Question

Derivation of photon momentum - just need help understanding some intermediate steps

Answer 1

I think I understand up to step 1.61 and 1.62 and then I start to get lost

Answer 2

@sillybilly123 whenever you get a spare moment

Answer 3

sure

1.62 is the end so you done?!

Answer 4

sorry I meant to get to this earlier

can you explain steps 1.61 and 1.62, I'm a little lost

Answer 5

I'm gonna guess that it's the \(\int v ~ d(mv)\) that is unhelpful. it's lazy.

well:

\(d(mv) = d\left( \dfrac{m_o v}{\sqrt{1 - (v/c)^2}}\right) \)

Quotient Rule:

\(= m_o \dfrac{ dv \sqrt{1 - (v/c)^2} - v (1/2) (1 - (v/c)^2)^{- 1/2} (- 2v /c^2) dv}{1 - (v/c)^2 } \)

\(= m_o \dfrac{ 1 }{(1 - (v/c)^2)^{3/2} } ~dv\)

So:

\(\int v ~ d(mv) \equiv m_o \int \dfrac{ v }{(1 - (v/c)^2)^{3/2} } ~dv\)

You can now pattern match into their suggested standard integral, it works; BUT that is what Wolfram is for .... AND this is already very do-able if you notice this differentiation pattern:

\(\dfrac{d}{dv} \left(\dfrac{ 1 }{(1 - (v/c)^2)^{1/2} }\right) = \dfrac{1}{c^2} \dfrac{ v }{(1 - (v/c)^2)^{3/2} }\)

SO:

\( m_o \int_0^v \dfrac{ v }{(1 - (v/c)^2)^{3/2} } ~dv = m_o \int_0^v c^2 \dfrac{d}{dv} \left( \dfrac{ 1 }{(1 - (v/c)^2)^{1/2} } \right)~dv\)

\( = m_o c^2 \left. \dfrac{ 1 }{(1 - (v/c)^2)^{1/2} } \right|_0^v\)

which falls out as:

\( = m_o c^2 \left( \dfrac{ 1 }{(1 - (v/c)^2)^{1/2} } - 1 \right) \qquad \star\)


For reducing to classical result bit, I would use a Taylor Expansion, AND we can just do it as a Binomial, same thing:

\((1 - z)^{-1/2} = 1 + (-1/2) (-z) + (-1/2)(-3/2) (1/ 2!) (-z)^2 + ... \)

\(= 1 + \frac{z}{2} + \frac{3z^2}{8} + \mathcal O (z^3)\)

So with \(z = (v/c)^2\):

\(\star = m_o c^2 \left(1 + \dfrac{v^2}{2c^2} + \dfrac{3v^4}{8 c^4} + ... - 1 \right) = m_o \left( v^2 + \dfrac{3v^4}{8 c^4} + ... \right)\)

I hope I got that right, check the arithmetic if it actually matters, but the principle is established...

For the very last bit, expand star:

\(\star = \dfrac{m_o c^2 }{(1 - (v/c)^2)^{1/2} } - m_o c^2 = mc^2 - m_oc^2 = ....\)

Answer 6

There's a much cooler way to do this, found it in Eisberg's QM book, if you're interested.

It copies from Feynman's lectures.