﻿ Solution: Galilean Transformations | Primer on Geometric Algebra | David Hestenes

## Solution: Galilean Transformations

Problem: Apply the Principle of Relativity to Newton’s 2nd Law to prove that
any two inertial frames are related by a Galilean transformation

x  →  x′ = Rut                    where Rand are constant.

Derive therefrom the Galilean velocity addition theorem:    vu.

Solution: In one inertial frame, every free particle has constant velocity, its acceleration is zero.
Therefore, by Newton's 2nd Law, there are
no net forces acting on it. By the Principle of
Relativity, there can be no net forces acting on the particle in any other inertial frame, either.
Therefore, no matter how one frame is displaced from the other, if that displacement is inertial:

$\Large{{\bf{\ddot x'}}\,\,\, = \,\,\,0\,\,\, = \,\,\,R\left[ {0\,\, + \,\,2W{\bf{\dot x}}\,\, + \,\,\left( {\dot W\,\, + \,\,{W^2}} \right)\,{\bf{x}}} \right]\,\, + \,\,{\bf{\ddot a}}}$

Since this must be true for any free particles in the two frames, no matter what their velocity or position,

$\Large{\begin{array}{l}1:\,\,\,\,\,\,RW\,\,\, = \,\,\,0\,\,\, = \,\,\,\dot R(t)\,\,\,\, \Rightarrow \,\,\,\,R\,\,\, = \,\,\,{\rm{constant}}\,\,\,\,\left( {{\rm{and}}\,\,\,\,W\,\, = \,\,\,0} \right)\\2:\,\,\,\,\,\,{\bf{\ddot a}}(t)\,\,\, = \,\,\,0\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\bf{a}}(t)\,\,\, = \,\,\,{\bf{u}}\,t\,\, + \,\,{\bf{c}}\\\\{\sf{Therefore}}:\,\,\,\,\,\,\,\,{\bf{x'}}(t)\,\,\, = \,\,\,R\,{\bf{x}}(t)\,\, + \,\,{\bf{u}}\,t\,\, + \,\,{\bf{c}}\end{array}}$

The
Galilean velocity addition theorm is usually described as coming from his thought experiment
of a uniformly moving ship (calm sea, no buffeting) where below deck, in a windowless cabin,
seamen perform experiments (tossing balls, jumping, watching fish in a tank, watching butterflies
and fireflies). His conclusions are now called his Principle of Relativity. We saw how the equations above
follow. For a ball rolling on the floor of the cabin with constant velocity
v with respect to the ship
(our unprimed frame) and if the ship is moving with constant velocity u with respect to the shore
(
our primed frame), we may as well choose our constant frame rotation to be R=1. Then v' = v+u
follows from the time derivative of our last equation above. Since
v and u are constants, the ball's
velocity
v' as seen from the shore is also constant, meaning we still see it as free particle motion.

Once we've established the relationship between inertial frames written above, there's no need to restrict
ourselves to only free particles. We can easily write the
relating particle velocites in two different inertial frames.

$\Large{{\bf{x'}}(t)\,\,\, = \,\,\,R{\bf{x}}(t)\,\, + \,\,{\bf{u}}t\,\, + \,\,{\bf{c}}\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{\bf{v'}}(t)\,\,\, = \,\,\,R{\bf{v}}(t)\,\, + \,\,{\bf{u}}}$

Exercise:
Inside a cable car climbing a slope with constant velocity v0 an object is dropped from rest.
Derive equations for the trajectory within the car and with respect to the earth outside.

Solution: As usual, there are several ways we could define our two frames. We may as well call the angle of the
slope
α with respect to the locally "horizontal" earth. But, we could define the earth and cable car frames
both aligned with the earth ground, or both aligned with the slope, or one of each. Which gives neater
equations? When should we start the clock, when should we drop the object?

What does our
intuition say about the trajectory as seen in the two frames? Here are two guesses:

A.  In the earth frame, g points straight down, so an object dropped freely in the car will accelerate
in a straight line down. Then, as the cable car moves up the slope, the falling object will
seem to drift towards the back of the car, making
a backward parabolic path typical of
projectiles. After all, the people standing in the car have to lean forward to keep balanced
because
g is tilted backwards from the vertical with respect to their floor.

B.  If the cable car is moving smoothly at a constant velocity, we have the kind of inertial system
Galileo described in his moving ship. If people in the car closed their eyes, in principle they
shouldn't be able to detect their motion with respect to the earth. Sure, they can tell
g is tilted,
but the laws of physics should work the same, so we expect the dropped object to accelerate
along a straight line following the direction of g, tilted at the same angle at which the riders
lean from the vertical with respect to their floor (
α). Then, since the car is moving up the
slope, people standing outside would see
a parabolic projectile path going up the slope.

Of course these two guesses are drastically different, so at least one of them is incorrect. The trick is to
The object is dropped from rest with respect to which frame?" Before dropping it, the object
is at rest with respect to the person's hand who will drop it, and that person is at rest in the cable car.
Therefore we expect guess B to be correct (of course it's always a good idea to think about experimental
testing schemes). In Galileo's cabin moving at constant speed with respect to the shore, rolling a ball
on a level billiard table should show no differences depending on which direction it is rolled. It would be
only slightly more complicated in the cable car, because
g is not perpendicular to the floor, but if we
tilt the billiard table to be perpendicular to g, we could not detect our motion by rolling the ball in any
direction.

Probably the simplest choice of frames would be to align them both with the climbing slope and choose
the frames to be coincident at
t=0 at the bottom of the hill. Then the constant rotor is R=1, the
transformation equation is simply
x'(t) = x(t) + v0t, and both frames agree on the same value for g.

However, we'll try a more instructive choice, summarized in the following diagram:

These frames can never be coincident, but their origins coincide at time
tc, which is negative.
We've chosen the time when the object is dropped to be t
=0. The unit bivector is CCW. The object
is dropped from initial position x0 in the cable car frame. Notice that g' and g are identical vectors,
yet
g' points in the negative y'-axis direction, while g is tilted from the negative y-axis direction.

We can calculate the paths as seen from both frames and see if they are consistent with our

general inertial frame planar transformation:

$\Large{{\bf{x'}}(t)\,\,\, = \,\,\,R\,{\bf{x}}(t)\,\, + \,\,{\bf{a}}(t)\,\,\, = \,\,\,{e^{ - {\bf{i}}\alpha }}\,{\bf{x}}(t)\,\, + \,\,{{\bf{v}}_0}\,\,\left( {t - {t_c}} \right)}$

The rotor
R rotates the cable car and vectors expressed in unprimed coordinates CCW by α.
Notice that same rotor takes
g as seen in the unprimed frame into g' as seen in the primed frame.
In
the primed earth frame, the object has the initial velocity of v0 when dropped, so its path is:

$\Large{{\bf{\ddot x'}}(t)\,\,\, = \,\,\,{\bf{g'}}\,\,\, = \,\,\,{e^{ - {\bf{i}}\alpha }}\,{\bf{g}}\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{\bf{x'}}(t)\,\,\, = \,\,\,\frac{1}{2}\,{\bf{g'}}\,{t^2}\,\, + \,\,{{\bf{v}}_0}\,t\,\, + \,\,{{\bf{x}}_0}^\prime }$

In
the unprimed cable car frame, the object is dropped from rest from position x0, so:

$\Large{{\bf{\ddot x}}(t)\,\,\, = \,\,\,{\bf{g}}\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{\bf{x}}(t)\,\,\, = \,\,\,\frac{1}{2}\,{\bf{g}}\,{t^2}\,\, + \,\,{{\bf{x}}_0}}$

Let's plug the last two equations into the transformation equation above to find x0':

$\Large{\begin{array}{l}\frac{1}{2}\,{\bf{g'}}\,{t^2}\,\, + \,\,{{\bf{v}}_0}\,t\,\, + \,\,{{\bf{x}}_0}^\prime \,\,\, = \,\,\,{e^{ - {\bf{i}}\alpha }}\,\left( {\frac{1}{2}\,{\bf{g}}\,{t^2}\,\, + \,\,{{\bf{x}}_0}} \right)\,\, + \,\,{{\bf{v}}_0}\,\,\left( {t - {t_c}} \right)\\\\\,\,\,\,\,\,\,\,~~~~~~~~\,\,\,\, \Rightarrow \,\,\,~~~~~\,\,\,\,\,\,\,\,\,{{\bf{x}}_0}^\prime \,\,\, = \,\,\,{e^{ - {\bf{i}}\alpha }}\,{{\bf{x}}_0}\,\, - \,\,{{\bf{v}}_0}\,{t_c}\end{array}}$

Checking this expression against the diagram above confirms its validity.

Below is a simulation of the results of this exercise. We've chosen a slope of  tan
α = 0.5, giving
an incline of about 26.6°. To simplify the simulation, we drop the object when the frame origins coincide
so that
tc = 0. We've chosen the cable car velocity with respect to the earth to be  v0 = 7 (cos α, sin α).

If we check the "View false path" box, we'll see the path of a green object matching guess A above.
Actually,
that guess would be correct if the object were dropped from rest in the earth frame,
that is if it were dropped outside the cable car by a man standing on the ground and not riding with the car.
You can either use the time slider or click the "Start Animation" button.

If we check the "View correct path" box, and don't yet click the "View earth path" button, we'll see a

blue object follow a straight line acceleration with respect to the cable car, towards the earth, at an
angle
α with respect to the cable car's vertical. Drag the slider slowly to make sure this looks correct.
Then, click on the "View earth path" button to see a
trace of the blue object's path. Then we see the

parabolic path as seen in the earth frame.

To clear the trace and take another look, click in the upper right hand corner of the graphics window
to
reset the simulation (the reset symbol may not be visible, but should still work).

Problem: Discuss invariance of Newton’s first and second laws with respect to
Galilean time translation and scaling:
t    t′ = αβ      (where α and β are constant).

We've assumed the existence and usage of universally synchronized clocks. Now we want to see what
happens if we
replace those old clocks with new universal clocks which don't agree with the
old clocks either on
when time equals zero (βor on the rate of ticking (α). If we have the
expression for the path of a particle,
x(t), in a given frame, we replace t everywhere in the expression
with
αβ and see what happens to velocities and accelerations along the path. Then we can think
about how Newton's first and second laws might be affected.

But, when we take a time derivative of position to get velocity, for example, we don't take the time
derivative of
x(t') with respect to t' because that would just be a renaming of symbols and nothing could
really change. Instead, we still take a time derivative with respect to
t of the altered function x(αβ).
In general, for any particle path:

$\Large{\begin{array}{l}{\bf{x}}(t)\,\,\,\,\,\,\,\,\,\, \xrightarrow[{t'~~~=~~~\alpha~~t~~+~~\beta}]{~~Clock~~Change~~}\,\,\,\,\,\,\,\,\,\,{\bf{x}}(t')\\\\\\\\{\bf{\dot x}}(t)\,\,\, = \,\,\,\frac{d}{{dt}}\,\,{\bf{x}}(t)\,\,\,\,\,\,\,\,\,\, \xrightarrow[{t'~~~=~~~\alpha~~t~~+~~\beta}]{~~Clock~~Change~~}\,\,\,\,\,\,\,\,\,\,\frac{d}{{dt}}\,\,{\bf{x}}(t')\,\,\, = \,\,\,\frac{{dt'}}{{dt}}\,\,\frac{d}{{dt'}}\,\,{\bf{x}}(t')\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \,\,\,\alpha \,\,{\left[ {\frac{d}{{dt}}\,\,{\bf{x}}(t)} \right]_{t = t'}}\,\,\, = \,\,\,\alpha \,{\bf{\dot x}}(t')\\\\\\\\{\bf{\ddot x}}(t)\,\,\, = \,\,\,\frac{d}{{dt}}\,\,{\bf{\dot x}}(t)\,\,\,\,\,\,\,\,\,\, \xrightarrow[{t'~~~=~~~\alpha~~t~~+~~\beta}]{~~Clock~~Change~~}\,\,\,\,\,\,\,\,\,\,\frac{d}{{dt}}\,\,\left[ {\alpha \,{\bf{\dot x}}(t')} \right]\,\,\, = \,\,\,{\alpha ^2}\,{\bf{\ddot x}}(t')\end{array}}$

Notice that velocities have been magnified by
α after the clock change. How does this affect Newton's laws?

Newton's 1st Law says that a free particle has a constant velocity and is described in any inertial frame
by
v' = Rv + u, as summarized earlier. The form of the Galilean transformation between two inertial frames
doesn't change by replacing
t everywhere by t'. And, when a time derivative is taken to get the velocity addition
theorem relating the constant velocities, we simply get
αv' = αRv + αu, which is valid and the same formula
using either the old clocks or the new clocks. We conclude Newton's first law
is invariant with respect to
Galilean time translation and scaling.

Newton's 2nd Law is a bit trickier.

$\Large{{\bf{F}}(t)\,\,\, = \,\,\,m\,{\bf{\ddot x}}(t)\,\,\,\,\,\,\,\,\,\, \xrightarrow[{t'~~~=~~~\alpha~~t~~+~~\beta}]{~~Clock~~Change~~}\,\,\,\,\,\,\,\,\,\,{\bf{F}}(t')\,\,\, = \,\,\,{\alpha ^2}m\,{\bf{\ddot x}}(t')}$

Since the value of the acceleration has been altered, Newton's 2nd Law is not generally invariant under this
clock change. Still, there are some cases where the 2nd Law is invariant under this time transformation:

1.  β = 0  and  F(t) = kt2, where k is a constant (a bit of an oddball case).

2.  α = 1  and  F is independent of time. This is a more likely case. This transformation involves

time translation, but not scaling. In other words, the old and new clocks tick at the same
rate, but are not synchronized. Many forces are time independent, but we must remember
this also means the net force must, for example, be independent of velocity, too, since
velocity generally depends on time.

3.  The only net force is a drag force depending on v2. Since we've seen velocity gets magnified
by
α, there will be an α2 on both sides of the equation.