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

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

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



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.

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


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

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

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


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.


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 \(A’\).

\(A’=\int_{t_0}^{t_1} L[q(t)+\epsilon f(t), \dot {q(t)}+\epsilon f’(t)]dt\)

\(A’=\int_{t_0}^{t_1} L[q(t),\dot {q(t)}]+\epsilon [f\dfrac{\delta L}{\delta q}+f^.\dfrac{\delta L}{\delta \dot q}]+\epsilon^2 [...]dt\)


\(\dfrac{A’-A}{\epsilon }=\dfrac{1}{\epsilon}\int_{t_0}^{t_1}L[q(t),\dot {q(t)}]+\epsilon [f\dfrac{\delta L}{\delta q}+f^.\dfrac{\delta L}{\delta q^.}]+\epsilon^2 [...]-L[q,\dot q]dt\)

\(\dfrac{A’-A}{\epsilon }=\int_{t_0}^{t_1}[f\dfrac{\delta L}{\delta q}+f^.\dfrac{\delta L}{\delta \dot {q}}]+\epsilon [...]dt\)

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

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

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

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


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:


\((dv )^TMdv =(dv)^T dv = d x^2+d y^2 + d z^2\)

\(Action = \int \sqrt {dx^2+dy^2 + dz^2}\)

\(Action = \int \sqrt {\dot x^2+\dot y^2 + \dot z^2}d t\)

\(Action = \int vd t\)

What are symmetries here? Gallilean group and?

Euclidian rotations

The Euclidan group

The Galilean group

Examples from Euler-Lagrange


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

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

If \(M=I\), then:

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

In 1 dimension, Euclid:

\(L=\sqrt {\dot q^2}\) \(L=\dot q\)


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

\(\dfrac{\delta L}{\delta \dot q}=\dfrac{\delta }{\delta \dot q}\dot q\) \(\dfrac{\delta L}{\delta \dot q}=1\)

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:

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


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

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



We define the momentum as:

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

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