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.

 

[Vincf]

Upon further reflection, I think that my previous answer is correct except in some particular case. Let us consider the case where :
$$
L(v^2) = \sqrt{v^2} = \sqrt{v_x^2 + v_y^2 + v_z^2}.
$$
Then
$$
\frac{d L}{d v^2} = \frac{1}{2\sqrt{v^2}}.
$$
Therefore,
$$
\frac{\partial L}{\partial \mathbf{v}} = \frac{\mathbf{v}}{\sqrt{v^2}}.
$$
In this case, $\frac{\partial L}{\partial \mathbf{v}} = \text{const}$ only implies that the direction of $\mathbf{v}$ is fixed.
For example, if $\mathbf{v} = f(t)\,\mathbf{e}_x,$ then $\frac{\partial L}{\partial \mathbf{v}} = \mathbf{e}_x$
which is indeed constant. Am I mistaken?

 

[PeroK]

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?

I don't have Landau, but I'm not sure why this needs to be complicated. In this case:
$$L = v_x^2 + v_y^2 + v_z^2$$So:
$$\frac{\partial L}{\partial v_x} = 2v_x$$And, by Euler-Lagrange:
$$a_x = \frac{dv_x}{dt} = 0$$Likewise:
$$a_y = a_z = 0$$The vector derivative notation is ultimately a shorthand for this. If you try to take a shortcut, then it is easy to confuse yourself.

 

[Rick16]

I don't have Landau, but I'm not sure why this needs to be complicated. In this case:

I just wanted to ask why on earth the Lagrangian would be $L=\sqrt {v_x^2 +v_y^2 +v_z^2}$, when I discovered that you had changed that. I spent quite a while this morning trying to work through the problems at the end of chapter 1, and my head was already spinning. This Lagrangian really pushed me over the brink.

The way you wrote it now makes it all look very simple. Thanks a lot.

And to answer your question, why this needs to be complicated: this is what inexperienced people do.

 

[Vincf]

Perhaps I'm missing something, but Landau's aim is to show that the Lagrangian is of this form for a free particle. He does this in the following paragraph 1.4.
In paragraph 1.3, he uses very general arguments to show that the Lagrangian can only depend on the square of the velocity. Nothing more.

 

[Rick16]

Perhaps I'm missing something, but Landau's aim is to show that the Lagrangian is of this form for a free particle. He does this in the following paragraph 1.4.
In paragraph 1.3, he uses very general arguments to show that the Lagrangian can only depend on the square of the velocity. Nothing more.

Yes, I know, but PeroK's suggestion is one way to see why the velocity would have to be constant (provided I already know the form of the Lagrangian). Landau must have had something else in mind.

 

[Vincf]

Landau's goal is to recover the Lagrangian form of a free particle using only translational invariance, spatial isotropy, and the principle of Galilean relativity. Mass appears as a multiplicative constant such that the ratio of masses remains invariant under a change of unit.

It is this level of generality that makes Landau's books so extraordinary and, at the same time, so difficult. The French edition of the mechanics volume comprises only 246 pages and covers an enormous field. Volume 2 is of the same order!

 

[Rick16]

Upon further reflection, I think that my previous answer is correct except in some particular case. Let us consider the case where :
$$
L(v^2) = \sqrt{v^2} = \sqrt{v_x^2 + v_y^2 + v_z^2}.
$$
Then
$$
\frac{d L}{d v^2} = \frac{1}{2\sqrt{v^2}}.
$$
Therefore,
$$
\frac{\partial L}{\partial \mathbf{v}} = \frac{\mathbf{v}}{\sqrt{v^2}}.
$$
In this case, $\frac{\partial L}{\partial \mathbf{v}} = \text{const}$ only implies that the direction of $\mathbf{v}$ is fixed.
For example, if $\mathbf{v} = f(t)\,\mathbf{e}_x,$ then $\frac{\partial L}{\partial \mathbf{v}} = \mathbf{e}_x$
which is indeed constant. Am I mistaken?

I also tried out several different functions for $L(v^2)$, but I found that the velocity was constant for all of them. You apparently found one for which this is not the case. When Landau writes that the Lagrangian can only be a function of the square of the velocity, he actually does not write that it can be any function that containts the square of the velocity.

 

[Vincf]

Yes, but since he reasons in general terms, his conclusion should be true without exception?

As I told you, Landau's reasoning is sometimes extremely rapid. And he was known to be quite exceptional himself in physics. So much so that, when we don't understand something, we always wonder if it's because our brains are too slow.

Note that, even in Feynman's work, with all due respect to that course, we were able to detect quite a few arguments that were at least "incomplete."

 

[Rick16]

Landau's goal is to recover the Lagrangian form of a free particle using only translational invariance, spatial isotropy, and the principle of Galilean relativity. Mass appears as a multiplicative constant such that the ratio of masses remains invariant under a change of unit.

This would be my next problem: I don't understand the paragraph before equation 4.1. But like I already wrote, my head is already spinning, mostly because of problem 3, and it looks like I won't be able to understand anything more today.

 

[PeroK]

Yes, I know, but PeroK's suggestion is one way to see why the velocity would have to be constant (provided I already know the form of the Lagrangian). Landau must have had something else in mind.

Let's allow $L$ to be any function of $v^2$. Euler-Lagrange gives:
$$\frac d {dt}\big (\frac{\partial L}{\partial v^2}\frac{\partial v^2}{\partial v_x} \big ) = 0$$Hence:
$$\frac d {dt}\big (\frac{\partial L}{\partial v^2}v_x\big ) = 0$$And
$$\frac d {dt}\big (\frac{\partial L}{\partial v^2} \big )v_x + \frac{\partial L}{\partial v^2}a_x =0$$We can know work backwards and find the condition (if any) on the function $L(v^2)$ where the solution is a constant $v_x, v_y, v_z$, with $a_x =a_y = a_z = 0$? In this case, $v^2$ is also constant in time, hence any function of $v^2$ is constant in time.

That means that for any function $L(v^2)$, constant velocity motion is a solution to the E-L equations.

 

[Vincf]

Yes, it works in that direction. What I found more difficult was proving that it's the only possibility regardless of the Lagrangian that depends only on the square of the velocity.

 

[PeroK]

Yes, it works in that direction. What I found more difficult was proving that it's the only possibility regardless of the Lagrangian that depends only on the square of the velocity.

Maybe it isn't the only solution. Any function of $v$ is also a function of $v^2$. That said, you could specify that the Taylor expansion for $L(v)$ had only even powers. Perhaps that would work?

 

[kparchevsky]

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?

Let's allow $L$ to be any function of $v^2$. Euler-Lagrange gives:
$$\frac d {dt}\big (\frac{\partial L}{\partial v^2}\frac{\partial v^2}{\partial v_x} \big ) = 0$$Hence:
$$\frac d {dt}\big (\frac{\partial L}{\partial v^2}v_x\big ) = 0$$And
$$\frac d {dt}\big (\frac{\partial L}{\partial v^2} \big )v_x + \frac{\partial L}{\partial v^2}a_x =0$$We can know work backwards and find the condition (if any) on the function $L(v^2)$ where the solution is a constant $v_x, v_y, v_z$, with $a_x =a_y = a_z = 0$? In this case, $v^2$ is also constant in time, hence any function of $v^2$ is constant in time.

That means that for any function $L(v^2)$, constant velocity motion is a solution to the E-L equations.

 

[kparchevsky]

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?

I took a look at this chapter. It is really simple. You just did not get it. What Landau meant is
d/dt (dL/dv) = 0
This follows from the general Lagrange's equation (2.6), because our L does not depend on coordinates q (or r). Now, ANYTHING which time derivative is zero AT ALL TIMES MUST be constant (does not depend on t). Hence, dL/dv (which full time derivative is zero, which follows from the Lagrange's equation) must be constant.

 

[Rick16]

Hence, dL/dv (which full time derivative is zero, which follows from the Lagrange's equation) must be constant.

The fact that dL/dv is constant was never the question. The question was why v is constant.

 

[Rick16]

If L(v) is ABSOLUTELY arbitrary and depends on magnitude v only, I afraid that this statement is not true. Look at the following counter example

@Vincf already had a counter example in #5.

 

[Rick16]

In any case, this is my first reading of the text, and I am satisfied that I can show that $\vec v$ is constant when I assume that the Lagrangian is known. I don't need to show it for an arbitrary function $L(v^2)$ at this point. I may come back to it later, but now I must move on. Thank you everybody for your insights.