Integration simplification

1. May 29, 2012

sparkle123

I simply don't understand this conversion. I've only worked with du, dx, dy, etc. before. How do you change a d(γmu) into some kind of du?
(This is in the contest of relativity, although this info is not essential i think)

Thanks! :)

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2. May 29, 2012

SammyS

Staff Emeritus

What is $\displaystyle \frac{d}{du}\left(\frac{mu}{\sqrt{1-u^2/c^2}}\right)\ ?$

3. May 29, 2012

dimension10

Let $$x = \frac{{mu}}{{\sqrt {1 - \frac{{{u^2}}}{{c_{0}^2}}} }}$$

Then solve for u.

4. May 29, 2012

sparkle123

Figured it out thanks!

5. May 29, 2012

sparkle123

Wait dimension10, how do you solve it your way, after getting u = x/sqrt(m^2+x^2/c^2)?

6. May 29, 2012

dimension10

$$\begin{array}{l} {x^2} = \frac{{{m^2}{u^2}}}{{1 - \frac{{{u^2}}}{{{c_0}^2}}}}\\ {x^2} = \frac{{{m^2}}}{{\frac{1}{{{u^2}}} - \frac{1}{{{c_0}^2}}}}\\ \frac{{{m^2}}}{{{x^2}}} + \frac{1}{{c_0^2}} = \frac{1}{{{u^2}}}\\ {u^2} = \frac{1}{{\frac{{{m^2}}}{{{x^2}}} + \frac{1}{{c_0^2}}}}\\ {u^2} = \frac{{c_0^2{x^2}}}{{{m^2}c_0^2 + {x^2}}}\\ u = \frac{{{c_0}x}}{{\sqrt {{m^2}c_0^2 + {x^2}} }}\\ ...\\ \int {\frac{{{c_0}x}}{{\sqrt {{m^2}c_0^2 + {x^2}} }}{\rm{d}}x} = \int {m{{\left( {1 - \frac{{{x^2}}}{{c_0^2}}} \right)}^{ - \frac{3}{2}}}x{\rm{ d}}u} = \int {m{{\left( {1 - \frac{{{u^2}}}{{c_0^2}}} \right)}^{ - \frac{3}{2}}}u{\rm{ d}}u} \end{array}$$

7. May 29, 2012

sparkle123

Thanks a lot!