# The Naked Scientists Forum

### Author Topic: Doppler shift and its accuracy  (Read 5254 times)

#### engrByDayPianstByNight

• Sr. Member
• Posts: 105
##### Doppler shift and its accuracy
« on: 08/08/2007 21:45:38 »
Hello,

I have a basic idea of the Doppler effect/Doppler shift between a source (emitting electromagnetic waves) and a receiver (measuring the certain. However, as I look up some reference, I find that most books on this subject seem to discuss only two scenarios:

1)the source is moving and the receiver is stationary;
2)the source is stationary and the receiver is moving.

(Note: let "stationary" and "moving" be with respect to the ground/Earth), and I know that there are two different formulas, one for each of the above two scenarios (although if the relative velocity between the two nodes is much smaller compared with the speed of light c, both formulas yield approximately the same result). What about the scenario where both the source and receiver are moving? What would be the Doppler shift then? Somehow, I haven't been able to find that in the literature that I've read so far.

Also, in the industry, I'm wondering if anyone can provide me with some information about how accurately measuring the velocity of, say a car, can be achieved by the idea of Doppler shift using today's technology. For example, when two vehicles are 50[m] apart and the relative speed is 5[m/s], would today's technology be able to achieve a precision of less than 1% of error (i.e., 4.95--5.05[m/s])? Does anyone have any knowledge about that?

Thanks.

#### another_someone

• Guest
##### Doppler shift and its accuracy
« Reply #1 on: 08/08/2007 23:04:24 »
I am perplexed as to how one can have the notion of whether the source or the receiver is moving.  If there is no third body to relate to, then whether the source or the emitter is moving should, as far as I understand relativity, be indistinguishable.

http://en.wikipedia.org/wiki/Doppler_shift
Quote
For waves which do not require a medium, such as light or gravity in special relativity, only the relative difference in velocity between the observer and the source needs to be considered.

Clearly, where you have a medium (such as with sound waves), then you have to take into account the speed relative to that medium; but that is not pertinent to EM waves.

As for two vehicles being able to detect a 1% difference in speed, where both vehicles are moving, but at different speeds - the main criteria is the frequency of the signal you are using (higher frequency signals will have a greater doppler shift for the same speed).

The equation for calculating doppler effects for EM waves is:

Thus, for a 1GHz signal, for a vehicle moving 5 m/s, the frequency shift will be around 16.67Hz, while a speed of 5.05 m/s 16.83, a difference of 0.16Hz.

On the other hand, the values for a 10GHz signal would be 166.67Hz and 168.33 Hz respectively, giving you a 1.6Hz separation.

I would imagine it not too difficult to device a system that can discriminate those kind of differences.

#### engrByDayPianstByNight

• Sr. Member
• Posts: 105
##### Doppler shift and its accuracy
« Reply #2 on: 09/08/2007 01:50:53 »
I am perplexed as to how one can have the notion of whether the source or the receiver is moving.  If there is no third body to relate to, then whether the source or the emitter is moving should, as far as I understand relativity, be indistinguishable.

You're right. I should have said that both "source and receiver are moving with respect to the ground." Another website I went to distinguished the different scenarios in which the formulas are slightly different from the wikipedia website:

http://library.thinkquest.org/C006027/html-ver/op-dop.html

In it it discusses the two scenarios:

1)either the source is moving (towards/away from) the receiver, and the receiver is stationary, or
2)the receiver is moving (towards/away from) the stationary source.

So, I wonder, what if both are moving (again, w.r.t. the ground)? What's the appropriate formula to compute the shift then?

another_anyone gives an illustrative example in which a tiny Hz shift results from a 1GHz transmitted signal. Which is what I'm wondering: Does today's technology able to discern such a tiny difference in frequency, given all the interference and noise in the radio environment? How accurate are the commercial products using the Doppler shift? Anybody with any industry specs?

Thanks.

#### another_someone

• Guest
##### Doppler shift and its accuracy
« Reply #3 on: 09/08/2007 02:41:01 »
http://library.thinkquest.org/C006027/html-ver/op-dop.html

In it it discusses the two scenarios:

1)either the source is moving (towards/away from) the receiver, and the receiver is stationary, or
2)the receiver is moving (towards/away from) the stationary source.

So, I wonder, what if both are moving (again, w.r.t. the ground)? What's the appropriate formula to compute the shift then?

I am not surprised you are confused after reading that site - it really does not explain things well.

The ground is not at all relevant to any of the equations (not unless you are dealing with sound waves travelling through the ground).  The fact that the ground is nearby would make no difference than if the ground was not there at all.

The equation from Wikipedia is the correct one to use for EM waves (not for sound waves, or other waves travelling through a medium; but it is the correct one to use for EM waves because EM waves always travel at the same speed relative to the observer, no matter what speed the observer is travelling at).

another_anyone gives an illustrative example in which a tiny Hz shift results from a 1GHz transmitted signal. Which is what I'm wondering: Does today's technology able to discern such a tiny difference in frequency, given all the interference and noise in the radio environment? How accurate are the commercial products using the Doppler shift? Anybody with any industry specs?

I can't quote you what the state of the art is with regard to how accurate contemporary devices may be (others on this forum probably could answer that), but as for the level of noise having a significant effect, not on discriminating differences in frequency (which is why FM radio has so much less noise than AM radio).  What higher noise levels will give you is a problem in actually receiving any signal at all, so it may mean you need ramp up the strength of the signal in order to detect the signal at all, but if you can detect it, it will not effect your ability to determine its frequency.

#### engrByDayPianstByNight

• Sr. Member
• Posts: 105
##### Doppler shift and its accuracy
« Reply #4 on: 09/08/2007 05:08:52 »
another_anyone,

thanks. I re-read the Wikipedia post and realized that indeed, when the emitted wave is electromagnetic, it doesn't matter whether the source is moving and the receiver stationary, or vice versa, or both are moving. It's only when the emitted wave is, e.g., sound, would there be some differences in calculating the Doppler shift that has to involve specify the mobility/stationarity of the receiver and the source. Am I right? The presentation of the Wikipedia post (the General and Analysis sections) was a little misleading because the Analysis section appears to treat only sound waves, though, and I thought it was for both sound and electromagnetic...

Also, I saw that post claims that "These can be generalized into a single equation with both the source and receiver moving," which is what I've been wondering (although admittedly, I was thinking about electromagnetic waves). But perhaps someone can enlighten me on how this generalization can be formulated?

And I appreciate anyone's input on the industry specs about the accuracy of measuring Doppler shift...

#### engrByDayPianstByNight

• Sr. Member
• Posts: 105
##### Doppler shift and its accuracy
« Reply #5 on: 09/08/2007 05:11:24 »
Oops, I just realized I kept referring to you as "another_anyone." Sorry.

#### lightarrow

• Neilep Level Member
• Posts: 4586
• Thanked: 7 times
##### Doppler shift and its accuracy
« Reply #6 on: 09/08/2007 15:48:45 »
thanks. I re-read the Wikipedia post and realized that indeed, when the emitted wave is electromagnetic, it doesn't matter whether the source is moving and the receiver stationary, or vice versa, or both are moving. It's only when the emitted wave is, e.g., sound, would there be some differences in calculating the Doppler shift that has to involve specify the mobility/stationarity of the receiver and the source. Am I right?
Yes.

Quote
The presentation of the Wikipedia post (the General and Analysis sections) was a little misleading because the Analysis section appears to treat only sound waves, though, and I thought it was for both sound and electromagnetic...

Also, I saw that post claims that "These can be generalized into a single equation with both the source and receiver moving," which is what I've been wondering (although admittedly, I was thinking about electromagnetic waves). But perhaps someone can enlighten me on how this generalization can be formulated?

I think you should take the product of the two equations:

f = f0(1 - v0/v)*[1 - vs/(v + vs)]

v = sound's speed relative to the medium

vs = source's speed

v0 = observer's speed.

f' = f*Sqrt[(c+v)/(c-v)]  in the case of source approaching the observer (or viceversa)

f' = f*Sqrt[(c-v)/(c+v)]  in the case of source going away of the observer (or viceversa)

f  = light's frequency when the observer is stationary with respect to the source
f' = new light's frequency seen when they move with respect to each other
v  = relative speed -->  Edit. it means: speed of the source relative to the observer.
c  = light's speed.

(I imagine you could understand the reason for those two colours I gave to the 2 equations)
« Last Edit: 09/08/2007 18:44:46 by lightarrow »

#### another_someone

• Guest
##### Doppler shift and its accuracy
« Reply #7 on: 09/08/2007 17:14:54 »
I think you should take the product of the two equations:

f = f0(1 - v0/v)*[1 - vs/(v + vs)]

v = sound's speed relative to the medium

vs = source's speed

v0 = observer's speed.

f' = f*Sqrt[(c+v)/(c-v)]  in the case of source approaching the observer (or viceversa)

f' = f*Sqrt[(c-v)/(c+v)]  in the case of source going away of the observer (or viceversa)

f  = light's frequency when the observer is stationary with respect to the source
f' = new light's frequency seen when they move with respect to each other
v  = relative speed
c  = light's speed.

(I imagine you could understand the reason for those two colours I gave to the 2 equations)

I am sorry, I cannot see how this can be right?

If v is the speed relative to the medium, and the medium is light, then no matter what speed the observer and source are travelling to each other, the speed relative to the speed of light will always be the same, it will be the speed of light (i.e. light travels at the same speed for all observers).

#### lightarrow

• Neilep Level Member
• Posts: 4586
• Thanked: 7 times
##### Doppler shift and its accuracy
« Reply #8 on: 09/08/2007 18:41:00 »
I think you should take the product of the two equations:

f = f0(1 - v0/v)*[1 - vs/(v + vs)]

v = sound's speed relative to the medium

vs = source's speed

v0 = observer's speed.

f' = f*Sqrt[(c+v)/(c-v)]  in the case of source approaching the observer (or viceversa)

f' = f*Sqrt[(c-v)/(c+v)]  in the case of source going away of the observer (or viceversa)

f  = light's frequency when the observer is stationary with respect to the source
f' = new light's frequency seen when they move with respect to each other
v  = relative speed
c  = light's speed.

(I imagine you could understand the reason for those two colours I gave to the 2 equations)

I am sorry, I cannot see how this can be right?

If v is the speed relative to the medium, and the medium is light, then no matter what speed the observer and source are travelling to each other, the speed relative to the speed of light will always be the same, it will be the speed of light (i.e. light travels at the same speed for all observers).

I wrote: v  = relative speed
I thought it was clear that I intended: "speed of the source relative to the observer"; if it wasn't, I apology. I've corrected it in my previous post.
« Last Edit: 09/08/2007 18:45:26 by lightarrow »

#### engrByDayPianstByNight

• Sr. Member
• Posts: 105
##### Doppler shift and its accuracy
« Reply #9 on: 12/08/2007 00:43:06 »
Thanks, lightarrow, for the equations.

#### engrByDayPianstByNight

• Sr. Member
• Posts: 105
##### Doppler shift and its accuracy
« Reply #10 on: 12/08/2007 01:19:48 »
Actually, lightarrow, your color-coding the two equations inspires me to ask another question on astronomy (or is it cosmology?), something I'm not an expert of.

From our observation of the light we know the universe is expanding (due to the so-called redshift). Presumably, the light that we've observed was originally emitted some billions of years ago, right? If that's the case, we can claim that billions of years ago, the universe was in an expansion mode. How can we say it's still expanding today? Why could it not be contracting mode now?

If, on the other hand, the light that we observed and used to determine the universe is expanding emitted only a few hundred years ago, that implies that the source of the light is only a few hundred lightyears away from Earth, which is quite a local area on the cosmological scale. Then we could only claim that this area of the universe is expanding currently. How then can we apply this claim to the universe as a whole?

I know I must have missed something here. I would have moved this question to the Astronomy section if I knew how. So, I appreciate your answer or anyone else's. And feel free to treat me as an astronomy novice and explain in layman's terms.

#### lightarrow

• Neilep Level Member
• Posts: 4586
• Thanked: 7 times
##### Doppler shift and its accuracy
« Reply #11 on: 12/08/2007 11:18:38 »
Actually, lightarrow, your color-coding the two equations inspires me to ask another question on astronomy (or is it cosmology?), something I'm not an expert of.

From our observation of the light we know the universe is expanding (due to the so-called redshift). Presumably, the light that we've observed was originally emitted some billions of years ago, right? If that's the case, we can claim that billions of years ago, the universe was in an expansion mode. How can we say it's still expanding today? Why could it not be contracting mode now?

If, on the other hand, the light that we observed and used to determine the universe is expanding emitted only a few hundred years ago, that implies that the source of the light is only a few hundred lightyears away from Earth, which is quite a local area on the cosmological scale. Then we could only claim that this area of the universe is expanding currently. How then can we apply this claim to the universe as a whole?

I know I must have missed something here. I would have moved this question to the Astronomy section if I knew how. So, I appreciate your answer or anyone else's. And feel free to treat me as an astronomy novice and explain in layman's terms.
Not easy questions for me. I imagine theoretical models of the universe tell us it's expanding, but we have no idea of what the fartest part of the universe (beyond the actual observable limit) is actually doing. I leave this question to more expert people on the subject.

#### The Naked Scientists Forum

##### Doppler shift and its accuracy
« Reply #11 on: 12/08/2007 11:18:38 »