Recent content by Tangent87

Thanks all. Think it makes sense to me now.

After reading the wikipedia article and looking at many other threads on this forum I am still having a hard time understanding the difference between the Observable universe and the entire universe... Why is the entire universe not observable to us? The Big Bang happened 13.8 billion years...

Ah I see thanks. I've worked it through now and it does come out, the key is to not try and differentiate out the brackets but just deduce that the stuff inside the brackets must be a constant.

Hmm okay, so now on the next line I get: \left(\frac{dv}{du}u^2\dot{u}\right)_{,u}\dot{u} + \left(\dot{v}u^2\right)_{,v}\dot{v}=\frac{d^2v}{du^2}u^2\dot{u}^2 + 2\frac{dv}{du}u\dot{u}^2 + \frac{dv}{du}u^2\ddot{u} + \ddot{v}u^2=0Is that correct? Have you worked through it to the end sgd37?

It is, I got it from 2010 Part II paper 2 page 22: http://www.maths.cam.ac.uk/undergrad/pastpapers/2010/Part_II/PaperII_2.pdf

Thank you for replying. Ok so when we put k^c=(0,1) in we get: 0=\left(\dot{x}^ag_{av}\right)_{,b}\dot{x}^b=\left(\dot{v}g_{vv}\right)_{,b}\dot{x}^b=\left(\dot{v}u^2\right)_{,b}\dot{x}^b=\left(\dot{v}u^2\right)_{,u}\dot{u}+\left(\dot{v}u^2\right)_{,v}\dot{v} So now are you saying that...

Hi, I'm stuck on the last bit the attached question where we're given the metric ds^2=-du^2+u^2dv^2 and have to use equation (*) to find the geodesic equations. They tell us to use V^a=\dot{x}^a the tangent vector to the geodesic and presumably we use the three killing vectors they gave us, so...

If you're not happy with separating variables you can always use an integrating factor: \LARGE \frac{d}{dt}\left(U(t)e^{iHt/{\hbar}}\right) = 0 So then: \LARGE U(t)e^{iHt/{\hbar}} = U(0) U(0) can be normalised to 1 therefore \LARGE U(t) = e^{-iHt/\hbar}

Ah okay, that clears everything up, thanks again.

So in order to have the wavefunction being antisymmetric would I need to have something like \frac{1}{\sqrt{2}}(\psi_1(A)\psi_2(B)-\psi_2(A)\psi_1(B))? The only trouble I have with this is that the wavefunction now involves terms from the second energy level whilst we're only dealing with the...

Hi, I am doing question 32D on page 18 here: http://www.maths.cam.ac.uk/undergrad/pastpapers/2005/Part_2/PaperII_3.pdf and I am stuck on the second paragraph where we have to explain how to construct the two-particle states of lowest energy for (i). identical spin-1 particles with...

Ahhh thanks, done it now. It seems I was definitely going about it the wrong way.

Can anybody help me out with deriving the identity V_{a;b;c}=V_eR^e_{cba}? Forget about the EM stuff I don't care so much about that but I'd be very grateful for some help in deriving that identity. Thanks

Because the residue should be pi, not 2pi because when you close the semi-circular contour in the upper half plane you only go round the origin on an infinitesimal semi-circle so it's pi, it would be 2pi if we went all around the origin on a full circle but we don't in this case. (It might help...

The problem is part of a larger question (see page 59 here http://www.maths.cam.ac.uk/undergrad/pastpapers/2004/Part_2/list_II.pdf ) and I'm having trouble with the last bit as well where it goes into the EM stuff, I know it must obviously somehow relate to everything we've done above but I just...