# Thread: Einstein's Special Theory of Relativity Quesion

1. ## Einstein's Special Theory of Relativity Quesion

I have no idea how to start this problem, We barely learned alittle part of this in class today but the teacher just said read the book and it is easy to undertand. I read the book over and over but I don't know....

Einstein's Special Theory of Relativity says that mass m is related to v velocity by the formula

Here, m0 is the rest mass and c is the velocity of light. Use differentials to determine the percent increase in mass of an object
when its velocity increases from 0.8c to 0.82c.

Approximate percent increase:_______

2. Originally Posted by killasnake
I have no idea how to start this problem, We barely learned alittle part of this in class today but the teacher just said read the book and it is easy to undertand. I read the book over and over but I don't know....

$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$

I'm going to rewrite this as:

$m = m_0 \left ( 1 - \frac{v^2}{c^2} \right ) ^{-1/2}$

Now:

$dm = m_0 \cdot \frac{-1}{2} \left ( 1 - \frac{v^2}{c^2} \right ) ^{-3/2} \cdot \frac{-2v}{c^2} dv$

$dm = m_0 \frac{v}{c^2} \left ( 1 - \frac{v^2}{c^2} \right ) ^{-3/2} dv$

We have v = 0.8c and dv = 0.02c, so

$dm = m_0 \frac{0.8c}{c^2} \left ( 1 - \frac{(0.8c)^2}{c^2} \right ) ^{-3/2} \cdot 0.02c$

$dm = m_0(0.016) (1 - 0.64)^{-3/2}$

$dm = 0.074074m_0$

Now, we need this as a percentage increase in mass:

$\frac{dm}{m_0} \times 100$ %

= $\frac{0.074074m_0}{m_0} \times 100$ %

= $(0.074074) \times 100$ %

= 7.4074 %

-Dan