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My vector calculus homework is the next:

If $T = a_{\alpha\beta}(q^1,...,q^n)') \dot{q^\alpha} \dot{q^\beta}$ show that $2T = \dfrac{\partial T}{\partial \dot{q^\alpha} }\dot{q^\alpha}$

I think that T is the kinetic energy, $a_{\alpha\beta}$ the metric tensor and i got to show that T is a homogeneous function of degree 2, but i donĀ“t really know how to proceed since this is my first time dealing with tensors.

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It is actually fairly easy, but I will show it in many little steps. First, realize that the indices can always renamed if done properly. Secondly we use the fact that $q$ and $\dot{q}$ are independent. We start by taking the partial derivative of $T$ :

$$\frac{\partial T}{\partial \dot{q^\gamma}} = a_{\alpha\beta}\frac{\partial \dot{q}^\alpha}{\partial \dot{q}^\gamma} \dot{q}^\beta + a_{\alpha\beta}\dot{q}^\alpha\frac{\partial \dot{q}^\beta}{\partial \dot{q}^\gamma} = a_{\alpha\beta}\delta^\alpha_\gamma \dot{q}^\beta+ a_{\alpha\beta}\dot{q}^\alpha\delta^\beta_\gamma$$

In the next step we multiply the partial derivative of $T$ with $\dot{q}^\gamma$. Realize that it is not just a multiplication, but contains also a summation over the index $\gamma$:

$$\frac{\partial T}{\partial \dot{q^\gamma}} \dot{q}^\gamma = a_{\alpha\beta}\delta^\alpha_\gamma \dot{q}^\beta \dot{q}^\gamma +a_{\alpha\beta}\dot{q}^\alpha\delta^\beta_\gamma \dot{q}^ \gamma$$

We use the properties of the Kronecker-symbol $\delta^\alpha_\gamma$ which are applied on $a_{\alpha\beta}$ (however, they could be also applied on the $\dot{q}$s.)

$$\frac{\partial T}{\partial \dot{q^\gamma}} \dot{q}^\gamma = a_{\gamma\beta}\dot{q}^\beta \dot{q}^\gamma + a_{\alpha\gamma}\dot{q}^\alpha\dot{q}^ \gamma = a_{\gamma\beta}\dot{q}^\gamma\dot{q}^\beta+ a_{\alpha\gamma}\dot{q}^\alpha\dot{q}^ \gamma =a_{\alpha\beta}\dot{q}^\alpha\dot{q}^\beta+ a_{\alpha\beta}\dot{q}^\alpha\dot{q}^ \beta = 2T $$

Remains to rename the index $\gamma$ to $\alpha$. That's it.

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  • $\begingroup$ Thank you very much, what i did was: $T(\lambda\dot{q^\alpha}, \lambda\dot{q^\beta}) = a_{\alpha\beta} \lambda\dot{q^\alpha} \lambda\dot{q^\beta} = \lambda^2 T(\dot{q^\alpha}, \dot{q^\beta})$ for some $\lambda \in \mathbb{R}$, which implies that T is a homogeneous function of degree 2 and using euler's theorem for homogeneous functions we have $2T = \dfrac{\partial T}{\partial \dot{q^\alpha} }\dot{q^\alpha}$, but i don't know if what i did it's correct. $\endgroup$ – Daniel Teran Nov 16 '19 at 22:30

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