Why $\sum\limits_{i} \frac{\partial L}{\partial \dot{q_i}} \dot{q_i} = \sum\limits_{i} \frac{\partial T}{\partial \dot{q_i}} \dot{q_i} = 2T$? [closed]

From Landau and Lifschitz's "Mechanics"; section 6.

I understand up to this point

$$E \equiv \sum\limits_{i} \dot{q_i}\frac{\partial L}{\partial \dot{q_i}} - L$$

Then the author states:

Using Euler's theorem on homogeneous functions, we have $$\sum\limits_{i} \dot{q_i} \frac{\partial L}{\partial \dot{q_i}} = \sum\limits_{i} \dot{q_i} \frac{\partial T}{\partial \dot{q_i}} = 2T.$$

Can someone aid me and explain how to get from former equation to latter? How Euler's theorem implies the second equation?

Help will be appreciated.

closed as off-topic by sammy gerbil, Chris♦, Kyle Kanos, Jon Custer, Cosmas ZachosFeb 11 '18 at 3:47

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• Can you state Euler's theorem? In particular, for $T=T(\dot q)$ (for fixed $q$). Is this function homogeneous? of what degree? – AccidentalFourierTransform Feb 5 '18 at 21:58
• Hint: L&L apparently assume $T$ is quadratic in $\dot{q}$. – Qmechanic Feb 5 '18 at 22:04
• If you just based on a physical example of kinetic energy $T$ being a quadratic form, take $L = (1/2)\dot{q}^2 - U(q)$, you can see that this result holds immediately. So, now just generalize it, right? – Dr. Ikjyot Singh Kohli Feb 5 '18 at 22:50
• This is one instance in which the famous brevity of Landau and Lifshitz might not be particularly helpful. – probably_someone Feb 6 '18 at 15:09

The kinetic energy of a holonomic scleronomous particle system consisting of $n$ particle (described by the $N$ coordinates $q_i$, $i=1,\dots,N$) can be written as a quadratic form in $\dot{q}$: $$T= \frac{1}{2} \sum_{k,r=1}^{N} a_{kr}(q)\dot{q_k}\dot{q_r}.$$ This can be seen by putting $$\vec{v_j}=\frac{d}{dt}\vec{r_j}=\sum_{k=1}^{N}\frac{\partial \vec{r_j}}{\partial q_k}\dot{q_k}$$ (with $\vec{r_j}$ being the position vector of the $j$-th particle) in the kinetic energy definition formula $$T=\frac{1}{2} \sum_{j=1}^{n} m_j \vec{v_j}\cdot\vec{v_j}$$ and by defining $$a_{kr}=\sum_{j=1}^{n}m_j\frac{\partial \vec{r_j}}{\partial q_k}\cdot \frac{\partial \vec{r_j}}{\partial q_r}.$$

Recall that potential energy does not depend on $\dot{q}$, so its derivative is zero, leaving only the derivative of the kinetic energy. Using the above equation for $T$ the result you're looking for can be easily obtained from Euler's Homogeneous Function Theorem, since T is quadratic in $\dot{q}$.

An alternative method: straightforward calculation

The result can be computed also without using Euler's theorem, using the formula above for $T$ and noticing that $a_{kr}$ is symmetric. In fact, \begin{align} \sum_{i} \dot{q_i}\frac{\partial T}{\partial\dot{q_i}}&=\frac{1}{2}\sum_{i} \dot{q_i}\left(\sum_{r,k} a_{kr}(q) \frac{\partial}{\partial\dot{q_i}}(\dot{q_k}\dot{q_r})\right)\\ &=\frac{1}{2}\sum_{i}\dot{q_i}\left(\sum_{r,k} a_{kr}(q) (\delta^i_k \dot{q_r}) + \sum_{r,k} a_{kr}(q)(\delta^i_r\dot{q_k})\right)\\ &=\frac{1}{2}\sum_{i}\dot{q_i}\left(\sum_{r} a_{ir}(q) \dot{q_r} + \sum_{k} a_{ki}(q)\dot{q_k}\right)\\ &=\frac{1}{2}\sum_{i}\dot{q_i}\left( 2\sum_{r} a_{ir}(q)\dot{q_r}\right)\\ &=\sum_{r,i}a_{ir}(q)\dot{q_i}\dot{q_r}\\ &=2T \end{align}

Since you mention Euler's theorem, it is the following result:

Let $f : \mathbb{R}^n\to \mathbb{R}$ be a $C^1$ function which is homogeneous of degree $k$, i.e., satisfies

$$f(\lambda x)=\lambda^k f(x),$$

then it holds

$$kf(x)=\sum_{i=1}^n x^i D_if(x),$$

where $D_i$ is the $i$-th partial derivative.

Proof: Define $g(\lambda)=f(\lambda x)$ for fixed $x$. Take the derivative with respect to $\lambda$ using the chain rule:

$$g'(\lambda)=\sum_{i=1}^n D_if(\lambda x) \dfrac{d}{d\lambda}(\lambda x^i)=\sum_{i=1}^n x^i D_i f(\lambda x).$$

On the other hand, since $f(\lambda x)=\lambda ^k f(x)$ we have $g(\lambda)=\lambda^k f(x)$ and hence $$g'(\lambda)=k \lambda^{k-1}f(x)$$

equating the results obtained we have

$$k\lambda^{k-1}f(x)=\sum_{i=1}^n x^i D_i f(\lambda x),$$

finally compute the above expression at $\lambda = 1$, and the result follows.

Now, the authors assume $T$ is homogeneous of degree $2$ on the variables $\dot{q}^i$. This can be motivated by looking at the case of particle in three dimensions (where $q^1=x,q^2=y,q^3=z$) :

$$T(\dot{q}^i)=\dfrac{1}{2}m\sum_{i}(\dot{q}^i)^2$$

From where $T(\lambda \dot{q}^i)=\lambda^2 T(\dot{q}^i)$. With this hypothesis the result above can be used. Euler's theorem gives

$$T=\sum_i \dfrac{\partial T}{\partial \dot{q}^i}\dot{q}^i.$$

Notice that even if $T$ depends on $q^i$ isn't a problem. Suppose still $T$ is homogeneous of degree $2$ on the velocities. Consider $q^i$ fixed and define the function $f(\dot{q}^i)=T(q^i,\dot{q}^i)$ and apply Euler's theorem to it, you get the same result.

But $L = T-V$ and usually one assumes $V$ to beindependent of the velocities, so that

$$\dfrac{\partial T}{\partial \dot{q}^i}=\dfrac{\partial L}{\partial \dot{q}^i}$$

and hence

$$2T=\sum_i \dfrac{\partial L}{\partial \dot{q}^i}\dot{q}^i$$

and usually from this one concludes that $H = T+V$. The procedure however requires: (1) that $T$ be homogeneous of degree $2$ to Euler's theorem apply, (2) that $V$ be independent of the $\dot{q}^i$ in order to state that $\partial T/\partial \dot{q}^i = \partial L/\partial \dot{q}^i$, and (3) that $L = T - V$ as usually done in Classical Mechanics.