## Temporal distortion, geodesics(?), and My Little Pony

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Aedl Foxe
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### Temporal distortion, geodesics(?), and My Little Pony

[](/b00 "All right, here we go.")

There is a magical land wherein lies a certain landmark. This landmark (called henceforth the Spire) has the curious property of distorting time about itself, according to the function [imath]\tau = C r^{-n} + 1[/imath], where C and n are constants (in practice, C ≈ 11.5 and n ≈ 0.486) and r is the distance away from the Spire (i.e. the length of a circular arc concentric with the planet upon which the Spire is located drawn from the point where the Spire touches the planet's surface to a line passing through the Earth's surface and the point in question... or the 'map distance'; not taking the z-coordinate into account), and [imath]\tau[/imath] is the coefficient of temporal distortion with respect to "normal" time - which itself obviously does not exist in this world, due to the asymptotic nature of the temporal distortion function.

In case I need to explain what is meant by 'temporal distortion',
Spoiler:
suppose that you lived in a place far, far away from the Spire. If you had access to a watch, and looked up at the sun to track it, you'd notice that it came up about once every twenty-four hours - ignoring, of course, seasonal shifts. Now, suppose that you moved to a place much closer to the Spire. Here, if you kept track of the Sun, you'd notice that it was moving much faster than it used to - twice the speed, three times, or even faster. If you had a long-range viewing device of some kind, and watched your friends far away from the Spire, their actions would seem to you to be greatly sped up, and your actions, to them, would seem to be greatly slowed down.

Now. Suppose you want to travel from point A to point B in this world. Due to the distortion field, you might not want to travel in a straight line, like you would were all else equal. If point B is directly opposite point A from the Spire, for example, then walking in a straight line would actually take forever. The heat death of the universe would occur before you reached the Spire itself (though, as the old joke goes, you'd probably get close enough for all practical purposes)!

My question is this. Assuming a constant rate of travel, how can I find the equation of the path which will deliver our wayward traveler to his destination in the shortest amount of time (with respect, obviously, to the baseline [imath]\tau = 1[/imath])?

I have discovered - barring any error on my part, and I'd appreciate a second opinion on this - that if one takes a path defined by the vector-valued function [imath]r(t) = x(t)\hat{i} + y(t)\hat{j}[/imath], the total time taken from t = a to t = b will be equal to

$\int_a^b \left[ C \left( \left[ x(t) \right]^2 + \left[ y(t) \right]^2 \right)^{ - \frac{n}{2} } + 1 \right ] \sqrt{ \left[ x'(t) \right]^2 + \left[ y'(t) \right]^2 } \,\mathrm{d}t$

So I just need a function [imath]r(t)[/imath] that minimizes that.

I'd be grateful for any help, from full solutions to research recommendations.
Last edited by Aedl Foxe on Sun Feb 05, 2012 3:27 am UTC, edited 2 times in total.
"Maybe there are stupid happy people out there... And life isn't fair, and you won't become happier by being jealous of what you can't have... You can never achieve that degree of ignorance... you cannot unknow what you know." -E. Yudkowsky

Qaanol
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### Re: Temporal distortion, geodesics(?), and My Little Pony

You want to look into the calculus of variations.
wee free kings

Aedl Foxe
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### Re: Temporal distortion, geodesics(?), and My Little Pony

Functionals are often formed as definite integrals involving unknown functions and their derivatives. The interest is in extremal functions that make the functional attain a maximum or minimum value – or stationary functions – those where the rate of change of the functional is precisely zero.
Perhaps the simplest example of such a problem is to find the curve of shortest length, or geodesic, connecting two points.

..... yup, I do. Thanks. xP
"Maybe there are stupid happy people out there... And life isn't fair, and you won't become happier by being jealous of what you can't have... You can never achieve that degree of ignorance... you cannot unknow what you know." -E. Yudkowsky

Aedl Foxe
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### Re: Temporal distortion, geodesics(?), and My Little Pony

I would appreciate, were it possible, an example of the usage of the Euler-Lagrange equation on a function similar to the one stated above. The generic example given in Wikipedia is a bit over my head.
"Maybe there are stupid happy people out there... And life isn't fair, and you won't become happier by being jealous of what you can't have... You can never achieve that degree of ignorance... you cannot unknow what you know." -E. Yudkowsky

z4lis
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### Re: Temporal distortion, geodesics(?), and My Little Pony

I never had much experience with calculus of variations, and it's been a long time since I've done anything with it, but I believe you want to do something like the following:

1) Define F[p] to be that integral you have over a path p. Now, we want a path that minimizes this function, so if we could somehow compute F'[p], we would want F'[p] = 0.

2) Let c be a path that starts and ends at 0. Then c + p starts and ends at the same places p does, so we can define a function f(h) = F[p+hc]. Here, h is just a small real number. We're just slightly perturbing the original path p by hc. If p were a minimizer, then f(h) would have a local minimum at h=0, and so f'(0) = 0.

3) Compute f' for an arbitrary path, set it equal to 0, and then hope that the information given by that equation helps you figure out some conditions on p.

I'm too busy/lazy to do that calculation, but if you want to post results then people might be able to help you figure out how to move forward or what properties p needs to have to satisfy the equations you cook up.

Amusing side note: You could actually experimentally determine the paths by building a complicated apparatus to have the proper index of refraction corresponding to the modified time (velocity) and then shining light beams through it. It turns out light is way better than we are at the calculus of variations.

Edit: Have you considered switching to polar coordinates?
What they (mathematicians) define as interesting depends on their particular field of study; mathematical anaylsts find pain and extreme confusion interesting, whereas geometers are interested in beauty.

Timefly
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### Re: Temporal distortion, geodesics(?), and My Little Pony

I fail to see what this has to do with 'My Little Pony'.

legend
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### Re: Temporal distortion, geodesics(?), and My Little Pony

First of all, I'd definitely use polar coordinates here. Also because we're not interested in the parametrisation of the path, just it's shape, you can use [imath]\phi(r)[/imath] as the function you're looking for, instead of [imath]\phi(t),r(t)[/imath].
Those two things will simplify the problem quite a bit. Now let us define:
$L[\phi',\phi,r]:=\tau\sqrt{(r\phi')^2+1}$
So the equation we're solving is just:
$\delta \int L dr=0$
Now this is a classical calculus of variations problem (the wikipedia article on the subject, linked above is quite good imo) and we can apply the Euler-Lagrange equation:
$\frac{\mathrm{d} }{\mathrm{d} r}\frac{\partial L}{\partial \phi'}=\frac{\partial L}{\partial \phi}$
The important thing here is that we're looking at [imath]\phi',\phi,r[/imath] as independant variables.
In our case this gives us:
$\frac{\mathrm{d} }{\mathrm{d} r}\frac{\tau\phi'r^2}{\sqrt{\phi'^2r^2+1}}=0$.
This can be integrated once, solved for [imath]\phi'[/imath] and integrated again. (Although you probably won't be able to exactly solve the second integral in the general case.)

Aedl Foxe
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### Re: Temporal distortion, geodesics(?), and My Little Pony

z4lis wrote:I never had much experience with calculus of variations, and it's been a long time since I've done anything with it, but I believe you want to do something like the following:

1) Define F[p] to be that integral you have over a path p. Now, we want a path that minimizes this function, so if we could somehow compute F'[p], we would want F'[p] = 0.

2) Let c be a path that starts and ends at 0. Then c + p starts and ends at the same places p does, so we can define a function f(h) = F[p+hc]. Here, h is just a small real number. We're just slightly perturbing the original path p by hc. If p were a minimizer, then f(h) would have a local minimum at h=0, and so f'(0) = 0.

3) Compute f' for an arbitrary path, set it equal to 0, and then hope that the information given by that equation helps you figure out some conditions on p.

I'm too busy/lazy to do that calculation, but if you want to post results then people might be able to help you figure out how to move forward or what properties p needs to have to satisfy the equations you cook up.

Amusing side note: You could actually experimentally determine the paths by building a complicated apparatus to have the proper index of refraction corresponding to the modified time (velocity) and then shining light beams through it. It turns out light is way better than we are at the calculus of variations.

Edit: Have you considered switching to polar coordinates?

Or I could use A* to do the same thing for an arbitrary precision.

I've considered switching to polar coordinates, but I'd still have to solve the Euler-Lagrange equation on a functional of two functions of t; they'd just be r and theta. Except theta wouldn't affect the... oh. Oh! Oh. I might try that. Thanks!

And if you don't see what this has to do with My Little Pony, then this has nothing to do with My Little Pony and I added that as an attention-getter. If you do, then shhh. [](/b10)

EDIT: I've found a primer on the use of the Euler-Lagrange equation; I'll try to work through it on my own, and post back here if I get stuck. Thanks!
"Maybe there are stupid happy people out there... And life isn't fair, and you won't become happier by being jealous of what you can't have... You can never achieve that degree of ignorance... you cannot unknow what you know." -E. Yudkowsky

Aedl Foxe
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### Re: Temporal distortion, geodesics(?), and My Little Pony

legend wrote:First of all, I'd definitely use polar coordinates here. Also because we're not interested in the parametrisation of the path, just it's shape, you can use [imath]\phi(r)[/imath] as the function you're looking for, instead of [imath]\phi(t),r(t)[/imath].
Those two things will simplify the problem quite a bit. Now let us define:
$L[\phi',\phi,r]:=\tau\sqrt{(r\phi')^2+1}$
So the equation we're solving is just:
$\delta \int L dr=0$
Now this is a classical calculus of variations problem (the wikipedia article on the subject, linked above is quite good imo) and we can apply the Euler-Lagrange equation:
$\frac{\mathrm{d} }{\mathrm{d} r}\frac{\partial L}{\partial \phi'}=\frac{\partial L}{\partial \phi}$
The important thing here is that we're looking at [imath]\phi',\phi,r[/imath] as independant variables.
In our case this gives us:
$\frac{\mathrm{d} }{\mathrm{d} r}\frac{\tau\phi'r^2}{\sqrt{\phi'^2r^2+1}}=0$.
This can be integrated once, solved for [imath]\phi'[/imath] and integrated again. (Although you probably won't be able to exactly solve the second integral in the general case.)

I didn't see your post before, sorry! x3 Thanks a lot for the step-by-step. I'll walk through it, make sure I understand it.
"Maybe there are stupid happy people out there... And life isn't fair, and you won't become happier by being jealous of what you can't have... You can never achieve that degree of ignorance... you cannot unknow what you know." -E. Yudkowsky

tomtom2357
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### bounds on the correct path

Okay, how about we make C and n equal to 1 and a and b initially a distance of 1 away from the spire (and opposite each other), and try to figure that out first. Also, I am going to assume that the path is in a plane, so that nothing weird happens. My first thought is to try a circle path. This gets from a to b in time 2*pi, so anything that takes longer than that is a waste of time. Therefore, the optimal path must pass through only the points C such that |AC|+|BC|<=2*pi, this defines an ellipse with the spire at its center. Therefore, the optimal path must not pass through a point that is more than pi away from the spire. Now we know that time must always be slowed down by a factor of 1+(1/pi), or about 1.318. We can now use this to improve our estimate of the optimal path. Iterating this procedure, I have found that the optimal path can only pass through points C such that |SC|<pi-1 (S is the spire), and therefore that the temporal distortion is always at least 1+(1/(pi-1)). I hope that you can use this to find the optimal solution.

Edit: actually, you can also use my method to find the maximum temporal distortion (and therefore minimum distance from the spire) allowed without taking more time than my (not necessarily optimal) solution,and so the minimum distance from the spire is W(exp(2-pi)) (W is the lambert W function, and the value is approximately 0.2489), and therefore the maximum temporal distortion is 1+(1/(W(exp(2-pi)))), or about 5.0171. Using the bounds, I have calculated that any path will take longer than 2*(1+(1/(pi-1)))*(sqrt(1+(W(exp(2-pi)))^2), which is a tad hairy, but numerically this is about 3.0234.

Edit 2: Actually, I haven't yet found a path that goes the distance from a to b faster than my semi-circle, so I challenge anybody on the forum to find a more optimal path.
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antonfire
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### Re: Temporal distortion, geodesics(?), and My Little Pony

It's easy to check whether your semicircle path satisfies the Euler-Lagrange equation. If it does, it's probably a unique solution by differential equations stuff, so it's optimal. (Well, you have to be a bit more careful than that, but whatever.) If it doesn't, some nearby paths are shorter and it's not hard to construct one.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

Aedl Foxe
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### Re: bounds on the correct path

tomtom2357 wrote:Okay, how about we make C and n equal to 1 and a and b initially a distance of 1 away from the spire (and opposite each other), and try to figure that out first. Also, I am going to assume that the path is in a plane, so that nothing weird happens. My first thought is to try a circle path. This gets from a to b in time 2*pi, so anything that takes longer than that is a waste of time. Therefore, the optimal path must pass through only the points C such that |AC|+|BC|<=2*pi, this defines an ellipse with the spire at its center. Therefore, the optimal path must not pass through a point that is more than pi away from the spire. Now we know that time must always be slowed down by a factor of 1+(1/pi), or about 1.318. We can now use this to improve our estimate of the optimal path. Iterating this procedure, I have found that the optimal path can only pass through points C such that |SC|<pi-1 (S is the spire), and therefore that the temporal distortion is always at least 1+(1/(pi-1)). I hope that you can use this to find the optimal solution.

Edit: actually, you can also use my method to find the maximum temporal distortion (and therefore minimum distance from the spire) allowed without taking more time than my (not necessarily optimal) solution,and so the minimum distance from the spire is W(exp(2-pi)) (W is the lambert W function, and the value is approximately 0.2489), and therefore the maximum temporal distortion is 1+(1/(W(exp(2-pi)))), or about 5.0171. Using the bounds, I have calculated that any path will take longer than 2*(1+(1/(pi-1)))*(sqrt(1+(W(exp(2-pi)))^2), which is a tad hairy, but numerically this is about 3.0234.

Edit 2: Actually, I haven't yet found a path that goes the distance from a to b faster than my semi-circle, so I challenge anybody on the forum to find a more optimal path.

I had in fact noticed that a path which leaves the distance constant was likely to be the shortest path between two points equidistant from the Spire, but I hadn't proven it. I also had been working under the assumption that this could be generalized to any two points by letting the distance from the Spire vary directly with the change in theta (thus tracing out a portion of an Archimedean spiral), but again, I hadn't proven anything. But thank you for your contribution.
"Maybe there are stupid happy people out there... And life isn't fair, and you won't become happier by being jealous of what you can't have... You can never achieve that degree of ignorance... you cannot unknow what you know." -E. Yudkowsky

Aedl Foxe
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### Re: Temporal distortion, geodesics(?), and My Little Pony

legend wrote:First of all, I'd definitely use polar coordinates here. Also because we're not interested in the parametrisation of the path, just it's shape, you can use [imath]\phi(r)[/imath] as the function you're looking for, instead of [imath]\phi(t),r(t)[/imath].
Those two things will simplify the problem quite a bit. Now let us define:
$L[\phi',\phi,r]:=\tau\sqrt{(r\phi')^2+1}$
So the equation we're solving is just:
$\delta \int L dr=0$
Now this is a classical calculus of variations problem (the wikipedia article on the subject, linked above is quite good imo) and we can apply the Euler-Lagrange equation:
$\frac{\mathrm{d} }{\mathrm{d} r}\frac{\partial L}{\partial \phi'}=\frac{\partial L}{\partial \phi}$
The important thing here is that we're looking at [imath]\phi',\phi,r[/imath] as independant variables.
In our case this gives us:
$\frac{\mathrm{d} }{\mathrm{d} r}\frac{\tau\phi'r^2}{\sqrt{\phi'^2r^2+1}}=0$.
This can be integrated once, solved for [imath]\phi'[/imath] and integrated again. (Although you probably won't be able to exactly solve the second integral in the general case.)

Forgive me, but I'm a bit confused. If I'm using polar coordinates, then I want my path to be expressed in the form [imath]r(\theta)[/imath], don't I? So then I would construct my functional as [imath]L[\theta, r, r'][/imath]... am I incorrect in this assumption?

As I understand the problem, I'm trying to find the function [imath]r(\theta)[/imath] such that the line integral of [imath]C r^{-n} + 1[/imath] with respect to [imath]r(\theta)[/imath] is minimized. I am assuming (danger signs! No idea if this is correct!) that the equation for solving a line integral translated without trouble into polar coordinates, in which case the integral to be minimized would be
$\int_{a}^{b} \left[ C \left[ r(t) \right]^{-n} + 1 \right] \sqrt{ \left[ r'(t) \right]^2 + \left[ \theta'(t) \right]^2 } \,\mathrm{d}t$
which simplifies, since [imath]\theta[/imath] is equal to t, to
$\int_{a}^{b} \left[ C \left[ r(\theta) \right]^{-n} + 1 \right] \sqrt{ \left[ r'(\theta) \right]^2 + 1 } \,\mathrm{d}\theta$

Then I would use that integrand as my function for the Euler-Lagrange equation, right? Or actually, since [imath]\theta[/imath] doesn't actually affect the function, I would use the Beltrami identity.
"Maybe there are stupid happy people out there... And life isn't fair, and you won't become happier by being jealous of what you can't have... You can never achieve that degree of ignorance... you cannot unknow what you know." -E. Yudkowsky

legend
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### Re: Temporal distortion, geodesics(?), and My Little Pony

Using [imath]r(\theta)[/imath] makes indeed somewhat more sense than my approach. The only reason I did it the other way around is that my gut feeling told me the equations would get simpler that way (although I was too lazy to really calculate it through ). However in your approach you can indeed make use of Beltrami identity, which should make your life somewhat easier.
The only thing I have to object in your calculation so far is that the line segment formula in polar coordinates is
$ds^2=dr^2+(r d\theta)^2$
You can intuitively understand the extra r because the further away from the origin you get the larger your displacement gets for the same change of the angle. (Depending on how much you know about calculus/coordinate transformations it might be a good exercise to try deriving this yourself. I think understanding this gives you a much better understanding of the polar coordinates in general.)
In your case this means that in your final integral you have [imath]\sqrt{r'^2+r^2}[/imath].

Aedl Foxe
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### Re: Temporal distortion, geodesics(?), and My Little Pony

legend wrote:Using [imath]r(\theta)[/imath] makes indeed somewhat more sense than my approach. The only reason I did it the other way around is that my gut feeling told me the equations would get simpler that way (although I was too lazy to really calculate it through ). However in your approach you can indeed make use of Beltrami identity, which should make your life somewhat easier.
The only thing I have to object in your calculation so far is that the line segment formula in polar coordinates is
$ds^2=dr^2+(r d\theta)^2$
You can intuitively understand the extra r because the further away from the origin you get the larger your displacement gets for the same change of the angle. (Depending on how much you know about calculus/coordinate transformations it might be a good exercise to try deriving this yourself. I think understanding this gives you a much better understanding of the polar coordinates in general.)
In your case this means that in your final integral you have [imath]\sqrt{r'^2+r^2}[/imath].

So all told, the integral becomes
$\int^{a}_{b} \left[ C \left[ r(\theta) \right]^{-n} + 1 \right] \sqrt{\left[ r(\theta) \right]^2 + \left[ r'(\theta) \right]^2}\,\mathrm{d}\theta$

Is that what you're saying? And thanks for the correction; I'll try deriving that at a later time. I understand polar coordinates, I think, but not so much the precise isomorphism between the two ideas... hence naive mistakes like the one above. x3

If my understanding is correct, I'll proceed from that point.

EDIT: Procedure from that point:

Define:

$f := \left[ C r^{-n} + 1 \right] \sqrt{r^2 + r'^2}$

Therefore (letting K stand for Konstant, since we're already using C), by the Beltrami Identity:

$f - r' \frac{\partial f}{\partial r'} = K$

Finding the partial derivative:

\begin{align} \frac{\partial}{\partial r'} \left[ \left[ C r^{-n} + 1 \right] \sqrt{r^2 + r'^2} \right] &= \left[ C r^{-n} + 1 \right] \frac{\partial}{\partial r'} \left[ \sqrt{r^2 + r'^2} \right] \\ {} &= \left[ C r^{-n} + 1 \right] \frac{1}{2} \left[ r^2 + r'^2 \right]^{-\frac{1}{2}} 2 r' \\ {} &= \frac{\left[ C r^{-n} + 1 \right] r'}{\sqrt{r^2 + r'^2}} \end{align}

Therefore our equation becomes:

$\left[ C r^{-n} + 1 \right] \sqrt{r^2 + r'^2} - \frac{\left[ C r^{-n} + 1 \right] r'^2}{\sqrt{r^2 + r'^2}} = K$

or, nicely enough,

$\frac{\left[ C r^{-n} + 1 \right] r^2}{\sqrt{r^2 + r'^2}} = K$

Now I just have to solve the differential equation! Of course, I haven't taken Diff. Eq. yet, so it's time to hit Khan Academy. >>; Finding analytic solutions to nonlinear differential equations isn't difficult, right?
"Maybe there are stupid happy people out there... And life isn't fair, and you won't become happier by being jealous of what you can't have... You can never achieve that degree of ignorance... you cannot unknow what you know." -E. Yudkowsky

Sagekilla
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### Re: Temporal distortion, geodesics(?), and My Little Pony

Aedl Foxe wrote:Now I just have to solve the differential equation! Of course, I haven't taken Diff. Eq. yet, so it's time to hit Khan Academy. >>; Finding analytic solutions to nonlinear differential equations isn't difficult, right?

Sure, if you don't mind solving it in terms of a power series. Generally speaking these things are extremely difficult to do, especially with the one you have.
http://en.wikipedia.org/wiki/DSV_Alvin#Sinking wrote:Researchers found a cheese sandwich which exhibited no visible signs of decomposition, and was in fact eaten.

Aedl Foxe
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### Re: Temporal distortion, geodesics(?), and My Little Pony

Sagekilla wrote:Sure, if you don't mind solving it in terms of a power series. Generally speaking these things are extremely difficult to do, especially with the one you have.

Oh, good. x x Such challenges I set for myself.

EDIT: Oh! But upon observation of the equation I notice that r = C is a solution! So tomtom2357, your intuition was correct: a path which keeps the distance from the Spire constant is in fact the optimal solution when the two points are equidistant from the solution. Kudos to you.
"Maybe there are stupid happy people out there... And life isn't fair, and you won't become happier by being jealous of what you can't have... You can never achieve that degree of ignorance... you cannot unknow what you know." -E. Yudkowsky

Aedl Foxe
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### Re: Temporal distortion, geodesics(?), and My Little Pony

$r(\theta) = A*\int \sqrt{r(\theta)^4 + 23 r(\theta)^{3.514} + 132.25 r(\theta)^{3.028} - B r(\theta)^2}\,\mathrm{d}\theta$

See, this is why I don't like treadmills. - -
"Maybe there are stupid happy people out there... And life isn't fair, and you won't become happier by being jealous of what you can't have... You can never achieve that degree of ignorance... you cannot unknow what you know." -E. Yudkowsky

mfb
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### Re: Temporal distortion, geodesics(?), and My Little Pony

Aedl Foxe wrote:
Sagekilla wrote:Sure, if you don't mind solving it in terms of a power series. Generally speaking these things are extremely difficult to do, especially with the one you have.

Oh, good. x x Such challenges I set for myself.

EDIT: Oh! But upon observation of the equation I notice that r = C is a solution! So tomtom2357, your intuition was correct: a path which keeps the distance from the Spire constant is in fact the optimal solution when the two points are equidistant from the solution. Kudos to you.

Only for that specific r=C. It should be the r where going around the spire (full circle) is the quickest.
For equidistant points with r>C, the ideal path will between them be bent towards the spire (the path will have a minimal distance <r from it).
For equidistant points with r<C, the ideal path will between them be bent away from the spire (the path will have a maximal distance >r from it).

Aedl Foxe
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### Re: Temporal distortion, geodesics(?), and My Little Pony

mfb wrote:
Aedl Foxe wrote:
Sagekilla wrote:Sure, if you don't mind solving it in terms of a power series. Generally speaking these things are extremely difficult to do, especially with the one you have.

Oh, good. x x Such challenges I set for myself.

EDIT: Oh! But upon observation of the equation I notice that r = C is a solution! So tomtom2357, your intuition was correct: a path which keeps the distance from the Spire constant is in fact the optimal solution when the two points are equidistant from the solution. Kudos to you.

Only for that specific r=C. It should be the r where going around the spire (full circle) is the quickest.
For equidistant points with r>C, the ideal path will between them be bent towards the spire (the path will have a minimal distance <r from it).
For equidistant points with r<C, the ideal path will between them be bent away from the spire (the path will have a maximal distance >r from it).

Umm... no. r = K is a solution to the differential equation, where K is any constant. Check the math.
"Maybe there are stupid happy people out there... And life isn't fair, and you won't become happier by being jealous of what you can't have... You can never achieve that degree of ignorance... you cannot unknow what you know." -E. Yudkowsky

mfb
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### Re: Temporal distortion, geodesics(?), and My Little Pony

In the limit of C->0, the shortest path is a straight line.
The paths are continuous with C (unless I am really wrong), so for very small C (and an angular difference not close to pi), the direct path is close to a straight line.
Check the limits.

Aedl Foxe
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### Re: Temporal distortion, geodesics(?), and My Little Pony

mfb wrote:In the limit of C->0, the shortest path is a straight line.
The paths are continuous with C (unless I am really wrong), so for very small C (and an angular difference not close to pi), the direct path is close to a straight line.
Check the limits.

What limit? I'm sorry, but you're not making any sense. Are you sure you understand the problem?
"Maybe there are stupid happy people out there... And life isn't fair, and you won't become happier by being jealous of what you can't have... You can never achieve that degree of ignorance... you cannot unknow what you know." -E. Yudkowsky

legend
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### Re: Temporal distortion, geodesics(?), and My Little Pony

The problem is that your equation is in general form and should be valid for all possible values of C and n. If you set C=0 (or n=0) the spire "disappears" and the solution should obviously be a straight line, not a circle. This is strong indication that the r=const. isn't actually a solution.
My guess here would be that by using Beltrami identity you indirectly assume that your r' isn't zero.
Another remarkable thing here is that theta=const. should actually be a solution, but we're missing it, because it can't be written as r(theta).

Aedl Foxe
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### Re: Temporal distortion, geodesics(?), and My Little Pony

legend wrote:The problem is that your equation is in general form and should be valid for all possible values of C and n. If you set C=0 (or n=0) the spire "disappears" and the solution should obviously be a straight line, not a circle. This is strong indication that the r=const. isn't actually a solution.
My guess here would be that by using Beltrami identity you indirectly assume that your r' isn't zero.
Another remarkable thing here is that theta=const. should actually be a solution, but we're missing it, because it can't be written as r(theta).

Oh. Oh! I apologize, mfb; in retrospect your observation is perfectly clear.

..... that's puzzling.
"Maybe there are stupid happy people out there... And life isn't fair, and you won't become happier by being jealous of what you can't have... You can never achieve that degree of ignorance... you cannot unknow what you know." -E. Yudkowsky

Qaanol
The Cheshirest Catamount
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### Re: Temporal distortion, geodesics(?), and My Little Pony

The Euler-Lagrange equation provides a way to find functions for which the integral in question is locally stable. These can be maxima, minima, or inflection points in the space of possible function. In other words, the critical points. Further analysis is needed to identify actual maxima and minima, as well as to distinguish between local and global extrema.
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