How can I show that z = v/c for small velocities

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Homework Help Overview

The discussion revolves around demonstrating the relationship \( z = v/c \) for small velocities, starting from the equation \( 1+z=\sqrt{\frac{1+v/c}{1-v/c}} \). The subject area is likely related to special relativity or kinematics.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the expansion of the right-hand side of the equation and consider the implications of small velocities on the relationship between \( z \) and \( v/c \). There is a suggestion to simplify the expression by substituting small values for \( v \).

Discussion Status

The discussion includes attempts to manipulate the equation algebraically, with some participants confirming the correctness of the derived expressions. There is also a suggestion that a simpler approach might exist, indicating a variety of perspectives on the problem.

Contextual Notes

Participants are working under the assumption that \( v \) is small, which influences their reasoning and the methods they consider. There is a playful comment about the number of posts from one participant, which does not contribute to the problem-solving process.

Logarythmic
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How can I show that

[tex]1+z=\sqrt{\frac{1+v/c}{1-v/c}}[/tex]

becomes [tex]z \simeq v/c[/tex] for small velocities? Please give me a hint.
 
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Try writing the right hand side as [tex](1+v/c)^{1/2}(1-v/c)^{-1/2}[/tex]. Can you expand this?
 
So

[tex]1+z=\frac{\sqrt{1+\beta}}{\sqrt{1-\beta}}\simeq \left( 1+\frac{1}{2}\beta-\frac{1}{8}\beta^2+... \right) \left( 1+\frac{1}{2}\beta+\frac{3}{8}\beta^2+... \right) = 1+\beta+\frac{5}{8}\beta^2+...=1+\frac{v}{c}[/tex]

Correct?
 
Yup, that's correct.
 
cristo scroll up, lol
 
I would think there was an easier way...v is approaching zero, sub v=0 into the right hand side and its easy to see z is also approaching zero...

PS: Cristo has 666 posts...ooh
 

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