Ultrarelativistic approximation

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Discussion Overview

The discussion revolves around the mathematical demonstration of the relationship between the expressions sqrt(1-β²) and sqrt(2*(1-β)) in the context of ultrarelativistic speeds, where v approaches the speed of light (c). The focus is on exploring the mathematical steps and reasoning involved in this approximation.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about how to mathematically show that sqrt(1-β²) equals sqrt(2*(1-β)) when v is approximately equal to c.
  • Another participant suggests substituting β with 1 - ε, where ε approaches zero, and discusses taking limits and discarding higher-order terms in the process.
  • A third participant questions the presence of an extraneous β in the previous explanation but indicates understanding after the clarification.
  • One participant offers an alternative perspective, noting that 1 - β² can be factored as (1 + β)(1 - β) and suggests that since β is close to 1, the first term is nearly 2.

Areas of Agreement / Disagreement

There is no clear consensus on the best method to demonstrate the relationship, as multiple approaches are presented, and participants express varying levels of understanding and clarification.

Contextual Notes

The discussion includes assumptions about the behavior of β as it approaches 1 and the implications of higher-order terms in the mathematical expressions, which remain unresolved.

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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mps said:
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 \epsilon = 1 - \beta , then substitute \beta = 1 - \epsilon. You get

sqrt(1 - (1 -2 \beta \epsilon + \epsilon^2)) = sqrt(1 - (1 - 2 (1 - \epsilon) \epsilon + \epsilon^2)

Take the limit as \epsilon goes to zero. You might need a bit of calculus to do that. bit basically you keep all the terms proportioanl to \epsilon and trhow out all high order temrs proportional to \epsilon^2.
 
Last edited:
pervect said:
sqrt(1 - (1 -2 \beta \epsilon + \epsilon^2))

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

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