Recent content by chartery

@pines-demon, thanks for reply. If I plug in e+iωt , I get separation constant as -ħω. Does that mean it is a "negative energy" solution?

Hi Apologies for formatting, I can't get PF's new Tex to work for me. The (?most) general solution of the free Schroedinger eq. is e^{±i(kx-ωt)} , which implies e^{+iωt} should solve the time separated ODE: dψ/dt = -iωψ, but instead it satisfies dψ/dt = +iωψ which is not obviously (to me) the...

In many researches, for instance on Stack Exchange (https://physics.stackexchange.com) and even on Physics Forums (here!), I have never seen Carroll's book (Carroll, S. Spacetime and Geometry) needing a bibliographic reference, and so didn't imagine it would require inspired guesswork. It is...

Sorry. It didn't occur to me that such a widely recognised (respected?) source would have such a basic concept being contentious, so assumed it was everyone's approach (or at least knowledge), and wouldn't need explanation or specific referencing. The relevant passages are before equation 2.40...

My learning source; I thought widely so. Yes, I guess that was the implicit question in OP. I presumed that (as in Carroll) the format ##\text{d}r## indicated, or at least implied, that the latter version obtained in Visser's paper. It seemed sensible to understand any difference between...

Hmm. I had ##\partial_r## as basis for contravariant vectors and ##\text{d}r## (not ##dr##) as basis for covariants. And line element as ##ds^2 = g_{\mu\nu}\text{d}x^{\mu}\text{d}x^{\nu}##. Carroll seems to go to a lot of trouble to distinguish gradients from differentials. If not so, then my...

My inexpert studies had left me with the impression one should be careful about distinction between a basis vector and its associated differential (or partial). I wasn't certain about all the circumstances in which that care was required.

Also thanks. Reducing the dimension of a metric (by fixing an index) as a route to geometric insight got a bit lost for me trying to follow the abstruse coordinate manipulations!

Thanks. I was just uneasy about the simplicity of that deduction in circumstances where the metric was becoming singular and applying to a basis ##\text{d}r## (as opposed to the differential ##dr##).

My understanding was that the line element is equivalent to the metric, the (curvature) singularity of the metric is at ##\left( r=0\text{, } \theta=\frac{\pi}{2}\right)##, and is circular. The problem is with the ##\text{d}r## or cross terms in these coordinates. He is using Boyer-Lindquist...

From paper 'A brief introduction to Kerr spacetime' ( https://arxiv.org/pdf/0706.0622 ) setting m->0 in the line element in Kerr coordinates gives, equation 7 : $$ \text{d}s_0^{2} = -\left( \text{d}u + a\sin^{2}\theta \text{d}\phi \right)^{2}+2\left( \text{d}u + a\sin^{2}\theta \text{d}\phi...

Durr... Got fixated on second term of ## \eta^{\rho\lambda} {\partial_{ \mu}}\epsilon h_{\nu\lambda} - \epsilon h^{\rho\lambda} {\partial_{ \mu}}\epsilon h_{\nu\lambda}## (just in case someone of similar density looking up). Many thanks.

Sorry for gap. I can see Vanhees understands, though it seems to me if ##\partial h## is order ##1/\epsilon## then ##\epsilon h^{\rho\lambda} {\partial_{ \mu}}\epsilon h_{\nu\lambda}## is only order ##\epsilon## but needs to be order ##\epsilon^2## to be ignored in OP equation?

Sorry @vanhees71 I can't get the multiple quote insert to work!Yes, my problem was being sure that ##h^{\rho\lambda}{\partial_{ \mu}}h_{\nu\lambda}## terms were order ##h^2## It makes sense that ##\epsilon h^{\rho\lambda} {\partial_{ \mu}}\epsilon h_{\nu\lambda}## would be order ##\epsilon^2##...

Sorry, you've lost me. Were you referring to ##g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}##? My problem was how to know that the partial derivative (i.e. variation) of a small item was necessarily also small. if ##\frac{1}{1+x}## is how I should think of ##\partial_{\mu}## here, I'm afraid I need...