How do you find relative velocity of an electromagnetic wave?

for example: Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 1.1% greater than the frequency emitted by the source in the distant galaxy. What is the relative velocity of the galaxy relative to the earth?

1 Answer
Aug 4, 2016

Answer:

#0.011c#, rounded to three decimal places
where #c=3xx10^8ms^-1#

Explanation:

There is difference between the question heading and the main question.

We are required to find the relative velocity of the galaxy to the earth given is the frequency of the signal emitted by the source in the distant galaxy while observed on earth.
It is a question of application of Doppler effect as applicable to electromagnetic radiations.

We know from the postulate of Special Relativity that speed of light #c# is constant in vacuum and is independent of the speed of the source.
As such Doppler effect is observed in case of movement of source with respect to the observer as shown in the figure below which depicts Doppler redshift and blueshift.
wikipedia

It is given in the problem that The frequency of the signal observed on earth is #1.1%# greater than the frequency emitted by the source. Therefore, it is a case of blueshift and the source is moving towards the observer.

We know that for relativistic Doppler effect, observed frequency #nu_@# of e.m. radiation of frequency #nu_s# emitted by a source moving with a velocity #v# with respect to the observer is given by expression which needs to consistent with the Lorentz transformation

#nu_@=nu_ssqrt((1+beta)/(1-beta))# .......(1)
where #beta=v/c# and #v# is considered positive when the source is approaching the observer.
We have #nu_@/nu_s=1.011#. Equation (1) becomes
#1.011=sqrt((1+beta)/(1-beta))#
#=>1.011=sqrt((1+v/c)/(1-v/c))#
Squaring both sides and solving for #v#
#1.022121=(c+v)/(c-v)#
#=>1.022121xx(c-v)=(c+v)#
#=>1.022121c-1.022121v=c+v#
#=>2.022121v=0.022121c#
#=>v=0.022121/2.022121c#
#=>v=0.011c#, rounded to three decimal places
where #c=3xx10^8ms^-1#