Simple variable elimination

[tramar]

1. The problem statement, all variables and given/known data
From the equations:

h\nu - h\nu' = T= m_o c^2 (\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1)
\frac{h\nu}{c}=\frac{h\nu'}{c}cos\theta+\frac{m_o v}{\sqrt{1-\frac{v^2}{c^2}}}cos\phi
\frac{h\nu'}{c}sin\theta = \frac{m_o v}{\sqrt{1-\frac{v^2}{c^2}}}sin\phi
\alpha=\frac{h\nu}{m_oc^2}

(representing the energy and momentum conservation for Compton scattering, theta = photon scatter angle and phi = electron scatter angle)

Eliminate v and \theta to obtain:

T=m_oc^2 \frac{2\alpha^2cos^2\phi}{1+2\alpha+\alpha^2sin^2\ phi}

2. Relevant equations
See above


3. The attempt at a solution
This problem has been driving me insane. I know the physics of Compton scattering but when I try to eliminate the v and theta from these 3 equations I just get hopelessly lost in the math... I have about 5 pages of scribbles leading to nothing.

Desperate at this point, some math help would be appreciated.

[gabbagabbahey]

HInt: \sin^2\theta+\cos^2\theta=1:wink:

[tramar]

I know that.... the problem is when I try to make theta disappear, I have to bring in a v. If I want to make v disappear, I bring in a theta. The only way that it seems to "work" gives me the following mess:

\sqrt{1-\frac{v^2}{c^2}}= \left( \frac{T}{mc^2}+1 \right) ^{-1}
v=c\sqrt{\left( 1-\left(\frac{T}{mc^2}+1 \right)^{-2} \right)}
cos\theta = \sqrt{1-\left(\frac{c}{h\nu'} \right)^2 \frac{(mv)^2}{1-\frac{v^2}{c^2}} sin^2\phi}

As you can imagine, taking v and substituting it into the cos equation makes for a disgusting mess that doesn't lead to anything very productive.

[gabbagabbahey]

I know that.... the problem is when I try to make theta disappear, I have to bring in a v. .

I don't see why you say this... just solve the equation

\frac{h\nu}{c}=\frac{h\nu'}{c}cos\theta+\frac{m_o v}{\sqrt{1-\frac{v^2}{c^2}}}cos\phi

for \frac{h\nu'}{c}\cos\theta, square both sides, and then square both sides of the equation

\frac{h\nu'}{c}sin\theta = \frac{m_o v}{\sqrt{1-\frac{v^2}{c^2}}}sin\phi

and add the two resulting equations together...

[tramar]

Ok so that gives me:

\left(\frac{h\nu'}{c}\right)^2=p_e^2-2p_ecos\phi\frac{h\nu}{c}+\left(\frac{h\nu}{c}\rig ht)^2

where p_e=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}} (for simplicity)

Not sure how to proceed... if I solve for p I have to use a quadratic eqn giving another big mess?

[gabbagabbahey]

The next step would be to eliminate \nu' in favor of T by using your first equation...

[tramar]

So:

\frac{(h\nu)^2-2Th\nu+T^2}{c^2}=p_e^2-2p_ecos\phi\left(\frac{h\nu}{c}\right)+\left(\frac {h\nu}{c}\right)^2

My instinct says to solve for (cp_e)^2 and then substitute that into energy conservation equation...

h\nu+mc^2=h\nu'+\sqrt{(mc^2)^2+(cp)^2}

...but I'll still have a p_e in there. Not sure if that's the right course of action...

[gabbagabbahey]

Why not solve the energy conservation equation

T+m_0c^2=\sqrt{(m_0c^2)^2+p_e^2c^2}

for p_e and then substitute that into

\frac{(h\nu)^2-2Th\nu+T^2}{c^2}=p_e^2-2p_ecos\phi\left(\frac{h\nu}{c}\right)+\left(\frac {h\nu}{c}\right)^2

[tramar]

so

c^2p_e^2=T^2+2Tmc^2

After substituting...

(h\nu)^2-2Th\nu=2Tmc^2-2h\nu\sqrt{T^2+2Tmc^2}cos\phi+\frac{(h\nu)^2}{c}

And getting rid of the root gives me a massive equation with 12 terms on the right :S

[gabbagabbahey]

Why do you still have a factor of 1/c in the last term?

[tramar]

My bad, the last one doesn't have 1/c.

Subbing in I get:

2h\nu cos\phi \sqrt{T^2+2Tmc^2}=2Tmc^2+2Th\nu

I'll try to solve for T...

[tramar]

Awesome! FINALLY got it.

Thanks a million for your help!