In vector space \(\mathbb{R}^n\).
\(\mathbf{q} \in \mathbb{R}^n\)
Also known as a worldline.
Index to \(t\)
\(\mathbf{q} (t)\)
\(v=\dfrac{\delta \mathbf{q}}{\delta t}\)
\(a=\dfrac{\delta v}{\delta t}\)
\(a=\dfrac{\delta^2 \mathbf{q}}{\delta t^2}\)
We observe a particle moving in a path. We want to model the path that the particle takes.
The path is in a vector space, with coordinates \(\mathbf{q}\). These coordinates could refer to the \(x\), \(y\), \(z\) and \(t\) coordinates we are are familar with.
For the path we have a start point \(a\) and end point \(b\). We can define the length of the path as:
\(S=\int_a^b d\tau\)
We call \(S\) the action.
We use a linear metric.
\(\tau^2 = \mathbf q^T\mathbf M\mathbf q\)
So:
\(d\tau^2 =(d\mathbf q)^T\mathbf Md\mathbf q\)
\(S = \int_a^b \sqrt {(d\mathbf q)^T\mathbf Md\mathbf q}\)
\(S = \int_a^b \sqrt {\dfrac{1}{dt^2}(d\mathbf q)^T\mathbf Md\mathbf q}dt\)
\(S = \int_a^b \sqrt {(\dfrac{d\mathbf q}{dt})^T\mathbf M\dfrac{d\mathbf q}{dt}}dt\)
\(S = \int_a^b \sqrt {(\mathbf {\dot q})^T\mathbf M\mathbf {\dot q}}dt\)
We have:
\(S = \int_a^b \sqrt {(\mathbf {\dot q})^T\mathbf M\mathbf {\dot q}}dt\)
We can define:
\(L=\sqrt {(\mathbf {\dot q})^T\mathbf M\mathbf {\dot q}}\)
So we have:
\(S=\int_a^b L dt\)
\(\delta A=0\)
That is, the coordinates and their velocities are such that action is stationary.
We have \(q(t)\) which makes the action stationary. Consider adding proportion \(\epsilon\) of another function \(f(t)\) to \(q(t)\).
\(A’=\int_{t_0}^{t_1} L[q(t)+\epsilon f(t), \dot {q(t)}+\epsilon f’(t)]dt\)
\(\dfrac{A’-A}{\epsilon }=\dfrac{1}{\epsilon}\int_{t_0}^{t_1} L[q(t)+\epsilon f(t), \dot {q(t)}+\epsilon f’(t)]-L[q,\dot q]dt\)
We can do a Taylor expansion of \(A’\).
\(A’=\int_{t_0}^{t_1} L[q(t)+\epsilon f(t), \dot {q(t)}+\epsilon f’(t)]dt\)
\(A’=\int_{t_0}^{t_1} L[q(t),\dot {q(t)}]+\epsilon [f\dfrac{\delta L}{\delta q}+f^.\dfrac{\delta L}{\delta \dot q}]+\epsilon^2 [...]dt\)
So:
\(\dfrac{A’-A}{\epsilon }=\dfrac{1}{\epsilon}\int_{t_0}^{t_1}L[q(t),\dot {q(t)}]+\epsilon [f\dfrac{\delta L}{\delta q}+f^.\dfrac{\delta L}{\delta q^.}]+\epsilon^2 [...]-L[q,\dot q]dt\)
\(\dfrac{A’-A}{\epsilon }=\int_{t_0}^{t_1}[f\dfrac{\delta L}{\delta q}+f^.\dfrac{\delta L}{\delta \dot {q}}]+\epsilon [...]dt\)
We can now make the left side \(0\), by using the definition of stationary action.
\(\lim_{\epsilon \rightarrow 0} \dfrac{A’-A}{\epsilon }=\int_{t_0}^{t_1}[f\dfrac{\delta L}{\delta q}+f^.\dfrac{\delta L}{\delta \dot q}]dt\)
\(\int_{t_0}^{t_1}[f\dfrac{\delta L}{\delta q}+f^.\dfrac{\delta L}{\delta \dot q}]dt=0\)
\(\int_{t_0}^{t_1}[f\dfrac{\delta L}{\delta q}]dt +\int_{t_0}^{t_1}[f^.\dfrac{\delta L}{\delta \dot q}]dt=0\)
Note that
\(\int_{t_0}^{t_1}[f^.\dfrac{\delta L}{\delta \dot q}]dt=[f\dfrac{\delta L}{\delta \dot q}]_{t_0}^{t_1}-\int_{t_0}^{t_1}f \dfrac{d}{dt}\dfrac{\delta L}{\delta \dot q}dt\)
We assume that \(f(t_0)=f(t_1)=0\) and so:
\(\int_{t_0}^{t_1}[f^.\dfrac{\delta L}{\delta \dot q}]dt=-\int_{t_0}^{t_1}f \dfrac{d}{dt}\dfrac{\delta L}{\delta \dot q}dt\)
Plugging this back in we get:
\(\int_{t_0}^{t_1}[f\dfrac{\delta L}{\delta q}]-f \dfrac{d}{dt}\dfrac{\delta L}{\delta \dot q}dt]=0\)
\(\int_{t_0}^{t_1}f[\dfrac{\delta L}{\delta q}]-\dfrac{d}{dt}\dfrac{\delta L}{\delta \dot q}dt]=0\)
Since this applies to all possible functions we get:
\(\dfrac{\delta L}{\delta q}=\dfrac{d}{dt}\dfrac{\delta L}{\delta \dot q}\)
\(p=\dfrac{\delta L}{\delta \dot q}\)
\(L=\sqrt {(\mathbf {\dot q})^T\mathbf M\mathbf {\dot q}}\)
\(\dfrac{\delta L}{\delta q}- \dfrac{d}{dt}(\dfrac{\delta L}{\delta \dot q})=0\)
We have
\(S=\int_a^b L(q(t), \dot q(t))dt\)
\(\delta S=\delta \int_a^b L(q(t), \dot q(t))dt\)
\(J=\int_a^b L(t,q(t), \dot q(t)) dt\)
\(J=\sum_{k=0}^{n-1}\)
\(A=\sum L(x(t), \dot x(t)) \delta t\)
\(A=\sum L(\dfrac{x(t)+x(t-1)}{2}, \dfrac{x(t)-x(t-1)}{\delta t}) \delta t\)
\(\dfrac{\delta }{\delta x(t)}A = \sum \dfrac{\delta }{\delta x(t)}L(\dfrac{x(t)+x(t-1)}{2}, \dfrac{x(t)-x(t-1)}{\delta t}) \delta t\)
\(\dfrac{\delta }{\delta x(t)}A = \delta t [\dfrac{\delta }{\delta x(t)}L(\dfrac{x(t)+x(t-1)}{2}, \dfrac{x(t)-x(t-1)}{\delta t}) + \dfrac{\delta }{\delta x(t)}L(\dfrac{x(t+1)+x(t)}{2}, \dfrac{x(t+1)-x(t)}{\delta t})]\)
\(\dfrac{\delta }{\delta x(t)}A = \delta t [\dfrac{1}{2}L_x +\dfrac{1}{\delta t}L_{\dot x} + \dfrac{1}{2}L_x -\dfrac{1}{\delta t}L_{\dot x}]\)
\(A=\int_a^b L(q(t), \dot q(t)) dt\)
\(A=\sum L(q(t), \dot q(t)) \delta t\)
\(A=\sum L(\dfrac{q(t)+q(t-1)}{2}, \dfrac{q(t)-q(t-1)}{\delta t}) \delta t\)
\(\dfrac{\delta }{\delta q_i(t)}A = \sum \dfrac{\delta }{\delta q_i(t)}L(\dfrac{q(t)-q(t-1)}{2}, \dfrac{q(t)-q(t-1)}{\delta t}) \delta t\)
\(\dfrac{\delta }{\delta q_i(t)}A = \delta t [\dfrac{\delta }{\delta q_i(t)}L(\dfrac{q(t)+q(t-1)}{2}, \dfrac{q(t)-q(t-1)}{\delta t})+\dfrac{\delta }{\delta q_i(t)}L(\dfrac{q(t+1)+q(t)}{2}, \dfrac{q(t+1)-q(t)}{\delta t})]\)
\(\dfrac{\delta }{\delta q_i(t)}A = \delta t [\dfrac{1}{2}L_{q_i}+\dfrac{1}{\delta t}L_{\dot q_i}+\dfrac{1}{2}L_{q_i}-\dfrac{1}{\delta t}L_{\dot q_i}+\dfrac{\delta }{\delta q_i(t)}L(\dfrac{q(t+1)+q(t)}{2}, \dfrac{q(t+1)-q(t)}{\delta t})]\)
For the Euclidian metric:
\(M=I\)
\((dv )^TMdv =(dv)^T dv = d x^2+d y^2 + d z^2\)
\(Action = \int \sqrt {dx^2+dy^2 + dz^2}\)
\(Action = \int \sqrt {\dot x^2+\dot y^2 + \dot z^2}d t\)
\(Action = \int vd t\)
What are symmetries here? Gallilean group and?
\(\dfrac{\delta L}{\delta q}=\dfrac{d}{dt}\dfrac{\delta L}{\delta \dot q}\)
\(L=\sqrt {(\mathbf {\dot q})^T\mathbf M\mathbf {\dot q}}\)
If \(M=I\), then:
\(L=\sqrt {(\mathbf {\dot q})^T \mathbf {\dot q}}\)
In 1 dimension, Euclid:
\(L=\sqrt {\dot q^2}\) \(L=\dot q\)
So:
\(\dfrac{\delta L}{\delta q}=\dfrac{\delta }{\delta q}\dot q\) \(\dfrac{\delta L}{\delta q}=0\)
\(\dfrac{\delta L}{\delta \dot q}=\dfrac{\delta }{\delta \dot q}\dot q\) \(\dfrac{\delta L}{\delta \dot q}=1\)
Into Euler-Lagrange:
\(\dfrac{\delta L}{\delta q}=\dfrac{d}{dt}\dfrac{\delta L}{\delta \dot q}\) \(0=\dfrac{d}{dt}1\)
In three dimensions:
\(L=\sqrt {\dot x^2 + \dot y^2 + \dot z^2}\)
So:
\(\dfrac{\delta L}{\delta q}=\dfrac{\delta }{\delta q}\sqrt {\dot x^2 + \dot y^2 + \dot z^2}\) \(\dfrac{\delta L}{\delta q}=0\)
\(\dfrac{\delta L}{\delta \dot q}=\dfrac{\delta }{\delta \dot q}\sqrt {\dot x^2 + \dot y^2 + \dot z^2}\)
We define the momentum as:
\(p_j = \dfrac{\delta L}{\delta \dot q_j}\)
\(p_j = \dfrac{\delta }{\delta \dot q_j}\sqrt {(\mathbf {\dot q})^T\mathbf M\mathbf {\dot q}}\)