What's gamma for?

1. Nov 9, 2006

actionintegral

Does anyone have a brief explanation of why the gamma is necessary in the Lorentz transformation.

2. Nov 9, 2006

bernhard.rothenstein

gamma?

in order to account for relativistic effects like time dilation, doppler shift...
sine ira et studio

3. Nov 9, 2006

actionintegral

But the constancy of the speed of light is preserved simply by
x'=x-vt and t'=t-vx/cc. I'm trying to see where the scaling factor comes in.

4. Nov 9, 2006

nakurusil

Comes from preserving dx^2 -( c dt )^2

5. Nov 9, 2006

MeJennifer

Gamma, or sometimes called the Lorentz factor, represents the relationship between relativistic time and proper time or relativistic length and proper length.

Proper time and proper length are resp. the time and the length of an object in its restframe while relativistic time and relativistic length are resp. the time and length of an object that is moving relative to the frame of the observer.

The reason we need gamma to translate lengths and times from one frame to another is that the speed of light is always the same independent of motion. So, distance and duration cannot longer be absolute but instead depend on relative motion.

6. Nov 9, 2006

actionintegral

Fair enough - but why would I want to preserve THAT?

7. Nov 9, 2006

bernhard.rothenstein

gamma@

you are perfectly right. in the paper

arXiv.org > physics > physics/0409121

Physics, abstract
physics/0409121

Three levels of understanding physical relativity: Galileo's relativity, Up-to-date Galileo's relativity and Einstein's relativity: A historical survey
Authors: Bernhard Rothenstein, Corina Nafornita
Subj-class: Physics Education

We present a way of teaching Einstein's special relativity. It starts with Galileo's relativity, the learners know from previous lectures. The lecture underlines that we can have three transformation equations for the space-time coordinates of the same event, which lead to absolute clock readings, time intervals and lengths (Galileo's relativity), to absolute clock readings but to relative time intervals and lengths (up-to-date Galileo transformations) and to relative clock readings time intervals and lengths.
we have called the equations you mention "uptodate" Galileo transformations. Please have a critical look at it

8. Nov 9, 2006

nakurusil

Because it represents the line element.

9. Nov 9, 2006

robphy

There are experimental results [as well as theoretical/mathematical results] that effectively lead to this requirement.

10. Nov 9, 2006

robphy

For special relativity, the set of transformations one considers has to satisfy some conditions, including:
they are linear [lines map to lines] and they form a group [which preserves the metric]. [I don't think your transformations form a group.]

Especially for a theory of relativity, the transformations cannot have a timelike eigenvector [i.e. no distinguished observer].

Last edited: Nov 9, 2006
11. Nov 9, 2006

pervect

Staff Emeritus
The equations

x1 = x - v*t
t1 = t - (v/c^2) x

do have an inverse, so they will form a group. (A mapping must be associative and have an inverse to be a group, and IIRC associativity is a general property of any mapping so it shouldn't be an issue).

But the form of the inverse reveals why it's not the right choice:

t = (t1 + (v/c^2) x1) / (1 - v^2/c^2)
x = (x1 + v*t) / (1 - v^2/c^2)

We _also_ want the transforms to have the property that the inverse is generated just by changing the sign of v. This is justified by requiring that the transforms be isotropic. Isotropy is the missing element in the proposal.

12. Nov 9, 2006

actionintegral

Thanks, perv. I will investigate the lack of symmetry in the inverse.

13. Nov 9, 2006

neutrino

Because people don't want to keep writing $$\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$. :tongue:

Sorry...

14. Nov 9, 2006

robphy

If people wrote $$\cosh\theta$$ instead of $$\gamma$$ or $$\frac{1}{\sqrt{1-v^2/c^2 } }$$, these mysterious quantities might not seem as mysterious.