# Differentiation

• Feb 4th 2010, 07:40 AM
roshanhero
Differentiation
I am trying to prove that the group velocity is equal to the particle velocity by relativistic purpose
The relativistic energy is given by the relation
$E^2 = p^2c^2 + m_{0}^2c^4$
But I don't know how to differentiate it with respect to momentum "p"
• Feb 4th 2010, 10:45 AM
Aryth
Well, first you get E by itself:

$E = \sqrt{p^2c^2 + m_0^2c^4}$

Now, we take the derivative with respect to $p$ on both sides:

$\frac{dE}{dp} = \frac{d}{dp} \sqrt{p^2c^2 + m_0^2c^4}$

First we use the chain rule setting $u = p^2c^2 + m_0^2c^4$

$\frac{dE}{dp} = \frac{1}{2\sqrt{p^2c^2 + m_0^2c^4}}\frac{d}{dp}(p^2c^2 + m_0^2c^4)$

Now, you differentiate what is in the parentheses:

$\frac{dE}{dp} = \frac{1}{2\sqrt{p^2c^2 + m_0^2c^4}}(2pc^2)$

The two's cancel, and a c factors out of the radical in the denominator, which simplifies the derivative to:

$\frac{dE}{dp} = \frac{pc}{\sqrt{p^2 + m_0^2c^2}}$

And that's the derivative, maybe you can continue with this.