Cosmic Adventure 5.1 Relative Motion in Special Relativity

© ABCC Australia 2015 new-physics.com

FROM GALILEAN TO EINSTEINIAN

Cosmic Adventure 5.1


what is wrong with them?

They are not valid when the speed of light is considered

You know what is wrong with your

equations?


System x:

𝑥′ = 𝑥 − 𝑣𝑡

𝑦′ = 𝑦

𝑧′ = 𝑧

𝑡′ = 𝑡

System x’:

𝑥 = 𝑥′ + 𝑣𝑡

𝑦 = 𝑦′

𝑧 = 𝑧′

𝑡 = 𝑡′

Because in between frames, you should have

light speeding along

No Light Involved


To take the speed of light into consideration, you need to have a Lorentz transformation with the Lorentz factor! 𝛾 =

1

1 −𝑣2

𝑐2



𝑥′′ =𝑥′ − 𝑣𝑡

1 −𝑣2

𝑐2

𝑡′′ =𝑡′ − 𝑣𝑥′/𝑐2

1 −𝑣2

𝑐2

𝑥′ =𝑥′′ + 𝑣𝑡

1 −𝑣2

𝑐2

𝑡′ =𝑡′′ + 𝑣𝑥′′/𝑐2

1 −𝑣2

𝑐2

Relativistic Equations


𝛾 =1

1 −𝑣2

𝑐2

Where did you this funky Lorentz factor come from?


Normally I would not tell where I get these equations. But for this competition, I have to do so because Angela already knew it.


Beam

BBeam A

Viewer

Light Path Geometry [at Rest] on plan

These equations came from Michelson’s concept of the experiment itself.

In the absence of aether wind, the two lights will recombine at the viewer in sync. No fringe will be observed.


Bea

m B

Beam A

Viewer

Light Path Geometry According to Michelson

According to Michelson, the two lights will pursue different paths due to the influence of the aether wind.


Michelson’s Working Equation

𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝐵𝑒𝑎𝑚 𝐴

=2𝑙𝑜

𝑐 1 − 𝑣2/𝑐2

𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝐵𝑒𝑎𝑚 𝐵

=2𝑙𝑜

𝑐 1 − 𝑣2/𝑐2

In the experiment there are two durations of time for the two beams.


The ratio between the Times of the Beams

𝑇𝑖𝑚𝑒 𝑓𝑜𝑟 𝐵𝑒𝑎𝑚 𝐴

𝑇𝑖𝑚𝑒 𝑓𝑜𝑟 𝐵𝑒𝑎𝑚 𝐵

=2𝑙𝑜

𝑐 1 − 𝑣2/𝑐2÷

2𝑙𝑜

𝑐 1 − 𝑣2/𝑐2

=2𝑙𝑜

𝑐 1 − 𝑣2/𝑐2×𝑐 1 − 𝑣2/𝑐2

2𝑙𝑜

=1

1 − 𝑣2/𝑐2

1

1 − 𝑣2/𝑐2


This factor was later brought up by Albert Lorentz in his postulate of time dilation to explain the experiment. I got inspired and made use of it in my theory.

In its turn, the factor accounts for failure of the Michelson-Morley experiment to produce the desired results.

1

1 − 𝑣2/𝑐2


Looking at your equations, it is obvious the major variable is the relative velocity 𝑣. What if both systems are at rest?

𝑥′′ =𝑥′ − 𝑣𝑡

1 −𝑣2

𝑐2

𝑡′′ =𝑡′ − 𝑣𝑥′/𝑐2

1 −𝑣2

𝑐2

Relativistic Equations


Then 𝑣 will become zero and the equations will revert to the starting point.

𝑥′′ =𝑥′ − 𝑣𝑡

1 −𝑣2

𝑐2

→ 𝑥′

𝑡′′ =𝑡′ − 𝑣𝑥′/𝑐2

1 −𝑣2

𝑐2

= 𝑡′


What about when the observer and the object are at rest and are separated by a distance s?

They are not moving, so 𝑣 is again 0. 0’

𝑥′

P

𝑥′′

0’’ P

𝑠


VISONIC TRANSFORM WITHOUT MOTION

To be continued on Cosmic Adventure 5.2

Science

Cosmic Adventure 5.1 Relative Motion in Special Relativity