The Theory of the Transverse Doppler Effect

[Irrational]

I'm doing a night course in General Relativity and we're currently finishing off Special Relativity... We're working mainly off of D'Inverno.

We've just covered the relativistic doppler effect and some associated things like aberration.

When it came to talking about the transverse doppler effect,

\frac{\lambda}{\lambda_{0}}= \frac{1}{(1-v^{2}/c^{2})^{\frac{1}{2}}}

we were given a handout from C.J Davidson's letter "The Theory of the Transverse Doppler Effect" and the lecturer ran through the details of it.

The following bit I am having trouble working through though.

From the conservation of energy:

\frac{E_{1}}{(1 - \beta_{1}^{2})^{\frac{1}{2}}} = h\nu + \frac{E_{2}}{(1 - \beta_{2}^{2})^{\frac{1}{2}}}

From the conservation of momentum...

- along the X axis, we have

\frac{E_{1}\beta_{1}/c}{(1 - \beta_{1}^{2})^{\frac{1}{2}}} = \frac{E_{2}\beta_{2}/c}{(1 - \beta_{2}^{2})^{\frac{1}{2}}}cos{\gamma} + \frac{h\nu}{c}cos{\alpha}

- along the Y axis, we have

0 = \frac{E_{2}\beta_{2}/c}{(1 - \beta_{2}^{2})^{\frac{1}{2}}}sin{\gamma} - \frac{h\nu}{c}sin{\alpha}

where \beta_{1} = v_{1}/c, \beta_{2} = v_{2}/cFrom here, we eliminate \beta_{2} and \theta to get

\nu = \nu_{0}\frac{(1 - \beta_{1}^{2})^\frac{1}{2}}{(1 - \beta_{1} cos{\alpha})}

where:

\nu_{0} = \frac{(E_{1} + E_{2})}{2E_{1}}\frac{(E_{1} - E_{2})}{h}

and after this, it's plain sailing...

My problem is eliminating \beta_{2} and \theta. I just end up in a mess everytime I try to do it.

It's not homework but I wasn't sure whether to post it here or there. It's more filling in the missing gap. I know this is probably basic algebra and/or trigonometry but any help would be appreciated.

Also, for some reason my LaTex tags are bit weird in preview mode so let me know if something is wrong.

Thanks
Dave

 

[bcrowell]

Also, for some reason my LaTex tags are bit weird in preview mode so let me know if something is wrong.

Yeah, that's a known bug. I just submit my post and then go back and proofread my equations.

From here, we eliminate \beta_{2} and \theta to get

This is the first time theta was referred to...?

Whatever this paper is doing seems extremely cumbersome and inelegant. The basic idea of a transverse Doppler shift is very simple. It's just relativistic time dilation of the source, so it gives you an additional factor of gamma. If you want to derive a fully general formula, it seems like it ought to be possible to do it using four vectors in some nicer way.

 

[Irrational]

Apologies. that should have been eliminate \gamma, not \theta. my notes were dirrerent to the actual paper i was citing the equations from.

I'll include the background as well.

The Theory of the Transverse Doppler Effect

The ingenious and exceedingly precise observations made recently by Ives on the spectrum of hydrogen canal rays verify a long standing prediction of the spectral theory of relativity - the prediction that the frequencies of spectral lines will be found to vary with the velocity v of the particle producing them as (1 - v^{2}/c^{2})^{\frac{1}{2}}.

That this effect was to be expected could be shown by transforming the Expression for the phase of the train of spherical light waves from the space-time coordinates of rest to a second frame in which - that of the observer - with respect to which the first was moving with uniform velocity v. By this procedure it is found that the observer should perceive the ordinary Doppler effect with respect not to the "rest frequency" of the radiation v_{0} but with respect to a lower frequency v_{0}^{\prime} = v_{0}(1 - v^{2}/c^{2})^{\frac{1}{2}}. For an obeserver looking at right angles to the particle's velocity the ordinary Doppler effect vanishes, and this relativistic red shift should alone remain - hence the designation "transverse Doppler effect".

It is interesting, though not usrprising perhaps, to find that the prediction might have been based equally well on the relativistic mechanics of photon emission. Consider an excited atom of total rest energy E_{1}. It's rest mass is m_{1}=E_{1}/c^{2}. In motion with velocity v_{1} it's total energy is m_{1}c^{2}/(1 - v_{1}^{2}/c^{2})^{\frac{1}{2}} = E_{1}/(1 - \beta_{1}^{2})^{\frac{1}{2}}. Suppose now that it ejects a photon of energy h\nu in transition to a lower state of total energy E_{2}/(1 - \beta_{2}^{2})^{\frac{1}{2}}.

Conservation of energy requires that

<br /> h\nu + \frac{E_{2}}{(1 - \beta_{2}^{2})^{\frac{1}{2}}} = \frac{E_{1}}{(1 - \beta_{1}^{2})^{\frac{1}{2}}}<br />

Consider next the momenta. The initial momentum of the atom is m_{1}\nu_{1}/(1 - \beta_{1}^{2})^{\frac{1}{2}} = (E_{1}\beta_{1}/c)/(1 - \beta_{1}^{2})^{\frac{1}{2}}. After ejecting the photon it's momentum is (E_{2}\beta_{2}/c)/(1 - \beta_{2}^{2})^{\frac{1}{2}} with components parallel and normal to v_{1} proportional, respectively, to \cos{\gamma} and \sin{\gamma} where \gamma represents the angle between v_{1} and v_{2}. The corresponding components ofthe photon's momentum are (h\nu/c)\cos{\alpha} and (h\nu/c)\sin{\alpha}, where \alpha represents the angle between v_{1} and the direction in which the photon has been ejected. The three vectors are in a plane, and conservation of momentum requires that

<br /> \frac{E_{2}\beta_{2}/c}{(1 - \beta_{2}^{2})^{\frac{1}{2}}}cos{\gamma} + \frac{h\nu}{c}cos{\alpha} = \frac{E_{1}\beta_{1}/c}{(1 - \beta_{1}^{2})^{\frac{1}{2}}}<br />

and

<br /> \frac{E_{2}\beta_{2}/c}{(1 - \beta_{2}^{2})^{\frac{1}{2}}}sin{\gamma} - \frac{h\nu}{c}sin{\alpha} = 0<br />

If one uses these relations to eliminate \beta_{2} and \gamma, and writes \beta for \beta_{1}, one finds that

<br /> \nu = \nu_{0}\frac{(1 - \beta^{2})^\frac{1}{2}}{(1 - \beta cos{\alpha})}<br />

where

<br /> \nu_{0} = \frac{(E_{1} + E_{2})}{2E_{1}}\frac{(E_{1} - E_{2})}{h}<br />

I've worked through all of this and understand it all but can't do the basic elimination of \beta_{2} and \gamma to get the last equation.

 

[Irrational]

Finally managed to work through it and get the required ansswer. Was simply an exercise in algebra in the end. That and the most basic of trigonometric identities, \cos^{2}{\gamma} + \sin^{2}{\gamma} = 1.