# 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?

## 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:

$$p_j = \dfrac{\delta L}{\delta \dot q_j}$$

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