# Theory OF Relativity

Printable View

• May 9th 2009, 11:10 AM
Collegeboy110
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
• May 9th 2009, 11:15 AM
running-gag
Hi

For $\frac{v}{c} << 1$

$E = mc^2\left[\left(1-\frac{v^2}{c^2}\right)^{-\frac12}-1\right] = mc^2\left(1+\frac12\:\frac{v^2}{c^2}-1\right) = \frac12\:mv^2$
• May 9th 2009, 11:27 AM
Collegeboy110
Can you explain this please. Would that be number A.Thank you
• May 9th 2009, 11:48 AM
running-gag
When a is negligible with respect to 1 you can expand $\left(1+a\right)^n$ as $1+na$ at 1st order

Here $a=-\frac{v^2}{c^2}$ and $n = -\frac12$
• May 9th 2009, 11:53 AM
Collegeboy110
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..
• May 9th 2009, 12:09 PM
running-gag
Yes. If $\frac{v}{c} << 1$ then $\frac{v^2}{c^2} << 1$ and $\left(1-\frac{v^2}{c^2}\right)^{-\frac12} \simeq 1+\frac12\:\frac{v^2}{c^2}$
• May 9th 2009, 01:19 PM
Collegeboy110
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.