Susskind Classical Mechanics Lecture 2

Source

• YouTube

Overview

write this last

Consider:

possible links to other concepts

• what gaps in my knowledge does this expose?
  • need to revise basic rules of differentiation and integration

Summary

Susskind Classical Mechanics Lecture 2

0:00

Systems that change smoothly

0:00

Start by looking at Aristotle's (FALSE) laws

• Aristotle didn't know about friction
• thought object only moves if force applied
• $\overrightarrow{V}=\frac{\overrightarrow{F}}{m}$

0:00

Will show that Aristotle's Law fails the test of Reversibility

3:43

Restate as

$\overrightarrow{F}=m\overrightarrow{V}$

Simplify to 1-dimensional line

$F=m\dot{x}$

5:29

Start with a discrete model where time axis is divided into small steps $ϵ$

$F=m\dot{x}=m\frac{X(t+ϵ)-X(t)}{ϵ}$

Now make F a known function of t

$F(t)==m\frac{X(t+\mathrm{\Delta })-X(t)}{\mathrm{\Delta }}$

$\frac{\mathrm{\Delta }}{m}F(t)=x(T+\mathrm{\Delta })-X(t)$

$\frac{\mathrm{\Delta }}{m}F(t)+X(t)=x(T+\mathrm{\Delta })$

In other words we now have a law that tells us where the particle will be at time $T+\mathrm{\Delta }$

8:37

Another Aristotle approach to demonstrate non-reversibility

1D system, particle moving on a spring

$F(x)=-kx$

$-k\frac{\delta }{m}X(t)+X(t)=X(t+\delta )$

rearranging

$X(t)[1-\frac{k\delta }{m}]=X(t+\delta )$

11:18

Adapt this into a continuous model

$m\dot{x}=-\frac{k}{m}x$

jump through knowledge of exponential function:

$X=A{e}^{-\frac{k}{m}t}$

So the object moves towards ht e origin with exponentially decreasing speed

12:53

This is fundamentally not Reversible so is not a valid law in Classical Dynamics as we understand it today.

17:14

Newton's second law

$F=ma$

or

$F=m\ddot{x}$

Is this predictive?

Expanding acceleration via a discrete differentiation of velocity:

$a=\frac{X(t+\mathrm{\Delta })-X(t)}{{\mathrm{\Delta }}^2}-\frac{X(t)-X(t-\mathrm{\Delta })}{{\mathrm{\Delta }}^2}$

${\mathrm{\Delta }}^2\frac{F}{m}=X(t+\mathrm{\Delta })+(X)t-\mathrm{\Delta }-2X(t)$

${\mathrm{\Delta }}^2\frac{F}{m}+2X(t)-X(t-\mathrm{\Delta })=X(t+\mathrm{\Delta })$

So to know where the particle is next we need to know where it is now and where it weas in the past.

29:39

Initial conditions -> $X(t),\dot{X}(t)$

New property, momentum

$\overrightarrow{P}=m\overrightarrow{\dot{x}}$

$F=\dot{P}$ (1) $\overrightarrow{P}=m\overrightarrow{\dot{x}}$ (2)

33:32

$F=\frac{P(t+\mathrm{\Delta })-P(t)}{\mathrm{\Delta }}$

$P=\frac{X(t+\mathrm{\Delta })-X(t)}{\mathrm{\Delta }}$

So space of initial conditions is position and momentum

34:58

Now consider a particle in Phase Space P v X

38:13

Restate the spring problem with Newton's equation

$F=m\ddot{x}=-kx$

set mass and k to 1 for simplicity

$F=\ddot{x}=-x$

$X=C\cos (t-t_0)$

$P=\dot{X}=Csin(t-t_0)$

$X^2+P^2=C^2$

so in phase space this describes a circle

Deterministic and Reversible because predictable into future and past

The bigger C, the bigger the radius, C is the energy.

56:38

Newton's third law

whenever two objects interact, they exert equal and opposite forces on each other.

$\overrightarrow{F_{ij}}=-\overrightarrow{F_{ji}}$

Prove by avoidance of perpetual motion that forces must be aligned with the line of centres between the objects

1:01:46

Conservation of Momemtum

imagine collection of particles $i$ in a closed system

$m_i\frac{d^2\overrightarrow{r_i}}{d{t}^2}=\sum _{j\ne i}{F}_{ji}$

restating as momentum:

$\frac{d\overrightarrow{P_i}}{dt}=\sum _{j\ne i}{F}_{ji}$

summing over all particles

$\frac{d\overrightarrow{P_{TOTAL}}}{dt}=\sum _{ij}{F}_{ij}$

because the particle forces come in matched pairs this simplifies to

$\frac{d\overrightarrow{P_{TOTAL}}}{dt}=0$

1:07:04

Applying Newton's laws and conservation of energy to a point moving on a 1-d line

$F(x)=-\frac{dV(x)}{dx}$ where $V$ is the potential energy

$\int F(x)dx=-V(x)$

1:16:14

$T+V$ is conserved (T is Kinetic Energy, V is potential energy)

$F(x)=-\frac{dV(x)}{dx}=m\ddot{x}$

Take time derivative of $\frac{1}{2}m{\dot{x}}^2+V(x)=E$

$\frac{1}{2}m2\dot{x}\ddot{x}$

$\dot{x}[m\ddot{x}+\frac{dV}{dx}]=\frac{dE}{dt}$

using Newton 2nd law

$0=\frac{dE}{dt}$

so energy is conserved

1:24:22

Extending to arbitrary number of dimensions

$F_i(x)=m{\ddot{x}}_i=\frac{\partial V(x)}{\partial X_i}$

Cannot prove, but this is considered a Postulate - i.e. empirically true but not proven

Come back to harmonic oscillator - objects in phase space travel on contours of constant potential energy

See also