- #1
Converting d(γmu) to du: A Relativity Integration Simplification
- Thread starter sparkle123
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SUMMARY
This discussion focuses on the conversion of the differential form d(γmu) into du within the context of relativity. The participants explore the mathematical transformation involving the Lorentz factor and the relationship between variables u, m, and x. A specific solution is derived, leading to the expression for u in terms of x and m, ultimately facilitating the integration process. The conversation highlights the importance of understanding these transformations for solving problems in relativistic physics.
PREREQUISITES- Understanding of differential calculus and integration
- Familiarity with the Lorentz factor in relativity
- Knowledge of the variables involved in relativistic equations (u, m, x, c0)
- Ability to manipulate algebraic expressions and solve equations
- Study the derivation and applications of the Lorentz factor in relativity
- Learn about integration techniques in calculus, particularly for rational functions
- Explore the physical significance of the variables u, m, and x in relativistic contexts
- Investigate more complex transformations in differential forms within physics
Students and professionals in physics, particularly those focusing on relativity, mathematicians dealing with differential equations, and anyone interested in advanced calculus applications in theoretical physics.
- #2
SammyS
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sparkle123 said:
What is \displaystyle \frac{d}{du}\left(\frac{mu}{\sqrt{1-u^2/c^2}}\right)\ ?
- #3
dimension10
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Let x = \frac{{mu}}{{\sqrt {1 - \frac{{{u^2}}}{{c_{0}^2}}} }}sparkle123 said:
Then solve for u.
- #4
sparkle123
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Figured it out thanks! 
- #5
sparkle123
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Wait dimension10, how do you solve it your way, after getting u = x/sqrt(m^2+x^2/c^2)?
- #6
dimension10
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sparkle123 said:
\begin{array}{l}<br /> {x^2} = \frac{{{m^2}{u^2}}}{{1 - \frac{{{u^2}}}{{{c_0}^2}}}}\\<br /> {x^2} = \frac{{{m^2}}}{{\frac{1}{{{u^2}}} - \frac{1}{{{c_0}^2}}}}\\<br /> \frac{{{m^2}}}{{{x^2}}} + \frac{1}{{c_0^2}} = \frac{1}{{{u^2}}}\\<br /> {u^2} = \frac{1}{{\frac{{{m^2}}}{{{x^2}}} + \frac{1}{{c_0^2}}}}\\<br /> {u^2} = \frac{{c_0^2{x^2}}}{{{m^2}c_0^2 + {x^2}}}\\<br /> u = \frac{{{c_0}x}}{{\sqrt {{m^2}c_0^2 + {x^2}} }}\\<br /> ...\\<br /> \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} <br /> \end{array}
- #7
sparkle123
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Thanks a lot!