Understanding Einstein's Relativity: From Energy Equations to Series Development

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SUMMARY

Einstein's theory of relativity introduces a new expression for the energy of a mass m, defined as { m c^2 } / √{1 - { v^2 / c^2 }}. To derive the kinetic energy expression in series form, one utilizes Taylor's series expansion, specifically the McLaurin series, which simplifies the function (1 + x^2/c^2)^(-1/2) at x=0. The resulting series includes terms such as mc^2, m(v^2/2), and (3/8)m(v^4/c^2). This mathematical approach provides clarity on the transition from the energy equation to the series development.

PREREQUISITES
  • Understanding of Einstein's theory of relativity
  • Familiarity with Taylor's series and McLaurin series
  • Basic knowledge of calculus, including derivatives
  • Concept of kinetic energy in physics
NEXT STEPS
  • Study the derivation of the Lorentz factor in relativity
  • Learn about Taylor series and their applications in physics
  • Explore the implications of relativistic kinetic energy
  • Investigate advanced calculus techniques related to series expansions
USEFUL FOR

Physicists, mathematics students, and anyone interested in the mathematical foundations of Einstein's relativity and energy equations.

OneEye
In Relativity, (p. 45), Dr. Einstein says:

...the energy of a material point of mass m is no longer gievn by the well-known expression

[tex]m { v \over c^2 }[/tex]

but by the expression

[tex]{ m c^2 } \over \sqrt { 1 - { v^2 \over c^2 } }[/tex]

...If we develop the expression for the kinetic energy in the form of a series, we obtain

[tex]mc^2 + m { v^2 \over 2 } + { 3 \over 8 } m { v^4 \over c^2 } . . . .[/tex]

How does one get from the second equation to the third? What is meant by "develop[ing] the expression... in the form of a series"?

Any help would be more than appreciated.
 
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One way is to use a "Taylor's series" expansion.

If f is any function, analytic at x0 then
f(x)= f(x0)+ f '(x0)(x- x0)+ (f''(x0)/2)(x- x0)2+ (f'''(x0)/6)(x- x0)3+ ...
the general term is (f(n)(x0)/n!)(x-x0)n where f(n) is the nth derivative.

In particular, use the "McLaurin series" which is the Taylor's series with x0= 0 and f(x)= (1+ x2/c2)-1/2, set x= v, and then multiply by mc2.
 
HoI - Thanks - I suspected something like this, but lacked the mathematical toolkit to know what exactly was meant.

Most helpful!
 

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