1
$\begingroup$

For any integrand $G(\lambda, \mathbf{r}(\lambda), \dot{\mathbf{r}}(\lambda))$, where $\dot{\mathbf{r}} = \frac{d}{d\lambda}\mathbf{r}$, the Euler-Lagrange equations are $$ \frac{\partial G}{\partial \mathbf{r}} - \frac{d}{d\lambda}\frac{\partial G}{\partial \dot{\mathbf{r}}} = 0 $$

If $G = U\sqrt{\dot{\mathbf{r}}\cdot \dot{\mathbf{r}}}$, $U = U(\mathbf{r})$, show that a substitution into the Euler-Lagrange equation yields $$ U\ddot{\mathbf{r}} + (\dot{\mathbf{r}}\cdot \nabla U)\dot{\mathbf{r}} - (\dot{\mathbf{r}}\cdot \dot{\mathbf{r}})\nabla U=0 $$ under the assumption that $\mathbf{\dot r} \cdot \mathbf{\ddot r} = 0$.

I am new to the topic, may I please ask how the above substitution works out in detail? Thank you!

$\endgroup$
2
  • $\begingroup$ Is $U$ just a function of $\mathbf{r}$? $\endgroup$
    – paulinho
    May 27, 2020 at 14:25
  • $\begingroup$ @paulinho, yup! I've added that:) $\endgroup$ May 27, 2020 at 14:31

1 Answer 1

1
$\begingroup$

Let $\mathbf{r} = (r_1, r_2, \cdots, r_n)$. Let us just find the Euler-Lagrange equation for $r_1$. The Euler-Lagrange equations for any other $r_i$ will be symmetric, just with $r_1$ replaced by $r_i$.

Let $f(\mathbf{\dot r}) = \sqrt{\dot{\mathbf{r}} \cdot \dot{\mathbf{r}}} = \|\mathbf{\dot r}\|$. Then we wish to plug in $G(\mathbf{r}, \mathbf{\dot r}) = U(\mathbf{r}) f(\mathbf{\dot r})$ into the Euler-Lagrange equations. It is important to note that $\mathbf{r}$ and $\mathbf{\dot r}$ can be functions of $\lambda$, so $G$ also implicitly depends on $\lambda$.

The first term on the left-hand side of the Lagrange equation is easy to calculate: $$\frac{\partial G}{\partial r_1} = f(\mathbf{\dot r}) \frac{\partial U}{\partial r_1}$$ as the only term in $G$ explicitly depending on $r_1$ is $U$.

As for the second term, one useful identity that we can use is if $g(\mathbf{r}) = \sqrt{\mathbf{r} \cdot \mathbf{r}} = \|\mathbf{r}\|$, $$\frac{\partial g}{\partial \mathbf{r}} = \frac{\mathbf{r}}{\|\mathbf{r}\|} \implies \frac{\partial g}{\partial r_1} = \frac{r_1}{\|\mathbf{r}\|}$$

Hence, it follows that $$\frac{\partial f}{\partial \dot r_1} = \frac{\dot r_1}{\|\mathbf{\dot r}\|} \implies \frac{\partial G}{\partial \dot r_1} = \frac{U}{\|\mathbf{\dot r}\|} \dot r_1 = U(\mathbf{r}) \alpha(\mathbf{\dot r})$$

where $\alpha = \dot r_1 / \|\mathbf{\dot r}\|$. We must now take the total derivative with respect to $\lambda$ of this term. Note that $$\frac{d}{d \lambda} \frac{\partial G}{\partial \dot r_1} = \frac{d (U \alpha)}{d \lambda}= \frac{\partial (U \alpha)}{\partial \mathbf{r}} \cdot \mathbf{\dot r} + \sum_{i = 1}^n \frac{\partial (U \alpha)}{\partial \dot r_i} \cdot \ddot r_i = \alpha (\mathbf{\dot r} \cdot \nabla U) + U \sum_{i = 1}^n \frac{\partial \alpha}{\partial \dot r_i} \cdot \ddot r_i $$ The last equality holds because again the only $\mathbf{r}$ explicit dependence of $\partial G / \partial \dot r_1$ is in $U$ and the only explicit $\mathbf{\dot r}$ dependence is in $\alpha$. Since $\alpha$ is symmetric for all $r_i \neq r_1$, we delineate two cases when calculating $\partial \alpha / \partial \dot r_i$:

  1. If $i = 1$, then it can be verified that $$\frac{\partial \alpha}{\partial \dot r_i} = \frac{\partial \alpha}{\partial \dot r_1} = \frac{1}{\|\mathbf{\dot r}\|^3} \sum_{i = 2}^n \dot{r_i}^2 = \frac{\|\mathbf{\dot r}\|^2 - \dot{r_1}^2}{\|\mathbf{\dot r}\|^3}$$

  2. If $i \neq 1$, then it can be verified that $$\frac{\partial \alpha}{\partial \dot r_i} = - \frac{\dot r_1 \dot r_i}{\|\mathbf{\dot r}\|^3}$$

Plugging these results back into the expression we had, we obtain $$\frac{d}{d \lambda} \frac{\partial G}{\partial \dot r_1} = \alpha (\mathbf{\dot r} \cdot \nabla U) + \frac{U}{\| \mathbf{\dot r}\|^3} \left[\| \mathbf{\dot r}\|^2 \ddot r_1 - \dot r_1 \sum_{i = 1}^n \dot r_i \ddot r_i \right] = \alpha (\mathbf{\dot r} \cdot \nabla U) + \frac{U \ddot r_1}{\| \mathbf{\dot r}\|} - \frac{U \dot r_1 (\mathbf{\dot r} \cdot \mathbf{\ddot r})}{\| \mathbf{\dot r}\|^3}$$ Plugging everything back in, we get the following equation in $r_1$: $$\|\mathbf{\dot r} \| \frac{\partial U}{\partial r_1} - \frac{\dot r_1}{\|\mathbf{\dot r}\|} (\mathbf{\dot r} \cdot \nabla U) - \frac{U \dot r_1}{\|\mathbf{\dot r}\|} + \frac{U \dot r_1 (\mathbf{\dot r} \cdot \mathbf{\ddot r})}{\| \mathbf{\dot r}\|^3} = 0$$ Multiplying through by $\|\mathbf{\dot r}\|$ gives $$U \ddot r_1 + (\mathbf{\dot r} \cdot \nabla U) \dot r_1 -(\mathbf{\dot r} \cdot \mathbf{\dot r}) \frac{\partial U}{\partial r_1} - \frac{U \dot r_1 (\mathbf{\dot r} \cdot \mathbf{\ddot r})}{\| \mathbf{\dot r}\|^2} = 0$$ A similar equation holds for all $\dot r_i$, so putting everything together we get that $$U \mathbf{\ddot r} + (\mathbf{\dot r} \cdot \nabla U) \mathbf{\dot r} -(\mathbf{\dot r} \cdot \mathbf{\dot r}) \nabla U - \frac{U \mathbf{\dot r} (\mathbf{\dot r} \cdot \mathbf{\ddot r})}{\| \mathbf{\dot r}\|^2} = 0$$ However, the last term goes to zero by assumption (recall $\mathbf{\dot r} \cdot \mathbf{\ddot r} = 0$), so we are done. $\square$

$\endgroup$
2
  • $\begingroup$ I'm really thankful for your answer! There was indeed an assumption that $\dot{\mathbf{r}}\cdot \ddot{\mathbf{r}}=0$. Sorry I missed that out, I never got to that stage of derivation. This question arises from ray tracing in seismology, where $\mathbf{r}(\lambda)$ is the ray parameter for the wavefronts. Ray travel time is $\int_S^R U\sqrt{\dot{\mathbf{r}}\cdot \mathbf{r}} \, d\lambda$, where $U(\mathbf{r})$ is the reciprocal of velocity. Plugging the integrand into the Euler-Lagrange equation extremizes the integral and allows one to obtain the ray tracing equations :) $\endgroup$ May 28, 2020 at 9:14
  • $\begingroup$ Interesting stuff! I'm going to edit the answer and the question in case any one else stumbling on this question has the same confusion. $\endgroup$
    – paulinho
    May 28, 2020 at 14:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .