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=0
Is that correct? Have you worked through it to the end sgd37?
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}
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 [Broken]
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...
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 [Broken]) 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...
Also forgot to mention your integrand is EVEN so you can write the integral as:
\frac{1}{2}Re\left(\int_{-\infty}^{\infty}\frac{1-e^{ix}}{x^2}dx\right).
Hopefully you know what to do now.
Because e^{ix}=cosx+isinx and so your integral is just the real part of my integral. What level of maths are you at? Because if you don't know that identity your unlikely to know the complex methods that will allow you to do this integral.
I don't know how your book is explaining convolutions, but this book does it very well (scroll down to page 291 and page 292 for the exponential distribution example):
http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter7.pdf
That has everything you need.
1. Homework Statement
I am trying to show that for a vector field Va which satisfies V_{a;b}+V_{b;a} that V_{a;b;c}=V_eR^e_{cba} using just the below identities.
2. Homework Equations
V_{a;b;c}-V_{a;c;b}=V_eR^e_{abc}(0)
R^e_{abc}+R^e_{bca}+R^e_{cab}=0 (*)
V_{a;b}+V_{b;a}=0 (**)
3. The...
Re: Geodesics
If we parametrize the geodesic by \lambda then the velocity is given by \frac{dX^a(\lambda)}{d\lambda}=\dot{X}^a
Thus, \nabla_{\lambda}\dot{X}^a=\dot{X}^b\nabla_{b}\dot{X}^a=\dot{X}^b\left(\partial_b \dot{X}^a+\Gamma_{bc}^a...
You made a mistake in evaluating the integral, it should be:
\LARGE {\int_0^t {e^{i\omega t^'}}dt^'} = \frac{1}{i\omega}\left(e^{i\omega t}-1\right)
Do you see why?
So from what I can tell you're happy with writing the geodesic equations in different coordinates except you can't get the relation between the connections in the two coordinate systems? I read your pdf and you seem to be making a right mess of this. Here's the simplest way to do it:
\nabla_a...
The function Tan-1z is defined by:
Tan^{-1}z=\int_0^z \frac{dt}{1+t^2}
where the path of integration is a straight line. For what region of the complex plane is Tan-1z analytic?
I'm not too sure about this. I can see that the integrand is analytic everywhere except at t=+i and t=-i so...
Ok now we're getting somewhere, I can see why there must be a |10>|10> term because 0 and 0 is the other combination of the m_i that add to give 0, but according to that wikipedia table it has coefficient -\sqrt{\frac{1}{3}} how is this determined by inspection? Because in an exam I have neither...
Hi Andy, thanks for replying. I've made a bit more progress on this now as I've realized that to get to |1 0> I have to take the orthogonal combination of |2 1> so that we'll obtain the top state |1 1> and then apply J_ to get |1 0> !! But from this I'm still not sure as to how to get to |0 0>...