[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

Doppler effect

DotarSojat@aol.com writes:
 > In response to my 8/26 note, Steve wrote (also on 8/26)--
 > >...one can look at the relationship of received power to emitted
 > >power just as well in terms of Doppler shifting as by using the
 > >relationship between emitter time and receiver time (and in a
 > >fundamental sense, they are the same thing expressed in differ-
 > >ent terms; Doppler shifting is merely the result of different
 > >rates of elapsed proper time between an emitter and a
 > >receiver).
 > The last part, "Doppler shifting is merely...," bothered me.
 > I thought I had read something different.  In going back to
 > consult V.A. Ugarov, "Special Theory of Relativity," Mir, 1979,
 > I found on page 86--
 >   "...the Doppler effect is formed of two independent parts:
 >   (1) it is connected with the continuously changing distance
 >   between the observer and the source; (2) it is also connected
 >   with the transformation of time intervals between events on
 >   transition from one reference frame to another."
 > (He went on to note that the second effect can be detected when
 > there is no radial motion; he called it "the transverse Doppler
 > effect.")

In special relativity, if two observers do not move relative to each
other, then they also observe identical rates of passage of time.  In
essence the two "independent parts" referred to by Ugarov are actually
mutally dependent -- if two objects have continually changing distance
between them, then they are experiencing relative motion and they must
also have different rates of elapsed time.  Therefore transformations of
time intervals between them will result in discrepancies between when
each observes particular events to take place.

However, my statement is a bit of an oversimplification, as Doppler
shift actually results from the transformation of spacetime coordinates,
and not just from relative rates of elapsed time.  As a result Doppler
shifting is a somewhat more complex phenomenon when examined in more
detail; in the general case photons not only change wavelength but their
apparent directions of motion, and the amount of Doppler shift depends
on the angle between the direction of motion and the direction of

You may find it illuminating to derive the Doppler shift from the
Lorentz transformation.  I'll do the simple one-space-dimensional case

The spacetime path of a light ray in a one-dimensional space can be
represented by a vector [ t x ] where t = x (where we measure t and x in
the same units; use c*t = x if you like conventional units).  If we
consider a photon with wavelength lambda', then one wavelength could be
represented by the vector [ lambda' lambda' ] in the moving object's

The Lorentz transformation, with only one space dimension considered,
can be represented by the matrix

[ gamma    v*gamma ]
[ v*gamma  gamma   ]

where v is the velocity, and gamma is 1/sqrt(1 - v^2); multiplying this
matrix with a vector on the right representing a quantity in the moving
frame (pardon my looseness with column vs. row vectors) produces the
transformed vector in the stationary frame.  Therefore, in the
stationary frame one wavelength of the photon is seen as:

[ gamma    v*gamma ] [ lambda' ]
[ v*gamma  gamma   ] [ lambda' ]

= lambda' * gamma * [ (1 + v)  (1 + v) ]

(the last part is a 2-element vector, both of whose components are the
quantity 1+v).

The question then is, what does the stationary observer see as the
photon's transformed wavelength lambda?  The answer is relatively
simple: The photon in the observer's frame has the wavelength lambda' *
gamma * (1 + v), simply by taking the x-component of the transformed
vector, which can be simplified like this:

lambda = lambda' * (1 + v) / sqrt(1 - v^2)
       = lambda' * sqrt(1 + v) * sqrt(1 + v) / (sqrt((1 + v)*(1 - v))
       = lambda' * sqrt(1 + v) * sqrt(1 + v) / (sqrt(1 + v) * sqrt(1 - v))
       = lambda' * sqrt((1 + v) / (1 - v))

Note that positive v corresponds to a receding object, so when v is
positive observed photons have longer wavelengths and hence lower
energies than they do relative to the moving object.

The derivation is even more interesting in two or three dimensions, as
it demonstrates not only the transverse Doppler shift, but an even more
interesting phenomenon dubbed the "headlight effect" -- the angle
between the direction of motion and the direction of emission of a
photon is observed to be different in the moving object's frame and the
stationary observer's frame.  You can derive this using a vector
representing one wavelength of the photon in the moving object's frame,
where the photon is emitted with an angle theta' to the moving object's
x-axis (I'll use only two space dimensions for simplicity):

lambda' * [ 1  cos(theta')  sin(theta') ]

and the Lorentz transformation matrix for motion v in the x direction:

[ gamma    v*gamma  0 ]
[ v*gamma  gamma    0 ]
[ 0        0        1 ]

Leaving the math as an exercise to the reader, the relationship between
theta' (the angle of emission in the moving frame) and theta (the
observed angle in the stationary frame) is:

cos(theta) = (v + cos(theta')) / (1 + v*cos(theta'))

The relationship between lambda' (the wavelength in the moving object's
frame) and lambda (the observed wavelength in the stationary frame) is:

lambda = lambda' * gamma * (1 + v*cos(theta'))

Note that if theta' = 0, then theta = 0 and the relationship between
lambda' and lambda is the same as in the one-dimensional case above.

It's a little more tricky to talk about the transverse Doppler effect,
as we have to consider the difference between the angle of emission in
the moving frame and the angle of reception in the observing frame.
Let's consider the case where cos(theta) = 0, which means that the angle
of reception is +/- pi/2 (the observed photon is seen as perpendicular
to the direction of motion in the stationary frame).  In that case,
cos(theta') = -v, so:

lambda = lambda' * gamma * (1 + v * -v)
       = lambda' * (1 - v^2) / sqrt(1 - v^2)
       = lambda' * sqrt(1 - v^2)
       = lambda' / gamma

On the other hand, when the angle of emission in the moving frame is
+/- pi/2 and cos(theta') = 0, then cos(theta) = v, and

lambda = lambda' * gamma

In this case light emitted perpendicular to the object _as seen by the
object_ is observed to come from an angle less than pi/2 relative to the
direction of motion of the object.  For v approaching 1, light emitted
from the entire front half of the object is concentrated in a small cone
with half-angle theta = acos(v) -- hence the "headlight effect".

This should demonstrate that although we're using different terms and
methods, we really are talking about the same thing, as the formulas for
Doppler shift are easily derived from the Lorentz transformation.