Fourth Dimension Calculation?

[jinydu]

I checked your calculations of the acceleration at the surface of the 4D planet using a different method, and they do seem correct.

However, I'm not sure about the value of G being higher by a factor of 3pi/8. This would also increase the mass of the planet by 3pi/8, and hence end up increasing the acceleration at the surface of the planet by (3pi/8)^2, so not everything to do with gravity would increase by 3pi/8.

Bug Detected: For some reason, "[eight])" keeps getting replaced by a smilie.

Edit by BobXP: I fixed your "bug". Edit or quote to see how.

[pat]

I've started working more formally on the calculations that I attempted above. Rather than cram them into BBCode and HTML, I have been using LaTeX. You can check out my start here: http://www.csh.rit.edu/~pat/math/quickies/orbit4/orbit4.pdf

I'm to a point where I cannot find any way to simplify the equation. It doesn't immediately appear to me that the only solution is a circle. But, it is immediately apparent that there are circles which are solutions.

And, for the record, I'm male... but am not at all concerned with being thought female.... it's all fine... it's all good....

[PWrong]

I checked your calculations of the acceleration at the surface of the 4D planet using a different method, and they do seem correct.

Thanks.

However, I'm not sure about the value of G being higher by a factor of 3pi/8. This would also increase the mass of the planet by 3pi/8, and hence end up increasing the acceleration at the surface of the planet by (3pi/8)^2, so not everything to do with gravity would increase by 3pi/8.

Hey you're right, I got it all wrong . The pi*3/8 is just part of the extra mass. I was a bit too eager to invent a new constant.

Bug Detected: For some reason, "[eight])" keeps getting replaced by a smilie.

It's supposed to do that. An 8 next to a bracket ) looks like a sideways smily with sunglasses. Emoticons are always ruining my equations.

[Geosphere]

Emoticons are always ruining my equations.

Then be sure to check that little "disable smilies in my post" box above the 'submit' button.

[PWrong]

I've started working more formally on the calculations that I attempted above. Rather than cram them into BBCode and HTML, I have been using LaTeX. You can check out my start here: http://www.csh.rit.edu/~pat/math/quickies/orbit4/orbit4.pdf

I'm to a point where I cannot find any way to simplify the equation. It doesn't immediately appear to me that the only solution is a circle. But, it is immediately apparent that there are circles which are solutions.

Wow, tricky. I printed it out and had a go at it, but I don't think I can help. I can follow most of the reasoning, although I had to look up angular momentum. I can see how it would work with an extra r(t), or if it was a circle. I also worked backwards from the end and got back to the correct start. That's about as much as I can do.

I'm just wondering though: Is the ending you came to the same as the 3D equivalent except the r(t) doesn't cancel out? It seems like it should be, but your method is a bit different from the one on the Max Francis webpage.

Also, how come you don't mention the words like perihilion or use e for eccentricity anywhere? Would it help if you did?

[Keiji]

I've fixed the smiley bug

[pat]

I'm just wondering though: Is the ending you came to the same as the 3D equivalent except the r(t) doesn't cancel out?

That's correct. In the 3D case, the r(t) cancels. In the 4D case, there's an extra one.

It seems like it should be, but your method is a bit different from the one on the Max Francis webpage.

Hmm... I tried to follow along line-for-line, equation-for-equation, with his page. I'm not sure where I depart. My section headings match his page titles. Most of my commentary parallels his.

Also, how come you don't mention the words like perihilion or use e for eccentricity anywhere? Would it help if you did?

All of those words come into his development after he integrates the equation that I cannot integrate (because of the extra r(t)). I haven't tried skipping ahead in his development and seeing if I can get back to where I am. Maybe it will help. And, maybe ellipses will still work (though it seems doubtful).

[mghtymoop]

Integrate the whole thing, it is really that simple, don't remove variables because you believe they are over time because with time being the 4th dimension this makes no sence, only problem is you are left with unsloved constants but this gets you a lot closer to your goal

[pat]

Integrate the whole thing, it is really that simple, don't remove variables because you believe they are over time because with time being the 4th dimension this makes no sence, only problem is you are left with unsloved constants but this gets you a lot closer to your goal

The integrand on the left hand side is: r(t)^d²r/_dt². But, the bold r is a vector, the italic r its magnitude. I don't know how to integrate that. It's basically like trying to integrate f(t)g''(t). It's not as simple as F(t)g'(t). Am I missing some nice way to write things?

[pat]

The integrand on the left hand side is: r(t)^d²r/_dt². But, the bold r is a vector, the italic r its magnitude. I don't know how to integrate that. It's basically like trying to integrate f(t)g''(t). It's not as simple as F(t)g'(t). Am I missing some nice way to write things?

More exactly, what is the integral of:
sqrt{ x(t)² + y(t)² + z(t)</sup>2</sup> + w(t)² } x''(t)

If it were only x'(t) instead of x''(t), then I could make a decent stab at it. But, as it isn't.... I'm confused. I should probably go back and re-read my calculus books.... it's been 14 years.

[PWrong]

Oh, now I can help!
You can do it with integration by parts:

"the integral of [ f(x)g'(x) ] = f(x)g(x) - integral[ f'(x)g(x) ]"
It works for 2nd derivatives as well:
"the integral of [ f(x)g''(x) ] = f(x)g'(x) - integral[ f'(x)g'(x) ]"

Using this rule:
integral[r(t)r''(t)] = r(t)r'(t) - integral[r'(t)r'(t)]

So you can substitute that into the last equation. I tried to go on further, but I got stuck pretty quickly. It looks like you might have to integrate it all again. Anyway, I hope that helped

[pat]

Oh, now I can help!
You can do it with integration by parts

Aha! Thanks for the pointer. I had tried playing some other variants out with integration by parts. But, it kept getting messy on me. That looks pretty good though. I'll take a look at that.

Following the derivation in my college calculus book, I came up with:
^d²p/_dθ = - α² p

Where p = (1/r) and α = √ ( K² + GM ) where K is a constant, G is the universal gravitational constant, and M is the mass of the sun.

But, that doesn't make for any stable orbits since the solution to that differential equation is:
p(θ) = a * sin( α θ + θ₀ )
which would mean that p becomes zero twice each orbit. p being zero means that the planet is infinitely far from the sun. Hmmm...

But, I'll get back to work on the other derivation with the integration by parts.

[PWrong]

Following the derivation in my college calculus book, I came up with:
^d²p/_dθ = - α² p

Hey, that's the formula for Simple Harmonic Motion. I just leant that today in calc, although we only do it in one dimension. In this case, it looks like p is in SHM with respect to θ, which is a bit different from ordinary SHM.

Oooh, I think I just found the mistake. Normally, an object oscillates around the origin. It usually makes perfect sense to put the origin at 0. But this would make the inverse of the radius oscillate around 0 as θ increases, which doesn't make sense, and it leads to infinities.

But, that doesn't make for any stable orbits since the solution to that differential equation is:
p(θ) = a * sin( α θ + θ₀ )

My calc book has the same solution, assuming the origin is at zero.
But it shouldn't have to be at zero. So when we integrate the equation, we have to introduce the constant. How wonderful, it's everybody's favourite careless error, the constant of integration! Ah, the irony. Anyway, I think that might work.

The solution to the equation should be:
p(θ) = a * sin( αθ + θ₀ ) + c
To keep p above zero, we need a minimum value for c.

0 = a * sin( αθ - k ) + c where k is such that the right hand side is at the minimum.

So, k = -pi/2
0 = a * sin( αθ - pi/2 ) + c
0 = -a * cos( αθ ) + c

since a is the amplitude, we should have the following information.
p oscillates between c - a and c + a. c is the centre.

We also have the formula for p(θ):
p(θ) = a * sin( αθ + θ₀ ) + c
which we can turn into a formula for r(θ):
r(θ) = 1 / (a * sin( αθ + θ₀ ) + c)

I've graphed some functions like these, and they work as I expected. The radius won't escape to infinity as long as c>a. The graph of y = r(x) has a few short peaks in it. The greater the difference between c and a , the smaller the peaks are.

The surprising part is when you graph it in polar form. It looks like most 4D orbits are not elliptical, but "flowery"! The peaks in the radius make the graph look like a flower with petals. The number of petals is approximately equal to α:

I've probably made a mistake, but I hope not. This is just too ridiculous to be wrong. Oh well, if it is wrong, it's still funny, and it shouldn't take long to find the mistake.

[PWrong]

, as soon as I posted that I worked out what's wrong with it. You can get the same patterns with a sin function instead of a cosec function. So there must be a mistake somewhere. Oh well, the first bit might be right. I'm up too late anyway, so I have an excuse for making bad assumptions. *yawn*

[pat]

The secant doesn't work because it's second derivative isn't proporational to itself.

Also, x(t) = a sin( bt + c ) + d has second derivative: - a b² sin( bt + c ). But, in order to satisfy the differential equation: ^d²x/_dt² = - b² x, then a must be one and d must be zero. So, it cannot all be offset somewhere. It has to go through zero. Which means that the radius has to go through infinity. Yuck.

[PWrong]

Oh, right. Oh well, so much for my idea. There must be a mistake somewhere though. For one thing, the planet has to travel an infinite distance in a finite time, so there's at least one contradiction there. Oh well, the integration by parts might work out better.

Could you show me how to get from the last equation in your pdf to the SHM formula? I'm curious as to where p came from.

[pat]

Actually, I did have a sign error in calculation in one place. But, the big error that I had was conceptual. I was thinking that equations like:
r = a csc ( b θ + c ) were bad or wrong because the planet would have to go infinitely far away and still come back. But, the error in my thinking was that it would have to come back.

The real trick is that nothing in the above equation requires that θ increase without bound.

With that understanding, I finished up my derivation based upon the derivation in the calculus book: Calculus with Analytic Geometry by John B. Fraleigh (Second Edition, ISBN 0-201-12010-0).

The upshot is this. There are four types of "orbits". There is the very unstable circular orbit (though, technically, this is a special case of the next type). There are orbits where the planet spirals outward or inward a constant distance every revolution. There are orbits where the planet spirals in faster and faster with each revolution. And, there are "orbits" where the sun doesn't quite catch the planet but rather spins it around a few times before sending it off into space.

Full details are here: http://www.csh.rit.edu/~pat/lj/orbit4.pdf

[PWrong]

Aw, that's not fair. You've done all that work and you just got the same result as everyone else. Doesn't it violate all the laws that you've already proven?

For instance, if the planet spirals inward, how can it sweep out equal areas in equal times? Also, what's the difference between spinning around a few times then flying off, and simply spiralling outwards? :?

I haven't had a chance to go through the whole thing entirely, but I have a couple of questions.

1. Could you explain how you came to equation 6.3? You suddenly introduce e into the equation. It seems like a huge jump to make. Is that the maths constant e, or a variable? I can almost

2. Why is it that the equations for Beta=0, Beta>0 and Beta<0 are all completely different?

3. The graph of equation 6.4 is an ugly star-shaped orbit. Why is that?

[pat]

Doesn't it violate all the laws that you've already proven?

I don't think so, for two reasons... first, my previous calculations had a sign error... second... the only orbit that would last forever is a circle (of course, that's unstable, so there would have to be only two bodies in the Universe).

For instance, if the planet spirals inward, how can it sweep out equal areas in equal times?

I'm not sure it does. I don't think I've claimed that. But, I'm pretty sure that it's true. It can do that by spinning around the sun faster as it gets closer.

Also, what's the difference between spinning around a few times then flying off, and simply spiralling outwards? :?

In the "simply spiralling outwards" case, I was saying that it would get a constant distance further from the sun with each revolution. In the "flying off" case, it approaches the sun for a few revolutions and then starts getting further and further at a huge rate (per revolution). And, in the "flying off" case, there is an asymptote, I believe where its rate of revolution around the sun has slowed so tremendously that (while its angle will always increase, it will never pass the asymptote). For all purposes, it has ceased to spiral.

1. Could you explain how you came to equation 6.3? You suddenly introduce e into the equation. It seems like a huge jump to make. Is that the maths constant e, or a variable?

It is the mathematical constant e. The solution to the differential equation: ^d²y/_dx² = b²y is y = a e^bx.

2. Why is it that the equations for Beta=0, Beta>0 and Beta<0 are all completely different?

Because Beta represents the critical balance between whether the gravity of the sun is strong enough to catch the planet. If beta is small, it will be very close to a circular orbit.

3. The graph of equation 6.4 is an ugly star-shaped orbit. Why is that?

I don't think it should be a star-shaped graph. It should definitely go off to infinity. Any one planet will only follow one of those lines from infinity to infinity. The star was unable to keep the planet in orbit. The star's gravity was too weak.

[pat]

I should also note that the three cases for Beta less-than, equal-to, or greater-than zero would all be the same case if I felt like allowing complex exponentials. But, I thought it would be less confusing to those who haven't done much with complex numbers if I stuck to real numbers.

[PWrong]

Hmm. It all seems to make sense now. I see how the differential equation works, although it still seems odd to have the constant e pop up like that.

The graph only looks like a star when you graph the radius against theta. The planet keeps going to infinity and coming back from the other side of the universe to do something else. Now I understand that theta never reaches the asymptote, so the star shape is irrelevant.

In the file, what do you mean by, "this jives with our understanding of constant Beta"?

I'm fairly good with complex numbers, but I can't see why you would use them here.

Well, I guess I'll have to accept that orbits don't work properly in 4D. That's disappointing. Is there any other kind of stable system? We need to develop some kind of sensible universe that Emily can live in, even if we have to adjust things a bit. I think we should consider some of the following possibilities, no matter how far fetched they seem.

1. Perhaps three bodies can make a stable system.

2. We should test Swirl Gyro's idea, and see if 5D orbits work.

3. We could just set things up so that Beta is extremely small, so that planets take longer than the age of the universe to spiral off. It might not always work though.

4. It might turn out too difficult to calculate, but could Einstein's relativistic equations have a different outcome to using Newtonian mechanics? It's a long shot, but maybe 4D general relativity is sufficiently different from 4D gravity to allow elliptical orbits.

5. Maybe we could confine the planets into orbit by wrapping 4D space around some exotic 5D curve.

[pat]

I'm fairly good with complex numbers, but I can't see why you would use them here.

The solution to the differential equation:
^d²y/_dx² = b² y
is, technically:
y = a e^bx
for some complex constant a regardless of whether b² is positive, negative, or zero. That reduces all three cases for β down to one (where the appearance of e may be even more shocking).

As for "jives with our understanding of β"... that constant represents the balance between the angular momentum of the planet and the force of the sun's gravity. If the force of gravity is weak, we'd expect the planet to go flying off somewhere. If the force of gravity is strong, we'd expect the planet to crash into the sun. If the force of gravity is just right, we'd expect a circular orbit.

[jinydu]

Well, I guess I'll have to accept that orbits don't work properly in 4D.

If you had to explain, in one sentence, why orbits are unstable in 4D but not in 3D, what would the sentence be, please? (Sorry, I just graduated from high school. Although math has always been my favorite subject, I haven't learned enough yet to really understand the details.)

[pat]

If you had to explain, in one sentence, why orbits are unstable in 4D but not in 3D, what would the sentence be, please?

Gravity dissipates too much. --shrug-- I will have to try the 2-D case and see if gravity's too strong.

[jinydu]

Gravity dissipates too much. --shrug--

And this cannot be compensated by placing the planets at a shorter distance to the sun? Oh well, I guess I really will have to wait until I complete a few university courses...

[pat]

It's not that gravity is too weak. It's that it falls off too rapidly.

Gravity falls off as the cube of the radius. The momentum is proportional to the square of the radius. Any small change in the radius makes one or the other take over. If you get too close, it's the sun. If you get too far, it's the momentum.

In three-dimensions, gravity falls off as the square of the radius. The momentum is proportional to the square of the radius. If you change the radius, the two things stay in balance.

[RQ]

For B^2y=ae^bx, isn't that assuming x=2?

[pat]

Isn't what assuming that x = 2? But, I don't think so.

Let y = a e^bx . Then, ^dy/_dx = a b e^bx and ^d²y/_dx² = a b² e^bx = b² y.

[pat]

And, we know that if b = θ i where i = √ (-1), then e^bx = e^{θ i x} = cos θ + i sin θ So, we get the harmonic case when b² is negative.

[Geosphere]

Gravity falls off as the cube of the radius.

Is this a fact? I've never heard this. What ratio at the cube of radius?