Recent content by LAHLH

yep, already done that (I think it's now known as brightrecruits.com), not much in my field unfortunately-seems to be mostly condensed matter or more experimental stuff, as I say I believe hyperspace, academicjobsonline, spires-jobs to be the best places for theory postdocs I have found. Just...

I'm in the UK. Thanks for the APS career website, I will look there too, but previously this hadn't been mentioned to me. I have of course talked about this with supervisor and my primary applications are to places he has recommended, however at this stage I am trying to broaden my search also...

there is also spires-jobs. Does no one else have anything?

Where to people usually look for postdoc adverts? I am aware of hyperspace and academicjobsonline but are these the only places to check? (I am interested in theoretical ads, specifically gravity etc)

must be a silly mistake anyone know where?

Sorry that was a typo, should be L=T (from c=1). Then G=1, means M=L^3/T^2, and combinging with this L=T, gives M=L=T. But now if we want hbar=1, this implies M=T/L^2, which combing with what we learned from c=1,G=1 (such as L=T) seems to suggest M=1/L (as oppose to M=L). All my equals are...

If we set c=G=1, then c=1 leads to L=1/T , the G=1 means that M=L^3/T^2 and combining the two means that M=L=1/T so far so good. But say if I also want \hbar=1 this seems to imply M=T/L^2 but combining this with c=1 which gave the L=1/T now suggests that M=1/L=T . Which seems...

So I should have really been using \bar{R}=\omega^{-2}R+2(n-1)g^{\alpha\beta}\omega^{-3}(\nabla_{\alpha}\nabla_{\beta}\omega)+(n-1)(n-4)g^{\alpha\beta}\omega^{-4}(\nabla_{\alpha}\omega)(\nabla_{\beta}\omega) for the (+----..) conv. Is my conformal transformation for the box still valid? In...

I'm working from the book Birrell and Davies and they have the plus sign \Box \phi +\frac{1}{4}\frac{n-2}{n-1} R \phi , they also use (+----) convention however. Actually I now realize that I've been using the transformations for the Ricci etc from Carroll who uses the opposite sign convention...

I'm not sure I agree with that. I do accept that you could do the problem the way you posted with that identity, and will try to verify that when I get chance next, but I don't buy that you can't do it using the covariant derivs as long as you are careful. I'm starting in the conformal frame and...

The identity you posted for box phi you mean? If so yes I agree that that's correct. I started off doing the calculation using this identity too but haven't finished the algebra yet. I can't see however why doing it the way I initially set out too would not be an equivalent way however, unless I...

where did I make this mistake? I was tring to be very conscious of not doing that. omega and phi are scalars so the \nabla_{\nu} cov deriv acts as a partial deriv and this is how I generate the second equality in my second line of equations in my post prev to yours. Or are you thinking of...

\bar{\Box}\bar{\phi}=\omega^{-2}\Box\bar{\phi}+(n-2)g^{\alpha\beta}\omega^{-3}(\nabla_{\alpha}\omega)(\nabla_{\beta}\bar{\phi} ) What I did was, for example in the first term of this on RHS :\omega^{-2}\Box\bar{\phi}, I wrote...

I didn't, I just used \bar{\Box}\bar{\phi}=\omega^{-2}\Box\bar{\phi}+(n-2)g^{\alpha\beta}\omega^{-3}(\nabla_{\alpha}\omega)(\nabla_{\beta}\bar{\phi}) and...

I'm trying to prove the conformal invariance (under g_{\mu\nu}\to\omega^2 g_{\mu\nu}) of \bar{\Box}{\bar{\phi}}+\frac{1}{4}\frac{n-2}{n-1}\bar{R}\bar{\phi} I've found that this equation is invariant upto a quantity proportional to...