__Return to current section__**: ****Planar Rigid Body Motion and Reference Frames****.**

**Problem: **Apply the ** Principle of Relativity** to Newton’s 2

^{nd}Law to prove that

any two inertial frames are related by a

*Galilean transformation*** x **→ **x**′ = *R***x **+ **c **+ **u***t** ** *where *R*, **c **and **u **are constant**.**

Derive therefrom the **Galilean ****velocity addition theorem:**** **** ****v****′** = **v **+ **u.**

*Solution:*** **In one inertial frame, every ** free particle** has

**constant velocity**, its

**acceleration is zero**.

Therefore, by Newton's 2

^{nd}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:

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

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

(

**), we may as well choose our constant frame rotation to be**

*our primed frame**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 *generalized* Galilean velocity addition theorem

relating particle velocites in two different inertial frames.

**Exercise: **Inside a cable car climbing a slope with constant velocity

**v**

_{0}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

**say about the trajectory as seen in the two frames? Here are two**

*intuition***:**

*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

ask "** 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*) +

**v**

_{0}

*t*, 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 *t*_{c}, 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 **x**_{0} 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**:

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

**v**

_{0}when dropped, so its path is:

In ** the unprimed cable car frame**, the object is dropped from rest from position

**x**

_{0}, so:

Let's plug the last two equations into the transformation equation above to find **x**_{0}**'**:

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 *t*_{c} = 0. We've chosen the cable car velocity with respect to the earth to be **v**_{0} = 7 (cos *α*, sin *α*).

If we check the "View false path" box, we'll see the path of a ** green object** matching

**above.**

*guess A*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

**. Then we see the**

*trace of the blue object's path***.**

*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*′ = *α**t *+ *β*** **** (**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 (**β*

**or 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

*α*

*t*+

*β*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**(*α**t *+ *β*).

In general, for any particle path:

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

** Newton's 1^{st} Law** says that

**and is described in any inertial frame**

*a free particle has a constant velocity*by

**v'**=

*R*

**v**+

**u**, as summarized earlier. The

**of the Galilean transformation between two inertial frames**

*form*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

*α*

**v'**=

*α*

*R*

**v**+

*α*u, which is valid and the same formula

using either the old clocks or the new clocks. We conclude Newton's first law

**with respect to**

*is invariant*Galilean time translation and scaling.

** Newton's 2^{nd} Law** is a bit trickier.

Since the value of the acceleration has been altered, Newton's 2^{nd} Law is not generally invariant under this

clock change. Still, there are some cases where the 2^{nd} Law is invariant under this time transformation:

**1.** *β* = 0 and **F**(*t*) = *kt*^{2}, 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 **v**^{2}. Since we've seen velocity gets magnified

by *α*, there will be an *α*^{2} on both sides of the equation.

__Return to current section__**: ****Planar Rigid Body Motion and Reference Frames****.**