Talk:Theory of relativity/Special relativity

Exercise

According to the classical theory, the kinetic energy of a body can be written as: ${\frac {1}{2}}mv^{2}$, where $m$ is the mass, and $v$ is the velocity of the body. According to the special theory of relativity, the kinetic energy of a body can be written as $(\gamma -1)mc^{2}$, where $\gamma ={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}$ is called the Lorentz factor. Prove that the classical theory would be correct, according to the special theory of relativity, if the speed of light would be infinite. In other words, prove that: $\lim _{c\to \infty }(\gamma -1)mc^{2}={\frac {1}{2}}mv^{2}$

[Solution]

   We have to prove that

$\lim _{c\to \infty }(\gamma -1)mc^{2}={\frac {1}{2}}mv^{2}$

By extending we get

$=\lim _{c\to \infty }{\frac {(\gamma +1)(\gamma -1)mc^{2}}{(\gamma +1)}}$

Siplifying gives

$=\lim _{c\to \infty }{\frac {(\gamma ^{2}-1^{2})mc^{2}}{\gamma +1}}$

And further

$=\lim _{c\to \infty }{\frac {(\gamma )^{2}mc^{2}-(1)^{2}mc^{2}}{\gamma +1}}$

By substituting $\gamma$ we get

$=\lim _{c\to \infty }{\frac {\mathbf {\left({\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}\right)} ^{2}mc^{2}-mc^{2}}{\gamma +1}}$

Simplification gives

$=\lim _{c\to \infty }{\frac {{\frac {mc^{2}}{\mathbf {\left({\sqrt {1-{\frac {v^{2}}{c^{2}}}}}\right)} ^{2}}}-mc^{2}}{\gamma +1}}$

And so we get

$=\lim _{c\to \infty }{\frac {{\frac {mc^{2}}{\mathbf {\left(1-{\frac {v^{2}}{c^{2}}}\right)} }}-mc^{2}}{\gamma +1}}$

Which gives

$=\lim _{c\to \infty }{\frac {{\frac {mc^{2}c^{2}}{c^{2}-v^{2}}}-mc^{2}}{\gamma +1}}$

Further

$=\lim _{c\to \infty }{\frac {{\frac {mc^{4}}{c^{2}-v^{2}}}{\frac {-(mc^{2})(c^{2}-v^{2})}{c^{2}-v^{2}}}}{\gamma +1}}$

And naturally

$=\lim _{c\to \infty }{\frac {\frac {mc^{4}-mc^{4}+mv^{2}c^{2}}{c^{2}-v^{2}}}{\gamma +1}}$

This gives

$=\lim _{c\to \infty }{\frac {\frac {mv^{2}c^{2}}{c^{2}(1-{\frac {v^{2}}{c^{2}}})}}{\gamma +1}}$

And so

$=\lim _{c\to \infty }{\frac {mv^{2}}{(1-{\frac {v^{2}}{c^{2}}})(\gamma +1)}}$

By substituting $\gamma$ again we get

$=\lim _{c\to \infty }{\frac {mv^{2}}{(1-{\frac {v^{2}}{c^{2}}})({\sqrt {1-{\frac {v^{2}}{c^{2}}}}}+1)}}$

By inserting $\infty$ we get

$={\frac {mv^{2}}{(1-{\frac {v^{2}}{\infty ^{2}}})({\sqrt {1-{\frac {v^{2}}{\infty ^{2}}}}}+1)}}$

The definition of infinity gives

$={\frac {mv^{2}}{(1-0)({\sqrt {1-0}}+1)}}$

Simplification gives

$={\frac {1}{2}}mv^{2}$

$Q.E.D.$

--Ofey 18:58, 24 October 2008 (UTC)]

Put all relativity theory resources in a single resource

I think it would be better, if the user comes to a resource called "Theory of Relativity", where there are sections about "General relativity", "Special relativity" and other subsections. Currently, there is no main page that links to all relativity theory resources. But I have not enough experience here to do it.--Sae1962 (discuss • contribs) 10:03, 30 June 2015 (UTC)