Years ago, the textbook I used for Intermediate Algebra mentioned relativity in the rational functions chapter. They gave an expression for observed speed (of galaxies separating). I wanted to help my students practice the algebraic steps involved in solving an equation (with lots of variables) for one variable, and made up a silly story about a galacto-cop chasing a possibly speeding pirate. I have a few questions.
The equation I used was equivalent to s = (u+v) / (1 + uv/c2), where u and v are the observed velocities of two objects moving in opposite directions. s is the speed at which they're separating. At earthly speeds, s would equal u+v. Simple. But near the speed of light things get complicated. If u and v are each over half the speed of light, u+v would give us an s value over the speed of light. That's apparently not possible. Einstein (and others?) came up with the equation above, which describes relativistic effects on velocity. I think. (Please correct my statements here if they're inaccurate, misleading, or confusing.)
What I wanted to do with my students was to solve for u. I got u = (s-v) * c^2 / (c^2 - sv). First question, is this legitimate?
My story was that the galacto-cop's 'radar' (what else should I call it?) gave her the pirate's speed of separation from her ship, but she wants to know the pirate's 'true' speed, and has to figure this version of the formula out. I claimed the galactic 'speed limit' was 1/4*c.
My understanding of relativity is weak, and I'd like to get this story down a little better before I tell it in class this semester. Help?
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I'm having trouble with your idea of the "true speed"--it's important in relativity that there's really no such thing as a frame-independent "true speed".
ReplyDeleteMaybe there's privileged reference frame with respect to which the speed limit is measured? But that's some pretty arbitrary law-making! :)
Yeah, what Dave said...
ReplyDeleteIn relativity, there are frames of reference issues to consider. The trick here is that in the velocity addition formula you gave, s, u, and v are not all in the same frame of reference. There are two frames involved (which we'll call unprimed and primed) and two objects who's velocity we are considering: u and v.
Rewritten in these terms, under Galilean relativity, you have if u'=0, the v=u+v'.
If a rail car is headed out of the station at speed u (relative to the station), and a passenger is walking forward on the train at speed v' (relative to the train), then the passenger is moving at speed v=u+v' (relative to the station).
In Einstein Relativity, the formula is if u'=0 then v = (u+v')/(1+uv'/c^2).
To use your Galactic Cop example, this formula works if u is the speed of the cop (relative to the Earth), v' is the speed of the pirate relative to the cop, and v is the speed of the pirate (relative to the Earth).
Does that help?
Yeah, i had a hunch I wasn't allowed to think of it as true speed (hence the quote marks). How about the speed relative to the center of our galaxy?
ReplyDeleteBlaise, thanks. I think I misunderstood what I read on Wikipedia (I just re-read it). I'm wondering if the context given in that algebra text I got the formula from was wrong.
I'm still wondering if there's ever a need to get it from the first version I wrote (s=...) to the other (u=...).
It all depends on what you want to figure out, and what information you have.
ReplyDeleteIn this case, you have two frames of reference (primed and unprimed), the velocity between the two (u), and the velocity of an object in one of them (v'), and you need to compute the velocity of the object in the other (v).
What you can do is solve for v' in terms of v (or v in terms of s, using the original notation), and show that the two formulas are the same, modulo a switch in sign of u (or at least, it should be).
Your solving for u case would be finding the velocity between the the frames of reference.
So take this: Ship U can't see the Station, but Ship V can see both. Ship V can tell U both v' and v (or v and s in the original notation), and U can therefore compute u.
The trick is that the formula should be symmetric: You've got (effectively) three reference frames, and the formula for velocity conversion between them should be the same.
To take your solution ( u = (s-v)*c^2/(c^2-sv) ), divide the left-side by (c^2/c^2) and you get u = (s-v)/(1-sv/c^2). If you note that changing reference frames also changes signs (if A is moving at v relative to B, then B is moving at -v relative to A), then the formulas can be manipulated to the same form.
A heck, here it is... We've got three reference frames, s, u, and v. If we represent the velocity of a in relation to b as a_b, the original formula can be written as:
v_s=(u_s+v_u)/(1+(u_s)(v_u)/c^2)
Solving for u_s, you get your formula
u_s=(v_s-v_u)/(1-(v_s)(v_u)/c^2)
Noting that v_u = -u_v, and doing the appropriate substitution, you get:
u_s=(v_s+u_v)/(1+(v_s)(u_v)/c^2)
and from above:
v_s=(u_s+v_u)/(1+(u_s)(v_u)/c^2)
Note that the two formula are identical except for the swap of u and v.
We also have:
s_u = -(v_s+u_v)/(1+(v_s)(u_v)/c^2)
= (s_v+v_u)/(1+(s_v)(v_u)/c^2)
= (v_u+s_v)/(1+(s_v)(v_u)/c^2)
...which is the same form, etc.
That's the real lesson of relativity, that the formula are the same for all observers.
It's worth noting, btw, that this form of the velocity addition formula works when the three reference frames are all moving colinearly with each other -- i.e. s_u and s_v are in the same direction (or opposite directions) -- and has to be modified when s_u and s_v are at angles to each other.
ReplyDeleteI think the big point is that the formula you produced by solving for u is, in fact, the same formula you started with, if you pay attention to the directions of the various velocity vectors.
ReplyDeleteYou can have your speed limit by having a police station as the reference point for 0 velocity. Within one galaxy, everything's pretty much moving the same speed anyway, so someone cruising along at 1/4 c would definitely need to get pulled over.
Thanks, Josh! I had a nagging suspicion that might be what some of what's been said here would lead to. I don't see it yet, but I'll sit down with it one day, and figure it out.
ReplyDelete