Velocity of a free particle using Landau's approach

[Rick16]

TL;DR

trying to show that the velocity of a free particle is constant

I have begun reading Landau's Mechanics. In chapter 1, §3 he writes $\frac {\partial L} {\partial \vec v}=const.$, where $L$ is a function of $v^2$: $L=L(v^2)$. He then writes that from this it follows that $\vec v=const.$

I want to try to show formally that v is constant, but I am not sure if I am doing it right.

I begin by replacing partial derivatives with regular derivatives, because they are easier to write in Latex and I don't think that partial derivatives are needed here anymore. So $\frac {dL(v^2)}{d\vec v}=\frac {dL}{dv^2} \frac {dv^2}{d\vec v}=\frac {dL}{dv^2} \frac {d}{d\vec v}(\vec v \cdot \vec v)=2\vec v\frac {dL}{dv^2}=const.$

Now I take the term $\frac {dL}{dv^2}$ and replace $L$ with $T-U$. Since $U$ is not a function of $v^2$, it follows that $L(v^2)=T(v^2)=\frac{1}{2}mv^2$, and $\frac {dL}{dv^2}=\frac {m}{2}=const.$ Then $2\vec v\frac {dL}{dv^2}=\vec v \cdot const.=const. \Rightarrow \vec v=const.$ Can I do it like this?

 

[Rick16]

This question does not seem to evoke a lot of interest. Could somebody just tell me if it is legitimate to write $\frac{d}{d\vec v}(\vec v \cdot \vec v)=2\vec v$? I.e. I wonder if I can take the derivative directly with respect to the vector as a whole or if I must take the derivative componentwise? The result is the same.

 

[Vincf]

I am not sure you have correctly understood the notation used by Landau. There is a footnote explaining that

$$
\frac{\partial L}{\partial \mathbf{v}}
$$
is a vector whose components are
$$
\frac{\partial L}{\partial \mathbf{v}} =
\begin{pmatrix}
\frac{\partial L}{\partial v_x} \\
\frac{\partial L}{\partial v_y} \\
\frac{\partial L}{\partial v_z}
\end{pmatrix}
$$
Thus, the partial derivatives are indeed necessary ?

Since $L = L(\mathbf{v}^2)$ with $\mathbf{v}^2 = v_x^2 + v_y^2 + v_z^2,$ we have :
$$
\frac{\partial L}{\partial v_x}
=
\frac{\partial L}{\partial (\mathbf{v}^2)}
\frac{\partial (\mathbf{v}^2)}{\partial v_x}
$$
and therefore
$$
\frac{\partial L}{\partial v_x}
=
\frac{\partial L}{\partial (\mathbf{v}^2)} \cdot 2 v_x
$$
$$
\frac{\partial L}{\partial v_y}
=
\frac{\partial L}{\partial (\mathbf{v}^2)} \cdot 2 v_y
$$
$$
\frac{\partial L}{\partial v_z}
=
\frac{\partial L}{\partial (\mathbf{v}^2)} \cdot 2 v_z
$$

If we take the ratio of these expressions, we find that $v_y/v_x$ and $v_z/v_x$ are constants and therefore, we can express everything in terms of $v_x$. So, $v_x$ is a constant.

Finally, we conclude that $v_y$ and $v_z$ are also constant.

Perhaps we can conclude more quickly? To be honest, when I was a student, I sometimes spent quite a bit of time on the "it is obvious that" sections of Landau.

 

[Rick16]

If we take the ratio of these expressions

Thank you, that is an interesting trick with the ratios. It is not likely that I would have thought of it.

Perhaps we can conclude more quickly?

I suppose you know how to do it? Because I don't.