# Re: Galilean transformation equations

*From*: Alen <alen1@xxxxxxxxxxxxxxx>*Date*: Sun, 4 Jan 2009 04:52:13 -0800 (PST)

On Jan 4, 2:36 pm, rbwinn <rbwi...@xxxxxxxx> wrote:

On Jan 3, 7:37 pm, Alen <al...@xxxxxxxxxxxxxxx> wrote:

On Jan 1, 11:07 am, rbwinn <rbwi...@xxxxxxxx> wrote:

On Dec 31, 7:00 am, "Sue..." <suzysewns...@xxxxxxxxxxxx> wrote:

On Dec 30, 9:01 pm, rbwinn <rbwi...@xxxxxxxx> wrote:

x'=x-vt

y'=y

z'=z

t'=t

Robert B. Winn

~<< the famous Lorentz transformation ensures

that the velocity of light is invariant between

different inertial frames, and also reduces to

the more familiar Galilean transform in the

limit v << c. >>~http://farside.ph.utexas.edu/teaching/em/lectures/node109.html

Classical Electromagnetism:

An intermediate level coursehttp://farside.ph.utexas.edu/teaching/em/lectures/lectures.html

Sue...

Thank you, Sue. The reason why I like the Galilean transformation

equations is because, using velocity of light instead of speed of

light like scientists insist on doing, the velocity of light is

invariant between different inertial frames, and does not have to

reduce in any way to become the Galilean transformation equations.

They are already the Galilean transformation equations.

Now, suppose that we have a photon going in the +x direction

relative to frames of reference S and S'. According to the Galilean

transformation equations, x'=x-vt. The photon travels a distance of

x' in S' while it is traveling a distance of x in S. The Michelson

Morley experiment says that the photon has a velocity of c in S and a

velocity of c in S'.

So, if x'=x-vt, then if n' is time on a clock in S', then

cn'=ct-vt

The photon has a speed of c in both frames of reference.

Suppose that a photon is going in the -x direction relative to S

and S'. Then the photon has a velocity of -c relative to S and S'.

The Galilean transformation equation that shows the distance traveled

by the photon in S and S' is

-cn'= -ct-vt

The photon has a speed of c in both frames of reference.

You claim that the Lorentz equations show the velocity of light to

be invariant in S and S'. I would submit that the velocity of light

is not invariant in S and S' according to the Lorentz equations, but a

photon going in the +x direction has a velocity of c relative to S and

S', while a photon going in the -x direction has a velocity of -c

relative to S and S', the difference being that there is a length

contraction in the Lorentz equations and there is not a length

contraction in the Galilean equations. The speed of light is

invariant in the Lorentz equations, but the speed of light is also

invariant in the Galilean transformation equations, as I have shown

above. The velocity of light is not invariant in either set of

equations. -c does not equal +c.

Thank you for your response.

Robert B. Winn

I once came across this possibility, and also found that

it is possible to eliminate length contraction in the direction

of motion, but I also found that it gives rise to a problem

in that there will be an asymmetry between oscillations of

light in the direction of motion and in a transverse direction.

What length contraction does is to ensure that, in the

moving frame, from the perspective of the stationary frame,

light oscillates back and forth with the same frequency

in all directions, independently of the direction of relative

motion. Without length contraction, photons within a

clock mechanism will appear to oscillate more slowly

in the direction of relative motion than in the transverse

direction, unless you postulate length stretching in

transverse directions.

When Einstein derived the equations, he reconciled the

derivation in both the direction of motion and the transverse

directions.

Alen- Hide quoted text -

- Show quoted text -

Alen,

Thank you for your comment. This is exactly the subject that I

hoped to discuss with someone. Here is how I break the problem

down. We have two sets of coordinates with no length contraction.

x'=x-vt

y'=y

z'=z

t'=t

So Michelson and Morley run an experiment in which scientists say they

have proven that light is traveling at c= 186,000 mi/sec in both

frames of reference. That would mean that t' in the Galilean

transformation equations is not the time on a clock in frame of

reference S' by which the speed of light was measured to be c. The

Galilean transformation equations define t' to be t, time as shown by

a clock in frame of reference S which shows light to be c in that

frame of reference. Consequently, time on a clock in S' has to be

represented by some other variable than t' if the Galilean

transformation equations are to be used. So we will say that time on

a clock in S' is n'.

So if w is velocity of light, then the results of the Michelson-

Morley experiment show

x=wt

x'=wn'

We use velocity of light instead of speed of light like scientists

do because if x and x' are negative, then w=(-c), not +c .

x'=x-vt

wn'=wt-vt

n'=t(1-v/w)

This means that the clock in S' is running slower than the clock in

S, allowing both clocks to show the speed of light to be c=186,000 mi/

sec.

Now with regard with your question about frequency, the result I get

from these equations is that the wavelength and frequency will also be

the same in both frames of reference according to this interpretation

of the Galilean transformation equations.

Robert B. Winn

You are setting up a situation with light travelling along

x', in the direction of the relative motion. But consider also

setting up a situation with light travelling transversely,

such as along the y direction, while the relative motion

is along the x direction, with the velocity of light being

c in both frames, and see what kind of transformation

equations you can get!?

Alen

.

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