diff options
author | Preston Pan <preston@nullring.xyz> | 2024-05-29 19:24:55 -0700 |
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committer | Preston Pan <preston@nullring.xyz> | 2024-05-29 19:24:55 -0700 |
commit | c35d434272e09256030cb113ec62b72c3dbae84f (patch) | |
tree | f4cb8599abda34eaaf6cf89d8c603c25e5a9a517 /mindmap/Lagrangian mechanics.org | |
parent | e0a4aa29f8ccbc9e86ec202be199756aea2528b6 (diff) |
new files; more mindmap
Diffstat (limited to 'mindmap/Lagrangian mechanics.org')
-rw-r--r-- | mindmap/Lagrangian mechanics.org | 10 |
1 files changed, 8 insertions, 2 deletions
diff --git a/mindmap/Lagrangian mechanics.org b/mindmap/Lagrangian mechanics.org index d306be7..c3e078d 100644 --- a/mindmap/Lagrangian mechanics.org +++ b/mindmap/Lagrangian mechanics.org @@ -19,7 +19,10 @@ J[f] = \int_{a}^{b}L(t, f(t), f'(t))dt \\ Defines the actual relationship between $f(t)$ and its level of optimization, where $a$ and $b$ represent the start and end points for a certain curve. For example, if you wanted to minimize the surface area of something, $a$ and $b$ would be the starting and end points of the surface. -* Euler-Lagrange Equation +* Euler-Lagrange equation +:PROPERTIES: +:ID: aaba4bf0-3d82-4ede-8cf3-0a1ccddcd376 +:END: We first define some function: \begin{align*} g(t) := f(t) + \epsilon \nu(t) @@ -69,7 +72,10 @@ must be zero, we get the Euler-Lagrange equation: This is because the integral implies that for all selections for this function $\nu(t)$, $\nu(t)(\frac{dL}{df} - \frac{d}{dt}\frac{dL}{dg'}) = 0$. Because $\nu(t)$ can be any function assuming it satisfies the boundary conditions, this can only be the case if $\frac{dL}{df} - \frac{d}{dt}\frac{dL}{dg'} = 0$. In physics, we re-cast $f$ as $q$ and $f'$ as $\dot{q}$, where $q$ and $\dot{q}$ are the /generalized coordinates/ and /generalized velocities/ respectively. -* The Hamiltonian +* Hamiltonian +:PROPERTIES: +:ID: 3473dbbe-35b8-4aad-b08f-f02d9929c932 +:END: The Hamiltonian represents the total energy in the system; it is the [[id:23df3ba6-ffb2-4805-b678-c5f167b681de][Legendre Transformation]] of the Lagrangian. Applying the Legendre Transformation to the Lagrangian for coordinate $\dot{q}$: \begin{align*} |