How to break in a new chalkboard

Step 4 (more).Update: Wohooo, this pic made it to Lifehacker!

• WaTTacK 3y

Well, nice pic indeed. But why is the circled part so? I mean where does (1+(gamma²/2)+...) come from?
(I have not studied relativity that much, I am just curious. But then again I think it is more of a mathematical approssimation which anyway I don't get.)
• That's the Lorentz Factor (en.wikipedia.org/wiki/Lorentz_factor). What this development does is bringing a "classic" equation for the kinetic energy from a relativistic equation (top left). The only assumption is that velocity is way smaller than the speed of light (v << c, that is gamma << 1). Hope it helps.
• WaTTacK 3y

Ok, that's great. Thanks. I did read the link, without actually grasping a lot of it though (silly me).
The part of the picture that troubles me is when you go from [(1-γ²)^(-1/2)] to [(1+γ²/2)+...].
I do have some reminiscences from physics class of the first substitutions in the first two rows, but I don't get that particular approximation.
Is there any chance you could explain it simply to a physics noob like me?
• Not a problem. That's called a Taylor series approximation (en.wikipedia.org/wiki/Taylor_series). Basically, when you do such an approximation, the first few terms in the series account for most of the value of the function at a particular point. In our case, we're assuming that v (velocity) is way smaller than the speed of light (this is called "classic conditions"), and so v/c is very close to zero. Therefore, we can do a Taylor series approximation "at gamma = 0".

You can see the complete result from Wolfram Alpha (www.wolframalpha.com/input/?i=taylor%5B%281-x%5E2%29%5E%2...).
• WaTTacK 3y

Oh, that was it. That's why I couldn't understand it, I never studied the Taylor series.
Thanks a lot mate!
• My pleasure.
• Gud
• Alan Boy 2y

Great
• I'm so hold.
30,465 views
5 faves
Taken on July 15, 2010
• 4.3 mm
• Show EXIF

Beta