- #1

ian_dsouza

- 48

- 3

\vec{c}

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, 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) = x

_{1}(t) is a solution to the Euler-Lagrange equation corresponding to initial conditions x(t

_{1}) = X

_{1}and v(t

_{1}) = V

_{1}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) = x

_{1}(t)+c is also a solution to the equation corresponding to the initial conditions x(t

_{1}) = X

_{1}+c and v(t

_{1}) = V

_{1}, 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.