The Lagrangian of a free particle. (mv²/2)

[Coffee_]

Hello everyone, I've been having trouble with the following reasoning for a while. The book I use for learning is Landau and Lifschitz vol1.

When the concept of the Lagrangian is introduced in textbooks it is some abstract function of the position vector, velocity vector and time. Then they try to derive the form of the lagrangian for a particle as if it was in an empty universe. The form of the Lagrangian for such a particle from any frame that describes space as homogeneous and isotropic (inertial frame) is argumented as will follow.

Let's say the particle follows a certain path in space. According to our minimal action principle the action is minimal here. Now the question is, if the particle traveled EXACTLY the same path, only starting from a different point in our coordinate system, would the action suddenly be different? (So the path hasn't rotated or anything, just translated so that the starting point starts somewhere else)

The answer is 'no' because space is homogeneous. This ''path'' that the particle has made should be the minimal action path no matter where in my coordinate system the particle started. So the big conclusion that I agree with is:

1) Action being minimal for this path should not depend on the position of this path in my coordinate system

In most textbooks I've encountered this conclusion in bold letters isn't mentioned and they instantly go form space being homogeneous to ''Lagrangian independent on the position vector''.

I have a lot of trouble with this step because while it sounds intuitive that this should be the case there is still an integral operator between the action and the Lagrangian and I'm interested in a more formal treatment of this thought step.

The same argumentation goes for time and orientation of the velocity vector and eventually with some other arguments they find L=mv²/2.

 

[maajdl]

Try to figure out what would happen if the the assumption was removed.

 

[Coffee_]

Try to figure out what would happen if the the assumption was removed.

The assumption in bold letters? I totally agree with that.

Or are you talking about the assumption that the Lagrangian should be independent of r?

 

[WannabeNewton]

Space being homogenous means that $x^i \rightarrow x^i + dx^i$ is a symmetry. A free particle Lagrangian must respect this symmetry so $L(x^i + dx^i) = L(x^i)$ which implies $\frac{\partial L}{\partial x^i} = 0$ so $L = L(v^2,\hat{v}, t)$ where $\hat{v}$ is the unit vector in the direction of velocity.

Space being isotropic means that $\theta \rightarrow \theta + d\theta, \phi \rightarrow \phi + d\phi$ is a symmetry and again the free particle Lagrangian must respect this so $\partial_{\theta}L = \partial_{\phi}L = 0$ which implies $L = L(v^2, t)$. Similarly applying homogeneity in time reduces the free particle Lagrangian to $L(v^2)$.

 

[Coffee_]

Space being homogenous means that $x^i \rightarrow x^i + dx^i$ is a symmetry. A free particle Lagrangian must respect this symmetry so $L(x^i + dx^i) = L(x^i)$ which implies $\frac{\partial L}{\partial x^i} = 0$ so $L = L(v^2,\hat{v}, t)$ where $\hat{v}$ is the unit vector in the direction of velocity.

Space being isotropic means that $\theta \rightarrow \theta + d\theta, \phi \rightarrow \phi + d\phi$ is a symmetry and again the free particle Lagrangian must respect this so $\partial_{\theta}L = \partial_{\phi}L = 0$ which implies $L = L(v^2, t)$. Similarly applying homogeneity in time reduces the free particle Lagrangian to $L(v^2)$.

My problem with this reasoning is that space being homogeneous doesn't say anything about the Lagrangian. Although the lagrangian is going to decide how our particle moves our ''LAW'' is that the action is minimal for the correct path. So if we want to do the reasoning correctly we can't go from space being homogeneous to the Lagrangian instantly. It has to be space homogeneous -> our law is the same everywhere in space -> action is minimal independent of where the particle moves.

Now I'm interested how to formally go from this conclusion to the Lagrangian.

 

[jameswhite]

The lagrangian is different for different systems, the logic here is simply that if take it as a given that space is homogenous then it equivalent to say that the lagrangian cannot depend on the raw position. I think the problem you are having is that you are trying to go from the action principle to the lagrangian. It doesn't work this way, think of them as separate issues.

The principle of least action is a law, yes. But it applies to anything and everything. The lagrangian here is not being derived from the principle of least action. The statement is merely that the lagrangian cannot depend on the position with respect to some arbitrary coordinate system. That is what we mean when we say that space is homogenous.

The principle of least action (PLA) is the tool that we use to derive the Euler-lagrange equations from the general form of the lagrangian. I think what you need to do is to separate the two things in your mind. The PLA is one thing, the concept of a free particle is another. If you actually wanted to calculate the equation of motion of a free particle from it's lagrangian then you would employ the PLA. But for the sakes of determining the lagrangian we don't use it at all. The PLA is for determining the trajectory of a system, the symmetries come before that.