# How to show that kinetic energy is a homogeneus function of degree 2? [closed]

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 don´t really know how to proceed since this is my first time dealing with tensors.

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.

• 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. – Daniel Teran Nov 16 '19 at 22:30