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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.

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    $\begingroup$ 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, $\frac{\partial L}{\partial x} \neq 0$. $\endgroup$
    – Herr_Mitesch
    May 20, 2015 at 16:45
  • $\begingroup$ @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 $\endgroup$
    – facenian
    May 20, 2015 at 17:26

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As the counter example given by Herr_Mitesh shows it is not true and this is because the lagrangian is not uniquely determined. In physics sometimes you don't have to think like in mathematics and in this case you must content yourself thinking that if the lagrangian does not contain x as a variable that is enough for the condition of homogeneity to be fulfilled

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  • $\begingroup$ 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. $\endgroup$
    – Matrix23
    May 21, 2015 at 9:04
  • $\begingroup$ I was just wanted to check if the last part of the question: the logic of extending this from infinitesimal to finite displacements, was a valid one. $\endgroup$
    – Matrix23
    May 21, 2015 at 9:17
  • $\begingroup$ At least in on variable I think the extension to finite displacments is just one the funtamental theorems of calculus $\endgroup$
    – facenian
    May 21, 2015 at 9:46

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