The phenomenon of a Doppler Effect is fairly easy to visualize. If a series of consistent waves are traveling at a relatively constant velocity, then if you stand still (as described by the location of the source of the waves), then you will sense waves that arrive at very consistent times. We say that they have a constant frequency.

There are four major examples of this "wave propagation": water waves, sound waves, light waves and electromagnetic radiation. We can consider water waves as being the easiest to visualize. In this case, we need to assume that the depth of the water is relatively constant, because the speed of water waves changes with the depth of the water. (For the other three types of waves, that is not true, sound waves travel at the "speed of sound" and light and electromagnetic radiation travel at the "speed of light").

Imagine that you are in a large pond that has a constant depth, and there is a drip of water or some other mechanism that consistently disturbs the water's surface. Water waves will propagate outward (as circles of constantly increasing size) across the pond. IF you stand still, the waves will arrive where you are, at the same intervals they were formed at. Since our source makes one new wave every, say, 2.0 seconds, then you would feel a new wave get to you every 2.0 seconds. We would say that you sense a frequency of 1 (second)/2.0 seconds/wave, or 0.5 wave per second. That frequency would not change. If the source only created waves of that frequency, you, at a distance, might be able to figure out some things about the source.

Now, imagine that you decided to move TOWARD that source across the pond. How many wave crests would you encounter? Well, it would be MORE than before! If the waves could be "frozen in place" then you would encounter a new wave crest every few feet of movement. This effect gets ADDED to the waves that are passing you going the other way. This is called the Doppler Effect. The fact that you are MOVING (relative to the source) means that you have to ADD two different effects, the original propagation of the waves across the pond and your motion across the pond, to know how many waves you would actually encounter.

If you decided to move AWAY FROM the source, fewer of the waves could "catch up to you" each second, and you would sense a frequency that was LOWER than when you were standing still.

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The Doppler Effect is given by 1 / (1 - v/u) (for a stationary observer and no wind or other affecting causes), where v is the velocity (speed) along the line between the source and the observer, and u is the velocity of propagation. This is the multiplication factor between the actual frequency and what you sense. For example, if a REALLY fast train was heading toward you at around 350 mph (half the speed of sound), then v/u = 0.5 and the formula is then 1 / (1 - 0.5) or 2.0. This means that for that situation, twice (2.0 times) as many waves would get to you per second from the train's whistle. that would make it sound an entire octave higher than it normally sounds. You have certainly noticed that when a high speed train whistles as it approaches you, the pitch then drops as it passes you, and is noticeably lower in pitch as it goes away from you. This is the Doppler Effect, and it is given by this formula. A train that approaches you at 75 mph would calculate to 1 / (1 - 0.1) = 1.111. Therefore, about 1.11 times as many waves would be heard, which means the pitch would sound slightly over one full note higher.

The same formula works for after it has passed you and is moving away. In this case, v is negative. For the fast train, we would have 1 / (1 - (-0.5)) or 1 / 1.5 or 0.6667. You would hear a pitch a little less than one octave below its actual pitch.

Note that if you knew the speed of sound and accurately measured the pitch before and after, you would be able to calculate the SPEED of the train! Similarly, if you happened to know the pitch (note) that the train actually created, and accurately measured EITHER the approaching or receding pitch, you could also calculate the speed of the train.

Now consider light, from very distant stars. In most cases, we know what "pitch" of light that ACTUALLY gets emitted from glowing hydrogen gas, for example. When discussing light, we refer to frequency or its inverse, wavelength, to describe different "pitches" (colors) of light. Wavelength is convenient because if we take a spectrum of the light from such a source, it is often found to contain lines in that spectrum that can easily be identified as to wavelength. In a laboratory, we can get the same kind of spectrum, from Hydrogen that we know is not moving away from us! Now, say that we see that a line in the star spectrum is shifted toward the red by one part in 1000 of the wavelength. Following the above Doppler reasoning, we would say that 1.001 = 1 / (1 - v/c), where c is now the speed of light. We would calculate that v/c is very close to 0.001, so that would mean that the star was moving away from us (RED shift) at about 0.001 times the speed of light or 186 miles per second. If we found a shift toward the blue, it would mean the star is moving TOWARD us at that speed.

Einstein's Theory of Special Relativity causes a modification to the Doppler Effect equation. Once Einstein realized that the speed of light was the fastest possible speed in the Universe, he realized a problem. Say that we are receiving light waves from a star that is standing still. Yes, we would see the light as traveling at c, the speed of light. But now, if that star was approaching us at half the speed of light, and it sent light out ahead of it (at the speed of light), the first thought is that that light should be moving at 1.5c, the addition of the two speeds. Just like what happens with a train whistle when the train is moving toward you, simple addition of speeds.

But that would violate the maximum speed being the speed of light! Einstein solved this with his Special Relativity. He stated that distances and time can APPEAR different, RELATIVE to different viewpoints! A lot of math followed, but the result is that Einstein determined that the rate of passage of time has to seem NORMAL to a traveling person (or star) but slowed down to observers that are not moving at that high speed. (There has been a LOT of experimental proofs of this, so it is certainly true!)

Physicists call the change in the rate of the passage of time "Time Dilation". Since WE see the passage of time of a rapidly receding star to be apparently "slowed down" then we see, effectively, that the atoms and electrons in the Hydrogen gas to seem to be moving more slowly. So, if we counted the number of orbits of an electron, we would get a smaller count, exactly according to Einstein's equation of the Time Dilation effect. What that means is that the FREQUENCY of that oscillation will seem lower, or REDDER in a spectrum. So, for a rapidly receding star, we would have TWO effects, Doppler and Time Dilation, both of which would cause us to see the spectral lines to be shifted toward the red end of the spectrum. Einstein made a Relativistic adjustment to the Doppler Effect equation to account for this Time Dilation effect.

If the star was rapidly moving toward us, The Doppler shift would be toward the BLUE end of the spectrum, while the Time Dilation shift would still be toward the red. Einstein's Relativistic Doppler formula calculates this, too, and gets a resulting red shift that is smaller that the non-Relativistic Doppler red shift would have been. It turns out that the Doppler effect is always at least three times as great as the Time Dilation effect, so Doppler always prevails.

There are also Relativistic Gravitational Red Shift and the other known effects, which still remain valid, but those effects are generally MUCH smaller and usually can be neglected.

The relativistic Doppler effect is given by:

where z is given as usual in octaves of wavelength shift.

For an object ACTUALLY receding from us at v = 0.6c, this would give a Doppler Red Shift z = 1.0 (one octave)

For an object ACTUALLY receding from us at v = 0.8c, this would give a Doppler Red Shift z = 2.0 (two octaves)

The Time Dilation effect (time slowing proportion) is given by:

For an object ACTUALLY receding from us at v = 0.6c, this gives a Time Dilation factor of 0.8 (which represents a Dilation Red Shift of z = 0.32)

**Therefore, an object ACTUALLY receding from us at 0.6c would exhibit a Red Shift of z = (2.5 (non-relativistic) * 0.8 (time dilation factor) - 1 or 2.0 - 1.0 or z = 1.0, the relativistic red shift factor.**

For an object ACTUALLY receding from us at v = 0.8c, this gives a Time Dilation factor of 0.6 (which represents a Dilation Red Shift of z = 0.74)

**Therefore, an object receding from us at 0.8c would exhibit a Red Shift of z = (5.0 * 0.6 - 1.0 or 3.0 - 1.0 or z = 2.0.**

This is the mathematical explanation of how the Relativistic Doppler red shift is really a combined effect of the standard Doppler formula and the Time Dilation factor, as the same results are obtained either way.

[NOTE: If the Time Dilation factor is 0.5, then time appears to pass
at exactly half its rest-frame rate, which would represent a
1.0 octave shift of the perceived frequency.
The conversion is given by:
z = -(log(TD_{f}))/(log(2)) ]

This means that the lower apparent average atom velocity gives
us received **light that appears to come from a source of lower surface
temperature than it actually is.** The actual object surface temperature
can therefore be substantially higher. Consider an object that is moving away
from us at an ACTUAL velocity of 0.87c. For this recessional velocity,
Time Dilation makes time appear to pass at exactly half of its
normal rate. Molecular and atomic velocities would therefore be half,
as well. This means that effectively, we would see an object that
indicates a surface temperature of half its actual rest-frame value.

This discussion has an interesting effect on the Twins Paradox example that is commonly used in presenting Special Relativity. Say the traveling twin is moving at 0.87c (to us) and so we see everything occurring on the spaceship as being at exactly 1/2 the rates we would normally see. Say that he drinks a glass of cool water, which we would normally say is at 10°C (or 50°F) or 283°K. But we would now see that water (if we could watch its atoms moving around in it, the actual definition of temperature) to appear to be at 283/2 or 142°K or around -131°C. We should see him trying to drink a piece of really frozen solid ice!

Any text on thermodynamics tells that blackbody radiation rates go as the fourth power of the absolute temperature (the Stefan-Boltzmann law). This means that the object / star in question is actually radiating 16 times as much energy as previously thought, if it is ACTUALLY at twice the surface temperature that we see.

First Developed, Apr 1989,

First Published on the Web: Jun 8, 1997

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C Johnson, Theoretical Physicist, Physics Degree from Univ. of Chicago