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Author Message
cubivore
I'm late to the party.

AngusA's solution is convincing. I need to either derive or find a reference that talks about that tan (theta) = sqrt((c-v)/(c+v)) tan (theta). I'm sure it's true.

I was seeing Adamek as going faster because I was blaming the increase in apparent size of the disk on a length contraction in their z-axis. A's z was 5 times shorter than B's, so gamma would be 5 between them. That neglected the relativistic effect on the triangle's hypotenuse (edge of the disk to the observer), which is probably why I was having such trouble.

I also got hung up on the ratio of the perceived distances being tan(theta)/tan(5*theta). That is overcome when one takes into consideration that the disc is "small."

It's been a while since my special relativity classes at university. I seem to recall us avoiding problems like this, too. Ah well.

Always learning!

Posted: Sun Nov 05, 2006 10:59 pm
LoneWolf
baf wrote:
Blast! It looks like you're right.

But I also don't see anything wrong with my math. I suspect that the problem results from my simplifying assumption, and that if you throw that out, the resulting (more complicated) equations would give us a range of velocities for both A and B for which the numbers come out right and both ships are actually moving in the right direction. But I don't feel like doing that right now. Maybe tomorrow, when I'm fresher of mind.

The problem is your first equation. Based on your assumptions, it should be:

Spoiler (Rollover to View):
5*Lb = Lc
since Lc has to be 5 times further than Lb.

If you use this with all of your equations, the answer reduces to x = 12/13 (even the negative goes away).

Posted: Wed Jul 19, 2006 7:32 pm
baf
I don't mean a range of solutions. I just mean a range of velocities that each ship could have. The solution is the relative speed of the two ships -- that is, the difference in their individual speeds. I'll be surprised if the difference varies as the individual speeds do.

Posted: Fri Apr 28, 2006 12:23 am
Ashin
baf wrote:
Blast! It looks like you're right.

Oh

baf wrote:
the resulting (more complicated) equations would give us a range of velocities for both A and B for which the numbers come out right and both ships are actually moving in the right direction. But I don't feel like doing that right now. Maybe tomorrow, when I'm fresher of mind.

, I did the same thing initially. And you're entirely correct too, you end up with a range of answers. However, only one appears to actually solve the problem correctly with a rational number, and I think thats .98c (or close to, I scratched this out by hand).

This card still strikes me as odd...

Posted: Thu Apr 27, 2006 2:47 pm
baf
Blast! It looks like you're right.

But I also don't see anything wrong with my math. I suspect that the problem results from my simplifying assumption, and that if you throw that out, the resulting (more complicated) equations would give us a range of velocities for both A and B for which the numbers come out right and both ships are actually moving in the right direction. But I don't feel like doing that right now. Maybe tomorrow, when I'm fresher of mind.

Posted: Sun Apr 16, 2006 10:04 pm
trystero
Baf, it seems to me that your solution proves that A is moving away from the disk, in violation of the problem statement.
Spoiler (Rollover to View):
If I plug your answer (v = 12/13 * c) back into your equation (2), I get

La = Lb * sqrt(1-(v/c)^2) = Lb * sqrt (1 - (144/169)) = Lb * sqrt (25/169) = Lb * 5/13

And you've already stated that Lb = Lc * 5 (and that therefore Lc = Lb * 1/5) in your equation (1).

Now, 5/13 = 25/65, and 1/5 = 13/65, meaning that 5/13 > 1/5, meaning that La is greater than Lc. And if, as you state, Lc is the distance that A was from the disk some time t (= Lc/c) ago, and La is the distance A is from the disk now, and La is greater than Lc, then A is moving away from the disk as time increases. (Or, as you point out in your later post, from A's frame of reference he sees the disk as moving away from him. It's still the same issue.)
Am I missing something?

Posted: Sun Apr 16, 2006 11:59 am
Dust
Thanks very much.
Spoiler (Rollover to View):
Ok, that make sense to me.

Here is also the point you make the approximation I mentioned: you need to assume that the distance for the edge of the disc and the center is the same. Which should be a perfectly valid assumption.

Ok, I have my peace of mind

Posted: Fri Apr 07, 2006 4:37 am
baf
Certainly.
Spoiler (Rollover to View):

The equation in question deals with the relationship between La and Lc. These are both in A's frame of reference, so everything I say below assumes this frame. In A's frame of reference, A is stationary and the disk is moving towards him.

Now, the light that A is currently seeing makes it look like the disk is a distance of Lc away. That means it came from the disk at a moment when the disk was Lc from A. Since A is unmoving, the light has travelled a distance of Lc. Since it's light, it was going at speed c. So the time it took to travel that distance was Lc/c.

At the start of that time, the distance between the disk and A was Lc, as noted already. Right now, the distance between A and the disk is La. So the distance has changed by (Lc-La).

So the distance has changed by Lc-La during a timespan of Lc/c. That means it's changed at a speed of (Lc-La)/(Lc/c). This gives us the relative speed of the disk and A.

But in the beginning, we made the simplifying assumption that B is stationary relative to the disk. That means that B's speed relative to A is the same as the disk's speed relative to A. So the relative speed of the two ships (v) is also (Lc-La)/(Lc/c). And that's equation 3.

Posted: Thu Apr 06, 2006 7:54 pm
Dust
baf wrote:
I solved this card recently...

Hi baf, could you explain in more detail where your equation 3 comes from?
You lost me at this point.

Posted: Sat Apr 01, 2006 3:16 am
baf
I solved this card recently, and now that I've looked at the discussion here, I'm surprised to see explanations so different from, and in some ways contradictory to, the way I was thinking about it. I don't know this "headlight effect", and suspect that I've just re-derived it from the Lorentz equation. At any rate, if the above explanations give you trouble (as they do for me), here's one you might like better.

Spoiler (Rollover to View):

First, to simplify things a little, I'm going to assume that ship B is stationary. If it's possible to satisfy the conditions of the problem with one ship going at a speed of zero, we really don't need a more general solution. This way we only have to consider two frames of reference rather than three.

So, A sees five times the diameter that B sees. If it looks larger, that's because in A's frame of reference, it's closer (due to Lorentz contraction). This is a simple matter of perspective: to look five times as large, it has to be 1/5 as far away. (Not 1/25 as people have stated above; I don't know where that came from. Perspective makes linear size inversely proportional to distance.)

Complicating this is the fact that A is looking at old light. The distance between A and the disk has changed in the time that it took for the light that A is seeing right now to travel the distance. So in his frame of reference, the disk is even closer than it looks. This is easy to miss; I certainly missed it on my first try, as did several other people who posted here.

Now to quantify all this. Call the distance from the disk to the ships La in A's frame of reference and Lb in B's frame of reference. There's a third distance that's important: the distance, in A's frame of reference, between A and the disk at the moment when the disk gave off the light that A is currently seeing. Call this Lc. (We don't need a corresponding figure for B, because B's distance from the disk is unchanging.) A's speed relative to the disk we will call v; B's speed relative to the disk we have assumed to be 0. The speed of light is, of course, c.

So, the perspective on the disk tells us:
1) Lb = Lc * 5

The Lorentz transformation gives the relationship between the distances in the two frames of reference:
2) La = Lb * sqrt(1-(v/c)^2)

And, finally, the effects of the movement. In A's frame of reference, the distance has changed by (Lc-La) in the time that the light took to travel a distance of Lc at the speed of light (Lc/c), so:
3) v = (Lc-La) /(Lc/c) = c(Lc-La)/Lc

The figure we want to find is the relative velocity as a fraction of the speed of light, or v/c.

And the rest is just algebra. Substituting 1) into 2) gives us
4) La = Lc * 5 * sqrt(1-(v/c)^2)
Equation 3) can be put in terms of La:
5) La = Lc - (Lc*v/c)
Combining 4) and 5):
6) Lc - (Lc*v/c) = Lc * 5 * sqrt(1-(v/c)^2)
Since v now appears only as part of "v/c", let's just write this as x from now on:
6b) Lc - (Lc*x) = Lc * 5 * sqrt(1-x^2)
Divide both sides by Lc:
7) 1-x = 5 * sqrt(1-x^2)
Square both sides:
8) (1-x)^2 = 25 * (1-x^2) = 25 * (1-x)(1+x)
9) 1-x = 25 * (1+x) = 25 + 25x
10) -24 = 26x
11) x = -24/26 = -12/13

And, well, my answer is negative, but that just means I was facing the wrong direction when I wrote down equation 3.

Posted: Wed Mar 29, 2006 2:42 pm
Dust
et wrote:

( I still don't understand why the angle of the cone of light decreases with relativistic speeds. The web site referred to doesn't explicitly explain that.)

Spoiler (Rollover to View):

You can understand the general effect without any need for relativity - see for example http://en.wikipedia.org/wiki/Aberration_of_light
or http://www-spof.gsfc.nasa.gov/stargaze/Saberr.htm.
This effect was discovered by Bradley.

Relativity merely changes the size of the effect. Mathematically, you arrive at the formula by starting with the relativistic velocity transformation in polar coordinates. See for example http://arxiv.org/ftp/physics/papers/0505/0505025.pdf or Section 3.3 in Wolfgang Rindler, "Essential Relativity".

Posted: Mon Mar 20, 2006 6:48 am
Dust
Solution by AngusA

Thanks very much to AngusA for his post.
I was really stuck and his explanation helped me a lot.

I do, however, have two comments:

Spoiler (Rollover to View):

1) I don't think the observers emit any light. This is not mentioned in the card and if such an assumption would be required for the solution I would regard this as pretty unfair.

It isn't required, though: relativistic light aberration yields the desired result (at least approximately, see below), without any need to shine light on the disc, and if you read the link in AngusA's post carefully it says exactly this:

"So what does this mean in relation to Warp? Well, the important thing to note is that the torch isn't giving off any less light. The same amount of light is being focused into a smaller area. Therefore, an approaching torch will seem brighter than a stationary torch. Likewise, a receding torch will appear darker. This doesn't just apply to torches. It also applies to any object traveling towards an observer."

This is also the answer to a previous question from Magma.

2) The formula is not exactly correct. It should read:
(see http://www.mathpages.com/rr/s2-05/2-05.htm)

tan F'/2 = tan F/2 * sqrt( (c+v)/(c-v) )

so F/2 and F'/2 instead of F and F'.

F is the angle between the direction of movement (here: the axis of the disc) and a given feature (here: the edge of the disc).

Now, for a large distance from the disc and a small disc it is probably approximately true that

tan F/2 = 1/2 tan F

which is not generally the case. If we make that approximation we end up at the result posted by AngusA.

Posted: Sun Mar 19, 2006 12:58 pm
Magma
A very conclusive and elegant solution, nicely done.

Posted: Sun Feb 26, 2006 6:57 pm
AngusA
et,

I think the beam narrows because of space dilation in the direction of travel. I didn't find a website that ran through the proof but the equation I used looks like it comes from the Lorentz transform.

Posted: Tue Feb 21, 2006 2:32 pm
Stratman
May I echo that...
Thanks AngusA, now I understand.
javascript:emoticon('')

Posted: Tue Feb 21, 2006 8:02 am
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