Talk:Special Relativity

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[edit] 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 $(γ - 1) m c 2$ , 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

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

By substituting $γ$ we get

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

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

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

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

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

This gives

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

And so

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

By substituting $γ$ 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)]

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