hello world physics - part 1 classical mechanics

I've been struggling to understand General Relativity - the theory that describes gravity - for a long time now.  One thing I realized might help is a very simple example - solving a very simple, recognizable problem using general relativity.  In an analogy with software engineering, I'm looking for how write "hello world" in new programming language.

For the equivalent of programming's "hello world" in physics, I'm going to choose the problem of tossing a ball up into the air.  I'm going to solve this problem using 4 different types of physics:  classical/Newtonian, quantum mechanics, general relativity, and quantum electrodynamics.  I hope this will give me (and you perhaps) a better understanding of the differences and similarities between these theories, and some practical understanding of how one uses / applies them.

For all of these examples, we'll use these assumptions:

• no air resistance / friction
• perfectly elastic collisions
• the surface of the earth is immovable (does not recoil in collision)
• force of gravity is constant / acceleration of gravity is constant (9.8 $m/s^2$)

Classical mechanics

summary

velocity as a function of time	$ v = -gt + v_0 $
position aka height as a function of time	$ x = -\frac{1}{2}gt^2 + v_0t $
time to reach maximum height	$$ t_\text{max height} = \frac{v_0}{g} $$
maximum height	$$ x_\text{max} = \frac{1}{2} \frac{v_0^2}{g} $$
time to return to ground	$$ t_\text{return} = 2 \frac{v_0}{g} $$

derivation

We'll use Newton's second law of motion, in equation form:

$$ F=ma $$

• $F$ is the force applied to the ball
• $m$ is the mass of the ball
• $a$ is the acceleration of the ball

Acceleration is the second derivative of position - height above the Earth's surface - ($x$) with respect to time:

$$ a = \frac{d^2x}{dt^2} $$

Substituting the definition of acceleration into Newton's second law yields:

$$ m\frac{d^2x}{dt^2} = F $$

The only force applied to the ball is the force of gravity:

$$ F = -mg $$

Combining these equations yields our differential equation of motion:

$$ m\frac{d^2x}{dt^2} = -mg $$

Simplifies to:

$$ \frac{d^2x}{dt^2} = -g $$

To simplify solving this equation, let's first substitute in for the velocity

$$ v=\frac{dx}{dt} $$

$$ a=\frac{dv}{dt} $$

Then our differential equation of motion becomes:

$$ \frac{dv}{dt} = -g $$

Integrating this equation yields:

$$ v = -gt + v_0 $$

$v_0$ is the constant of integration of integration.  It is the initial velocity of the ball - we can see this by setting $t=0$ in the above equation:

$$ v(t=0) = -g\cdot0 + v_0 $$

$$ v(t=0) = v_0 $$

Now we can substitute the definition of velocity into this equation:

$$ \frac{dx}{dt} = -gt + v_0 $$

Integrating this equation yields:

$$ x = -\frac{1}{2}gt^2 + v_0t + x_0 $$

$x_0$ is the initial position of the ball which you can see by setting t=0 as we did for the velocity equation (left as an exercise for the reader ;).  For simplicity we'll set it equal to zero so our equation becomes:

$$ x = -\frac{1}{2}gt^2 + v_0t $$

This traces out the exact path of the ball over time.  Some questions we can answer with these equations:

• How high does the ball go?
• When does the ball hit the Earth again?

To determine how high the ball goes, we can use the fact that at its highest point the ball stops moving - only for an infinitesimal amount of time - before falling back to Earth.  That means at the highest point the velocity is zero. Therefore we use the velocity equation, set $v=0$ an solve for $t$.  Once we know $t$ we can substitute into the position equation.  First, using the velocity equation setting $v=0$:

$$ 0 = -gt + v_0 $$

$$ gt = v_0 $$

$$ t = \frac{v_0}{g} $$

In other words, the initial velocity divided by the acceleration due to gravity yields the time at which the ball reaches its maximum height.  Now we can solve for the position at this time to get the maximum height the ball reaches:

$$ x_{max} = -\frac{1}{2}g \left( \frac{v_0}{g} \right)^2 + v_0 \left( \frac{v_0}{g} \right) $$

$$ x_{max} = -\frac{1}{2} \frac{v_0^2}{g} + \frac{v_0^2}{g} $$

$$ x_{max} = \frac{1}{2} \frac{v_0^2}{g} $$

When does the ball hit the ground again?  We can use the equation for the position, set $x=0$ and solve for t:

$$ 0 = -\frac{1}{2}gt^2 + v_0t $$

$$ 0 = t \left( -\frac{1}{2}gt + v_0 \right) $$

One solution is $t=0$.  The other solution we get by solving:

$$ 0 = -\frac{1}{2}gt + v_0 $$

$$ \frac{1}{2}gt = v_0 $$

$$ t = 2 \frac{v_0}{g} $$