Relativistic Doppler effect on gamma rays I'm trying to solve the following problem :

An electron-positron pair annihilates, creating two photons. At what speed must an observer move along the line of the photons in order that the wavelength of one photon be twice that of the other?

So far, I've found using $E=m_0c^2$ that the energy of each photon is 511 keV, and using $E = \frac{hc}{\lambda}$ that the corresponding wavelength is $\lambda = 2.429\cdot10^{-12}$m.
The observed wavelength for the photon coming towards the observer is then given by the relativistic Doppler formula : $\lambda_o = \lambda_s \sqrt{\frac{1+\beta}{1-\beta}}$ where $\beta = \frac{v^2}{c^2}$. What I don't understand is how to calcutate the wavelength for the photon propagating away from the observer.
It seems to me that it'd be impossible to even observe a photon moving away from you... Also, since the two photons are emitted from the same source, and a photon's speed relative to the observer is always $c$ in all IRFs, shouldn't both photons experience the same redshift/blueshift anyway?
Thanks,
--hlouis
Edit : Thanks to Mr. dmckee, I was able to solve the problem. If the observer carries a long pole with detectors at both ends, and the emission of both photons occurs when both ends are on opposite sides of the source, then one detector is moving away from the source, while the other is moving towards it. Then the answer is the speed needed to shift down the wavelength by 25% for one detector, and up 25% for the other.
 A: [Note: Since the statement of the question, including the recent "Edit" after comments, already generally describes how to obtain the solution, my following answer is largely limited to pointing out some remaining mistakes; at least unless requested otherwise.]

The observed wavelength for the photon coming towards the observer

... more specificly: towards a detector whose motion wrt. the source is quentified by the real number $\beta$; with $\beta > 0$ for motion of detector and source "towards" each other, and $\beta < 0$ for motion of detector and source "away from" each other ...

is then given by the relativistic [Doppler] formula: $\lambda_o = \lambda_s \sqrt{ \frac{1 + \beta}{1 - \beta} }$

... yes; but: ...

where $\beta = \frac{v^2}{c^2}$

... no. Rather with $\beta = \sqrt{\frac{v^2}{c^2}}$; or plainly $\beta = \frac{v}{c}$, if $v$ is appropriately understood as "speed" of detector and source wrt. each other.

Edit: [...] a long pole with detectors at both ends, and the photon emission occurs when both ends are on opposite sides of the source, then one detector is moving away from the source, while the other is moving towards it.
At what speed must an observer move along the line of the photons

... or adapted to the "Edit":
At what speed must the pole with the two detectors move in line ("line of sight") with the source ...

in order that the wavelength of one photon be twice that of the other?

So given the formula above, the remainig mathematical task is to solve
$2 = \lambda_{\text{towards}} / \lambda_{\text{away}} = \left(\lambda_s \sqrt{ \frac{1 + \beta}{1 - \beta} } \right) / \left(\lambda_s \sqrt{ \frac{1 - \beta}{1 + \beta} } \right) = \left( \frac{1 + \beta}{1 - \beta} \right)$
for $\beta$.
The result is of course:
$\beta = 1/3$, i.e. $v = c/3$ (referring to the detector and source moving towards each other).

Then the answer is the speed needed to shift down wavelength by 25% for one detector, and up 25% for the other.

No: since
$\sqrt{ \frac{1 - 1/3}{1 + 1/3} } = \sqrt{ \frac{2/3}{4/3} } = \sqrt{1/2}$
and correspondingly
$\sqrt{ \frac{1 + 1/3}{1 - 1/3} } = \sqrt{ \frac{4/3}{2/3} } = \sqrt{2}$,
the "shifts" involved are rather:
"down by 29.289... %" and "up by 41.42... %", respectively.
