Ultrarelativistic approximation

  • Thread starter mps
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  • #1
mps
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I read that when v≈c,
sqrt(1-β2) = sqrt(2*(1-β)).
How do you show this mathematically? I have no idea. Thanks! :)
 

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  • #2
pervect
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I read that when v≈c,
sqrt(1-β2) = sqrt(2*(1-β)).
How do you show this mathematically? I have no idea. Thanks! :)

Let [itex]\epsilon = 1 - \beta[/itex] , then substitute [itex]\beta = 1 - \epsilon[/itex]. You get

sqrt([itex]1 - (1 -2 \beta \epsilon + \epsilon^2)[/itex]) = sqrt([itex]1 - (1 - 2 (1 - \epsilon) \epsilon + \epsilon^2)[/itex]

Take the limit as [itex]\epsilon[/itex] goes to zero. You might need a bit of calculus to do that. bit basically you keep all the terms proportioanl to [itex]\epsilon[/itex] and trhow out all high order temrs proportional to [itex]\epsilon^2[/itex].
 
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  • #3
mps
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sqrt([itex]1 - (1 -2 \beta \epsilon + \epsilon^2)[/itex])

I think you might have an extraneous β here, but thanks a lot!!! I get it now :)
 
  • #4
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It's even easier to see than that. Just remember that [itex]1-\beta^2 = (1+\beta)*(1-\beta)[/itex]. Then, since [itex]\beta[/itex] is very close to 1, the first term is only very slightly smaller than 2.
 
  • #5
mps
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Thank you! :)
 

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