The parallax distance to an object is the distance you get by moving your (single) eye's position by a transverse distance x, measuring the angular displacement θ of the object given that move, and dividing the transverse distance by the angle.
Problem-set problem: You are outside after the rain late in the day and you see a rainbow. What is the parallax distance to the rainbow?
(Thanks to Andrei Gruzinov at NYU.)
Undefined.
ReplyDeleteinfinite
ReplyDelete@Anonymous, @Foster: You are both wrong!
ReplyDeleteI'm not at all sure how to think about this problem!
ReplyDeleteMy first thought was that rainbows are virtual objects -- so, when two people see what they take to be the same rainbow, they are seeing distinct things. No real object. Hence, no real angle that lets you see the same object after moving. But, you said that Anonymous was wrong, so I thought some more.
Then I thought, "Maybe the angular displacement is infinite, and so the distance is zero." But that doesn't seem to make sense either, since an infinite angle doesn't seem to have any meaning.
Then I thought maybe the rainbow works like an image in a mirror. After all, the light we see is a sort of refracted reflection of the sun. In that case, I suppose the distance is the same as the distance from the sun to the rain (or whatever) plus the distance from the rain to where I am standing. I am leaning toward this answer, but it also doesn't seem right somehow (although I can't quite say why). More importantly, I'm not sure how I would calculate that distance from the details you've given. Or am I supposed to be able to do so?
So, yeah. Fun problem. I have no idea how to think about it, but I would really like to see how it is done. If you don't want to put up the answer on your blog, would you mind emailing me?
57.3(x/θ)
ReplyDeleteA piece of information I'm not immediately seeing how to use is "late in the day". Does it mean that the Sun is on the horizon, and that helps somehow? Or is there something else to be extracted from that fact? (Feeling slow this morning.)
ReplyDeleteWow, all of a sudden this post got exciting.
ReplyDeleteJonathan: Think about the position of the rainbow for different observers, relative to exceedingly distant stars (or quasars). That's how parallax works.
Iain: You can only see rainbows near sunrise or sunset for various reasons, so it is a red herring, really.
Going along the same lines as Jonathan and thinking of the rainbow as a reflection in a mirror perpendicular to the sun-observer line, if the observer also took measurements of the sun after turning 180 degrees at each point, she would record the same angular displacement as that of the rainbow, so the parallax distance would be the same as the distance from the sun to the observer (approx; I'm not exactly sure how the slight change in that distance when the observer moves transversely affects things).
ReplyDeleteThe only additional complication I can think of is that the angular displacement observed for a rainbow would have the opposite sign to that observed for a fixed object; does it make sense to talk of a negative parallax distance in these circumstances?
For distant objects (using the small angle approx.), the parallax distance is the distance.
ReplyDeleteI would think the rainbow is always centered around the sun. So 1 AU.
Good grief. I'm ashamed to say that I said 'infinity' without thinking, but 1 AU is right. What an excellent question, though! I'm now thinking about orbiting the Sun in a rain filled sphere just over 1 AU in radius....
ReplyDeleteactually, 1 AU is not right, at least not precisely... Compare the parallax of the rainbow to that of a hypothetical object along the same line of sight...!
DeleteI would imagine.... parallax distance = (your distance from the sun) + 2 x (your distance from the droplets)
ReplyDeleteUnless I'm forgetting something. Isn't that the correct answer Hogg?