Recent content by giraffe714

So I went down a bit of a rabbit hole and turns out, from what I could gather: a) The action variable behaves strangely at the separatrix. b) The transition between ## H_{osc} ## and ## H_{rot} ## need not be continuous, which is okay because I completely misinterpreted what kind of continuity I...

As said in the tl;dr: is the Hamiltonian necessarily differentiable (hence continuous) at the separatrix in the action-angle formalism? After all, the action variables are different depending on the type of motion. As far as I know the Hamiltonian H = H(J) can be found by inverting J for E, and...

Ohh, alright. That makes sense, thank you. But once again, what happens when we multiply the abstract P by some abstract b_1? Why don't we get b'_1? Because the inverse of some 2x2 P (I would assume it has to be invertible) is $$ \frac{1}{ab-cd} \begin{pmatrix} d && -c \\ -b && a \end{pmatrix}...

Yes, it does, from a concrete point of view at least. When we multiply ## A(1,0) = (1, 0)' ## what we're doing is taking the coefficients ## (1, 0) ## and ## (1, 0)' ##, correct? But then, for any basis B, the basis vectors in that basis would always "look like" the standard basis, and when we...

The formula my textbook provides for finding change of basis matrices is: $$b'_j = a_{1j} b_1 + \cdots + a_{nj} b_n$$ I assume, since that's the convention and also because Wikipedia itself uses this formula like this, that the first index of the c's is the row, and the second is the columns...

Oh, yes, that's correct, my apologies.

So firstly I calculated the partial derivatives of f to be: $$ \frac{\partial f}{\partial x} = \frac{2x}{2} (y'^2 - \frac{1}{3} y^6) + \frac{x^2}{2} (2y'y'' - 2y^5y') = xy'^2 - \frac{1}{3} y^6 + x^2y'(y'' - y^5) $$ $$ \frac{\partial f}{\partial y} = \frac{x^2}{2}*\frac{1}{3}*6y^5 = x^2y^5 $$ $$...

That actually makes sense, thank you. I think I was just missing that S is a function of q_k which are also functions of t so that clears it up. One question though, just to make sure: the fact that ## P_k = \beta_k = const. ## doesn't *follow* from ## Q_k = \alpha_k = const. ##, instead they're...

As stated in the TLDR, I don't understand why the derivative of the Jacobi complete integral with respect to the constant α must be another constant, and furthermore why that constant is negative. The textbook I'm following, van Brunt's The Calculus of Variations proves it by taking: $$...

Okay, yeah, that makes sense, so the ##F_x = Fcosθ = 2k(\sqrt(x^2+y^2+(z - S)^2) - l_0)cos(arcsin(\frac{S}{\sqrt(x^2+y^2+(z - S)^2)}))## but that... that seems like too much trigonometry to be correct, plus I don't think there's a trig identity that simplifies cos(arcsin(whatever))? Could be...

ohh, so it's just trigonometry now right? yea... yea I think that makes sense, so then $$ V = k*\sqrt(x^2+y^2+(z - S)^2) \Rightarrow F_x = -\frac{1}{2}k*2x = -kx $$ But that's even further from the expression we're trying to get?? Plus, how will I be able to get a general expression for the...

What is that diagram trying to say exactly? That the x component of the force is along the x direction..?

Ohh, well it'd be sqrt(x^2+y^2+(z - S)^2), right?

Oh, yea, that makes sense. So I don't even have to derive an expression for V in terms of ##\phi##? That's a relief. How far is it from the anchor point? Well it's a length z - S from the anchor point, because it's a distance z from the origin, and the origin is a distance S away from the...

What do you mean by this? Why don't the springs change length when the mass is being displaced from equilibrium? It can be at any length ##l_0## from the spring if it is relaxed (and the second one doesn't break, but we're not given any condition for breaking so I'll just assume that's not a...