Invariance of Euler-Lagrange Equations and Conserved Quantities

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In this article we discuss some crucial properties that arise from the homogeneity and isotropy of space and time. We will identify some quantities that as the direct consequence of homogeneity and isotropy are conserved under a certain symmetry transformation.

\begin{align} \text{homogeneity of time} &\rightarrow \text{Energy Conservation}\\ \text{homogeneity of space} &\rightarrow \text{Momentum Conservation}\\ \text{isotropy of space} &\rightarrow \text{Angular Momentum Conservation}\\ \end{align}

Contents

Homogeneity of Time and Energy Conservation

If you calculate the equations of motion for a Lagrangian starting for time $t_0$ to $t_1$, does the Lagrangian and hence the equations of motion change if we calculate the equations of motion for the same time span but starting at a later time? That is, if we shift time by $\Delta t$ and make the transformation $t \rightarrow t'=t+\Delta t$ does the Lagrangian change? To answer this question we analyze the total time derivative of the Lagrangian. \begin{align} \frac{d L}{dt}=\sum_i \frac{\partial L}{\partial q_i}\dot{q_i}+\frac{\partial L}{\partial \dot{q}_i}\ddot{q}_i \end{align} If the Lagrange function would be time dependent we would get an additional term with the partial derivative of the Lagrangian with respect to time $\frac{\partial L}{\partial t}$. By use of the Euler-Lagrange Equation, we replace $\frac{\partial L}{\partial q_i}$ by $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i}$. \begin{align} \frac{d L}{dt}=\sum_i \dot{q_i} \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i}+\frac{\partial L}{\partial \dot{q}_i}\ddot{q}_i \end{align} Now we see, that this is nothing but the derivative of the product \begin{align} \frac{d L}{dt}=\sum_i\frac{d}{dt} \left( \dot{q_i} \frac{\partial L}{\partial \dot{q}_i} \right) \end{align} Rearranging terms we have \begin{align} \frac{d}{dt} \left( \sum_i \dot{q_i} \frac{\partial L}{\partial \dot{q}_i} -L \right)=0 \end{align} Thus whatever is in the Brackets is conserved. \begin{align} \sum_i \dot{q_i} \frac{\partial L}{\partial \dot{q}_i}-L = const. \tag{1} \end{align} We define this constant as the Hamiltionian $H$. Since there is no explicit time dependence of the Lagrangian, the Lagrangian is invariant with respect to time transformations $t \rightarrow t'=t+\Delta t$. So far so good, but what does the conserved quantity represent? For a Lagrange function of the form $L=T(\dot{q}_i)-V(q_i)$ it is true that \begin{align} \sum_i \dot{q_i}\frac{\partial L}{\partial \dot{q}_i} = \sum_i \dot{q_i}\frac{\partial T}{\partial \dot{q}_i} = 2T \end{align} now plugging this into, equation (1) \begin{align} H=\sum_i \dot{q_i} \frac{\partial L}{\partial \dot{q}_i} - L \end{align} becomes \begin{align} H=2T-L=2T-T+V=T+V=E \end{align}

Time Symmetry: A time homogeneous (independent) Lagrangian is invariant under time translation. The energy of the system is conserved and equals the Hamiltonian.

The energy in Cartesian coordinates is \begin{align} E=\sum_i\frac{m_i v_i^2}{2} + U(r_1,r_2,\dots) \end{align}

Homogeneity of Space and Momentum Conservation

Homogeneity of space causes another invariance. If we translate every particle in a system by a small amount $\vec{\varepsilon}$, we claim that the Lagrangian doesn't change. Let's analyze the change in the Lagrangian in more detail. After transformation we have new coordinates $r'=r+\varepsilon$. If the Lagrangian doesn't change the following must hold \begin{align} \sum_i \frac{\partial L}{\partial r_i} =\sum_i \frac{\partial L}{\partial r'_i} \end{align} which is indeed true since $\frac{\partial r'}{\partial r_i}=1$ for all $i$ \begin{align} \sum_i \frac{\partial L}{\partial r_i} =\sum_i \frac{\partial L}{\partial r'_i} \frac{\partial r'_i}{\partial r_i} \end{align}

Translational Symmetry: A(n) (active) transformation of all particles or a (passive) transformation of the coordinate system does not alter the Lagrangian.

Thus the change in the Lagrangian \begin{align} \delta L = \sum_i \frac{\partial L}{\partial r_i} \delta r_i=\vec{\varepsilon} \sum_i \frac{\partial L}{\partial r_i} \end{align} must be zero. Since $\vec{\varepsilon}$ is arbitrary, $\delta L =0$ must be zero \begin{align} \sum_i \frac{\partial L}{\partial r_i} =0 \end{align} Rewriting this equation by use of the Euler-Lagrange Equation \begin{align} \sum_i \frac{d}{dt}\frac{\partial L}{\partial v_i} =\frac{d}{dt} \sum_i \frac{\partial L}{\partial v_i}=0 \end{align}

we find that the Momentum is conserved \begin{align} p=\sum_i \frac{\partial L}{\partial v_i}=\sum_i m_iv_i=const. \end{align}

The term $\frac{\partial L}{\partial r_i}$ can be interpreted as the the force acting on the $i$-th particle \begin{align} \frac{\partial L}{\partial r_i}=-\frac{\partial U}{\partial r_i} \end{align} For a closed system the sum of all forces is zero. \begin{align} \sum_i F_i=0 \end{align} In Cartesian coordinates the canonical momentum coincides with the components of the vector $P_i$. For other generalized coordinates can be a more complicated relation than just the product of mass and velocity.

Isotropy of Space and Angular Momentum Conservation

Isotropy means the property that rotational translation does not alter the Lagranian. An infinitesimal rotation $\delta \vec{\varphi}$ leaves the Lagrangian invariant, as it's demonstrated below \begin{align} \begin{pmatrix} x'\\ y' \end{pmatrix}=\begin{pmatrix} \cos \delta\varphi & \sin \delta\varphi \\ -\sin \delta\varphi & \cos \delta\varphi \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix} \end{align} For very small $\delta \varphi$ the approximations \begin{align} \cos(\delta \varphi) \approx 1\\ \sin(\delta \varphi) \approx \delta \end{align} apply and the rotation matrix for an infinitesimal rotation becomes \begin{align} \begin{pmatrix} \delta x'\\ \delta y' \end{pmatrix}=\begin{pmatrix} 1 & \delta \\ -\delta & 1 \end{pmatrix}\begin{pmatrix} \delta x\\ \delta y \end{pmatrix} \end{align} Likewise we have for the velocity \begin{align} \begin{pmatrix} \delta \dot{x}'\\ \delta \dot{y}' \end{pmatrix}=\begin{pmatrix} 1 & \delta \\ -\delta & 1 \end{pmatrix}\begin{pmatrix} \delta \dot{x}\\ \delta \dot{y} \end{pmatrix} \end{align} What we are interested in, is whether or not the rotation of the coordinate system does effect the Lagrangian, which for a particle in two dimensions is \begin{align} \mathcal{L}=\frac{m}{2} (\dot{x}^2+\dot{y}^2)-V(x^2+y^2). \end{align} From the Lagrangian we see immediately that if $\dot{x}^2+\dot{y}^2$ and $x^2+y^2$ are invariant under the rotational transformation, so is the Lagrangian. The change in radius $r^2=x^2+y^2$ is \begin{align} \delta (x^2+y^2)=2x \cdot \delta x + 2y \cdot \delta y=2x \cdot \delta x + 2y \cdot \delta y \end{align} From the rotation matrix we have \begin{align} x'&=x+y \cdot \delta \Rightarrow \delta x'=y \cdot \delta\\ y'&=-x \cdot \delta +y \Rightarrow \delta y'=-x \cdot \delta \end{align} thus the change in $r$ is \begin{align} \delta (x^2+y^2)=2x \cdot \delta x + 2y \cdot \delta y=2xy \cdot \delta + 2y(-x) \cdot \delta =0 \end{align}

The same calculation for radial velocity $\dot{r}^2=\dot{x}^2+\dot{y}^2$ gives \begin{align} \delta (\dot{x}^2+\dot{y}^2)=2\dot{x} \cdot \delta \dot{x} + 2\dot{y} \cdot \delta \dot{y}=2\dot{x}\dot{y} \cdot \delta + 2\dot{y}(-\dot{x}) \cdot \delta =0 \end{align}

Rotational Symmetry: The radius as well as the radial velocity do not change under rotational transformation thus the Lagrangian is invariant under rotatitional translation.

For a particle with position vector $\mathbf{r}$ from the origin. The change in $\delta \mathbf{r}$ due to the infinitesimal rotation is proportional to the distance $\mathbf{r}$. \begin{align} \delta \mathbf{r} = \delta \vec{\varphi} \times \mathbf{r} \end{align} the time derivative of this relation gives a similar expression for velocity \begin{align} \delta \mathbf{v} = \delta \vec{\varphi} \times \mathbf{v} \end{align} The change in the Lagrangian due to rotation is \begin{align} \delta L = \sum_i \left( \frac{\partial L}{\partial r_i}\delta r_i +\frac{\partial L}{\partial v_i}\delta v_i \right) =0. \end{align} These equations together with the definitions \begin{align} \frac{\partial L}{\partial r_i}=\dot{p_i} \end{align} \begin{align} \frac{\partial L}{\partial v_i}=p_i \end{align} give \begin{align} \delta L = \left( \dot{\mathbf{p}} \cdot \delta \vec{\varphi} \times \mathbf{r} +\mathbf{p} \cdot \delta \vec{\varphi} \times \mathbf{v} \right) =0\\ \end{align} we use that even permutations of $\dot{\mathbf{p}} \cdot \delta \vec{\varphi} \times \mathbf{r}$ are the same $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}), $ \begin{align} \delta L = \delta \vec{\varphi} \left( \mathbf{r} \times \dot{\mathbf{p}} +\mathbf{p} \times \mathbf{v} \right) =0\\ \delta L = \delta \vec{\varphi} \frac{d}{dt} ( \mathbf{r} \times \mathbf{p} ) =0\\ \end{align} $\delta \vec{\varphi}$ is arbitrary, thus for the Lagrangian to be invariant we need \begin{align} \frac{d}{dt} \sum_i \mathbf{r} \times \mathbf{p} =0\\ \end{align} to be zero.

Which means the angular momentum is conserved. \begin{align} \mathbf{L} = \mathbf{r} \times \mathbf{p}=const.\\ \end{align}

General Conserved Quantities

A general transformation of the form \begin{align} \delta q_i = f_i(q) \cdot \delta \tag{2} \end{align} doesn't change the Lagrangian. From this general symmetrie we get \begin{align} \delta L &= \sum_i \left( \frac{\partial L}{\partial q_i}\delta q_i +\frac{\partial L}{\partial \dot{q}_i}\delta \dot{q}_i \right)\\ &= \sum_i \left( \dot{p}_i\delta q_i + p_i \delta \dot{q}_i \right)\\ \sum_i \frac{d}{dt}(p_i \cdot \delta q_i) \end{align} and this is by (2) \begin{align} \delta L &=\frac{d}{dt} \sum_i p_i \cdot f_i(q) \cdot \delta=0 \end{align} and the quantity we call $Q$ is conserved \begin{align} Q &= \sum_i p_i \cdot f_i(q) \cdot \delta \end{align}




Further Reading:

  • L. D. Landau & E. M. Lifschitz - Lehrbuch der Theoretischen Physik I: Mechanik