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.

  • 2
    $\begingroup$ It's explained right in the thing you quote. (The part in the parentheses) $\endgroup$ Commented Jul 6, 2019 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
    Commented Jul 6, 2019 at 14:25
  • $\begingroup$ I don't understand your comment. $\endgroup$ Commented Jul 6, 2019 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
    Commented Jul 6, 2019 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. $\endgroup$ Commented Jul 6, 2019 at 14:35

1 Answer 1


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.


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