How Do You Solve This Special Relativity Integral?

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SUMMARY

The forum discussion centers on solving the integral related to special relativity, specifically the expression \(\int^1_{-1} \frac {dz} { [1- \frac {v^2} {c^2} + \frac {v^2} {c^2} z^2 ]^{\frac {3} {2}}}\) with \(c=10^8 \text{ m/s}\). The proposed solution involves the expression \(\frac 1 {( \frac {v^3}{c^3})} \frac {z} {( \frac {c^2}{v^2}-1 ) \sqrt{( \frac {v^2}{c^2} -1 +z^2})} |^{1}_{-1}\). A user suggested using the substitution \(z = \tan \theta \sqrt{(\frac {c^2}{v^2})-1}\), but encountered issues when \(v\) is small, leading to a zero result. Further advice included differentiating the expression \(z/\sqrt{a^2 + z^2}\) to aid in solving the integral.

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Homework Statement



I am trying to solve this given [tex]c=10^8m/s[/tex].

[tex]\int^1_{-1} \frac {dz} { [1- \frac {v^2} {c^2} + \frac {v^2} {c^2} z^2 ]^{\frac {3} {2}}}[/tex]

The answer is:

[tex]\frac 1 {( \frac {v^3}{c^3})} \frac {z} {( \frac {c^2}{v^2}-1 ) \sqrt{( \frac {v^2}{c^2} -1 +z^2})} |^{1}_{-1}[/tex]

The Attempt at a Solution



I tried substitude with [tex]tan \theta= \frac{z}{\sqrt{(\frac {c^2}{v^2})-1}}[/tex] but it won't work because if v is small, the answer is zero.

Please give me advice how to solve this.

Thanks

Alan
 
Last edited:
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hi yungman! :smile:

(you keep miscounting the {} :wink:)

try differentiating z/√(a2 + z2) :smile:
 
tiny-tim said:
hi yungman! :smile:

(you keep miscounting the {} :wink:)

try differentiating z/√(a2 + z2) :smile:


Yes I fixed the latex problems I have.

I'll try and come back later.

Thanks
 

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