Recent content by Favicon

Ah yes, that's a good point. Thanks very much for your help Tiny :)

Please bear with me as I'm quite rusty on my differential equation solving. By "the obvious linear substitution" I take it you mean p(t) = q(t) + \frac{gcos(\alpha)}{\omega^2sin^2(\alpha)} Then, since \alpha and \omega are both constant, we have \ddot{p} = \ddot{q} Substituting those into...

Homework Statement A bead is free to slide along a rigid, straight wire, whilst the wire is rotating at angular velocity ω about the z-axis and is tilted away from the z-axis at angle α. I have the equation of motion (EOM) and need to find an explicit solution for the distance of the bead along...

Excellent, thank you Fex. It seems that just after asking for help is when I usually solve a problem myself..

I may be being blindingly stupid here, but I don't see how that answers my original question. What is it about the position and velocity components of a free particle which means they cannot fit that equation?

Actually, after a few minutes pondering, I'd like to attempt an answer myself, based on symmetry. The Lagrangian cannot depend on individual velocity components (but it can depend on the overall magnitude of the velocity) because of rotational symmetry - with no external forces acting on the...

I've heard it said that the Lagrangian of a free particle cannot possibly be a function of any position coordinate, or individual velocity component, but it is a function of the total magnitude of velocity. Why is this the case? I'd be grateful for any pointers in the direction of either a...

Ah ha, it was the \frac{\partial }{\partial \dot{q_j}}\frac{dF}{dt} = \frac{\partial F}{\partial q_j} that I missed. That got rid of all the 3rd order derivatives, leaving only 2nd order one's which all cancel each other out in the end. Thanks very much for the help ehild!

Indeed, I was aware of the action as the simpler way to prove it, but I wanted to do it with derivatives too as the book I'm working from doesn't discuss action until later on. Thought it made a good exercise in partial differentiation and chain rule, which it seems I needed! In fact I'm still...

Ok, sorry to clutter the boards with another thread about it. I forgot to search for existing messages before I posted and thought it might have developed recently due to a server update or something similar. Is it possible for me to delete this thread to get it out of the way? (I'm on the...

I just discovered that when making in a new post, if I preview before submitting, the template for new posts gets appended onto the end of what I typed in the box for editing before submission. I presume that this is a bug, not a feature, as I did use the template in writing my post before I...

Homework Statement This isn't strictly a homework question as I've already graduated and now work as a web developer. However, I'm attempting to recover my ability to do physics (it's been a few months now) by working my way through the problems in Analytical Mechanics (Hand and Finch) in my...

I'm working through a proof that every linear operator, A, can be represented by a matrix, A_{ij}. So far I've got which is fine. Then it says that A(\textbf{e}_{i}) is a vector, given by: A(e_{i}) = \sum_{j}A_{j}(p_{i})e_{j} = \sum_{j}A_{ji}e_{j}. The fact that its a vector is fine...

Thanks for the latex tip. Yes it probably is a non-rotating black hole. I'm actually writing a program to produce the expected line spectrum from a black hole, but the description I've been given (to explain the physics of relativistic line smearing) isn't very clear so when it talked about...

Sort of - it depends on whether your photon gets within the event horizon or not. According to general relativity, massive objects cause space-time to warp in such a way that other objects (including photons) are pulled towards them, so a photon passing close to a black hole will always have...