Deriving the Equation for Relativistic Mass

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SUMMARY

The discussion focuses on deriving the equation for relativistic mass, specifically the expression d(γμ) = m(1 - u²/c²)^(-3/2) du. Participants clarify that γ is defined as γ = 1/√(1 - u²/c²) and emphasize the need to apply the product rule for differentiation. The conversation highlights the importance of recognizing du as an infinitesimal quantity, which leads to the correct interpretation of the derivative on the left-hand side. The final result is obtained by substituting the expression for γ and simplifying the equation.

PREREQUISITES
  • Understanding of relativistic mechanics and the concept of relativistic mass.
  • Familiarity with calculus, specifically differentiation and the product rule.
  • Knowledge of the Lorentz factor, γ, and its formula γ = 1/√(1 - u²/c²).
  • Basic understanding of differential notation and infinitesimals in physics.
NEXT STEPS
  • Study advanced calculus techniques, particularly the product rule and chain rule in differentiation.
  • Explore the implications of relativistic mass in high-velocity physics scenarios.
  • Learn about the Lorentz transformations and their applications in special relativity.
  • Investigate the relationship between energy, momentum, and relativistic mass in particle physics.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the mathematical foundations of relativistic mechanics and the derivation of equations related to relativistic mass.

digital19
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Homework Statement



Show that [tex]d(\gamma mu)=m(1- \frac{u^2}{c^2})^{-3/2} du[/tex]

Homework Equations



It is known that is
[tex]\gamma=\frac{1}{\sqrt{{1- \frac{u^2}{c^2}}}}[/tex]

The Attempt at a Solution



The question stated in part 1 is the precise question given in the textbook.

I'm not sure how to proceed here. I believe it's asking you to take the derivative using the product rule. I even wondered if it was a type and du was supposed to be on the left hand side.
 
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It's not a typo,
[tex]d(\gamma mu)=m(1- \frac{u^2}{c^2})^{-3/2} du[/tex]
is just physicists' notation for
[tex]\frac{d}{du} (\gamma mu)=m(1- \frac{u^2}{c^2})^{-3/2}[/tex]
(if you want, consider du as an infinitesimal quantity, dividing by it gives you a differential quotient aka derivative on the left hand side).
So indeed, you just plug in the expression for [itex]\gamma[/itex] you gave and differentiate w.r.t. u; then simplify to get the requested result
 
What text are you using?
[tex]u^2[/tex] stands for [tex]\vec u}\cdot{\vec u}[/tex].
This leads to an additional term in the derivative.
 

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