Recent content by csco

Bill: it isn't homework. I found this problem while studying and it made me rethink about a few things in hamiltonian mechanics. I only wanted to know of a general method to solve problems of this sort since I had never seen them before that's why the question on how to deal with them. I gave...

No answers? I thought finding a CT between two given hamiltonians would be a standard problem that I just didn't know how to solve.

Hello everyone, I am given the inital hamiltonian H = (1/2)*(px2x4 - 2iypy + 1/x2) and the transformed hamiltonian K = (1/2)*(Px2 + Py2 + X2 + Y2) and I'm supposed to show there exists a canonical transformation that transforms H to K and find it. I don't know how to solve problems of this sort...

I don't understand your question. The lemma allow us to remove the restriction on the range of f because that what the lemma is about. Uniqueness on arbitrary intervals follows from local uniqueness that's what the lemma says. There's a short proof of this fact. Let there be two solutions...

Any suggestions or comments on all of this anyone? I expected I would get a definitive answer at least for the unidimensional case but maybe the problem is too difficult and I'm just wasting my time trying to solve it? I'm hoping the question is being understood and if not please let me know and...

But that's not a problem because you can turn a higher order equation into a first order one by increasing the dimensions of the space. So I'm considering higher order ODEs too because I'm not fixing the n in Rn. I don't understand how are you relating the maximum of the solution with the...

Hi AlephZero and thanks for your answer. I don't understand very well where are you trying to get at with what you are saying but I wasn't very clear in my first post so I'll define the terms I'm using more clearly. I'm considering an ODE of the form x' = f(t,x) (I'm ignoring dependence on...

I would like to know if the blow up time of a ordinary differential equation with the lipschitz condition is a continuous function (in its domain whatever it might be) of the initial conditions and parameters. With blow up time I mean the length of the time interval to the future of the inital...

Take 3 chords that form a right triangle and intersect with each other in the circumference then if you add a new chord that intersects them you will necessarily have a triple intersection right? Ignoring that degenerate case in which chords intersect in the circumference this seems to be true...

By definition |a_n,b_n\rangle are eigenkets of A and B with eigenvalues an and bn: A|a_n,b_n\rangle = a_n|a_n,b_n\rangle B|a_n,b_n\rangle = b_n|a_n,b_n\rangle Using the previous equations and linearity of A and B you have: BA|a_n,b_n\rangle = Ba_n|a_n,b_n\rangle = a_nB|a_n,b_n\rangle =...

The flux density is \mathbf{F} \cdot \mathbf{\hat{n}} while \frac{\int \mathbf{F} \cdot \mathbf{\hat{n}}dA}{A} is the average flux density over the surface

The answer is no. You are looking for solutions of the equation: \frac {n(n+1)(2n+1)}{6} = \frac {n(n+1)(n+2)}{6} Cancelling terms we get 2n+1 = n+2 Solving for n you find n = 1 as the only solution

The expressions are not equal. Try z = 0, you get 1 in the first expression and -i in the second. The valid equation is: \frac {z-i} {-z-i} = \frac {z(i-1) +i +1} {z(1-i) +i +1} To check that just multiply and divide the LHS by (i - 1)