A Roller Coaster Ride through Relativity/The Doppler shift

From Wikibooks, open books for an open world
Jump to: navigation, search

The Doppler Shift[edit]

When you hear an ambulance go by, the pitch of its siren falls. Since the ambulance is moving through the air and it is you who is stationary, this is due to the moving source effect. It works like this.

Suppose that the ambulance is moving away from you and its siren has a period T0. During the time it takes for the siren to emit one complete wave, the wave front has moved towards you a distance cT0 but the ambulance has moved in the opposite direction a distance vT0. The wave is therefore stretched to a total distance (cT0 + vT0)

Since this wave reaches me travelling at a speed of c, the time T it takes for the wave to pass me is (cT0 + vT0)/c ie:

T=T_0(1+v/c)

The apparent time period has been increased by a factor of (1 + v/c).

(If you are a musician, you will know that a change of one semitone – which is 1/12th of an octave – is caused by a change in pitch of 12√2 or 1.06 This means that an ambulance travelling at 45 mph (which is 6% of the speed of sound) will have its siren Doppler shifted by a whole tone as it goes by – a semitone up as it approaches and a semitone down as it recedes.)

Now exactly the same thing happens with light. Hadn't you noticed that in addition to the siren sounding lower, all the lights on the ambulance look redder too? No? Well I am not too surprised as the ambulance may be travelling at an appreciable fraction of the speed of sound, but it is not travelling anything like as fast as the speed of light so the effect is not going to be very noticeable.

The argument in light is exactly the same except that, in addition to the increase due to the wavelength stretching effect, we must also include the increase due to time dilation. All we have to do is replace T0 by \gammaT0 ie:

T=\gamma T_0(1+v/c)={T_0(1+v/c) \over \sqrt{1-v^2/c^2}}

Multiplying top and bottom by c and cancelling a factor of √(c + v) leads finally to

T=T_0 \sqrt{c+v \over c-v}

But I thought you said that there was only one Doppler effect in light. You seem to have found a formula for the moving source effect – but what about the moving observer effect?

OK. Consider the normal effect (in sound) first. Suppose you are blowing a whistle to attract the attention of the ambulance driver (who is still travelling away from you). The waves you send (which have a wavelength cT0) are chasing the ambulance with a closing speed of (c - v). The time it takes for one wave to catch up the ambulance is going to be cT0 / (cv) so:

T={T_0 \over 1-v/c}

Again the time has been increased so the ambulance driver will hear a lower pitched whistle, but you will notice that the formula is not the same as the formula for the moving source effect which multiplies by (1 + v/c) rather than dividing by (1 - v/c).

Now we must be a little careful when applying the relativistic time dilation correction here. From our point of view, it is the ambulance man's clock which is going slow. So if it takes T of our seconds to reach the ambulance, it will take fewer of his. This means that we must divide by \gamma, not multiply. Hence

T={T_0 \over(1-v/c) \gamma} = {T_0 \sqrt{1-v^2/c^2} \over (1-v/c)}

which, surprise, surprise, leads us to exactly the same formula!

Astronomers are usually more interested in changes in wavelength than changes in time period, but the formula is essentially the same as wavelength is proportional to period. ie:

\lambda=\lambda_0\sqrt{c+v \over c-v}

I find it rather pleasing that the formula for the Doppler shift in light turns out to be the geometric mean of the two normal Doppler shift formulae. Plotting some graphs will help to sort out the differences.

Relativistic and non-relativistic Doppler shifts

From a physicists point of view, it is the Doppler shift factor here defined as \lambda / \lambda_0 which is of primary interest. From an astronomer's point of view, it is the change in wavelength \lambda - \lambda_0 which is generally measured and they usually quote a star's red-shift factor defined as (\lambda - \lambda_0) / \lambda_0. It is easy to see that Red-shift factor = Doppler shift factor - 1

If the speed of the source is much smaller than the speed of light, we can simplify the formula using the binomial theorem as follows:

{\lambda \over \lambda_0} = \sqrt{c+v \over c-v} = \sqrt{1+v/c \over 1-v/c}
{\lambda \over \lambda_0} \approx (1+ \tfrac{1}{2} v/c)(1- \tfrac{1}{2} v/c) \approx 1+v/c

hence:

 \text{Red-shift} \approx v/c

Note that this expression can only be used in the range where the speed of the galaxy is less than 20% of the speed of light. The Andromeda galaxy, for example, shows a red-shift of -0.001 (ie it is in fact blue-shifted) and is moving towards us at a speed of about 300 km s-1. Don't worry about a collision though – at 2.5 million light years away it is going to take 2.5 billion years to reach us!

Quasars with very large red-shift factors have been observed – for example: quasar PC1247+3406 has a red shift of 4.897 therefore a Doppler shift factor of 5.897 and a recession speed of 94% of the speed of light. In 1929, Edwin Hubble discovered that all the distant galaxies were receding from us and subsequent measurements have indicated that there is a simple proportional relationship between speed and distance, with the edge of the observable universe at about 13.6 billion light years. If PC1247's red shift is due to the Doppler effect and nothing else, it would seem to indicate that it lies a staggering 12.8 billion light years away.

Of course, if a galaxy were to travel away from us at a speed equal or greater than that of light, it would be invisible because all light from it would be red-shifted out of existence.

I suppose so – but surely the point is academic because you have already pointed out many times that nothing can travel faster than light.

True – but as it happens, it is not inconceivable that a galaxy should recede from us at such a speed in spite of what we have said about the addition of velocities etc. Cosmologists believe that the distant galaxies show large red-shifts, not so much because they are moving away from us in an otherwise fixed space; they prefer to think of the galaxies themselves as being stationary – but the space in between them is expanding.

What does that mean?

Well, imagine a whole lot of ants crawling around on a balloon which is being blown up.

The ant balloonn

The distance between the ants is increasing all the time – but the ants are not really moving at all. Also, at any given time, the rate at which the distance between two particular ants increases is directly proportional to the distance between them. Eventually, when the balloon gets large enough, there will be a pair of ants whose separation is increasing faster than the speed of light.

Our universe could be like this. There could indeed be galaxies out there which are moving away from us faster than the speed of light!

Well, as I said before, the idea is academic because we could never see them or go there – even in principle.

Yes – you are probably right, though there is a possibility that, if the expansion of the universe were to slow down, some of these galaxies might come into view again. The latest research seems to indicate that, in fact, the expansion of the universe is accelerating. If this is the case, some galaxies which we can see now will disappear over the horizon, never to be seen again. The truth is, we don't really know what kind of universe we are living in – mainly because astronomers are seriously divided on the question of how much matter it contains. But that is another story.

The third hill[edit]

The roller coaster, still travelling at high speed, emerges from the spiral and shoots up a gentle rise. Once again, as it approaches the summit, you have a chance to catch your breath and look around.

Hey! Look at those boys over there! That's really clever!

You're right, it is. The boys are tossing a pair of footballs to each other – but instead of just playing catch, they are bouncing the balls off the other one so that they catch the one they have just thrown.

That must be really difficult. you say. You have got to toss the ball with exactly the right speed so that it comes back to you. If you toss it too slowly it won't have enough momentum to bounce the other ball back to its owner.

But you don't have time to watch much longer. The roller coaster is starting its third big descent. Hang on!

To your surprise you discover a football in your lap, just like the ones the boys were throwing. On a whim you toss it sideways out of the coaster. Just then, ahead of you, you see one of the boys throwing his ball towards the tracks. To your astonishment you realize that the two balls are on a collision course, but your ball seems to be travelling faster than his. On the other hand, to the boy beside the tracks, it is you who are moving swiftly by and (because your actions are slowed down by time dilation) it seems to be his ball which is travelling faster than yours. Each of you predicts that the balls will bounce asymmetrically towards the other. What in fact happens is that the balls bounce off each other perfectly and the boy catches his ball while your ball lands in your lap! This all happens because of:

Bizarre consequence number 12
Moving objects increase their mass

Back to the introduction ...

Back to top ...

Next page ...