# Struggling with Doppler shift formulas

1. Jul 9, 2015

### JohnnyGui

Good day to you all,

I wanted to share a struggle with you all that I can't seem to get out of. It's about calculating, in the case of sound, the decrease or increase in frequency (thus increase or decrease in wavelength) when a sound source is respectively moving away or towards an observer with a velocity using the formula z = v / s (where s is the speed of sound). I know that this formula is only accurate for small velocities.

Scenario in which a sound source is moving away from the observer:
In this case, I'd automatically conclude that the observed wavelength (λObs) would be larger than the emitted wavelength (λEmit). The factor the emitted wavelength gets larger with would be the ratio of of the speed of sound (s) divided by its speed minus the velocity of the object that is moving away (v).

λEmit x (s / (s - v)) = λObs

The frequency would be inversely lower by a factor of (s - v) / s

fEmit x ((s - v) / s) = fObs

If one would want to calculate λEmit from the observed wavelength and the velocity of the object, the formula would look like this:

λObs / (s / (s - v)) = λEmit

The part (s / (s - v)) could be rewritten as 1 / (1 - (v / s)) (divide all parameters by s). Since z = v / s, then I could rewrite that part as 1 / (1 - z). Thus, the formula would be:

λObs / (1 / (1 - z)) = λEmit

This means that:

λObs / λEmit = 1 / (1 - z)

In principle, the z in this formula would be considered the redshift in the case of light, since the object is moving away from the observer.

Here's when thought I had the formula right, until I read the formula on the wiki page here: https://en.wikipedia.org/wiki/Redshift which shows that the formula should be:

λObs / λEmit = 1 + z

Instead of my concluded formula: λObs / λEmit = 1 / (1 - z)

My question is, what the heck am I doing wrong here?

Last edited: Jul 9, 2015
2. Jul 9, 2015

### nasu

Your formula for frequency does not look right.
Compare with the one here, for example:
http://hyperphysics.phy-astr.gsu.edu/hbase/sound/dopp.html

Note that for small values of the ratio v/s the two formulas are approximately the same.
And 1/(1-z) is approximately equal to 1+z for z<<1.

3. Jul 9, 2015

### JohnnyGui

Ah, so my concluded formula λObs / λEmit = 1 / (1 - z) is actually correct but for small velocities?

Would you happen to know how λObs / λEmit = 1 + z is concluded for larger velocities? Do they implement a Lorentz transformation or something similar?

4. Jul 15, 2015

### JohnnyGui

I think I've found the culprit

@nasu: If you say that λObs / λEmit = 1 / (1 - z) is the same as λObs / λEmit = 1 + z for small z values, then my concluded frequency formula fObs / fEmit = 1 - z would be also the same as the formula mentioned in your link: fObs / fEmit = 1 / (1 + z) for small z values.

The reason why there's a difference between my concluded formulas and the formulas in your given link is because it depends on which scenario you're derivating the formula from.

Here's why:

Scenario of a sound source receding from you:
In this scenario, you'd conclude that λObs would be larger than λEmit by a factor of (s / (s - v)); the ratio of the speed of sound and the speed of sound minus the source's receding speed (the speed of sound "gets slower" which causes a longer wavelength). This factor can be rewritten as (1 - (1 - z)). From this conclusion you'd conclude that if a sound source is approaching you, that the ratio would be inverted, thus ((s - v) / s) so that λObs would be smaller than λEmit by that factor. This ratio could be rewritten as 1 - z.
Thus, from a scenario in which the sound source is receding from you, you'd conclude that λObs / λEmit = (1 / (1 - z)) for receding sound sources and λObs / λEmit = 1 - z for approaching sound sources.

Scenario of a sound source approaching you
In this scenario, you'd conclude that λObs would be smaller than λEmit by a factor of (s / (s + v)); the ratio of the speed of sound and the speed of sound plus the source's velocity (the speed of sound "gets" an additional velocity which causes a shorter wavelength). This facor can be rewritten as (1 / (1 + z)). From this conclusion you'd conclude that if a sound source is receding from you, that the ratio would be inverted, thus ((s + v) / s) so that λObs would be larger than λEmit by that factor. This ratio could be rewritten as 1 + z.
Thus, from a scenario in which the sound source is approaching you, you'd conclude that λObs / λEmit = (1 / (1 + z)) for approaching sound sources and λObs / λEmit = 1 + z for receding sound sources.

See how I now have the 2 formulas for receding sound sources: λObs / λEmit = (1 / (1 - z)) and λObs / λEmit = 1 + z which gives the same result for small z values.
As well as 2 formulas for approaching sound sources: λObs / λEmit = 1 - z and λObs / λEmit = (1 / (1 + z)) which also give the same result for small z values.

The thing I can't understand is why these two pairs of formulas give the same result only for small z values while they are concluded in the same manner.

Last edited: Jul 15, 2015
5. Jul 15, 2015

### Khashishi

Because they are both just approximations for
$f=\left(\frac{c+v_r}{c+v_s}\right) f_0$
Obviously, two approximations will only (approximately) agree in the region where the approximation is valid.

6. Jul 15, 2015

### nasu

@JohnnyGuy

No, the reason is not different scenarios. Your formula for receding source is different when compared with the formula for the same scenario in hyperphysics.
So I think that you are simply using the wrong formula. The fact that in the limit of small speeds it gives the same numerical results does not mean that is the "correct" formula.

7. Jul 15, 2015

### JohnnyGui

@nasu : What I'm doing here is trying to conclude the correct formulas by myself, not using them. Unfortunately, I can't seem to understand how the formulas in the mentioned link are concluded.

@Khashishi: Is there a way to explain why both of my mentioned pairs of formulas are merely approximations? What part in the physics of Doppler shifts causes that my concluded formulas are just approximations?

8. Jul 15, 2015

### Khashishi

Sorry, I mean one of them is an approximation.
"Since z = v / s" is an approximation.
$z = \sqrt{\frac{1+\beta}{1-\beta}} -1$
λObs / λEmit = 1 - z is not approximate.

9. Jul 15, 2015

### nasu

Maybe this is what you try to do. But you did not show how did you get your first formula. You just assumed that it is right.
You can find the derivation in standard textbooks. For all scenarios.

10. Jul 15, 2015

### JohnnyGui

@nasu : I concluded my first formula based on other readings such as here: https://www.physicsforums.com/threads/prove-that-z-v-c.273160/ . From the readings I understood that the factor a wavelength gets larger or smaller by is the fraction of the original speed of sound compared to the velocity of its source. That's what I was trying to explain which lead me to conclude my formula.

@Khashishi: Hmm, so you meant only z = v / s is the approximate? How about (1 / (1 + z)) or (1 / (1 - z))? Are those approximates?

11. Jul 15, 2015

### nasu

"the factor a wavelength gets larger or smaller by is the fraction of the original speed of sound compared to the velocity of its source"

Well, that will translate into a formula like
Δλ/λ=v/c
which is not what you started with. Maybe this is the problem.

Last edited: Jul 15, 2015
12. Jul 15, 2015

### JohnnyGui

Shouldn't that sentence also translate into a fraction of (s / ((v - s)) or (s / (v + s)) (fraction of original speed of sound compared to the velocity of its source) like in my given link, which would be a factor that λEmit gets larger or smaller by?

I've written it down and was able to conclude from λObs / λEmit = (s / (s - v)) that (λObs - λEmit) / λObs = v / s
The conclusion is quite long though. I could put it on here if you want to.

Last edited: Jul 15, 2015
13. Jul 15, 2015

### nasu

Can you show the source for that quote, before we go further?

14. Jul 15, 2015

### JohnnyGui

15. Jul 23, 2015

### JohnnyGui

If not, how exactly should the formula then be concluded other than λEmit x (s / (s - v)) = λObs for recessing sound sources and λObs / λEmit = (s / (v + s)) for approaching sound sources?

16. Jul 23, 2015

### Nathanael

Why not just start from scratch and say the observed wavelength is the emitted wavelength plus/minus the distance traveled by the source in the time between emitting?

That would give you λobsemit almost straight away, and it's conceptually more understandable.

17. Jul 27, 2015

### JohnnyGui

Not sure how the formula would look like based on your description.

I've made my problem a bit clearer in this picture:

s = the speed of sound v = the velocity of the moving sound source in the direction of the given arrow

With "they" I mean the following sources: https://en.wikipedia.org/wiki/Redshift & http://formulas.tutorvista.com/physics/doppler-shift-formula.html
While this link is supporting my concluded formula for an approaching sound source: https://www.physicsforums.com/threads/prove-that-z-v-c.273160/

See how my formulas differ from them? I don't get why my conclusion is wrong and how "they" have concluded their formulas regardless of these formulas being the same as mine for very small v values.

Last edited: Jul 27, 2015
18. Jul 27, 2015

### Staff: Mentor

You are saying that for a source moving towards you, the observed frequency will be given by $f'= \frac{s + v}{s} f$. This is incorrect. nasu gave you a link to the correct formula and its derivation.

19. Jul 27, 2015

### JohnnyGui

Thanks for your answer. I'm aware that they're different. So what you are saying is that the explanation in post #2 here: https://www.physicsforums.com/threads/prove-that-z-v-c.273160/ is wrong since he's also saying that f' = (s + v) / s)) x f for an approaching sound source? Or did I read the explanation wrong?

20. Jul 27, 2015

### Staff: Mentor

You read it wrong. That post is discussing the case of a moving observer, not a moving source:

The formula for a moving source and stationary observer is different from that for a stationary source and moving observer.

21. Jul 30, 2015

### JohnnyGui

I'm amazed.. I didn't know that should be different. Shouldn't an observer observe the same frequency if the sound source approaches him at the same speed as him approaching the sound source instead? I can't seem to wrap my head around on how that should be different.

22. Jul 30, 2015

### A.T.

23. Jul 31, 2015

### Staff: Mentor

You need to review how the Doppler formula for sound is derived.

24. Jul 31, 2015

### olivermsun

A key thing to realize is that sound waves propagate at some speed $c$ relative to the medium. So it does actually matter whether the sound source is moving relative to the medium or the listener is moving or both.