Why Does the Lagrangian of a Free Particle Depend Only on Velocity Magnitude?

[Favicon]

I've heard it said that the Lagrangian of a free particle cannot possibly be a function of any position coordinate, or individual velocity component, but it is a function of the total magnitude of velocity. Why is this the case? I'd be grateful for any pointers in the direction of either a physical or a mathematical explanation

EDIT: I should make it clearer that this is assuming no prior knowledge of kinetic energy or Newton's laws

 

[FeX32]

The Lagrangian can be expressed about any coordinate that can be written in the form:
\begin{equation}
\frac{d}{dt}\left(\frac{∂L}{∂\dot{q}_i}\right)-\frac{∂L}{∂q_i}=Q_i
\end{equation}
Where L is the Lagrangian (kinetic energy - potential energy)

 

[Favicon]

Actually, after a few minutes pondering, I'd like to attempt an answer myself, based on symmetry.

The Lagrangian cannot depend on individual velocity components (but it can depend on the overall magnitude of the velocity) because of rotational symmetry - with no external forces acting on the particle, how could the direction of it's velocity have any impact on the physics of the situation?

As for the positional coordinates, I presume these can't affect the Lagrangian because of translational symmetry. (Again, with no external forces acting on the particle, the location has no bearing on its energy, which after all is what the Lagrangian describes)

Have I got the right idea here, or am I making false assumptions?

 

[FeX32]

Of course, the energies are usually much easier to express in certain coordinate systems.

 

[Favicon]

The Lagrangian can be expressed about any coordinate that can be written in the form:
\begin{equation}
\frac{d}{dt}\left(\frac{∂L}{∂\dot{q}_i}\right)-\frac{∂L}{∂q_i}=Q_i
\end{equation}
Where L is the Lagrangian (kinetic energy - potential energy)

I may be being blindingly stupid here, but I don't see how that answers my original question. What is it about the position and velocity components of a free particle which means they cannot fit that equation?

 

[FeX32]

the location has no bearing on its energy

There you have it

 

[FeX32]

I may be being blindingly stupid here, but I don't see how that answers my original question. What is it about the position and velocity components of a free particle which means they cannot fit that equation?

They can fit the equation. Just find any coordinate "q" that you can write the energies in.

 

[Favicon]

Excellent, thank you Fex. It seems that just after asking for help is when I usually solve a problem myself..

 

[FeX32]

To answer you question directly. There is an infinite amount of coordinate systems that you can express the Lagrangian in. The thing is, usually one or two of them are the simplest, mathematically.

 

[FeX32]

No problem. Yes, it seems you understand it essentially.