# I Radar Doppler effect: classical vs relativistic

1. Jul 15, 2016

### penelopa

I think the relativity provides the Doppler shift (in a radar - double version) in a form:
$\frac{1-v}{1+v}$

which is quite simple, as a square of a single Doppler: $f' = f\sqrt{\frac{1-v}{1+v}}$

But I never seen what is a classical version of this - what is a formula for a radar shift in the classical, real math?

2. Jul 15, 2016

### penelopa

I try to rediscovery the original classical version, in this way:

SR assumes the relative speed is equal to: $v=\frac{v_s-v_r}{1-v_s\cdot v_r}$

therefore if I simply insert this into the relativistic equation, then I should get just the classical version, I think!

$\Large\frac{1-v}{1+v}=\frac{1-\frac{v_s-v_r}{1-v_s\cdot v_r}}{1+\frac{v_s-v_r}{1-v_s\cdot v_r}}=\frac{1-v_s\cdot v_r-v_s+v_r}{1-v_s\cdot v_r+v_s-v_r}=\frac{(1-v_s)\cdot(1+v_r)}{(1+v_s)\cdot(1-v_r)}=\frac{1-v_s}{1-v_r}\frac{1+v_s}{1+v_r}$

Indeed it's good result!, because a single classical Doppler is equal to:
1. for a forward direction: $\frac{1-v_s}{1-v_r}$
2. for a backward direction: $\frac{1+v_s}{1+v_r}$

thus the double Doppler is simply a combination of the both!

Last edited: Jul 15, 2016
3. Jul 16, 2016

### penelopa

So, I can conclude finally: the speed's measurement, by the Doppler shift observation, is impossible in principle,
because the Doppler shift is a function of two variables in fact:
a speed of an measured object, and a speed of a radar itself, not a single one number,
which can be rediscovered by a one measurement of the bounced light frequency change.

4. Jul 16, 2016

### marcusl

The shift is expressed in terms of the relative range rate between radar and target.

5. Jul 16, 2016

### penelopa

What a range rate? The shift evidently depends on the two variables: vr and vo.

And there is no possibility to measure the real speed difference: vr-vo, in which we are interested!

6. Jul 16, 2016

### Ibix

I don't understand your experimental setup. What are $v$, $v_r$, $v_s$, vr and vo the velocities of, and measured with respect to what?

I should add that I'm not sure that there's an answer to your question. I suspect that in the rest frame of the radar emitter you will find that the non-relativistic case is the same as the relativistic one. In any other frame you will probably get nonsense because the non-relativistic case can't handle the propagation of light in a consistent way without proposing an ether that we know doesn't exist.

Last edited: Jul 17, 2016
7. Jul 17, 2016

### penelopa

My setup is a typical measurement of a speed of moving body, using a radar.

Radar-->vr <===singnal===> O - My car --->vo

The Radar can't measure the speed's difference, which is equal to: $v = v_o-v_r$, nor my car speed: $v_o$.

It can measure something like the relativistic speed only: $v' = \frac{v_o-v_r}{1-v_ov_r} <> v$

Probably exist some procedure to measure this speed: v, but we must make two measurements, not a single only,
because this quantity depend on the two independent variables: vr and vo.

8. Jul 17, 2016

### penelopa

I can provide a better example:

we shoot a rocket to the Moon, for example.
The rocket accelerates at the beginning, and we know what is the acceleration precisely;
and further it moves freely in a space with a speed which we know already, because we projected precisely the whole journey.

Now we measure the speed of our rocket, using radar, and... what we discovery?
The radar moves together with the Earth, and the Sun moves also in the Galaxy, ect.

So the radar's final result would be: $v' = \frac{v_o-v_r}{1-v_0v_r}$ instead of our predetermined precisely speed: $v = v_o-v_r$!!!

Last edited: Jul 17, 2016
9. Jul 17, 2016

### Ibix

So what you are saying is that the radar gun is moving at speed $v_r$ relative to some reference frame, and the target is moving at $v_o$ relative to that same reference frame, and you wish to know the value $v_o-v_r$, right?

The first question is, why? The radar gun is moving inertially, to a very good approximation, so is free to consider itself at rest. In fact, it would be difficult to build a radar gun that "naturally" did not consider itself at rest. So $v_r=0$ and your question resolves to what is $v_o$. All your comments about the speed of the Sun and the galaxy are irrelevant. Speed relative to what? The radar gun is free to say, at least for the ~0.25s duration of a radar pulse to the Moon and back, that they are all in motion relative to it, which is at rest.

Given all that, the result is simple. You can always use clocks and rulers in the rest frame of the radar gun to analyse the experiment, so the result is that the received frequency is $f'=f(c-v_o)/(c+v_o)$, as you said. This is true both in the relativistic and non-relativistic cases. Although you can only analyse the non-relativistic case at all as long as you assert that the speed of light is $c$ in the radar gun's rest frame and don't ask any questions about why that is.

A hint: Doppler radar is widely used by the police, the military, civilian meteorological work, and probably many other applications I haven't thought of. If your analysis suggests that something in such widespread use doesn't work, it's most likely a flaw in your analysis.

10. Jul 17, 2016

### penelopa

I understand what You are talking about, but I'm interested in the real speed of my rocket, which is determined perfectly to be: $v=v_o-v_r$, because the rocket's trajectory has been projected precisely to be just that, not any other.

But a radar gives me this: $v'=\frac{v_o-v_r}{1-v_ov_r}$, which is evidently different than the predetermined v! So, I can't belive in the results of any radar, because it's incorrect - wrong, in principle - always!

Last edited: Jul 17, 2016
11. Jul 17, 2016

### Ibix

Your problem here is that you appear to be using non-relativistic physics to calculate the "determined perfectly" speed $v$, and using relativistic physics to do the measurement, $v'$. Unsurprisingly, they don't agree. That's what happens when you don't use a self-consistent methodology. Either use relativistic physics to transform the projected velocity or don't use it to transform the measured velocity.

Edit: Or, possibly, you're trying to measure different things. The radar gives you the relative speed of the radar set and the target, but you appear to want the separation rate in some other frame (though why you'd call that the "real speed", I don't know).

Last edited: Jul 17, 2016
12. Jul 17, 2016

### penelopa

But You can see, I use the both: relativistic and classical together!
I'm discovered successfully the classical radar Doppler shift, using the relativistic speed composition idea only: $v'=\frac{v_o-v_r}{1-v_ov_r}$

One additional question: what is a rocket speed which is flying with a speed 10 km/s to the Moon, in the domain of the relativity?

13. Jul 17, 2016

### penelopa

I named the v as a real speed, because I know the distance to the Moon, and I have an clock, then I can determine a flying time: d = vt;

14. Jul 17, 2016

### Ibix

Except that you complain that they don't give the same answer. And the process you follow in post #3 doesn't make any sense to me. The answer should be the same in the relativistic and non-relativistic case, as I said before.

What you would do is use the Doppler formula, $f'=f(c-v)/(c+v)$, to measure the velocity $v$ of the rocket relative to the radar set. Then you add on the velocity, $v_r$, of the radar set relative to you. If you are happy with the Newtonian approximation that will be simply $v_o=v+v_r$. That will match your calculations, assuming you've done them correctly.

It isn't, which is one reason to suspect that you're doing something wrong. You can use relativistic physics if you want, but you shouldn't need to.

But this is where I'm confused: you seem to want the velocity of the rocket with respect to the radar set. Which is exactly what the radar set gives you. I've no idea why you're messing around with the velocity of the radar set as well.

15. Jul 17, 2016

### penelopa

I ignored Your Doppler formula earlier, without any comment, but it's evidently wrong.

The correct formula is in my post #2.
It's of two-variable function. So, it's impossible to recovery these two variables by one measurement only, and even a difference of the both: vs-vo,
because the function can't be reduced to the one dimension only: v = vo-vr.

16. Jul 17, 2016

### penelopa

And that is just a problem: a radar measurement gives only a... number: $v'=\frac{v_o-v_r}{1-v_ov_r}$ instead of the correct - desired: $v = v_o-v_r$.

17. Jul 17, 2016

### Staff: Mentor

18. Jul 17, 2016

### Staff: Mentor

This thread will remain closed since the OP's question has been answered, whether or not the OP wants to accept the answer.

@penelopa You were coming close to receiving a warning before I closed the thread. You appear to have some fundamental misconceptions that need to be corrected. To take just the main points:

You repeatedly describe the classical formula for the velocity difference as "real", "correct", etc. This is a serious misconception on your part. The "real", correct formula--the one that actually matches experiments--is the relativistic formula. The classical formula is only an approximation, which breaks down if the velocities are not small compared to the velocity of light.

SR does not "assume" this. SR derives this as a theorem from the fundamental postulates of the principle of relativity and the fact that the speed of light is the same in all inertial frames. And, as noted above, this derived theorem gives a formula that actually matches experiments. The classical formula does not.

Your description of what exactly you are trying to compute is also unclear, but I don't think it's worth trying to disentangle that at this point. You need to rethink your entire scenario in the light of the above.

If you have questions about the above, feel free to PM me.