Homogeneity of space and the form of the Lagrangian

[ian_dsouza]

I was reading that the homogeneity of space can lead to the conclusion that the lagrangian of a free particle is not explicitly dependent on its position. At the moment, this does not come very intuitively to me. By homogeneity, I understand that if you displace the initial position of a particle by a vector $$\vec{c}$$
, then all points on the trajectory of the particle are displaced by the same vector 'c'.

I am trying to work with the Euler lagrange equation. For now, consider the 1-dimensional case: $$L(x,\dot{x},t)$$ If x(t) = x₁(t) is a solution to the Euler-Lagrange equation corresponding to initial conditions x(t₁) = X₁ and v(t₁) = V₁ as in:
$$\frac{\partial L(x_1(t),\dot{x}_1(t),t)}{\partial x} - \frac{\mathrm{d} }{\mathrm{d} t} \frac{\partial L(x_1(t),\dot{x}_1(t),t)}{\partial \dot{x}}=0 \tag{1}$$then I'd like to prove that if x(t) = x₁(t)+c is also a solution to the equation corresponding to the initial conditions x(t₁) = X₁ +c and v(t₁) = V₁, as in:$$\frac{\partial L(x_1(t)+c,\dot{x}_1(t),t)}{\partial x} - \frac{\mathrm{d} }{\mathrm{d} t} \frac{\partial L(x_1(t)+c,\dot{x}_1(t),t)}{\partial \dot{x}}=0 \tag{2}$$then this must mean that $$\frac{\partial L(x,\dot{x},t)}{\partial x}=0 \tag{3}$$
I tried expanding the term $$L(x_1(t)+c,\dot{x}_1(t),t)$$ with Taylor's theorem. Assuming 'c' is not necessarily small, the series has an infinite number of terms. The LHS in eqn. (2) evaluated to zero using the identity in eqn. (1). But I am not sure how to prove eqn (3).

Any help is much appreciated.

 

[gtoguy97]

The homogeneity of space implies that the Lagrangian of a free particle is not explicitly dependent on its position. To see this, we can consider the Euler-Lagrange equation for a single particle:$$\frac{\partial L(x, \dot{x}, t)}{\partial x} - \frac{d}{dt}\left(\frac{\partial L(x, \dot{x}, t)}{\partial \dot{x}}\right) = 0$$If we assume that the Lagrangian is homogeneous of degree n in the position variable, then it can be written as$$L(x, \dot{x}, t) = x^n \tilde{L}(\dot{x}, t)$$where $\tilde{L}$ does not depend on x explicitly. Substituting this into the Euler-Lagrange equation yields$$x^n\frac{\partial \tilde{L}(\dot{x}, t)}{\partial x} - x^n \frac{d}{dt}\left(\frac{\partial \tilde{L}(\dot{x}, t)}{\partial \dot{x}}\right) = 0$$But since $x^n \neq 0$, we must have that $$\frac{\partial \tilde{L}(\dot{x}, t)}{\partial x} = 0.$$This implies that the Lagrangian does not depend on the position explicitly.