# Seeing Exoplanets Sideways

Given that I’m currently working on a book about relativity, I’m spending a lot of time idly thinking about various relativistic effects. Many of these won’t end up in the final book, but they’re fun to think about.

One thing that occurred to be earlier, while thinking about something else entirely, is the Doppler shift. In particular, I was thinking about the detection of planets around other stars, which is often done using the Doppler shift due to the orbiting planet’s tug on its star. If the orbit is more or less aligned with our line of sight to the star, then the star wobbles back and forth along our line of sight by a small amount, and this leads to a shift in the spectrum of light emitted by the star. We can use that shift to measure the speed, and by watching the variation in the speed, we can determine the orbital period of the star, and the mass of the planet, and fun stuff like that.

Of course, this leaves one small thing out. You’ll often hear people say that this only works for orbits aligned along our line of sight, because we need velocity along that axis to get a Doppler shift. That’s not completely true, though– there is, in fact, a Doppler shift of light emitted by an object moving at right angles to the observer. You can think of this as a relativistic effect: the frequency of the emitted light behaves like a sort of “clock,” and from relativity we know that moving clocks run slow. Thus, the frequency seen by a stationary observer looking at a moving source will be different than the frequency seen by an observer moving with the source of the light.

This Doppler shift due to transverse motion is a “second order” effect, which means that it depends on the square of the velocity. “But wait,” you say, “Things that are squared are bigger than things that aren’t squared! So why isn’t this the dominant effect that we see?” It’s because the thing that gets squared is the ratio of the speed of the object to the speed of light, which is always a number less than 1. And while numbers greater than one get bigger when you square them, numbers less than one get smaller. For motion that is toward or away from you, the first-order Doppler shift completely dominates the second-order shift. The second-order effect is the only Doppler shift for transverse motion, though, so that’s where we see it.

So, could you use the second-order Doppler shift to detect an extrasolar planet?

Your first guess might be “no,” because the planet is always in motion, so there wouldn’t be the change in velocity that is the real signature of Doppler shifts due to orbits along our line of sight. That’s not quite true, though, because planetary orbits are generally elliptical, not circular, so the speed of the planet in its orbit varies slightly. There would be a tiny difference in the shift.

So, how big a shift are we talking? Well, I can easily find values for the orbital speed of the Earth, so let’s use those as a starting point. Imagine that we set up a huge beacon on Earth, beaming out light with a very particular frequency, and an alien looking at us sideways was carefully monitoring the frequency of the light. How much variation would that alien see?

The quantity determining the shift of the frequency is:

1/γ = (1-β2)1/2

That’s the square root of 1 minus the speed of the Earth divided by the speed of light squared. So, what are the maximum and minimum values of this?

Using the maximum orbital speed of Earth from the link above, 30,287 m/s, the shift factor comes out to 0.999999994897. The minimum speed, 29,291 m/s gives a shift factor of 0.999999995227. That’s a difference of about 0.0000000003301. To put that in perspective, if the light used for the beacon was the microwave radiation used in cesium atomic clocks, with a frequency of 9,192,631,770 Hz, the shift between maximum and minimum orbital speeds would be about 3.3 Hz. That’s pretty tiny.

It gets worse, though, because what we actually look at when detecting extrasolar planets is not the speed of the planet, but the speed of the star. This is much, much lower, because the star is so much more massive than the planet. How much smaller? Well, for a very rough estimate, I would expect the speed of the star to be smaller by roughly the ratio of the masses. The Sun has a mass of about 1030kg compared to the Earht’s 1024, so that’s a factor of roughly 1,000,000. Which gets squared, so the transverse Doppler shift of light from the Sun due to the motion of the Earth would be 1012 times smaller than the light from the Earth itself.

That would mean the difference between the Doppler shifts at the maximum and minimum speeds of a star like the Sun due to a planet like the Earth would be on the order of 10-22. Since the very best atomic clocks on Earth are only good at the 10-18 level, I think it’s safe to say that we need the orbits of extrasolar planets to be more or less aligned with our line of sight to see a Doppler shift from the motion…

1. #1 ThirtyFiveUp
November 24, 2010

Just to give thanks for your blog which I read daily to learn stuff which is amazing.

The universe is one entity and I am thankful to be a part of it. May we all live in wonder and awe.

November 27, 2010

Is this the general or special relativity? Who is the target audience and how much math do they need to know?

3. #3 qbsmd
November 29, 2010

Is using numbers for earth really a fair test? I had thought most of the exoplanets found have been Jupiter sized and closer to their stars (causing a much more noticeable Doppler effect on their stars). Also, earth has a relatively low eccentricity compared to other planets (http://en.wikipedia.org/wiki/Table_of_planets_in_the_solar_system).

A planet the size of Jupiter in an earth or Venus distance orbit with an eccentricity of .04-.05 would probably be detectable according to your threshold. Which raises the question of what the threshold should be; I had the impression that we didn’t quite have the ability to detect small planets yet, which is why the ones detected are so large.

4. #4 llewelly
November 29, 2010

Is using numbers for earth really a fair test?

It’s a fair test if you want to search for earth-like planets. If you’d rather search for hot Jupiters, no.

5. #5 Craig Heinke
December 3, 2010

Planets in close orbits lose their eccentricities quickly. But let’s say the planet is maximally eccentric, and at its closest moves at the speed of a hot Jupiter in a 3.5 day orbit around a solar-like star. My students calculate the average speed of such a hot Jupiter, it’s ~135 km/s. Let’s make it a really massive hot Jupiter, almost a brown dwarf, at 10 Jupiter masses. Then the star moves at 1.35 km/s. Beta=4.5e-6, and the second-order shift is then ~1e-11, which gives you your delta lambda/lambda. So that’s a shift in the spectral line equivalent to 3 mm/s. Current best technology is reaching 1 m/s, and at this point we’re having to fight with chromospheric activity on the star; so no, we can’t detect any planets by this method now.

Since the first-order effect goes as sin(i)*beta_star, the first-order effect will dominate for inclinations larger than ~3*10^-4 degrees. I estimate that only one star in 10^11–i.e., maybe one star in our galaxy–will have an inclination that close to pole-on from our perspective.

Interesting post!