Schrödinger's hat

Well our goal is to make this: [math]m(\sqrt{(x - x_1)^2 + (y - y_1)^2}) = n(\sqrt{(x - x_2)^2 + (y - y_2)^2}) [/math] Look like this: [math] (x-x_3)^2 + (y-y_3)^2 = r[/math] Where [math]x_3, y_3 \mbox{and} r[/math] are some combinations of our original variables. We can start by shifting everything so that one of our circles is at the origin. This doesn't change the problem because it's just the same as re-labelling our axes (everywhere you see an x, you replace it with that number - x_1 and same w/ y). I put a little dash next to them so we know they're not the original x and y, but any results with these new coordinates hold for the old ones. It's a trick to get rid of some of the mess. This is my new set of variables: [math]x' = x-x_1,\quad y' = y - y_1[/math] So everywhere I see x,y I can replace them with [math]x = x'+x_1,\quad y = y' + y_1[/math] Also divide everything by n at the same time. [math]\frac{m}{n}\left(\sqrt{(x' + x_1 - x_1)^2 + (y' + y_1 - y_1)^2}\right) = \sqrt{(x' + x_1 - x_2)^2 + (y' + y_1 - y_2)^2} [/math] Next step, I cancel the numbers with opposite signs and replace some numbers with shorter names for them [math]a = x_2 - x_1,\quad b = y_2 - y_1,\quad c = \frac{m}{n}[/math] [math]c\left(\sqrt{(x')^2 + (y')^2}\right) = \sqrt{(x' - a)^2 + (y - b)^2} [/math] This following isn't really necessary, but if you square both sides and start rearranging you might see that it makes things a bit tidier. [math]x'' = (c^2 - 1)x'[/math] Then the last part is completing the square. This is a common method of dealing with quadratics. Read over what I said carefully, look up the derivation for the quadratic formula and/or ask for further help if you need it. None of these tricks are necessary to solve the problem, you can just have at it with algebra. It gets very big and messy doing it that way though and might be a little intimidating.

Any luck with that ratio of distances one Vastor? Imatfaal and I nutted it out in another thread. It gets pretty gruesome without a few tricks to simplify it. Some things that may help: Pick new coordinates so that the origin is at the centre of one of the circles. This is equivalent to shifting everything over. [math]x' = x-x_1,\quad y' = y - y_1[/math] Put frequently appearing variables in another variable: [math]a = x_1 - x_2,\quad b = y_1 - y_2,\quad c = \frac{m}{n}[/math] A re-scaling may also be useful later on. You should see where to do it if you need it. [math]x'' = (c^2 - 1)x'[/math] Remember the method of completing the square. If you see this: [math] x^2 + 2kx [/math] You can look at this: [math] (x + k)^2 [/math] And see that it expands to: [math] x^2 + 2kx + k^2[/math] So then you know: [math] x^2 + 2kx = x^2 + 2kx + k^2 - k^2 = (x + k)^2 - k^2[/math] Combine these methods carefully and you should be able to prevent it turning into a giant mess.

DH, Is it more that the supernova ejects mass into surrounding/remaining clouds, or that clouds condense more rapidly on the supernova remnant?

Please. If you do not understand the comic, then ask and someone will explain. Even without understanding the details of what the characters are saying the point is still there. The joke is that the light haired character is complaining about scientists taking the wonder out of everything by studying it. Then the dark haired character comes in, very excited about details that would be thought of as extremely boring by most people. She then goes on to say how beautiful something that is essentially a lump of slime is. The point it is raising is that the scientist, in studying all the details, sees beauty where others do not.

Stars don't burn all of the available hydrogen before they die. They just burn up enough that they can't burn what remains at a rate fast enough to stop the star from collapsing which could have merged with the expanding debris. There would have also been other gas clouds that had not collapsed.

Pah, I'll have no such superstitious nonsense.

Do not think for a minute that knowing how something works makes it less awe inspiring or wonderful. The more I study, the more often I find myself sitting and staring in amazement at some object or phenomenon I had previously taken completely for granted.

To know that, wouldn't we have to know how the density of opportunities for life changes with time. Perhaps it is much more likely in this comparatively young universe? On top of that, someone has to land near the beginning of the scale. A civilisation that was mediocre in every single respect would in many ways be less mediocre over all.

A heart is little use in figuring out the universe. Very good for gaping in wonder at what you've discovered though. There's beauty and elegance in pretty much every field of science, much of which can only start to be hinted at to the layman. I do not need a magic man for this. I actually find the idea of attributing everything to one somewhat belittling. The fact that we take the road of cold hard logic to get there doesn't lessen it in the slightest. The elegance of doing so has a beauty of its own.

Slight tangent: You can use science just as easily without believing it. Study the laws of logic, get a handle on the principle of induction, then take a set of axioms and possibly some hypothesis (whether or not you agree with any of them) and see where they lead. This is something you can quite easily choose to or not to do. You're right in that much of what you encounter in school or pop-science is analogous to being told about Gloria, but if you so desire you can meet Gloria and even dissect her into millions of tiny pieces to see how suitable she is. Not only that, you can get rid of the bad pieces, and then everyone is happier.

They are very different languages. HTML looks like this: <HTML> <head> Blah blah </head> <body> <p>Some stuff</p> </body> </HTML> [code] And so on. Javascript looks like this [code] var i, ham; ham = function(a,b,c) { for(var i = 0,ii =a; i<ii; i++) { b[i] = {Fish: c*i}; } }; You often see javascript embedded in an HTML file (inside <script> tags). You also frequently see bits of html in the javascript code, as its main use is building webpages.

I reiterate, what does the non-material world do? What effect does it have on me? If it has no effect, it can safely be discarded because there is no reason to believe it over any other claim of something with no effect. If it has an effect but is singular and not contingent on anything in the natural world, it is indistinguishable from randomness. We have plenty of things that act in this way (see atomic decay). If it is not contingent on anything in the natural world and applies everywhere, then it is natural law. If it has an effect and it is contingent on things in the natural world, then it can be examined with the tools of science. You keep bringing up mathematics. This is an entirely separate issue. Whether or not mathematics has its own existence, it is still completely reasonable (accessable by reason), and bound by rules. It either does not have its own existence (and is merely an abstraction on the natural world, like green -- green does not exist, it's a concept), or fits under the umbrella of natural (and we extend 'natural law' to include the way axioms interact with each other). There's also the issue of whether information does, in fact, have mass. If it is not bound by traditional logic, it is either bound by another formal reasoning system (there are plenty, look them up) or I can say whatever I like about it. Including that it is bound by logic. If you're going to pull the problem of induction card, you have bigger problems than whether science explains everything. Induction is internally consistent. Deduction is internally consistent. They are both methods of reasoning. Prove to me without using deduction, that deduction is valid. Exactly the same problem. Yet people only ever whine about induction. I can go one further and say that the only reason we accept deduction as a valid form of reasoning is that it is internally consistent, and it works. This is an inductive argument. So just what do you mean by supernatural/not governed by natural law? I cannot seem to find any concept/thing that they apply to, which isn't completely irrelevant to everything.

[math]\infty \mbox{infty} \div \mbox{div} \times \mbox{times} \star \mbox{star} \cdot \mbox{cdot}[/math] [math]S p\,a\;c\quad e\qquad s[/math] [math]21^a[/math] It's very finnicky about parentheses, and rarely gives intelligible error messages when you misplace one (I don't think thhe forum gives intelligible error messages at all)

Ah, in that case, you can totally get away with changing to a new coordinate system (centred on one of the circles and re-scaled by: [math]\left(\tfrac{m}{n}\right)^2 - 1[/math] so it's nice and pretty). [math]x'=d(x-j),\;y'=d(y-k)[/math] [math] \left(x' - b \right)^2 + \left(y' - c \right)^2 = a(b^2 + c^2) [/math] Feel free to copy anything I did over at the appropriate time.

I could put whatever I like into [math] \alpha, \;\beta,\;\gamma,\mbox{ and }\delta[/math] here: [math] x' = \alpha x + \beta t[/math] [math] t' = \delta t + \gamma x [/math] As long as it gives the suitably mirrored result when reversing v and still have a consistent set of equations. [math] x' = \frac{1}{\gamma} x[/math] [math]t' = \gamma t [/math] (with the traditional definition of gamma) Is one example. This produces the correct time dilation and length contraction formulae, it gives the same result in both frames, but it produces nonsense when we try to apply it anywhere else. So Principle of relativity + time dilation -> Lorentz transforms Doesn't hold without further data/constraints/assumptions Principle of relativity + time dilation + transformation consistent with measurements-> Lorentz transforms Does, but then you can just go Principle of relativity + consistent with measurements-> Lorentz transforms

The sun is a miasma of incandescent plasma. The sun's not simply made out of gas: no, no no The sun is a quagmire, it's not -- made of fire- Forget what you've been told in the past! Was so happy when I saw they corrected themselves. I didn't even realise they were still putting out albums until someone linked me to the more recent stuff. Although it's arguably less catchy than the original.

fractions are \frac{}{} with the top in the first parentheses, and the bottom in the second. Click on the images of the equations to see the source: [math] \oint_{s}\left(\iiint\limits_{\mbox{ham}}^{\mbox{fish}^a} \psi^{ \left[\begin{array}{ccc} \frac{a}{b} &\partial_s&\tfrac{\partial}{\partial \pi}\\ 1 &2 &3\\ 4 &5 &6 \end{array}\right]} \; d\mathcal{V}\right) d\mathbf{R}\rightarrow \pm \infty = \sqrt{3}[/math] Produces [math] \oint_{s}\left(\iiint\limits_{\mbox{ham}}^{\mbox{fish}^a} \psi^{\left[\begin{array}{ccc}\frac{a}{b}&\partial_s&\tfrac{\partial}{\partial \pi}\\1&2&3\\4&5&6\end{array}\right]} \; d\mathcal{V}\right) d\mathbf{R}\rightarrow \pm \infty = \sqrt{3}[/math] Most things in LaTeX are fairly straightforward. If I don't know how to do something, 4/5 times my first guess will be correct. There's a tutorial in the maths section, as well as a bunch of examples in the sandbox and maths section (remember, click the images). The forum often eats your formatting on the raw input. So if it's really complicated, paste it elsewhere (or get used to writing horrible unformatted messy latex ).

Many of the things from They Might Be Giants

May I enquire as to what your profession is such that it entailed doing this calculation by hand? Or is this merely your version of goofing off?

Also, now that I actually turn my brain on, this seems a lot simpler. [math] m\left(\sqrt{(x - j)^2 + (y - k)^2}\right) = n\left(\sqrt{(x - r)^2 + (y - s)^2}\right) [/math] Set some variables: [math] x' = x-j,\;y' = y-k,\; a = \left(\frac{m}{n}\right)^2\;b=j-r\;c = k-s[/math] Now: [math] a\left((x')^2 + (y')^2\right) = (x' + j - r)^2 + (y' + k - s)^2 [/math] [math] a\left((x')^2 + (y')^2\right) = (x' + b)^2 + (y' + c)^2 [/math] Set: [math]d=a-1[/math] And: [math] d(x')^2 - 2bx' + d(y')^2 - 2cy' = b^2 + c^2 [/math] [math] (x')^2 + \frac{-2b}{d}x' + (y')^2 + \frac{2c}{d}y' = \frac{b^2 + c^2}{d} [/math] [math] (x' - \frac{b}{d})^2 + (y' - \frac{c}{d})^2 = \frac{b^2 + c^2}{d} + \left(\frac{b}{d}\right)^2 + \left(\frac{c}{d}\right)^2 [/math] [math] (dx' - b)^2 + (dy' - c)^2 = db^2 + dc^2 + b^2 + c^2 = a(b^2 + c^2) [/math] Finally: [math] \left(x - j - \frac{b}{d}\right)^2 + \left(y - k - \frac{c}{d}\right)^2 = \frac{a}{d^2}(b^2 + c^2) [/math] Resubbing in variables could be done before dividing by d to simplify things somewhat further. And finally realise one is doing algebra for entertainment and go to find something productive to do...

I meant after that last step you wrote, but at any rate, I feel it is now time for a DON'T CROSS THE STREAMS Too lazy to find the image, I'll let your imagination do the work.

That's another dimension. You have to move perpendicular to your two directions on the surface of the ball to get there. Analogous in 3d Euclidean would be moving in a direction that is not up, down, left, right, forward, back or any combination. There is no outside to the closed 2d space on the ball. It's embedded in a higher dimension, but that is a distinct concept.

Lots of that junk on the RHS will cancel down or turn into neatly bracketed multiplication of stuff if you care. Turning it into [math](nr\pm mj)[/math] type of thing before combining the brackets should help

Vectors don't make too much sense for a photograph, unless your camera works more like an eye I guess. 3-D is already stored in much the same way as 2-D vector graphics (except it's 3-D data). There's actually been a bit of interest going the other way (storing your 3-D object as a bunch of small particles rather than larger geometric shapes, although it's more of a hybrid, because the larger shape is stored too for when it's far away.)

Because I'm bored/curious/out of practise with algebra: [math] (m^2 - n^2)x^2 + (m^2 - n^2)y^2 -x(2m^2j -2n^2r) -y(2m^2k -2n^2s) = n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2[/math] Tidy up a bit. Get some stuff we're not interested in on the rhs. [math] (m -n)(m + n)x^2 + (m -n)(m + n)y^2 -x2j(m^2j - n^2r/j) -y2k(m^2 -n^2s/k) = n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2[/math] [math] (m - n)(m + n)x^2 + (m - n)(m + n)y^2 -x2j(m -n)(m + nr/j) -y2k(m - n)(m + ns/k) = n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2[/math] [math] (m + n)x^2 + (m + n)y^2 -x2(km + nr) -y2(km + ns) = \frac{n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2}{m-n}[/math] ^Edit: Mistake. Have fun. Now get the quadratics in a nice form: [math] x^2 + y^2 -x\frac{2(km + nr)}{(m + n)} -y\frac{2(km + ns)}{(m + n)} = \frac{n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2}{m^2-n^2}[/math] For convenience, set: [math]a = \frac{-(km + nr)}{(m + n)}[/math] [math]b = \frac{-(km + ns)}{(m + n)}[/math] Now: [math] x^2 + y^2 +2ax + 2by = \frac{n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2}{m^2-n^2}[/math] Note that: [math] (x+a)^2 = x^2+2ax + a^2 \rightarrow x^2 + 2ax = (x+a)^2 - a^2[/math] [math] \left(x - \frac{(km + nr)}{(m + n)}\right)^2 + \left(y - \frac{km + ns)}{(m + n)}\right)^2 = \frac{n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2}{m^2-n^2} + \left(\frac{(km + nr)}{(m + n)}\right)^2 + \left(\frac{(km + nr)}{(m + n)}\right)^2 [/math] And you're left with a horrific mess on the RHS which can be simplified, and a k I substituted instead of a j on line 4.