4
$\begingroup$

I'm in trouble with comprehending derivation of Lagrangian of a particle in non-inertial , translational and rotational frame of reference by Landau's Mechanics.
More precisely I don't understand why the following can happen.

$$ L^\prime=\frac{1}{2}m{v^\prime}^2-m\mathbf{W}(t)\cdot\mathbf{r}^\prime-U \tag{39.4} $$

The velocity $\mathbf{v}'$ of the particle relative to $K'$ is composed of its velocity $\mathbf{v}$ relative to $K$ and the velocity $\mathbf{\Omega}\times\mathbf{r}$ of its rotation with $K$: $\mathbf{v}^\prime=\mathbf{v}+\mathbf{\Omega}\times\mathbf{r}$ (since the radius vectors $\mathbf{r}$ and $\mathbf{r}'$ in the frames $K$ and $K'$ coincide). Substituting this in the Lagrangian (39.4), we obtain

$$ L=\frac{1}{2}mv^2+m\mathbf{v}\cdot\mathbf{\Omega}\times \mathbf{r}+\frac{1}{2}m(\mathbf{\Omega}\times \mathbf{r})^2-m\mathbf{W}\cdot \mathbf{r}-U \tag{39.6} $$

(in p.127, L.D. Landau and E.M. Lifshitz Mechanics )

The problem is the rectlinear term. Why the second term in (39.4) $m\mathbf{W}\cdot\mathbf{r}^\prime$ simply converts into the third term in (39.6) $m\mathbf{W}\cdot\mathbf{r}$. Radius vectors $\mathbf{r}$ and $\mathbf{r}'$ have rotational relation, hence I guess it can't be just replaced each other.

$\endgroup$
  • 2
    $\begingroup$ It's explained right in the thing you quote. (The part in the parentheses) $\endgroup$ – Aaron Stevens Jul 6 at 14:13
  • $\begingroup$ @AaronStevens I think the thing they share is only their origin and the watching direction of $K$ and $K'$ are different. How can they be identical? $\endgroup$ – darkspider Jul 6 at 14:25
  • $\begingroup$ I don't understand your comment. $\endgroup$ – Aaron Stevens Jul 6 at 14:27
  • $\begingroup$ @AaronStevens Sorry, English is not my usual language. I meant why does it happen that $r$ and $r^\prime$ coincide in spite that their frame are different. $\endgroup$ – darkspider Jul 6 at 14:33
  • 1
    $\begingroup$ Because it is the position vector. It points from the origin to the object in question. This doesn't depend on the reference frame. $\endgroup$ – Aaron Stevens Jul 6 at 14:35
2
$\begingroup$

Expanding on AaronStevens's above comment: The reference frames $K$ and $K^{\prime}$ share the same origin. Therefore position vectors ${\bf r}={\bf r}^{\prime}$ (measured relative to the origin) are the same.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.