Galilean Transformations

Date: 09/01/99 at 00:19:55
From: Erika Kurt
Subject: Modern Physics - Galilean transformation

Could you please explain to me the basic principles of Galilean 
transformation? I understand the definition, but I don't understand 
how to do problems that involve it. Plus, it doesn't seem to work for 
things like a frame traveling at the speed of light. I tried looking 
up the math that it involves in my Physics textbook, but the notation 
is so different from what I used in my Calculus BC in high school that 
I can't tell what is going on. I also haven't been able to find any 
in-depth (yet basic) explanations on the Internet. Could you please 
help me? I personally think that it just involves adding or 
subtracting velocity vectors... but this seems too simple for the 
complex terminology.

Erika

Date: 09/01/99 at 10:47:38
From: Doctor Rick
Subject: Re: Modern Physics - Galilean transformation

Hi, Erika.

You're quite right: the Galilean transformation is a simple thing 
expressed in complex (or at least unfamiliar) terminology, and it 
doesn't work right for frames traveling near the speed of light.

What's going on here is that you are dressing up old, standard, 
everyday physics (the kind that Isaac Newton and Galileo developed, 
and that works fine for things moving at ordinary speeds) in a form 
that lets you compare it with relativistic physics.

The Galilean transformation, underneath its disguise, is old, familiar
stuff, _not_ relativistic physics - which is why it doesn't work right 
(in relativistic terms) near the speed of light. The reason you are 
learning it now is so you will have something to compare with the 
Lorentz transformation, which is the basis of special relativity.

The Galilean transformation is:

     (1)  x' = x - vt
     (2)  t' = t

where v is the relative speed between two reference frames (x, t) and 
(x',t'). Equation (1) is the same in the Galilean transformation 
(Newtonian physics) or in the Lorentz transformation (relativistic 
physics); it's just the old familiar rate equation. But in Newtonian 
physics, time is the same in all reference frames (equation 2) - 
something you probably never questioned until you learned something 
about relativity.

In relativity, the time at which an event occurs is not universal, but 
instead it depends on your reference frame. The Lorentz transformation 
is:

     (3) x' = x - vt
     (4) t' = -vx/c^2 + t

I could tell you more about the consequences of equation (4), but 
you'll be learning all that. I'll just point out that if v is much 
less than the speed of light, the first term on the right is very 
small, and equation (4) is essentially the same as equation (2). 
That's why Newton's laws (and the Galilean transformation) are all we 
need for ordinary physics, at ordinary speeds.

If you have particular problems involving the Galilean transformation, 
I would be glad to take a look and see how I could help you. Try to 
write down whatever part of the solution you are able to do, so I 
don't waste time explaining parts that you do understand.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   

Date: 09/06/99 at 22:33:08
From: Doctor Douglas
Subject: Re: Modern Physics - Galilean transformation

Hi Erika,

Here's an explanation that complements the answer that you have 
received earlier from Dr. Rick. Below we will work through an example 
that I hope will help clarify things.

Your explanation of the Galilean transformation is completely correct. 
In fact, the Galilean transformation is nothing more than careful 
addition and subtraction of velocity vectors. As you say, it doesn't 
work for frames whose velocities differ by a substantial fraction of 
c, the speed of light (for such problems, you need the Lorentz 
transformation.) The Lorentz transformation reduces to the Galilean 
transformation when all of the velocities involved are slow compared 
with c.

So, with the basic understanding that the Galilean transformation is 
simply addition/subtraction of vectors, let's try working through a 
problem to see if we can make the notation clearer. I'm going to 
choose a problem that involves conservation of momentum and 
conservation of kinetic energy, since these are usually concepts that 
are treated before the transformation equations, but if you haven't 
seen these before, please write back and I'll try to come up with a 
different example.

Let's consider an elastic collision of two masses (m = 1 kg, M = 2 kg) 
along the x-direction. Let m have initial velocity v1 = 4 m/s and M 
have initial velocity V1 = 0 m/s. Now, momentum is conserved:

     m*v1 + M*V1 = m*v2 + M*V2

and so is kinetic energy (since the collision is elastic):

     (1/2)*m*v1^2 + (1/2)M*V1^2 = (1/2)m*v2^2 + (1/2)M*V2^2

In numbers these equations yield (I multiplied through by 2 in the 
kinetic energy equation):

     momentum:          1*4 + 2*0 = 1*v2 + 2*V2, or v2 = 4 - (2*V2)
     kinetic energy:   1*16 + 2*0 = 1*v2^2 + 2*V2^2

Combining these enables us to solve for v2 and V2, and you can easily 
check that the final velocities are v2 = -4/3 m/s and V2 = +8/3 m/s.

All well and good so far? Now let's imagine we view this collision as 
taking place on a moving vehicle. Let's imagine that I did this 
collision experiment on a moving train, and that I measured all of the 
numbers above: m, M, v1, V2, etc. Let's say that you are on the ground 
and watching the same experiment, but that the train on which I am 
riding is moving at a relative velocity b = +5 m/s. That is, you see 
the train (and me) moving to the right at 5 m/s. Incidentally, I see 
_you_ moving at -5 m/s.

The whole idea is what quantities do _you_ measure as I and the 
experiment whiz by. Well, the masses don't change:

     m' = m  = 1 kg
     M' = M  = 2 kg

The velocities measured by you do get shifted by the Galilean 
transformation.

     v1' = v1 + b = 4+5 = 9 m/s 
     V1' = V1 + b = 0+5 = 5 m/s

You see how I've used primes to indicate all quantities measured in 
_your_ frame. This is often the standard notation (and part of the 
reason why things could get confusing), but you can use any variable 
names you like, such as q and Q for your masses, and u1 and U1 for the 
initial velocities in your frame, etc. Now, we've calculated the final 
velocities v2 and V2 in my frame, how about your frame? For example, 
will V2' = V2 + b? The answer is yes, and that is a valid way to solve 
the problem (and it's easier since we've already done the algebra in 
my frame). But let's verify it by directly computing everything:

     momentum:       (m')(v1') + (M')(V1') = (m')(v2') + (M')(V2')
     kinetic E:  (m')(v1')^2 + (M')(V1')^2 = (m')(v2')^2 + (M')(V2')^2

in numbers, these are:

     momentum:     1*9 + 2*5 = 1*(v2') + 2*(V2')
     kinetic E:  1*81 + 2*25 = 1*(v2')^2 + 2*(V2')^2

and solving for the two unknowns (the first equation gives 
v2' = 19-2*V2', and substituting for v2' in second equation) yields 
v2' = +11/3 m/s and V2' = +23/3 m/s. So a quick check does show that 
v2' = v2 + b and V2' = V2 + b. That is reassuring - our algebra is 
probably correct.

What this means is that the physics is the same (conservation of 
momentum or kinetic energy, or Newton's force law, or whatever) in any 
frame in which you choose to do the calculation. For example, 
sometimes it is particularly convenient to work in the "center-of- 
mass" frame of the various bodies. In our example above, that would 
have been at b = +2 m/s. You might find it interesting to verify 
conservation of momentum and kinetic energy in that frame as well.

Good luck, and write back if you have any trouble with this 
explanation.

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/