Recent content by tetris11

@HoplessAstro - Yup of the 80 percent of my class who did Bsc's, over 1/2 of them went to banking (check UCL' alumni page). In the masters years the rest of us pretty much panicked with the best of us going on to do PhD's at Oxbridge and the other just calling it a day and looking into other...

Cheers man, that actually makes complete sense - but just for the record: gij gij = δii = n, where n is number of dimensions? I'm just wondering how you knew it was four without knowing how dimensions it was. Tensors aren't all 4-d, right?

Well, g^{i j} u_i u_j = 1 g^{i j} g_{i j} = ?? uh... g? or 0? Might need to help me out here, maths isn't my first language...

Okay so I have: Eqn1) Tij=\rhouiuj-phij = \rhouiuj-p(gij-uiuj) Where Tij is the energy-momentum tensor, being approximated as a fluid with \rho as the energy density and p as the pressure in the medium. My problem: Eqn2) Trace(T) = Tii = gijTij = \rho-3p My attempt: Tr(T) = Tii...

Homework Statement Hi there, really easy question but I can't get a straight answer online. What's \intx ef(x) dx The Attempt at a Solution I get (ef(x) x)/f'(x) - \int (ef(x))/f'(x) dx And when I try to do parts on the last component it will proably just go one forever. Help?

Hey there, I'm having a real hard time understanding exactly how to manipulate tensors (let alone know what they actually are). I learn brilliantly from example and repetition (the understanding comes later) but the internet and my lecture notes seem to be void of any kind of worked example...

ln(\frac{t^{2}}{t'}) = ln(R) t^{2} = R\frac{dt}{dx} \int\frac{1}{t^{2}} dt =\int -R dx \frac{-1}{t} +k = -Rx +c t = \frac{A}{Rx+j} thanks dude!

So I then get: ln(t') +c = 2ln(t) +k ? 2ln(t) - ln(t') -R = ln(\frac{t^{2}}{t'}) = ln(R) (my maths is pretty rusty)

Homework Statement t''[x] = \frac{2}{t} t'[x]^{2} The Attempt at a Solution This is not your basic 2ODE, since I can't separate the components into y'', y' and y. Help? I've so far tried: \frac{d^{2}t}{dx^{2}}=\frac{2}{t}(\frac{dt}{dx})^{2} \frac{dt}{dx}=\frac{2}{t}(\frac{dt}{dx})^{2}dx...

Since: V'^{a} = \frac{dX'^{a}}{dX^{b}}V^{b} W'_{b} = \frac{dX^{c}}{dX'^{b}}W_{c} \frac{dC^{d}}{dX^{b}} *\delta_{'b}C^{'d} = \frac{dC^{d}}{dX^{b}}* \frac{dC^{'d}}{dX^{'b}} = \frac{dC^{d}}{dX^{'b}}* \frac{dC^{'d}}{dX^{b}} = \frac{W'_{b}}{W_{b}}*\frac{V'^{d}}{V^{b}} = ? I'm still pretty...

C^{'d} = \frac{dX^{'a}}{dX^{b}}C^b not to sure about the other one...

Homework Statement How does \delta_{b}C^{d} transform? Also compute \delta^{'}_{b} C^{'d}The Attempt at a Solution \delta_{b} C^{d} = \frac{dC^{d}}{dX^{b}} ?I think I am supposed to prove that its a scalar, but I really have no starting point. Any extensive help would be really great.

Ok, so I'm fairly sure that the value of k = \sqrt{1 - (1/9)^2} = \sqrt{80/81} Tricky part now is the expectation value:<A> = \sum |Cn|2 an = |C1|2 a1 + |C2|2 a2 I'm told that the corresponding eigenvalues are +1 and -1. So <A>= |1/9|2 (+1) - |(80/81|(-1) = 81/81 = 1 ? So for a normalised...

Homework Statement |O> = k |R1> + 1/9 |R2> a) Find k if |O> has already been normalized, and b) then the expectation value. The Attempt at a Solution a) To Normalise: |(|O>)|2 = (1/9 |R2> + k |R1>).(1/9 |R2> - k|R1>) = 1/81|R2>2 - k2|R1>2 = 1 I just assumed that |k| = (1-(1/81))0.5, but...

I don't really follow. How does that contract? In your example, would Fuv contract to something like Jk And then would Bz contract to L (no upper indices?) Would it be valid to say DuvExyFuv would contract to something like Jkab?