# SR and classical Doppler shift

1. Jul 1, 2009

### nutgeb

Alice and Bob are each in their own spaceship, and are travelling radially, initially away from each other, and subsequently toward each other, at some highly relativistic relative speed, say 99c. Each ship emits light flashes toward the other. Alice observes the light from Bob's outbound flight to be redshifted, and the light from his inbound flight to be blueshifted. Ditto for Bob.

I was surprised that, both coming and going, the SR time dilation factor $$\gamma$$ causes the received light spectrum to be more blueshifted (or less redshifted) than the spectrum shift caused by the classical Doppler effect alone.

This is effect is apparent in the SR redshift equation, which multiplies the classical Doppler shift:

$$\lambda_{o} = \lambda_{e} / (1 - v/c)$$

by the $$\gamma$$ factor:

$$\sqrt{ (1-v/c) (1+v/c) }$$

which always results in a decrease in wavelength from the classical Doppler shifted wavelength.

The result is shown graphically in the chart at the bottom of this http://www.fourmilab.ch/cship/doppler.html" [Broken] at the Fourmilab website.

Intuition might suggest (incorrectly) that a slower (time dilated) clock at the emitter would shift the classical Doppler effect in the red direction, analagous to how the slowing of the emitter's clock by gravitational time dilation affects wavelength.

Presumably the physical explanation for why SR time dilatation instead shifts incoming light in the blue direction is that the distance to the emitter becomes Lorentz contracted due to its velocity, enabling wave crests to arrive at a more rapid rate than they are emitted. In other words, the Lorentz contraction of the intervening distance crowds the incoming waves radially into a smaller space, thereby decreasing their wavelength.

But then, why isn't the shift toward blue caused by the Lorentz contraction exactly offset by the shift toward red caused by the emitter's clock running slower? The emitter's slower clock should cause the frequency originally emitted to decrease (and therefore the wavelength to increase), as measured by the observer's faster clock.

Evidently the increased crowding of wave peaks due to Lorentz contraction must be a larger effect (potentially much larger) than the decreased frequency due to time dilation at the emitter.

Last edited by a moderator: May 4, 2017
2. Jul 2, 2009

### Staff: Mentor

Hi nutgeb,

Here is a spacetime diagram of the Doppler shift from a 1 Hz source moving at 0.6c to the right (Doppler shift factor of 2). Hopefully it helps clear things up, everything is there.

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3. Jul 2, 2009

### nutgeb

Thanks for the diagram DaleSpam.

Please correct me if I'm wrong, but I believe that the diagram shows the total SR Doppler redshift, but it doesn't separate that redshift into its classical and SR time dilation components.

I'm trying to dissect the 'physical' explanation for why the classical Doppler effect is always more redshifted than the SR Doppler redshift; that is, why introducing SR time dilation always causes the classical Doppler redshift to shift toward the blue spectrum, never toward the red.

Last edited: Jul 2, 2009
4. Jul 2, 2009

### Staff: Mentor

The physical explanation is simply that the classical formula is wrong, it is an approximation that is only valid for v<<c. When that assumption is violated it biases the answer blue (i.e. neglects time dilation which always reddens).

Note that length-contraction is not an issue here, study the diagram to convince yourself of that. The speed of light together with the frequency determines the wavelength, it is not independent. Remember what length contraction is: a reduction in length from the rest frame to some other frame. There is no rest frame for light.

5. Jul 2, 2009

### nutgeb

As I said, it's not true that classical Doppler shift biases blue, instead it biases red, which is quite counterintuitive. Please look at the graph I linked to in post #1 of this thread. Try the math I set out.
Surely the distance between the emitter and the observer is Lorentz contracted.

EDIT: I printed your diagram and measured the purely vertical distance between the t'=0 and t'=2 lines at ~ 1.25 inches. The same for the purely horizontal distance between the x'=0 and x'=2 lines. By comparison, the vertical distance between the t (unprimed) lines is ~ 1.5 inches, same as the horizontal distance between the x (unprimed) lines. So your diagram seems to indicate that t' and x' are time contracted and distance contracted respectively, relative to t and x. The time contraction seems counterintuitive, which is my point.

Last edited: Jul 2, 2009
6. Jul 2, 2009

### JesseM

Put it this way, suppose we lived in a universe with totally Newtonian laws of physics, with no natural length contraction or time dilation; then if you have a clock moving along with an emitter measuring the frequency of pulses being sent out, and you compare that with the frequency of pulses being received by an observer who sees the emitter in motion (as measured by his own clock, which of course naturally remains in synch with the emitter clock in a Newtonian world), then you have the classical Doppler shift formula. Now, suppose in this Newtonian world we artificially slow down the clock moving along with the emitter by a factor of $$\sqrt{1 - v^2/c^2}$$, where v is the velocity of the emitter relative to the observer; if you now compare the frequency of the pulses being sent from the emitter as measured by the artificially-slowed down clock with the frequency of the pulses being received by the observer as measured by his normal clock, you get back the relativistic Doppler shift formula.
Not relevant, because the only thing you need to calculate the relativistic Doppler shift is the difference between the distance of the emitter when one pulse is sent and the distance when the next pulse is sent, and just as in Newtonian physics this is simply the velocity of the emitter in the observer's frame times the time interval between pulses being sent in the observer's frame, length contraction doesn't figure into it because nowhere are you concerned with the distance between things in the emitter's frame.

It might help to look at this numerical example I posted on another thread:

suppose a clock is traveling away from me at 0.6c, and it's programmed to send out a flash every 20 seconds in its own rest frame. In my frame, because of time dilation the clock is slowed down by a factor of $$1/\sqrt{1 - 0.6^2}$$ = 1.25, so it only flashes every 1.25*20 = 25 seconds in my frame. But that doesn't mean I see the flashes every 25 seconds, the gap between my seeing flashes is longer since each flash happens at a greater distance. For example, suppose one flash is emitted when the clock is at a distance of 10 light-seconds from me, at time t=50 seconds in my frame. Because we assume the light travels at c in my frame, if the flash happens 10 light-seconds away the flash will take 10 seconds to reach me, arriving at my eyes at t=60 seconds. Then, 25 seconds after t=50, at t=75, the clock emits another flash. But since it was moving away from me at 0.6c that whole time, it's increased its distance from me by 0.6*25 = 15 light-seconds from the distance it was at the first flash (10 light-seconds away), so it's now at a distance of 10 + 15 = 25 light-seconds from me, so again assuming the light travels at c in my frame, the light will take 25 seconds to travel from the clock to my eyes, and since this second flash happens at t=75 in my frame, that means I'll see it at t=100 seconds. So, to sum up, the clock flashes every 20 seconds in its own rest frame, and once every 25 seconds in my frame due to time dilation, but I see the first flash at t=60 seconds and the second at t=100 seconds, a separation of 40 seconds. This means the frequency that I see the flashes (1 every 40 seconds) is half that of the frequency the clock emits flashes in its own frame (1 every 20 seconds), which is exactly what you predict from the relativistic Doppler equation if you plug in v=-0.6c (negative because the clock is moving away from me): $$\sqrt{\frac{1 - 0.6^2}{1 + 0.6^2}} = \sqrt{0.25} = 0.5$$.

Last edited: Jul 2, 2009
7. Jul 2, 2009

### nutgeb

Jesse, I understand how the SR Doppler shift equation works numerically, and I understand how to combine the classical Doppler shift equation with the SR gamma equation to get SR Doppler shift. So you don't have to explain that to me.

I think you're missing my point, because you're not reading my posts carefully enough. The point that's counterintuitive is that the SR gamma clock correction shifts the classical Doppler equation in BLUE direction (shorter wavelength), not in the RED direction (longer wavelength) as one has every right to expect when the emitter's clock is running slower than the observer's.

8. Jul 2, 2009

### JesseM

If you understood that, why did you suggest length contraction plays a role, when (as I explained) it is clearly irrelevant?
No, I understood that your main point was about looking for some intuitive explanation for why the shift goes in the blue direction, I just didn't choose to address that larger point in my post. I was just addressing the sub-argument you made that it might have something to do with length contraction.
Why would you have "every right to expect" that? Like I said, even in a purely Newtonian universe if the emitter's clock is slowed down then you get the same formula as the relativistic Doppler shift, which is blueshifted compared to what happens if the emitter's clock runs normally in a Newtonian universe. You haven't really given a clear explanation of why you think it should be the other way around.

Look at my numerical example again, but this time thinking in terms of the Newtonian world before and after we artificially slow the emitter's clock down. Either way, assume the true rate at which the emitter is emitting flashes is once every 25 seconds (since different frames won't disagree on this rate in Newtonian physics). If the emitter is moving away from me at 0.6c, then by the calculations I already gave, I'll receive the flashes once every 40 seconds. So, the ratio of frequency received over frequency emitted is (1 every 40 seconds)/(1 every 25 seconds) = 0.625, exactly what you'd expect from the classical formula. Now, keep everything exactly the same but slow down the clock moving with the emitter, so that it only ticks forward by 20 seconds in the 25 seconds of "actual" time between the emitter sending out flashes. In this case the ratio of frequency received over frequency-emitted-as-measured-by-slowed-clock is (1 every 40 seconds)/(1 every 20 seconds) = 0.5, which is the same as what's predicted by the relativistic Doppler formula. I guess you could summarize this by saying that 1) the relation between the frequency the signals are received by the observer and the frequency they are being emitted in the observer's frame is the same in both Newtonian physics and relativity, since in either case the calculations only involve quantities in the observer's frame, and 2) given some particular emission frequency in the observer's frame, the emission frequency will be larger when measured by the clock moving along with the emitter (because the clock is ticking slower so the period it measures between signals is smaller), and so the ratio (frequency signals are received by observer)/(frequency signals are sent by emitter as measured by emitter's clock) must be smaller.

9. Jul 2, 2009

### nutgeb

OK, I'm glad we agree that the SR gamma correction always changes the classical Doppler shift in the blue direction - shortening the wavelength - regardless of whether the emitter is coming or going.

Intuitively I expect it to be the 'other way around' because, if the emitter's clock runs slower than the observer's, that means that the time interval between wave emissions is increased, as measured by the observer's clock. An increased interval between wave emissions means a lower observed frequency and a correspondingly longer wavelength: i.e. a change in the red direction.

The same logic is often used to explain the gravitational redshift: The clock runs slower in a gravity well, so the light it emits is redshifted when received by an observer outside the gravity well, whose clock runs relatively faster. There also is an alternative explanation that photons 'lose energy climbing out of the gravity well', so I don't need to hear about that please. I just want to understand the clock explanation for the SR Doppler shift.

10. Jul 2, 2009

### JesseM

By the way, note that in my argument above I am considering what happens when you keep the emission rate in the observer's frame fixed (which as I pointed out also ensures that the rate that the observer receives signals is fixed) and turn time dilation on and off for the clock moving along with the emitter. Maybe you're taking a different approach conceptually, imagining keeping the emission rate in the emitter's frame fixed and considering what happens when you turn time dilation on and off. In this case, since turning time dilation "on" means that the emission rate in the observer's frame goes down, and this means the rate the observer receives signals goes down as well (a redshift, not a blueshift), then since we were keeping the emission rate in the emitter's frame fixed, it makes perfect sense that the ratio (frequency signals are received by observer)/(frequency signals are sent by emitter as measured by emitter's clock) would get smaller as compared to the classical case, since the numerator of the fraction has gotten smaller while the denominator has stayed fixed. So maybe the problem is that you are using "redshift" and "blueshift" confusingly, although the ratio above (which is what is given by the Doppler shift formula) gets smaller, this is compatible with the idea that if you keep the emission rate as measured in the emitter's frame fixed, then when you turn time dilation "on" the frequency the signals are received by the observer gets smaller, meaning the period between the observer receiving signals is larger, which is what I would call a "redshift".

11. Jul 2, 2009

### JesseM

Not if by "changes" you mean keeping the emitter frequency fixed in its own frame and then looking at the frequency/wavelength of signals hitting the observer both with and without time dilation--in this case, as I explained in the comment I just posted above, the frequency of signals hitting the observer goes down when you turn time dilation "on", which is equivalent to saying the wavelength expands rather than shortens as compared with when time dilation was "off".

But I don't know if this is what you meant when you talked about redshift/blueshift, the problem is you aren't being specific about what you want to keep fixed and what you want to vary when we compare the classical case to the relativistic case.

Last edited: Jul 2, 2009
12. Jul 2, 2009

### nutgeb

I'm referring to the common-sense situation where Bob and Alice synchronize their clocks, then Alice flies away somewhere and then flies back radially past Bob and away at a constant velocity. She transmits light at a pre-agreed frequency, but due to SR Doppler shift resulting from their relative motion, Bob receives that light at a different frequency. Light emitted on Alice's way in toward Bob is received blueshifted by Bob; light emitted as Alice moves radially away from Bob is redshifted when received by Bob.

I want to focus exclusively on one question: Why does the SR gamma correction always change the classical Doppler shift in the BLUE direction - shortening the wavelength and increasing the frequency - regardless of whether the emitter is coming or going?

Please look at the chart I linked on the Fourmilab website in my first post, and think about what it means. The green 'relativistic' line is always BELOW (or equal to) the red 'classical' line, clearly showing that the SR gamma correction shortens the wavelength as compared to the classical Doppler shift. A shorter wavelength means a higher frequency. For example, when Alice approaches Bob at a velocity of just slightly below c, the light Bob receives from her is nearly infinitely blueshifted as shown by the relativistic line, while it only approaches 1/2 of the emitted wavelength as shown by the classical line.

Last edited: Jul 2, 2009
13. Jul 2, 2009

### JesseM

What do you mean "change"? What is being compared and what variable are you examing to see the change? Presumably you are in some sense imagining repeating the Alice/Bob experiment in both a non-relativistic universe and a relativistic universe and comparing some aspect of the result, but what do you want to keep fixed in both universes? As I said, if we imagine Alice is sending out pulses at the same frequency as measured by her own clock in both universes, and we imagine her speed relative to Bob is the same in both universes, and we compare the frequency that the pulses hit Bob in both universes, in this case regardless of whether Alice is traveling away from or towards Bob the frequency that the pulses hit Bob will be smaller in the relativistic universe than the non-relativistic one, meaning the wavelength he measures is lengthened in the relativistic universe as compared to the non-relativistic one, not shortened. But if this is not the comparison you want to make, you really need to be specific. If you have trouble explaining it in words, please give me a specific numerical example where you repeat the calculation in both the relativistic and non-relativistic universe and show that the wavelength of something is shorter in the relativistic universe than the non-relativistic one.
I'm confused by that graph because it doesn't seem to agree with the Doppler shift equations on this page. If you pick a given velocity for the source like 0.6c away from the observer, you find that according the the classical formula, $$\nu_{observed} = \nu_{source}\frac{1}{1 + 0.6}$$ which means $$\nu_{observed} = 0.625 \nu_{source}$$; and since frequency = c/wavelength, this means $$\lambda_{observed} = 1.6 \lambda_{source}$$. On the other hand, with the source having the same velocity of 0.6c away from the observer, the relativistic formula says that $$\nu_{observed} = \nu_{source} \sqrt{\frac{1 - 0.6}{1 + 0.6}}$$ which means $$\nu_{observed} = 0.5 \nu_{source}$$; and again, using frequency = c/wavelength this means that $$\lambda_{observed} = 2 \lambda_{source}$$. So here it seems that for the same velocity of 0.6c, the redshift should be greater in the relativistic case than the classical case. These numbers also agree with my numerical example below...have I made a mistake in the math somewhere? If not, an idea that just occurred to me is that the fourmilab graph might be assuming that in the "classical" case the light is always moving at c relative to the source rather than relative to the observer as I assumed above, implying if we treat light as a wave in a classical medium like the "aether", they would be assuming the source was always at rest in the medium while the observer was moving relative to it (in contrast the equation on the page I linked to above seems to be assuming the observer is at rest in the medium in the classical case).

edit: No, I think I just realized my actual mistake...although the classical formula $$\nu_{observed} = 0.625 \nu_{source}$$ is correct for the scenario above, I can't just use frequency=c/wavelength on both sides because classically, if the source is moving at 0.6c relative to the medium in which light waves move at c, then in the source's frame the waves are only moving at 0.4c, so in this frame the equation should be frequency = 0.4c/wavelength. So, plugging in we should have $$\frac{c}{\lambda_{observed}} = 0.625 \frac{0.4c}{\lambda_{source}}$$ which gives $$\lambda_{observed} = 4 \lambda_{source}$$. The relativistic formula for a source moving away from the observer at 0.6c was $$\lambda_{observed} = 2 \lambda_{source}$$, so this does seem to show that if in both the classical and relativistic universe we assume the source is emitting light at a certain fixed wavelength and moving at a certain fixed speed, then the wavelength measured by the observer will be smaller in the relativistic universe than the wavelength measured by the observer in a classical universe (more blueshifted), although this depends on the assumption that in the classical universe the observer is the one at rest relative to the medium who measures light waves to measure at c (if you instead assumed the source was at rest in the classical case the answer might be different).

Last edited: Jul 2, 2009
14. Jul 2, 2009

### Staff: Mentor

Don't forget that the classical Doppler formula is not relativistic. In other words, it is not simply the relative velocity which is important. Instead, the classical Doppler formula makes a distinction between a moving source and a moving receiver:

$$f_{r} = \left ( \frac {v+v_{r}}{v + v_{s}} \right ) f_{s}$$

The relativistic Doppler formula is always redder than the classical formula for a stationary receiver (vr=0), which is the usual case. The relativistic Doppler formula is always bluer than the classical formula for a stationary source (vs=0). The link you posted must have used the classical formula for a stationary source, although they didn't specifically say so.

15. Jul 2, 2009

### JesseM

Based on what I said above, doesn't this depend on whether you keep the frequency constant as measured in the emitter's frame, or whether you keep the wavelength constant? If you use the same wavelength in the emitter's frame for both the classical and relativistic cases, and also keep the velocity of the emitter relative to the observer constant, then if you also assume it is the observer who's stationary relative to the medium in the classical case, it seems that the wavelength seen by the observer is actually greater in the classical case than in the relativistic case (so he sees the light as redder in the classical case, not the relativistic one).

Just as an example, in my numerical example earlier I assumed the emitter was sending out pulses every 20 seconds in its own frame, and that it was moving away from the observer at 0.6c, and showed that this meant the observer would receive pulses every 40 seconds. In the relativistic case, this obviously means the wavelength measured by the observer is 40 light-seconds, and the wavelength measured by the emitter is 20 light-seconds. Now let's carry over that wavelength of 20 light-seconds as measured by the emitter to a classical case where the observer is at rest in the aether and the emitter is moving at 0.6c relative to the aether. In this case the emitter will only measure the signals to be traveling at 0.4c in its own rest frame, so if the wavelength in its frame is 20 light-seconds the period must be 20/0.4 = 50 seconds in its frame. And in classical physics all frames agree on time-intervals so the period must also be 50 seconds in the observer's frame. In the observer's frame, if pulse #1 is emitted at distance X from himself, then 50 seconds later the light from that pulse will be 50 light-seconds closer than X, while the emitter will be 0.6*50 = 30 light-seconds farther than X and emitting the next pulse at that moment, so the distance between pulses (i.e. the wavelength) as measured by the observer should be 50 + 30 = 80 light-seconds, double what it was in the relativistic case (and four times the wavelength in the emitter's frame, matching the formula I got in the previous post).

16. Jul 3, 2009

### JesseM

Grrr, just realized I messed this one up, if the observer is at rest in the aether and the emitter is moving away from the observer at 0.6c, then in the emitter's frame the signals are moving at 1.6c, not 0.4c. So, if the wavelength is 20 light-seconds the period must be 20/1.6 = 12.5 seconds, which means in the observer's frame if pulse #1 is emitted at a distance X, then 12.5 seconds later the light from that pulse will be 12.5 light-seconds closer than X, and the emitter will be 0.6*12.5 = 7.5 light-seconds farther than X and emitting the next pulse, so the wavelength in the observer's frame is 12.5 + 7.5 = 20 light-seconds, exactly the same as in the emitter's frame. In fact it's obvious the wavelengths should be the same in both frames in classical physics, since there is no disagreement about simultaneity or about the length of rulers, so if in one frame two pulses are lined up with either end of a 20 light-second ruler at a single instant, all classical frames agree on this.

And as before, in the relativistic case if the emitter is sending out pulses with a wavelength of 20 light-seconds in its own frame, then in the observer's frame the wavelength is 40 light-seconds, more redshifted than in the classical case where the emitter was sending out pulses of the same wavelength. So it looks like if we fix (wavelength of emitted pulses in emitter's frame) and include the assumption that the observer is the one at rest in the aether, the result is that in the relativistic universe the pulses being received by the observer will be more redshifted than they would be in the classical universe. And if we stick to the assumption that the observer is stationary but fix (frequency of emitted pulses in emitter's frame), then since this was one pulse every 20 seconds in the relativistic case it would have to be be the same in the classical case, and since the pulses are moving off at 1.6c in the emitter's frame according to the classical case, the wavelength is 20*1.6 = 32 light-seconds, and according to the argument at the end of the last paragraph this would be the same as the wavelength in the observer's frame, so with these assumptions it still turns out that the wavelength seen by the observer in the relativistic universe (40 light-seconds) is also larger than in the classical universe.

Furthermore, if we fix either (wavelength of emitted pulses in emitter's frame) or (frequency of emitted pulses in emitter's frame) but now change the classical case so that the emitter is the one at rest in the aether, it will still be true that the wavelength seen by the observer in the relativistic universe is more redshifted than the than the wavelength in the classical universe; keeping the numbers the same for the relativistic case (1 pulse/20 second frequency in emitter's frame, 20 light-second wavelength in emitter's frame), it works out that in the classical universe the observer will measure a wavelength of 20 light-seconds in both of these cases too. So I really have no idea what the fourmilab page could have been calculating that would indicate the observed wavelength more blueshifted in the relativistic case than the classical case!

17. Jul 3, 2009

### Staff: Mentor

Careful, this is a classical formula and only applies in the "luminiferous aether" frame. Remember, it is wrong and non-relativistic, so you cannot logically change reference frames without running into problems.

EDIT: Nevermind, I see that you figured that out in your next post.

18. Jul 3, 2009

### nutgeb

Thanks DaleSpam, you hit the nail on the head. The Fourmilab chart shows the classical formula for a moving observer and stationary emitter, but in the cosmological context we're normally interested in a stationary observer and moving emitter. The classical Doppler formulas for wavelength are as follows (where outward relative velocity has a positive sign):

Moving emitter, stationary observer: $$\lambda _{o} = \lambda _{e} (1 + v_{e})$$

Moving observer, stationary emitter: $$\lambda _{o} = \lambda _{e} / (1 - v_{o})$$

As you say, when the emitter is moving, the classical shift is biased red as compared to the relativistic shift. When the observer is moving, the classical shift is biased blue as compared to the relativistic shift. If the moving emitter line were plotted on Fourmilab chart, it would be the same as the red line except inverted in both the x and y axes.

Interestingly, the relativistic shift is nothing other than the geometric mean of the two classical formulas. See http://www.mathpages.com/rr/s2-04/2-04.htm" [Broken]. Calculate the geometric mean by multiplying the two classical formulas together and taking the square root.

Another way to look at it is that the relativistic formula utilizes the relativistic velocity sum of the emitter's and observer's respective velocities relative to the stationary 'medium':

$$v_{sum} = \frac{v_{e} + v_{o} }{ 1 + v_{e} v_{o} }$$

The classical formula assumes there is a stationary medium.

Last edited by a moderator: May 4, 2017
19. Jul 3, 2009

### JesseM

Those formulas are incorrect. It is true that the classical formulas relating frequencies would look similar:

Moving emitter, stationary observer: $$\nu_{o} = \nu_{e} \left ( \frac {v}{v + v_{e}} \right )$$

Moving observer, stationary emitter: $$\nu_{o} = \nu_{e} \left ( \frac {v + v_{o}}{v} \right )$$

However, when it comes to wavelength both the emitter and the observer automatically agree on the wavelength in the classical case as I mentioned earlier, since in the classical case there are no disagreements between frames about simultaneity or distances between wave peaks. So, in the classical case $$\lambda_{o} = \lambda_{e}$$ regardless of whether the observer or the emitter is moving relative to the medium. It is true that if the emitter is sending out waves at a constant frequency then the wavelength will change depending on its speed relative to the medium (and the direction the waves are being emitted), but this is not the same as a disagreement between classical frames about the wavelength, as long as you fix the velocity of the emitter relative to the medium and the direction the waves are emitted, then all frames will agree on the wavelength.

20. Jul 3, 2009

### nutgeb

Yes Jesse, I already fixed the wavelength formulas in an edit.

Your second point is trivial if the observer and emitter agree on what frame the medium is stationary in, but if they can't detect the medium's motion, they will get confused and disagree on each other's relative speed, as calculated from the classical Doppler shift.

Last edited: Jul 3, 2009