Show that the redshift can be defined in terms of frequencies rather than wavelengths by z=(ve-v

step 1 Hello everyone, welcome to special activity again. And now let us move on to a doubler effect. D

Show that the redshift can be defined in terms of...

Key Concepts

Redshift

Redshift is a measure of the change in the frequency or wavelength of light from an astronomical object, indicating that the object is moving away from the observer. It is defined as the fractional change between the emitted and observed quantities, and it encodes important information about the relative motion or expansion of the universe.

Doppler Shift

The Doppler shift describes how the observed frequency or wavelength of light (or any wave) changes due to the relative motion between the source and the observer. When the source is moving away, the light is redshifted; when approaching, it is blueshifted. This effect is fundamental in astrophysics for determining the velocities of celestial objects.

Relativistic Doppler Effect

For objects moving at speeds close to the speed of light, the classical Doppler shift must be modified by special relativity. The relativistic Doppler effect incorporates a square root factor involving the velocity divided by the speed of light, ensuring the transformation accords with the principles of relativity and providing accurate results even for high-speed receding sources.

Small Velocity Approximation

When the speed of the receding object is much less than the speed of light, a Taylor expansion can be employed to simplify the relativistic Doppler formula. In this limit, the redshift becomes linearly proportional to the recessional velocity (z ? v/c), which makes the analysis and interpretation of small shifts more straightforward.

Frequency-Wavelength Relationship

The relationship between frequency and wavelength, mediated by the constant speed of light, allows redshift to be defined interchangeably in terms of frequency or wavelength. This concept is critical because it provides the flexibility to analyze spectral shifts using the most appropriate variable for a given observational context.

Transcript

00:02 Hello everyone, welcome to special activity again.

00:04 And now let us move on to a doubler effect.

00:08 Dobler effect is when you have a source that generates wave and either the observer, the one who received the signal or the source itself is moving.

00:22 So there is a motion associated with the generation of these waves.

00:27 So you don't know as an observer for this for this you do not get the same frequency of the source for example if you have a sound wave that makes 20 hertz for example so the and it's moving with a velocity let's say five meter per seconds or whatever the sound that you perceive will not be with the same frequency and there is a really actually between the observed frequency and the source of frequency we can write it this way one minus plus i will say now why this is minus plus over one over one plus minus v over so this is the case for for sorry light waves so, and of course, this is time f .s.

01:31 So the observed frequency, there is related to the source frequency with this factor here.

01:40 And you may note that if we have the negative sign, that means if we have the negative sign here and the plus sign here, that means this factor here and the numerator will be smaller than, this factor here and we get here a fraction less than one which means the observer frequency will be less than the source frequency and you can visualize this as follows if you if you have a source a source here that's generate some waves around it and here's the observer and he is moving away away from this source so what happened this is that this original wavelength lambda between two bigs or two minimums or two similar points here on this wave will will be at will have an additional distance here associated with that which is the vt here so somehow with this wavelength gets a bigger value than the original value it becomes higher than the original lambda here and higher wavelength higher lambda means a frequency that is smaller you know lambda times f equal to c so r to v for the case of the speed of sound that's fine however the the two here are inversely proportional to each other so once lambda gets higher the frequency gets shorter and vice versa so in this case with the preceding observer observer who is moving away from the source we get a frequency that's that's a higher observed frequency that's higher the other way around the other case is when you have a lambda f sorry lambda that's smaller with me which means this observer now is moving toward the source.

04:01 So lambda here gets us shorter and the wave and the frequency gets higher.

04:08 And this is what i was referring to by the minus plus here.

04:14 By the way, minus plus plus minus means the first minus here is associated with the plus here.

04:20 And the plus here is connected with this minus here.

04:24 So they are packaged together.

04:26 They came together come together here so so for a source that's moving away sorry for observers observer that's moving away or a source actually that's moving away we have a wavelength we have a wavelength that's higher and frequency that's smaller this will be the case with a negative sign sorry a frequency will will be smaller right so we are correct here this minus sign here will be the the one in charge in this case and if you have a source that's moving toward the observer or the other way or around the observer is moving toward the source the plus sign here will be the one in charge so plus minus me that let me let me highlight this let me highlight this so plus minus means this factor here is higher so the frequency is higher the observer's frequency is higher which means approaching so yellow here means approaching and for the minus sign with the minus sign that is highlighted was blue here minus plus means but going far away proceeding okay so percy let me let me write it down very seeding and now we have our foundation for for this problem and and let us now see how how can we do this and and let us consider part a of this problem here part a of this problem requires us to to get an approximation an approximation for this expression in the case where v is much smaller than c and we can do this we can use this to approximate this expression here first let let us rewrite this expression in a nicer format here if observed is one and by the way it mentions that we are dealing with a preceding guess so this observer here is moving away from the source for example so the minus sign here will will be the one that will work out and the plus sign in the denominator so let me write it this way and see how can we deal with it.

07:38 So 1 plus v or c to the power minus 1 half.

07:46 Okay.

07:48 And all this times a source, of course.

07:54 And we need to take this one step further.

07:58 We are given the final expression in the problem statement that we will have.

08:05 Have the final expression as the f observed over sorry delta f over f source will be equal to minus v r c however we need to prove it we need to prove it here so let me start first by taking the binomial extension the first term of the binomial expansion for the square root here and the square root here and if you do this, you do this, you may recall that one minus...

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