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 Forum index » Diversions » Perplex City Puzzle Cards » PXC: Silver Puzzle Cards
[Puzzle] #245 - Silver [Maze Set] - Relativity
Moderators: AnthraX101, bagsbee, BrianEnigma, cassandra, Giskard, lhall, Mikeyj, myf, poozle, RobMagus, xnbomb
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Author Message
Ashin
Veteran


Joined: 22 Nov 2005
Posts: 140

 		Awesome guys, and without using a Lorentz transformation. *bows down* Definitely a much more elegant solution that the nasty string of equations I was using.

Also, I didn't mean to imply that I think working together is bad, or for that matter sharing hints and ideas. I guess it's just frustrating when I click on a forum page and the second or third post down has the answer, kinda takes the fun out of it. But I agree about the cards not being the overall main challenge. Guess I just need to loosen up a little Very Happy

Oh, and the Wave 2 cards are only available in the UK right now (as far as I know), so I'm waiting on shipping right now from Firebox.

PostPosted: Thu Dec 01, 2005 9:50 pm
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neuromancer
Decorated


Joined: 04 Aug 2005
Posts: 168
Location: Birmingham, UK

 		Well I got my solve accepted now. Unfortunately since the answer was put up everyone that had this card and looks at these forums solved the card, so I didn't get into the first 10 solves list Evil or Very Mad . Damn MC for locking the cards, why the hell do they do that!!

Just for the sake of completeness, the answer is:

Spoiler (Rollover to View):
12/13C. But you don't need to enter the 'C'.

_________________
The car in front.....it's mine!!!

Windows, the most successful computer virus in history...

PostPosted: Fri Dec 02, 2005 7:50 pm
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TheBozzball
Boot

Joined: 29 Nov 2005
Posts: 12

 		When I said that the picture on the card was a clue, that is because the solution uses the:

Spoiler (Rollover to View):
Headlight Effect


PostPosted: Wed Dec 07, 2005 2:09 pm
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Stratman
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Joined: 03 Feb 2006
Posts: 81
Location: Kettering UK

 		I realise this has been solved and the answer is correct.
But there is one part of the explanation I just can't seem to get my head around!
Why does Bajdan have to be 25x further away to see the diameter of the disc five times smaller?
Every way I look at this it seems to me you would see the disc five times smaller if you were five times further away.
Am I missing something obvious here?

PostPosted: Sat Feb 04, 2006 2:25 pm
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Magma
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Joined: 10 Jan 2006
Posts: 119

 		I can't figure it out either. I understand now that:

Spoiler (Rollover to View):
By the wording of the question it rules out the use of the equation to work out spatial contraction - Contraction only happens in the direction of motion, and as they are facing the axis of the disc, i.e. facing it head on, the apparent diameter would be unaffected by the speed of travel, only the apparent thickness of the disc.


But what I can't figure out is the 25 times distance. If I draw an object 10m tall, view it from the ground at 2m away, and calibrate against a big ruler stuck in the ground 1m away from me, the object is 5m tall. If I move 10m away and do the same with a big ruler 1m away from me, the object measures 1m tall. Is there some freakish relativistic reasoning going on here?

Also, from what my brain (such as it is) could fathom about the solution, it goes like this:

Spoiler (Rollover to View):
Call Adamek "Ship A" and Bajdan "Ship B".

Ship A is 1 unit distance away from the disc and is stationary.
Ship B is 25 units from the disc, 24 units from Ship A, and is moving towards both.

Ship B gives off light as it moves, and to see the disc that light must reach the disc, bounce off and reach the ship again.
We know that the light and the Ship B start at the same point, and finish at the same point.
However, the light has to bounce off the disc and return in addition to the distance the ship moves.
This means Ship B moves 24 units to the position of Ship A, and the light moves 25 units to the disc, then 1 unit back to Ship A's position.
So Ship B is moving 24 units in the time that it takes light to move 26 units. This means Ship B is moving at 24/26ths the speed of light. Simplified as a fraction it becomes 12/13.


Now this surely assumes that Ship B is seeing the disc using its own lights?
What happens if the disc is illuminated by another light source? If it makes no difference, why not?

I already wondered if the distant small disc is a star, and the diameter is indicative of its brightness, but intensity falls off as a square of distance while here the size falls off as the square root of the distance.

PostPosted: Sat Feb 18, 2006 2:22 pm
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AngusA
Boot


Joined: 02 Jan 2006
Posts: 44
Location: London, UK

 		I agree with what you're saying Magma. The solution as given just doesn't make sense to me, as this idea of how big something is seems entirely subjective (or maybe relative to how far away you are from it). Bozzball mentioned the Headlight effect which, from a quick look at Wikipedia, relates to relativistic effects on the intensity of light and when I have some time this weekend I'll look into it. Maybe how bright something looks versus what you'd expect is where the answer to this lies? Unfortunately, I'm two weeks behind on a Web course at the moment...

PostPosted: Sat Feb 18, 2006 2:56 pm
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et
Greenhorn

Joined: 02 Jan 2006
Posts: 3

 		I ask the same question as Stratman, Magma and Angus A. Why should the second person, B, be twenty five times the distance that Person A is from the disc object?

HELP!

Are we ignoring relativity because the diameter is perpendicular to direction of travel?

Am I wrong in assuming that the angle subtended by the disc will be five times less for B than A when it is further away?
If you can explain the thinking process, then I'll be a happier woman.

PostPosted: Sat Feb 18, 2006 6:31 pm
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AngusA
Boot


Joined: 02 Jan 2006
Posts: 44
Location: London, UK

 		I have a new solve for this puzzle. It's pretty complicated and you may want to take a look at http://en.wikipedia.org/wiki/Special_relativity_for_beginners to get you started.

Spoiler (Rollover to View):
The set up:

Adamek (Starship A) and Bajdan (Starship B) are travelling towards a disk. At the point they pass they are the same distance from the disk travelling at different velocities V a & V b respectively.

To measure the size of the disk they are both shining a cone of light on the disk so that just the disk is illuminated across the whole diameter. The cone of light has an angle F to the direction of travel (see http://www.adamauton.com/warp/lesson5.html) for a basic diagram.

Using Pythagoras: tan F= radius of the disk (R)/distance from the disk (d).

The Headlight Effect causes the angle F to narrow at relativistic (very fast) speeds according to following equation:

tan F' = tanF [(c-v)/(c+v)]^0.5 {square root of (c-v)/(c+v)}

where tan F' = the observed width of disk (R')/distance from the disk (d), giving:

R' = tanF [(c-v)/(c+v)]^0.5
d

or

R' = d*tanF [(c-v)/(c+v)]^0.5

So, we have the situation where A sees the disk to be 5 times bigger than B or
R' a = 5*R' b

d*tanF [(c-V a )/(c+V a )]^0.5 = 5*d*tanF [(c-V b )/(c+V b )]^0.5

Dividing both sides by d*tanF gives:

[(c-V a )/(c+V a )]^0.5 = 5*[(c-V b )/(c+V b )]^0.5

We know that B is travelling at a much higher velocity than A because he sees the disk to be much smaller. So, I'll take this from A's Frame of Reference, so that V a = 0. The left hand side of the equation resolves to 1, giving:

1 = 5*[(c-V b )/(c+V b )]^0.5

Squaring both sides gives:

1 = 25*(c-V b )/(c+V b )

Multiply both sides by (c+V b )

c+V b = 25c - 25V b

Rearranging:

26V b = 24c

or V b = 12c/13


PostPosted: Sun Feb 19, 2006 8:46 pm
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et
Greenhorn

Joined: 02 Jan 2006
Posts: 3

 		Laughing Many Many thanks Angus A. You've given an answer I am happy with. I only hope that Angus B is relatively as quick as you!

( 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.)

PostPosted: Mon Feb 20, 2006 8:39 pm
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Stratman
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Joined: 03 Feb 2006
Posts: 81
Location: Kettering UK

 		May I echo that...
Thanks AngusA, now I understand.
javascript:emoticon('Idea')

PostPosted: Tue Feb 21, 2006 10:02 am
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AngusA
Boot


Joined: 02 Jan 2006
Posts: 44
Location: London, UK

 		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.

PostPosted: Tue Feb 21, 2006 4:32 pm
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Magma
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Joined: 10 Jan 2006
Posts: 119

 		A very conclusive and elegant solution, nicely done.

PostPosted: Sun Feb 26, 2006 8:57 pm
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Dust
Greenhorn

Joined: 19 Mar 2006
Posts: 7

 		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.



PostPosted: Sun Mar 19, 2006 2:58 pm
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Dust
Greenhorn

Joined: 19 Mar 2006
Posts: 7
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".


PostPosted: Mon Mar 20, 2006 8:48 am
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baf
Boot

Joined: 29 Mar 2006
Posts: 46

 		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.



PostPosted: Wed Mar 29, 2006 4:42 pm
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