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

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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 travelled 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.

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