# Proving independence of the lagrangian on position of a free particle using the euler-lagrange equation

I asked a similar question some time back but am trying to work this from another angle.

In deriving the lagrangian of a free particle, we use the homogeneity of space to conclude that the lagrangian does not depend on its position vector $\vec{x}$. By homogeneity of space, I understand that if you displace the initial position of the particle by a vector $\vec{c}$, then all points on the trajectory of the particle are displaced by the same vector $\vec{c}$. Looking at the euler lagrange equation, considering a one degree of freedom case:

If $x_1(t)$ is a solution to the E-L equation corresponding to the initial condition $x(t_1)=X_1$, $$\frac{\partial L(x_1(t),\dot{x}_1(t),t)}{\partial x} - \frac{\mathrm{d} }{\mathrm{d} t} \frac{\partial L(x_1(t),\dot{x}_1(t),t)}{\partial \dot{x}}=0 \tag{1}$$ and $x_1(t) + c$ is also a solution to the E-L equation corresponding to the initial condition $x(t_1)=X_1 + c$, where $c$ is an infinitesimal displacement, $$\frac{\partial L(x_1(t)+c,\dot{x}_1(t),t)}{\partial x} - \frac{\mathrm{d} }{\mathrm{d} t} \frac{\partial L(x_1(t)+c,\dot{x}_1(t),t)}{\partial \dot{x}}=0 \tag{2}$$ then I'd like to prove that $$\frac{\partial L(x,\dot{x},t)}{\partial x}=0 \tag{3}$$

So far, I tried expanding $L(x_1(t)+c,\dot{x}_1(t),t)$ as a taylor series in powers of $c$. Since $c$ is very small, the linear terms in $c$ dominate. I can then reduce eqn. $(2)$ to: $$[L_{xx}-L_{xx\dot{x}}\,\dot{x} - L_{x\dot{x}\dot{x}}\,\ddot{x} - L_{x\dot{x}t}]\,c=0\tag{4}$$

I am not sure how I can proceed from here.

If I can prove that $\frac{\partial L(x,\dot{x},t)}{\partial x}=0$ for an infinitesimal displacement, I can imagine an infinity of such successive displacements $\vec{c}$, making $\frac{\partial L(x,\dot{x},t)}{\partial x}=0$ valid for finite displacements.

• It is simply not true. Take for example $L = x\dot x$, where the Euler-Lagrange-Equation is fulfilled for arbitrary $x(t)$, because it is just a total derivative. Still, $\frac{\partial L}{\partial x} \neq 0$. – Herr_Mitesch May 20 '15 at 16:45
• @Herr_Mitesch. It is true and that's because the Lagrangian is not uniquely determined, however what is really meant is that it is possible to find a Lagrangian that does not depend on x explictly – facenian May 20 '15 at 17:26

• I see your point. If the lagrangian is reformulated as $L(x,\dot{x},t)=\ell(x,\dot{x},t)+\frac{\mathrm{d} F(x,t)}{\mathrm{d} t}$, such that $\frac{\partial \ell(x,\dot{x},t)}{\partial x}=0$, then assuming eqn $(1)$ is true, I can prove the homogeneity of space, i.e., eqn $(2)$. At least I can then say that $\frac{\partial \ell(x,\dot{x},t)}{\partial x}=0$ is a sufficient condition for the homogeneity of space to be true. – IanDsouza May 21 '15 at 9:04