# Thread: annoying relativity

1. ## annoying relativity

I am having problems with an equation involving relativity, but my question is directly math related currently the equation is in the form of e=mc^2(blah blah blah), but I need it in the form e=mv^2(blah blah blah).

the current equation is "E=mc^2((1/(sqrt(1-v^2/c^2)))-1"

where sqrt(x) signifys the square root of x. please help.

what I really need is the answer, but I would love to know the steps taken.

2. Originally Posted by dented42
I am having problems with an equation involving relativity, but my question is directly math related currently the equation is in the form of e=mc^2(blah blah blah), but I need it in the form e=mv^2(blah blah blah).

the current equation is "E=mc^2((1/(sqrt(1-v^2/c^2)))-1"

where sqrt(x) signifys the square root of x. please help.

what I really need is the answer, but I would love to know the steps taken.
I do not understand you want to solve for $\displaystyle v$?

3. Originally Posted by dented42
I am having problems with an equation involving relativity, but my question is directly math related currently the equation is in the form of e=mc^2(blah blah blah), but I need it in the form e=mv^2(blah blah blah).

the current equation is "E=mc^2((1/(sqrt(1-v^2/c^2)))-1"

where sqrt(x) signifys the square root of x. please help.

what I really need is the answer, but I would love to know the steps taken.
What you want to do is expand the gamma function ($\displaystyle \gamma=1/\sqrt{1-v^2/c^2}$) in a Taylor series about v=0. So $\displaystyle E = mc^2(1+1/1!*\gamma'(v=0)+1/2!*\gamma''(v=0)+...)$. You should get $\displaystyle E=mc^2(1+1/2*v^2/c^2+3/8*v^4/c^4+...)$.

-Dan