Paths

Describing paths

Describing events

In vector space \(\mathbb{R}^n\).

\(\mathbf{q} \in \mathbb{R}^n\)

Describing the path of a particle

Also known as a worldline.

Index to \(t\)

\(\mathbf{q} (t)\)

Describing the velocity of a particle

\(v=\dfrac{\delta \mathbf{q}}{\delta t}\)

Describing the acceleration of a particle

\(a=\dfrac{\delta v}{\delta t}\)

\(a=\dfrac{\delta^2 \mathbf{q}}{\delta t^2}\)

Action

Action

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.

Linear metrics

We use a linear metric.

\(\tau^2 = \mathbf q^T\mathbf M\mathbf q\)

So:

\(d\tau^2 =(d\mathbf q)^T\mathbf Md\mathbf q \)

\(S = \int_a^b \sqrt {(d\mathbf q)^T\mathbf Md\mathbf q}\)

Time and velocity

\(S = \int_a^b \sqrt {\dfrac{1}{dt^2}(d\mathbf q)^T\mathbf Md\mathbf q}dt\)

\(S = \int_a^b \sqrt {(\dfrac{d\mathbf q}{dt})^T\mathbf M\dfrac{d\mathbf q}{dt}}dt\)

\(S = \int_a^b \sqrt {(\mathbf {\dot q})^T\mathbf M\mathbf {\dot q}}dt\)

The Lagrangian

Lagrangians

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:

\(S=\int_a^b L dt\)

Principle of stationary action

\(\delta A=0\)

That is, the coordinates and their velocities are such that action is stationary.

Euler-Lagrange

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:

\(\dfrac{\delta L}{\delta q}=\dfrac{d}{dt}\dfrac{\delta L}{\delta \dot q}\)

Definition: Momentum

\(p=\dfrac{\delta L}{\delta \dot q}\)

EL v2

\(L=\sqrt {(\mathbf {\dot q})^T\mathbf M\mathbf {\dot q}}\)

\(\dfrac{\delta L}{\delta q}- \dfrac{d}{dt}(\dfrac{\delta L}{\delta \dot q})=0\)

We have

\(S=\int_a^b L(q(t), \dot q(t))dt \)

\(\delta S=\delta \int_a^b L(q(t), \dot q(t))dt \)

\(J=\int_a^b L(t,q(t), \dot q(t)) dt\)

\(J=\sum_{k=0}^{n-1}\)

EL v3

\(A=\sum L(x(t), \dot x(t)) \delta t \)

\(A=\sum L(\dfrac{x(t)+x(t-1)}{2}, \dfrac{x(t)-x(t-1)}{\delta t}) \delta t \)

\(\dfrac{\delta }{\delta x(t)}A = \sum \dfrac{\delta }{\delta x(t)}L(\dfrac{x(t)+x(t-1)}{2}, \dfrac{x(t)-x(t-1)}{\delta t}) \delta t \)

\(\dfrac{\delta }{\delta x(t)}A = \delta t [\dfrac{\delta }{\delta x(t)}L(\dfrac{x(t)+x(t-1)}{2}, \dfrac{x(t)-x(t-1)}{\delta t}) + \dfrac{\delta }{\delta x(t)}L(\dfrac{x(t+1)+x(t)}{2}, \dfrac{x(t+1)-x(t)}{\delta t})]\)

\(\dfrac{\delta }{\delta x(t)}A = \delta t [\dfrac{1}{2}L_x +\dfrac{1}{\delta t}L_{\dot x} + \dfrac{1}{2}L_x -\dfrac{1}{\delta t}L_{\dot x}]\)

\(A=\int_a^b L(q(t), \dot q(t)) dt\)

\(A=\sum L(q(t), \dot q(t)) \delta t\)

\(A=\sum L(\dfrac{q(t)+q(t-1)}{2}, \dfrac{q(t)-q(t-1)}{\delta t}) \delta t\)

\(\dfrac{\delta }{\delta q_i(t)}A = \sum \dfrac{\delta }{\delta q_i(t)}L(\dfrac{q(t)-q(t-1)}{2}, \dfrac{q(t)-q(t-1)}{\delta t}) \delta t\)

\(\dfrac{\delta }{\delta q_i(t)}A = \delta t [\dfrac{\delta }{\delta q_i(t)}L(\dfrac{q(t)+q(t-1)}{2}, \dfrac{q(t)-q(t-1)}{\delta t})+\dfrac{\delta }{\delta q_i(t)}L(\dfrac{q(t+1)+q(t)}{2}, \dfrac{q(t+1)-q(t)}{\delta t})]\)

\(\dfrac{\delta }{\delta q_i(t)}A = \delta t [\dfrac{1}{2}L_{q_i}+\dfrac{1}{\delta t}L_{\dot q_i}+\dfrac{1}{2}L_{q_i}-\dfrac{1}{\delta t}L_{\dot q_i}+\dfrac{\delta }{\delta q_i(t)}L(\dfrac{q(t+1)+q(t)}{2}, \dfrac{q(t+1)-q(t)}{\delta t})]\)

The Euclidian metric

The Euclidian metric

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?

Euclidian rotations

The Euclidan group

The Galilean group

Examples from Euler-Lagrange

Outcome

\(\dfrac{\delta L}{\delta q}=\dfrac{d}{dt}\dfrac{\delta L}{\delta \dot q}\)

\(L=\sqrt {(\mathbf {\dot q})^T\mathbf M\mathbf {\dot q}}\)

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:

\(\dfrac{\delta L}{\delta q}=\dfrac{\delta }{\delta q}\sqrt {\dot x^2 + \dot y^2 + \dot z^2}\) \(\dfrac{\delta L}{\delta q}=0\)

\(\dfrac{\delta L}{\delta \dot q}=\dfrac{\delta }{\delta \dot q}\sqrt {\dot x^2 + \dot y^2 + \dot z^2}\)

Other

Momentum

We define the momentum as:

\(p_j = \dfrac{\delta L}{\delta \dot q_j}\)

\(p_j = \dfrac{\delta }{\delta \dot q_j}\sqrt {(\mathbf {\dot q})^T\mathbf M\mathbf {\dot q}}\)

Force