
Theory OF Relativity
Im having trouble with this homework problem, Its just that the formua throws me off, and I cant figure out what is asking.
#7. According to the Theory of Relativity the energy E of a body with mass "m" is given as a function of its speed "v" by
http://i49.photobucket.com/albums/f2...Relativity.jpg
Where c is constant, the speed of light.
A. Assuming v< c, expand E as a series in (V/C) to the second nonzero term.
B. Explain why (v/c) is very small, then E can be well approximated by
E=1/2 mv^2

Hi
For $\displaystyle \frac{v}{c} << 1$
$\displaystyle E = mc^2\left[\left(1\frac{v^2}{c^2}\right)^{\frac12}1\right] = mc^2\left(1+\frac12\:\frac{v^2}{c^2}1\right) = \frac12\:mv^2$

Can you explain this please. Would that be number A.Thank you

When a is negligible with respect to 1 you can expand $\displaystyle \left(1+a\right)^n$ as $\displaystyle 1+na$ at 1st order
Here $\displaystyle a=\frac{v^2}{c^2}$ and $\displaystyle n = \frac12$

So the second non zero term would be 1/2 mv^2?... As for be (v/c) is very small because V is always less than C, automatically thats less than 1.. it further becomes small when u square it..

Yes. If $\displaystyle \frac{v}{c} << 1$ then $\displaystyle \frac{v^2}{c^2} << 1$ and $\displaystyle \left(1\frac{v^2}{c^2}\right)^{\frac12} \simeq 1+\frac12\:\frac{v^2}{c^2}$

Ok So I get part B of this question Thank You. Im still having a little trouble expanding The series (v/c) to the second non zero term as in part A.