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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!

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    $\begingroup$ First, $p_{i} = \partial L / \partial \dot{u}_{i}$ not $\partial L / \partial u_{i}$. Secondly, the kronecker delta appears from the product rule and using \begin{align} \sum_{j=1}^{N} \partial_{\dot{u}_{i}}(\dot{u}_{j}) &= \partial_{\dot{u}_{i}}(\dot{u}_{1} + \dot{u}_{2} + \dots + \dot{u}_{i} + \dots +\dot{u}_{N}) \\ &\underbrace{= 0 + 0 + \dots + 1 + \dots + 0}_{\text{$1$ in the $i^{th}$ position}} \\ &= 1 \text{ when $i = j$} \\ &= \delta_{ij} \end{align} $\endgroup$ – mattos Oct 3 '18 at 3:01
  • $\begingroup$ @Mattos The mistake on the momentum was a typo, my apologies. $\endgroup$ – Kosta Oct 3 '18 at 3:04
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    $\begingroup$ What do you mean by collapse two parts together? $\endgroup$ – edm Oct 3 '18 at 3:35
  • $\begingroup$ @edm I mean that the two separate sums become one. The final answer should only contain one sum, and because of the symmetry argument, they get added together. $\endgroup$ – Kosta Oct 3 '18 at 3:37
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    $\begingroup$ Could you include the final expression the professor wrote? $\endgroup$ – edm Oct 3 '18 at 3:40
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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.

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  • $\begingroup$ This is excellent, thank you. I knew the metric tensor was symmetric in that sense, but I did not understand how it pertained to this problem. $\endgroup$ – Kosta Oct 3 '18 at 4:26
  • $\begingroup$ For the last part, wouldn't $$ m\sum_{\alpha=1}^3 g_{i \alpha} \dot u_\alpha $$ also be an acceptable result since the sums are equivalent? $\endgroup$ – Kosta Oct 3 '18 at 4:27
  • $\begingroup$ @Kosta In the summation, $\alpha,\beta$ are just dummy variables. You can change it arbitrarily. So your sum is the same as my sum. $\endgroup$ – edm Oct 3 '18 at 4:35
  • $\begingroup$ Right. I was asking because of the Hamiltonian calculation. The final answer has both $p_\alpha$ and $p_\beta$, so I was just wanting to make sure. $\endgroup$ – Kosta Oct 3 '18 at 4:39

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