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We have the Lagrangian equation:

$$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial\dot{q_J}}\right)-\frac{\partial L}{\partial\dot{q_J}}=0$$

For the Motion of one particle: using Cartesian coordinates:

$$L=T=\frac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2)$$

Thus we have:

$$\frac{\partial L}{\partial\dot{x}}=m\dot{x}\qquad\qquad\frac{\partial L}{\partial\dot{y}}=m\dot{y}\qquad\qquad\frac{\partial L}{\partial\dot{z}}=m\dot{z}$$

I have having trouble deriving the following: $$\frac{\partial T}{\partial x}=\frac{\partial\left(\frac{1}{2}m\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right)\right)}{\partial x}=\frac{1}{2}m\frac{\partial\left(\left(\dot{x}^2\right)\right)}{\partial x}=\,\,\,...$$

I do not know how to solve the equation

The final answer given in the example is:

$$\frac{\mathrm{d}}{\mathrm{d}t}(m\dot{x})=F_x\qquad\qquad\frac{\mathrm{d}}{\mathrm{d}t}(m\dot{y})=F_y\qquad\qquad\frac{\mathrm{d}}{\mathrm{d}t}(m\dot{z})=F_z$$

jng224
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Jack Jack
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    Related: https://physics.stackexchange.com/q/885/2451 and links therein. – Qmechanic Feb 15 '21 at 15:44
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    $x$ and $\dot x$ are independent variables when comes to the derivatives of the Lagrangian. The derivative of an independent variable by another independent variable is simply zero. So $\frac{\partial \dot x}{\partial x} = 0$. – Hans Wurst Feb 15 '21 at 15:54

1 Answers1

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There isn't much to solve actually. The Euler-Lagrange equation is $$ \frac{d}{dt}\frac{\partial L(r,\dot r,t)}{\partial \dot r_i} = \frac{\partial L(r,\dot r,t )}{\partial r_i} $$ where $r$ stands for the collection of all coordinates $r_i \in r$. In your case $r=(x,y,z$). The Lagrange function for a free particle is simply $$ L(r,\dot r,t) = \frac{1}{2}m(\dot x^2+ \dot y^2+ \dot z ^2 ) = L(\dot r) $$ which is a special case of a Lagrange function that does not depend on $\dot r $ and $t$.

The Euler-Lagrang equation tells us now how to derive the differential equations that need to be solved to obtain the trajectories of the system.

In the derivatives $\frac{\partial L}{\partial r}$ and $\frac{\partial L}{\partial \dot r}$ we treat $r$ and $\dot r$ like independent variables following normal differentiation rules as you know them from normal functions. This leads to

$$ \frac{\partial L(\dot r)}{\partial x} = \frac{\partial }{\partial x}\frac{1}{2}m(\dot x^2+ \dot y^2+ \dot z ^2 ) = 0 \\ \frac{\partial L(\dot r)}{\partial y} = \frac{\partial }{\partial y}\frac{1}{2}m(\dot x^2+ \dot y^2+ \dot z ^2 ) = 0\\ \frac{\partial L(\dot r)}{\partial z} = \frac{\partial }{\partial z}\frac{1}{2}m(\dot x^2+ \dot y^2+ \dot z ^2 ) = 0 $$ The derivatives with respect to $\dot x, \dot y, \dot z$ are $$ \frac{\partial L(\dot r)}{\partial \dot x} = \frac{\partial }{\partial \dot x}\frac{1}{2}m(\dot x^2+ \dot y^2+ \dot z ^2 ) = m\dot x \\ \frac{\partial L(\dot r)}{\partial \dot y} = \frac{\partial }{\partial \dot y}\frac{1}{2}m(\dot x^2+ \dot y^2+ \dot z ^2 ) = m\dot y\\ \frac{\partial L(\dot r)}{\partial \dot z} = \frac{\partial }{\partial \dot z}\frac{1}{2}m(\dot x^2+ \dot y^2+ \dot z ^2 ) = m \dot z $$

Now we can plug in the derivatives into the Euler-Lagrange equation, $$\begin{aligned} \frac{d}{dt}\frac{\partial L(\dot r)}{\partial \dot x} &= \frac{\partial L(\dot r)}{\partial x} \\ \frac{d}{dt}m\dot x &= 0 \end{aligned}$$ which leads to $$ m\ddot x = 0 $$

This is the same as $$ma = F$$ with $F=0$, as expected for a free particle.

During the formalism, we sort of use $x, \dot x$ with two different meanings. In the derivatives they are simply independent variables of the Lagrange function. So treat the Lagrange function like you would treat any other function with different variables, for example $$ f(a,b) = b\\ \frac{\partial f}{\partial a} = 0 $$ Once you are done with the derivatives, you substitute the independent variable $x$ with $x(t)$ before evaluating the time derivative to obtain the differential equation for the trajectory $x(t)$.

Hans Wurst
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