Timeline for Why the free Lagrangian does not dependent on the velocity vector direction, only its speed?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Aug 22, 2023 at 13:27 | vote | accept | Giorgi | ||
| Aug 21, 2023 at 19:57 | comment | added | Buzz♦ | @Chemistry The Lagrangian is simply not a vector, any more than your weight is a vector; that's just not how it is defined in the formalism. If the Lagrangian had multiple components (like a vector), each component would produce a separate Euler-Lagrange equation, and they would not generally be consistent. The only want to form a quantity that is linear in $\vec{v}$ and a scalar is to take the dot product of $\vec{v}$ with another vector. | |
| Aug 21, 2023 at 7:13 | comment | added | Giorgi | Thanks for the great analysis. Imagine now that in the thought process, Landau might have come up with linear dependence on velocity, but we don't have $A$ at all. If so, $L$ would be a vector. For example: $L= \frac{1}{2}m(\vec v) = \frac{1}{2}m(v_{x\vec i} + v_{y\vec j} + v_{z\vec z})$. If so, $L$ ends up being a vector. can't euler lagrange be used here ? I know if L is a vector, action would be a vector, but by varional calculus, wouldn't we still arrive at euler lagrange ? if we could, $\frac{\partial L}{ \partial v_x} = \frac{1}{2}m$ and $\frac{d}{dt}(\frac{1}{2}m = 0$ => 0=0. Incorrect? | |
| Aug 21, 2023 at 1:01 | history | answered | Buzz♦ | CC BY-SA 4.0 |