0
$\begingroup$

I'm a bit stuck with the tensor analysis for the following problem. It was just introduced to me and I've never seen this before. All I'm looking for in a place to start, because I'm unsure where to even begin with this.

The distance squared between two infinitesimally close points in Cartesian coordinates is $ds^2 = dx_1^2 + dx_2^2 + dx_3^2$. So using the chain rule, the distance squared in general coordinates is \begin{align} ds^2 = \sum_{\alpha = 1}^3 \sum_{\beta = 1}^3 g_{\alpha \beta} du_{\alpha} du_{\beta} \end{align} where the metric tensor $g$ is \begin{align} g_{\alpha \beta} = \sum_{\mu = 1}^3 \dfrac{\partial x_{\mu}}{\partial u_\alpha}\dfrac{\partial x_{\mu}}{\partial u_\beta} \end{align} Note the metric tensor is a function of $u_1$, $u_2$, and $u_3$.

Write the Lagrangian for these coordinates, calculate the conjugate momenta in terms of the velocities, and from these calculate the Hamiltonian. Your result should be \begin{align} H = \dfrac{1}{2m} \sum_{\alpha = 1}^3 \sum_{\beta = 1}^3 p_{\alpha} g_{\alpha \beta}^{-1} p_{\beta} \end{align} where $g^{-1}$ is the inverse (matrix) of $g$

I know what to do. i.e., find the Lagrangian, take $p_i = d\mathcal L / d \dot q$, and use $H = \sum_i \dot q_i p_i - \mathcal{L}$. I'm just unsure on how to do this with the tensors. I know the first step, where $$ \mathcal L = \dfrac{m}{2}\dfrac{ds^2}{dt} = \dfrac{m}{2} \sum_{\alpha = 1}^3 \sum_{\beta = 1}^3 g_{\alpha \beta} \dot u_{\alpha} \dot u_{\beta}$$ But that's as far as I can get with my math skills. Can someone point me in the right direction? Preferably to a source that does not use Einstein notation.

Thank you!

$\endgroup$
1
$\begingroup$

You should have $\mathcal{L}=\frac{m}{2}g_{\alpha\beta}\dot{u}^\alpha\dot{u}^\beta$ so $p_\alpha=mg_{\alpha\beta}\dot{u}^\beta=m\dot{u}_\alpha$ and $H+\mathcal{L}=p_\alpha\dot{u}^\alpha=\frac{1}{m}p_\alpha p^\alpha$, and the Hamiltonian and Lagrangian are each half of that. Bear in mind the index height is important; you need a metric tensor to change it. If you're shaky on tensors use matrices instead, viz. $\mathcal{L}=\frac{m}{2}\dot{u}^T g\dot{u}$ etc.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.