Laundau's derivation of the law of inertia

[aloyisus]

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:

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

and hence

<br /> \frac{\partial \L(v^{2}) }{\partial \mathbf{v}} = constant<br />

From here, Landau immediately goes on to say that "since \frac{\partial \L }{\partial \mathbf{v}} 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. \frac{\partial \L }{\partial \mathbf{v}} is not only a function of the magnitude of velocity, but a function of all the components. Surely the statement f(v) = f(v_{x},v_{y},v_{z}) = constant does not imply (v_{x},v_{y},v_{z}) = constant?

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

<br /> \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<br />

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

<br /> \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}}}<br />

Now, if all the components of the velocity were to scale by the same factor, \mathbf{v} \rightarrow \alpha \mathbf{v}, \alpha 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 v^{2}. He has not assumed it is a linear function of v^{2}. 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.

 

[kcdodd]

Just stabbing in the dark here, but regardless of what velocity does, p=dL/dv is the momentum which is a constant of motion for each coordinate. Looking at the Hamiltonian version, then velocity is v=dH/dp. If H only depends on p, and the p's are constant, then the v's must be constant too. Although, p=dL/dv would have to be invertible to get the Hamiltonian in terms of p's only, putting a restriction on the form of the Lagrangian.

 

[aloyisus]

Thanks for your reply, kcdodd. Any stab is appreciated.

Landau doesn't introduce Hamiltonian formation until much later in the book, so I'm not sure that's where he was going.

His statement "since \frac{\partial \L(v^{2}) }{\partial \mathbf{v}} is a function of the velocity only, it follows that v = constant" is made as though self-evident, and to me it isn't!

Has anybody worked through Landau and can clarify?

Thanks again.

 

[stevenb]

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

<br /> \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<br />

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

<br /> \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}}}<br />

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

In that section of the book, there is the statement that an inertial frame is considered and then a further statement about an inertial frame that "The homogeneity of space and time implies that the Lagrangian cannot contain explicitly either the radius vector (position) of the particle or the time, i.e. L must be a function of the velocity only".

In your counter-example, you are basically saying that velocity is dependent on time. However, since you assumed that the Lagrangian is the magnitude of velocity, you are automatically saying that the velocity can not change in time, nor in position, based on the meaning of Homogeneous Space and Time".

I think the reason that he does not explain, and the reason why you got confused are both the same. It's trivial. Dont' you hate that? Sometimes the most trivial things are the hardest to make sense of.

 

[kcdodd]

I googled this up: http://www.strw.leidenuniv.nl/~icke/ps/SymmetryLagrangian.pdf

Basically, in addition to position and time symmetry, it must also obey gallilean symmetry. So, according to that explanation, there is an additional assumption to get the right lagrangian. It is not enough to simply say it has a dependence only on velocity, but also one which does not alter the lagrangian under translations of the velocity. Your counter example likely does not do that.

 

[aloyisus]

kcdodd, thanks for the link. That's a good companion to the first chapter of Landau, but he doesn't provide any more information at the place where I'm stuck.

If we ignore my counter-example, and I try to state my problem really simply, we have a function \L which is a function of v^{2} only.

v^{2} is defined as v^{2} = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}

We also have, true for each component, \frac{\mathrm{d} }{\mathrm{d} t} \frac{\partial \L}{\partial v_{i}} = 0

Does it follow from just this information that the v_{i} are constant, w.r.t time? That's what the man Landau, and the pdf kcdodd linked to, appear to me to be saying.

stevenb, I'm very happy for this to be a trivial problem - I'm just not seeing it yet!

 

[stevenb]

Does it follow from just this information that the v_{i} are constant, w.r.t time? That's what the man Landau, and the pdf kcdodd linked to, appear to me to be saying.

I don't read either source as saying that. I think you need to take the full context of the discussion into it. The factors that seem relevant are as follows

1. All inertial frames are equivalent and the laws of physics are the same in all inertial frames.

2. The physics for a free particle is homogeneous is space.

3. The physics for a free particle is homogeneous in time.

stevenb, I'm very happy for this to be a trivial problem - I'm just not seeing it yet!

Yes, I agree it's not easy to see, and the Author wants you to do exactly what you are doing: question and prove the result. I just think it's ironic that often these very basic physical concepts are in fact difficult to come to terms with. Often solving a very complex problem using the end-resulting physics is much easier.

HINT:
You have focused on the interesting case to look at. If the Lagrangian is proportional to the magnitude of the velocity, you show that a scale factor of alpha can be considered. If alpha is just a changed constant, then all you are doing is using different units to describe the physics, which is no problem. However, if alpha is a function of time, then you can change the interpretation of the physics such that velocity is constant, but the frame of reference is time dependent. In other words you could put the alpha factor into the coordinate frame scaling. Now your frame is not an inertial frame, but the physics in the frame is still the same (same Lagrangian form). Shouldn't the use of a time dependent frame of reference change the form of the Lagrangian? In particular, wouldn't you expect the Lagrangian to have explicit time dependence, or some fictitious force? Let's say you can answer no to this last question for a point particle, then how would the theory extend to systems with more than one particle? Even without a formal proof, you can see an issue here.

By the way, can you find another counter-example for L that displays this issue?

 

[aloyisus]

Thanks stevenb,

I'm going to go away and think about what you just said ;)

 

[aloyisus]

OK, I think I've finally got it. If anyone's still listening, does my reasoning below make sense?

We have \frac{\mathrm{d} }{\mathrm{d} t} (\frac{\partial \L}{\partial \mathbf{v}}) = 0

which means 2v_{i} \frac{\mathrm{d} \L}{\mathrm{d} v^{2}} = constant for each of the three components.

Now \frac{\mathrm{d} \L}{\mathrm{d} v^{2}} is a function of the magnitude of the velocity only, so for all three equations to be simultaneously satisfied, it is strongly implied that v = constant.

The only way this might not be true would be if \frac{\mathrm{d} \L}{\mathrm{d} v^{2}} \propto \frac{1}{v_{i}}

which would also satisfy the equations. This would then have to be true for all three components simultaneously, which, due to the Lagrangians dependence on v^{2} would imply that v_{x} = v_{y} = v_{z}.

However, we can always choose a different inertial reference frame which breaks this relationship (discounting the trivial solution v = 0).

Therefore \frac{\mathrm{d} \L}{\mathrm{d} v^{2}} cannot be proportional to \frac{1}{v_{i}} and hence v is indeed constant.

Any thoughts? And thanks for reading.

 

[stevenb]

OK, I think I've finally got it. If anyone's still listening, does my reasoning below make sense?

We have \frac{\mathrm{d} }{\mathrm{d} t} (\frac{\partial \L}{\partial \mathbf{v}}) = 0

which means 2v_{i} \frac{\mathrm{d} \L}{\mathrm{d} v^{2}} = constant for each of the three components.

Now \frac{\mathrm{d} \L}{\mathrm{d} v^{2}} is a function of the magnitude of the velocity only, so for all three equations to be simultaneously satisfied, it is strongly implied that v = constant.

The only way this might not be true would be if \frac{\mathrm{d} \L}{\mathrm{d} v^{2}} \propto \frac{1}{v_{i}}

which would also satisfy the equations. This would then have to be true for all three components simultaneously, which, due to the Lagrangians dependence on v^{2} would imply that v_{x} = v_{y} = v_{z}.

However, we can always choose a different inertial reference frame which breaks this relationship (discounting the trivial solution v = 0).

Therefore \frac{\mathrm{d} \L}{\mathrm{d} v^{2}} cannot be proportional to \frac{1}{v_{i}} and hence v is indeed constant.

Any thoughts? And thanks for reading.

Still listening here.

I agree with 99% of what you wrote here. The only thing that I would modify is the statement that v_{x} = v_{y} = v_{z}.

I would instead say {{v_{x}(t)}\over{v_x(t_0)}} = {{v_{y}(t)}\over{v_y(t_0)}} = {{v_{z}(t)}\over{v_z(t_0)}}. Basically, the velocities don't have to be equal, but they would need to scale by the same ratio from a starting value, for example. I believe this is what you were implying (or even said explicitly) in your previous posts. The bottom line is that this relationship can not hold in ALL inertial frames.

I think there are a number of ways to get to your conclusion, which is why I didn't want to force any particular method on you (and probably why the other more knowledgeable people here didn't say more). It's better to come to grips with it on your own terms. I don't know if you've read ahead in the book yet, but his next step is to show you that not only is the velocity constant, but the Lagrangian must be proportional to v^2. So, in effect he automatically bypasses the concern about any other Lagrangian, and what it might imply. This amounts to the same type of argument.

Good work.