I'm stuck on some of the math behind this problem and could use some help working this out. I'm trying to calculate the Lagrangian, find the conjugate momenta, and finally calculate the Hamiltonian given the following information.
In cartesian coordinates, we have that $ds^2 = dx_1^2 + dx_2^2 + dx_3^2$. This gives the following tensor: $$ ds^2 = \sum_{\alpha=1}^3 \sum_{\beta=1}^3 g_{\alpha \beta}du_{\alpha} du_{\beta} $$ Where the metric tensor $g$ is $$ g_{\alpha \beta} = \sum_{\mu = 1}^3 \dfrac{\partial x_{\mu}}{\partial u_{\alpha}}\dfrac{\partial x_{\mu}}{\partial u_{\beta}} $$
I found the Lagrangian easy enough. This is $$ \mathcal{L} = \dfrac{m}{2}\sum_{\alpha=1}^3 \sum_{\beta=1}^3 g_{\alpha \beta} \dot u_{\alpha} \dot u_{\beta} $$
The conjugate momenta calculation confuses me once I get to a certain point. I will show my work up until there. \begin{align} p_i = \dfrac{\partial \mathcal L}{\partial \dot u_i} &= \dfrac{m}{2}\sum_{\alpha=1}^3 \sum_{\beta=1}^3 g_{\alpha \beta} \dfrac{\partial }{\partial \dot u_i} \bigg ( \dot u_{\alpha} \dot u_{\beta} \bigg ) \\ &=\dfrac{m}{2}\sum_{\alpha=1}^3 \sum_{\beta=1}^3 g_{\alpha \beta} \bigg ( \dot u_{\alpha} \delta_{i\alpha} + \dot u_{\beta}{\delta_{\beta i}} \bigg ) \end{align}
This is one source of my confusion. I'm not sure why the Kroneker delta's pop up in the last line. My professor tried to explain it to me, but I wasn't able to follow his logic there.
Continuing the problem, we have that \begin{align} p_i &=\dfrac{m}{2}\sum_{\alpha=1}^3 \sum_{\beta=1}^3 g_{\alpha \beta} \bigg ( \dot u_{\alpha} \delta_{i\alpha} + \dot u_{\beta}{\delta_{\beta i}} \bigg ) \\ &=\dfrac{m}{2}\sum_{\alpha=1}^3 \sum_{\beta=1}^3 g_{\alpha \beta} \dot u_{\alpha} \delta_{i\alpha} + \dfrac{m}{2}\sum_{\alpha=1}^3 \sum_{\beta=1}^3 g_{\alpha \beta} \dot u_{\beta}{\delta_{\beta i}} \end{align} From here, my professor invoked some symmetry argument on the metric tensor to collapse the two parts together. He was rushed and stopped here. I was completely lost at this point and I'm not sure how to proceed from there.
Once I get this, I should be able to find the Hamiltonain easy enough. I'm just new to tensor math and this is a bit confusing.
Thank you!
You made a mistake when computing $$\frac{\partial}{\partial \dot u_i}\dot u_\alpha\dot u_\beta.$$ By a simple application of product rule, it should be $$\frac{\partial}{\partial \dot u_i}\dot u_\alpha\dot u_\beta=\frac{\partial\dot u_\alpha}{\partial \dot u_i}\dot u_\beta+\dot u_\alpha\frac{\partial\dot u_\beta}{\partial \dot u_i}=\delta_{i\alpha}\dot u_\beta+\dot u_\alpha\delta_{\beta i}.$$ Kronecker delta comes up because $$\frac{\partial\dot u_\alpha}{\partial \dot u_i}=\delta_{i\alpha}.$$
As for "collapsing the expression", I presume your professor did the following.
The numbers $g_{\alpha\beta}$ have the property $g_{\alpha\beta}=g_{\beta\alpha}$. This is known as symmetry of the tensor. You may verify this by looking at their definition, but the tensor is actually symmetric intrinsically.
With this property, we can compute the following:
$$\begin{align}\frac{m}{2}\sum_{\alpha=1}^3\sum_{\beta=1}^3g_{\alpha\beta}(\delta_{i\alpha}\dot u_\beta+\dot u_\alpha\delta_{\beta i})&=\frac{m}{2}\sum_{\alpha=1}^3\sum_{\beta=1}^3g_{\alpha\beta}\delta_{i\alpha}\dot u_\beta+\frac{m}{2}\sum_{\alpha=1}^3\sum_{\beta=1}^3g_{\alpha\beta}\dot u_\alpha\delta_{\beta i}\\ &=\frac{m}{2}\sum_{\alpha=1}^3\sum_{\beta=1}^3g_{\alpha\beta}\delta_{i\alpha}\dot u_\beta+\frac{m}{2}\sum_{\color{red}{\beta}=1}^3\sum_{\color{red}{\alpha}=1}^3g_{\color{red}{\beta\alpha}}\delta_{\color{red}{i\beta}}\dot u_\alpha \end{align}$$ The two sums are equal and so $$\begin{align}\frac{m}{2}\sum_{\alpha=1}^3\sum_{\beta=1}^3g_{\alpha\beta}\delta_{i\alpha}\dot u_\beta+\frac{m}{2}\sum_{\beta=1}^3\sum_{\alpha=1}^3g_{\beta\alpha}\delta_{i\beta}\dot u_\alpha &=m\sum_{\alpha=1}^3\sum_{\beta=1}^3g_{\alpha\beta}\delta_{i\alpha}\dot u_\beta\\&= m\sum_{\beta=1}^{3}g_{i\beta}\dot u_\beta\end{align}$$ The last equality holds because $\delta_{i\alpha}=0$ for $\alpha\not=i$ and $\delta_{ii}=0$, so that when you sum over $\alpha$, only the term $\alpha=i$ remains.