below equation (3.19) he writes that epsilon is an arbitrary function of the coordinates.
 
I see. Thanks.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Back
Top