1
$\begingroup$

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.

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – sammy gerbil, Chris, Kyle Kanos, Jon Custer, Cosmas Zachos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ Can you state Euler's theorem? In particular, for $T=T(\dot q)$ (for fixed $q$). Is this function homogeneous? of what degree? $\endgroup$ – AccidentalFourierTransform Feb 5 '18 at 21:58
  • 3
    $\begingroup$ Hint: L&L apparently assume $T$ is quadratic in $\dot{q}$. $\endgroup$ – Qmechanic Feb 5 '18 at 22:04
  • $\begingroup$ 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? $\endgroup$ – Dr. Ikjyot Singh Kohli Feb 5 '18 at 22:50
  • $\begingroup$ This is one instance in which the famous brevity of Landau and Lifshitz might not be particularly helpful. $\endgroup$ – probably_someone Feb 6 '18 at 15:09
3
$\begingroup$

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}

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.