Recent content by Irrational

you have no idea how thick i feel right now. thanks for filling in the gap.

edit... now I'm finished stating the question.

Hi, I'm trying to work through something and it should be quite simple but somehow I've gotten a bit confused. I've worked through the Euler Lagrange equations for the lagrangian: \begin{align*} \mathcal{L}_{0} &= -\frac{1}{4}(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu}) \\ &=...

Finally managed to work through it and get the required ansswer. Was simply an exercise in algebra in the end. That and the most basic of trigonometric identities, \cos^{2}{\gamma} + \sin^{2}{\gamma} = 1.

Apologies. that should have been eliminate \gamma, not \theta. my notes were dirrerent to the actual paper i was citing the equations from. I'll include the background as well. I've worked through all of this and understand it all but can't do the basic elimination of \beta_{2} and \gamma...

I'm doing a night course in General Relativity and we're currently finishing off Special Relativity... We're working mainly off of D'Inverno. We've just covered the relativistic doppler effect and some associated things like aberration. When it came to talking about the transverse doppler...

http://en.wikipedia.org/wiki/Dirichlet_boundary_condition" "When imposed on an ordinary or a partial differential equation, it specifies the values a solution needs to take on the boundary of the domain" http://en.wikipedia.org/wiki/Neumann_boundary_condition" "When imposed on an...

ah. you said 'properly signed kronecker delta'... i presume you mean the minkoswki metric? :) yes, you would get minkowski space then alright.

perhaps I'm wrong but i'd imagine that if we include time, then kronecker delta = (1,1,1,1) mikowski metric = (-1,1,1,1) so no, you don't end up with minkowski space. if you take time out of the equaton, then yes, the two are the same but this is not really any help is it?

well if y = \ln x, then i guess that dy/dx = 1/x so evaluting the limit of this would give 0... which makes sense if you interpret what \ln x looks like graphically. as expected, the rate of change of the function would slow down to a point where it's basically not changing as x goes to infinity.

from your wiki link, the speed at which a convergent sequence approaches its limit is called the rate of convergence... ln(x) isn't a convergent series, it's a function. and as mentioned before, the series expansion for ln(x) only converges for a small range of x.

as you sure you're not trying to find the limit of ln(x) as x approaces infinity? the limit as x goes to infinity of ln(x) is infinity. what are you hoping ln(x) converges on? if you're talking about the rate of convergence of the taylor series expansion of ln(x), the series only converges in...

the minkowski metric is (-1,1,1,1) (with c taken to be 1) in general, the line element is d\tau^2 = -g_{\mu\nu}dx^{\mu}dx^{\nu} so in minkowski space, the line element becomes d\tau^2 = -dt^2 + dx^2 + dy^2 + dz^2 (which is what i think you are referring to in the equation above involving r)...

?? zeros in a digital world are kind of essential when the digital world is comprised mainly of ones and zeros... and i wouldn't go bringing god into anything. praise science.

what are you hoping your function converges on? i would have imagined you are trying to find the rate of convergence of a sequence approximating the function.