Atkisnon's derivation of \gamma factor.

  • #1
MathematicalPhysicist
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So I am rereading David Atkinson's QFT book:
http://books.google.co.il/books/about/Quantum_Field_Theory.html?id=vbAnQAAACAAJ&redir_esc=y

And I am puzzled by what is written in pages 5-6 section 1.3 Special Theory of Relativity.

He writes down the transformation between two inertial frames, one moving at speed v compared to the other.

[tex] x'=\gamma (x-vt) \ y'=y \ z'=z[/tex]

Now he writes on page 6:
What is gamma?
From the uniformity of space, we see that \gamma may not depend on the coordinates (...).
In the absence of gravity, \gamma is independent of t,x,y,z but it may depend on v.
...
Rotational invariance therefore means that \gamma must be independent of the sign of v, i.e, it's a function only of v^2.

My problem is with the last conclusion, why can't we have \gamma to be any even function of v, like v^4 and so forth?
 

Answers and Replies

  • #2
qbert
185
5
any even function of v is a function of v^2.

v^4 = (v^2)^2
 
  • #3
MathematicalPhysicist
Gold Member
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Feel so stupid now... :-D

Thanks.
 
  • #4
qbert
185
5
don't feel bad. everybody misses stuff like that at times.

in fact you could come back and ask. really? every even function
of v is a function of v^2? those are two different things in principle,
even functions f(v) = f(-v) and f(v) = h(v^2) for some h.

it's not that hard to prove either.
 

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