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aloyisus

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I'm coming back to physics after a long absence of fifteen years. Starting with classical mechanics, I thought I'd read Landau & Lifgarbagez, since I already have it. I know it's famously concise, but I didn't think I'd be stuck on page 5. If anyone can help me unravel the following, I'll be most grateful.

Landau derives the law of inertia by considering a free particle moving in an inertial frame of reference. Homogeneity of space & time means the Lagrangian can only depend on the magnitude of the velocity, and so it is some function of the velocity squared.

From Lagrange's equation, we have:

[tex]

\frac{\mathrm{d} }{\mathrm{d} t} (\frac{\partial \L(v^{2}) }{\partial \mathbf{v}}) = 0

[/tex]

and hence

[tex]

\frac{\partial \L(v^{2}) }{\partial \mathbf{v}} = constant

[/tex]

From here, Landau immediately goes on to say that "since [tex]\frac{\partial \L }{\partial \mathbf{v}}[/tex] is a function of the velocity only, it follows that v = constant", and there we have the law of inertia in an inertial reference frame.

It is that step I don't follow. [tex]\frac{\partial \L }{\partial \mathbf{v}}[/tex] is not only a function of the magnitude of velocity, but a function of all the components. Surely the statement [tex]f(v) = f(v_{x},v_{y},v_{z}) = constant[/tex] does not imply [tex](v_{x},v_{y},v_{z}) = constant[/tex]?

To kick it around a bit further, let me consider one component of the equation in more detail:

[tex]

\frac{\partial \L(v^{2})}{\partial v_{x}} = \frac{\mathrm{d} \L(v^{2})}{\mathrm{d} v^{2}} . \frac{\partial v^{2}}{\partial v_{x}} = 2v_{x} \frac{\mathrm{d} \L(v^{2})}{\mathrm{d} v^{2}} = constant

[/tex]

One possible solution is indeed that [tex]v_{x}[/tex], and by extension [tex]v_{y}[/tex] and [tex]v_{z}[/tex] are constant. However, it's possible to propose a form for the Lagrangian [tex]\L (v^{2}) = (v^{2})^{\frac{1}{2}}[/tex] which will give us:

[tex]

\frac{\partial \L}{\partial v_{x}} = \frac{v_{x}}{(v^{2})^{\frac{1}{2}}} = \frac{v_{x}}{(v_{x}^{2}+v_{y}^{2}+v_{z}^{2})^{\frac{1}{2}}}

[/tex]

Now, if all the components of the velocity were to scale by the same factor, [tex]\mathbf{v} \rightarrow \alpha \mathbf{v}[/tex], [tex]\alpha[/tex] would cancel and this would also be an allowed velocity - i.e. constant direction but changing magnitude.

Clearly the Lagrangian I've proposed is plain wrong, and if we were to plug-in the actual Lagrangian, it is clear that the velocity must be constant. But the point is that

Can anyone explain Laundau's reasoning to me? Or where my own reasoning is broken?

Many thanks in advance.

Landau derives the law of inertia by considering a free particle moving in an inertial frame of reference. Homogeneity of space & time means the Lagrangian can only depend on the magnitude of the velocity, and so it is some function of the velocity squared.

From Lagrange's equation, we have:

[tex]

\frac{\mathrm{d} }{\mathrm{d} t} (\frac{\partial \L(v^{2}) }{\partial \mathbf{v}}) = 0

[/tex]

and hence

[tex]

\frac{\partial \L(v^{2}) }{\partial \mathbf{v}} = constant

[/tex]

From here, Landau immediately goes on to say that "since [tex]\frac{\partial \L }{\partial \mathbf{v}}[/tex] is a function of the velocity only, it follows that v = constant", and there we have the law of inertia in an inertial reference frame.

It is that step I don't follow. [tex]\frac{\partial \L }{\partial \mathbf{v}}[/tex] is not only a function of the magnitude of velocity, but a function of all the components. Surely the statement [tex]f(v) = f(v_{x},v_{y},v_{z}) = constant[/tex] does not imply [tex](v_{x},v_{y},v_{z}) = constant[/tex]?

To kick it around a bit further, let me consider one component of the equation in more detail:

[tex]

\frac{\partial \L(v^{2})}{\partial v_{x}} = \frac{\mathrm{d} \L(v^{2})}{\mathrm{d} v^{2}} . \frac{\partial v^{2}}{\partial v_{x}} = 2v_{x} \frac{\mathrm{d} \L(v^{2})}{\mathrm{d} v^{2}} = constant

[/tex]

One possible solution is indeed that [tex]v_{x}[/tex], and by extension [tex]v_{y}[/tex] and [tex]v_{z}[/tex] are constant. However, it's possible to propose a form for the Lagrangian [tex]\L (v^{2}) = (v^{2})^{\frac{1}{2}}[/tex] which will give us:

[tex]

\frac{\partial \L}{\partial v_{x}} = \frac{v_{x}}{(v^{2})^{\frac{1}{2}}} = \frac{v_{x}}{(v_{x}^{2}+v_{y}^{2}+v_{z}^{2})^{\frac{1}{2}}}

[/tex]

Now, if all the components of the velocity were to scale by the same factor, [tex]\mathbf{v} \rightarrow \alpha \mathbf{v}[/tex], [tex]\alpha[/tex] would cancel and this would also be an allowed velocity - i.e. constant direction but changing magnitude.

Clearly the Lagrangian I've proposed is plain wrong, and if we were to plug-in the actual Lagrangian, it is clear that the velocity must be constant. But the point is that

*Landau is not using any knowledge of the actual Lagrangian beyond it's dependency on*[tex]v^{2}[/tex]. He has not assumed it is a linear function of [tex]v^{2}[/tex]. So how is he able to make the statement that v = constant?Can anyone explain Laundau's reasoning to me? Or where my own reasoning is broken?

Many thanks in advance.

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