Recent content by Hamiltonian

ok so for the LHS of the EL equation, we have, $$\frac{d}{dt}\left (\frac{\partial\mathcal{L}}{\partial\dot{\phi}^a} \right)=\frac{d}{dt}(g_{ab}\dot{\phi}^b)=\frac{d}{dt}(g_{ab})\dot{\phi}^b+g_{ab}\ddot{\phi}^b$$ $$ \frac{d}{dt} g_{ab} = \frac{d\phi^c}{dt} \frac{\partial g_{ab}}{\partial \phi^c}...

Ok, so since ##g_{ab}## is not generally constant I will have on the LHS of my EL equation, $$\frac{d}{dt}\left (\frac{\partial\mathcal{L}}{\partial \dot{\phi}^a} \right )=\frac{d}{dt}(g_{ab}\dot{\phi}^b)$$ from here would it make sense to use the product rule? although I wouldn't really know...

ok so in Einstein summation convention, my final equation reads, $$g_{ab}\ddot{\phi}^b=-\frac{\partial}{\partial \phi^a}\left (V(\phi^1,...,\phi^n) \right )$$ is it fair to say that I have solved the given problem? Have I shown this equation that I have obtained by plugging the Lagrangian into...

$$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^1}=g_{11}\dot{\phi}^1+g_{12}\dot{\phi}^2$$ $$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^2}=g_{22}\dot{\phi}^2+g_{21}\dot{\phi}^1$$ $$\implies \frac{\partial\mathcal{L}}{\partial \dot{\phi}^a}=\sum_{b=1}^n g_{ab}\dot{\phi}^b$$ now if we plug this...

ok, I see the mistake. $$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^1}=g_{11}\dot{\phi}_1+g_{12}\dot{\phi}_2$$ $$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^2}=g_{22}\dot{\phi}_2+g_{21}\dot{\phi}_1$$ so how could I combine these to write a more general formula for the ##n\times n## metric...

they do in the ##2\times 2## case right? $$\sum_{a=1}^{2} g_{1a}\dot{\phi}^a = g_{11}\dot{\phi}_1+g_{12}\dot{\phi}_2$$ or are you trying to say that this pattern changes, and becomes more complicated when we deal with a higher-order matrix(##g_{ab}##) so that's why this equation is incorrect?

In the past, I have shown relatively easily that if we have a lagrangian of the form ##\mathcal{L}=\frac{1}{2}\dot{\mathbf{q}}^2-V(\mathbf{q})## simply plugging this into the EL equation gives us newtons second law: ##\ddot{\mathbf{q}}=-\frac{\partial V}{\partial \mathbf{q}}##. I am unfamiliar...

I was originally thinking that once we derive the two ODEs from the EL equation we must somehow prove that the solutions to these ODEs will be straight lines, by solving the ODEs. But I think I understand now that is exactly what I have proved(without solving them ofcourse), the question states...

Yeah, that's what the question says, "thus deriving the ODEs for straight lines in polar coordinates". I do not understand why every straight line must satisfy these ODEs when written in polar coordinates is it because the path that minimizes the action for this particular lagrangian must be...

I am a little confused as to what you mean by ##x^2+y^2=r^2## looks like a straight line, in cartesian coordinates that's a circle with radius ##r##. I looked at the link you provided and it talks about how the equation for a straight line in polar coordinates is...

In polar coordinates, ##x=rcos(\theta)## and ##y=rsin(\theta)## and their respective time derivatives are $$\dot{x}=\dot{r}cos(\theta) - r\dot{\theta}sin(\theta)$$ $$\dot{y}= \dot{r}sin(\theta)+r\dot{\theta}cos(\theta)$$ so the lagrangian becomes after a little simplifying...

Are you implying, ##A=0## and ##B=\frac{1}{2\omega_0}## Edit: I shall not show my face around these parts of town henceforth.

I initially thought of $$A=sin(\omega_0 t)$$ and $$B=cos(\omega_0 t)$$ using the trig identity, ##cos(2\theta) = cos^2(\theta)-sin^2(\theta)## we get, $$2\omega_0 cos(2\omega_0 t) = cos(\omega_0 t)$$ but even this doesn't work.

Finding the first and second derivative of out ansatz, $$\dot x(t)=A(cos(\omega_0 t) - t\omega_0 sin(\omega_0 t)) + B(sin(\omega_0 t) + t\omega_0 cos(\omega_0 t))$$ $$\ddot x= A(-2\omega_0 sin(\omega_0 t) - t{\omega_0}^2cos(\omega_0 t)) + B(2\omega_0 cos(\omega_0 cos(\omega_0 t)...

The above theorem is trying to find the pdf of a transformed random variable, it attempts to do so by "first principles", starting by using the definition of cdf, I don't understand why they have a ##f_X(x)## in the integral wouldn't ##\int_{\{x:r(x)<y\}}r(X) dx## be the correct integral for the...