# Simple variable elimination

1. Oct 30, 2009

### 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.

2. Oct 31, 2009

### gabbagabbahey

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

3. Nov 2, 2009

### 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.

4. Nov 2, 2009

### gabbagabbahey

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...

5. Nov 2, 2009

### tramar

Ok so that gives me:

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

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?

6. Nov 2, 2009

### gabbagabbahey

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

7. Nov 2, 2009

### 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...

8. Nov 2, 2009

### 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

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

9. Nov 2, 2009

### 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

10. Nov 2, 2009

### gabbagabbahey

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

11. Nov 2, 2009

### 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...

12. Nov 2, 2009

### tramar

Awesome! FINALLY got it.

Thanks a million for your help!