## Miscellaneous Science Questions

**Moderators:** gmalivuk, Moderators General, Prelates

### Re: RELATIVITY QUESTIONS! (and other common queries)

I recently found out about Hamiltonian mechanics and so have been playing around with it a fair bit. In doing so, I decided to have a go at a relativistic version (although it possibly isn't correct as I am in no way sure that Hamilton's equations should still hold (as I assumed they did) in a relativistic context) by using a relativistic form for T(p).

[math]T(p)=\sqrt{p^2c^2+m^2c^4}[/math]

I checked that this returned the velocity when the derivative with respect to p was taken and it did, correctly (I believe), return the momentum divided by [imath]m\gamma[/imath] with [imath]\gamma[/imath] being given as a function of p.

I then thought I'd have a look at a relativistic harmonic oscillator using [imath]V(x)=\frac{1}{2}kx^2[/imath]

I was able to reduce this to the equation (for some reason, the equations seemed much easier to solve for the momentum than the position):

[math]\ddot p^2 + \frac{p^2\ddot p^2}{m^2c^2} - \frac{k^2p^2}{m^2}=0[/math]

Which bears an obvious resemblance to the classical form with a relativistic correction of [imath]\frac{p^2 \ddot p^2}{m^2c^2}[/imath] which will, as expected, produce large deviations for large momenta as well as high jerk situations and, according to wolfram alpha, the solutions appear to be approximately sinusoidal (which is good as it should approximate the classical sinusoidal harmonic oscillator at low momenta).

That said, I am unable to reduce this to a closed form (and, as the equation is non-linear, it appears that in all likelihood, no such form exists) and, not being familiar with the techniques of differential equations at all, am unable to plot the solutions (I am particularly interested in what happens to the family of [imath]\ddot p(0)=0[/imath] curves as [imath]p(0)[/imath] becomes close to [imath]m\gamma c[/imath].

All of this led to me to wonder the following:

1. Do hamilton's equations still hold for relativistic physics (with an appropriate hamiltonian)?

2. If anyone could give me a clue as to how to plot the solutions to the differential equation (thinking about it physically, I think that it should be triangle wave in the limit of large [imath]p(0)[/imath] and a sine wave in the limit of small [imath]p(0)[/imath]).

3. If anyone had any idea why it might be (or at least seem) easier to solve for momentum than position.

[math]T(p)=\sqrt{p^2c^2+m^2c^4}[/math]

I checked that this returned the velocity when the derivative with respect to p was taken and it did, correctly (I believe), return the momentum divided by [imath]m\gamma[/imath] with [imath]\gamma[/imath] being given as a function of p.

I then thought I'd have a look at a relativistic harmonic oscillator using [imath]V(x)=\frac{1}{2}kx^2[/imath]

I was able to reduce this to the equation (for some reason, the equations seemed much easier to solve for the momentum than the position):

[math]\ddot p^2 + \frac{p^2\ddot p^2}{m^2c^2} - \frac{k^2p^2}{m^2}=0[/math]

Which bears an obvious resemblance to the classical form with a relativistic correction of [imath]\frac{p^2 \ddot p^2}{m^2c^2}[/imath] which will, as expected, produce large deviations for large momenta as well as high jerk situations and, according to wolfram alpha, the solutions appear to be approximately sinusoidal (which is good as it should approximate the classical sinusoidal harmonic oscillator at low momenta).

That said, I am unable to reduce this to a closed form (and, as the equation is non-linear, it appears that in all likelihood, no such form exists) and, not being familiar with the techniques of differential equations at all, am unable to plot the solutions (I am particularly interested in what happens to the family of [imath]\ddot p(0)=0[/imath] curves as [imath]p(0)[/imath] becomes close to [imath]m\gamma c[/imath].

All of this led to me to wonder the following:

1. Do hamilton's equations still hold for relativistic physics (with an appropriate hamiltonian)?

2. If anyone could give me a clue as to how to plot the solutions to the differential equation (thinking about it physically, I think that it should be triangle wave in the limit of large [imath]p(0)[/imath] and a sine wave in the limit of small [imath]p(0)[/imath]).

3. If anyone had any idea why it might be (or at least seem) easier to solve for momentum than position.

Last edited by eSOANEM on Wed Jan 11, 2012 4:49 pm UTC, edited 1 time in total.

my pronouns are they

Magnanimous wrote:(fuck the macrons)

- Carlington
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### Re: RELATIVITY QUESTIONS! (and other common queries)

doogly wrote:At this level people are talking cosmologicly, and there, we are gonna assume the universe is homogenous and isotropic. The same for all points (in space, anyway - expanding in time, and if you have drunk enough kool aid at the GR font to get upset by picking out a preferred time direction, we just imagine that all matter that exists in the world can be treated like a single fluid, and it has a rest frame, so that does make things rather special), and the same in all directions.

So, if you are homogeneous and isotropic, there isn't a lot you can vary. You are going to be a space of constant curvature. Positive, negative or zero. Sphere, hyperbola or flat.

yurell wrote:I suppose the easiest two dimensional analogues are:

- Positive: A sphere
- Zero: A plane
- Negative: A saddle

Picturing curvature in more than two dimensions is ... really difficult for me, so I just stick with what the maths says.

Okay, positive is a sphere, and zero is flat, those two work. The one that I've never understood is negative curvature. If I follow my intuition, it tells me that making the curvature negative should just fold the universe back into a sphere, only this time one with us on the inside. But to us, it should be indistinguishable from the positive curvature sphere. Why is this not the case?

Kewangji: Posdy zwei tosdy osdy oady. Bork bork bork, hoppity syphilis bork.

Eebster the Great: What specifically is moving faster than light in these examples?

doogly: Hands waving furiously.

Eebster the Great: What specifically is moving faster than light in these examples?

doogly: Hands waving furiously.

Please use he/him/his pronouns when referring to me.

- shawnhcorey
**Posts:**42**Joined:**Sun Jan 08, 2012 2:08 pm UTC

### Re: RELATIVITY QUESTIONS! (and other common queries)

Carlington (The Aussie) wrote:Okay, positive is a sphere, and zero is flat, those two work. The one that I've never understood is negative curvature. If I follow my intuition, it tells me that making the curvature negative should just fold the universe back into a sphere, only this time one with us on the inside. But to us, it should be indistinguishable from the positive curvature sphere. Why is this not the case?

Here's a picture of negative curvature. The really tricky part is figuring out how a surface that has negative curvature at every point could curve back onto itself. It doesn't seem possible in three dimensions but might be in four (or more).

- doogly
- Dr. The Juggernaut of Touching Himself
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### Re: RELATIVITY QUESTIONS! (and other common queries)

OK, 2 dimensions.

You're on the surface of a sphere, like the earth. The curvature is +R^2. So you say to yourself, hmmm, what if I just changed my perspective, and declare that I am inside this sphere. What happens to the curvature now? Ah, well the principal curvature along this here latitude direction is now -R! And along this longitude is -R! Negative waaaaait..... The curvature is still +R^2!

In terms of arithmetic, this fact is just due to the product of negative numbers being a positive number.

In terms of geometry, what happens is that you haven't actually *done* anything. The sphere doesn't care where you're standing.

Higher dimensions remain a bit trickier, naturally.

You're on the surface of a sphere, like the earth. The curvature is +R^2. So you say to yourself, hmmm, what if I just changed my perspective, and declare that I am inside this sphere. What happens to the curvature now? Ah, well the principal curvature along this here latitude direction is now -R! And along this longitude is -R! Negative waaaaait..... The curvature is still +R^2!

In terms of arithmetic, this fact is just due to the product of negative numbers being a positive number.

In terms of geometry, what happens is that you haven't actually *done* anything. The sphere doesn't care where you're standing.

Higher dimensions remain a bit trickier, naturally.

LE4dGOLEM: What's a Doug?

Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.

Keep waggling your butt brows Brothers.

Or; Is that your eye butthairs?

Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.

Keep waggling your butt brows Brothers.

Or; Is that your eye butthairs?

- shawnhcorey
**Posts:**42**Joined:**Sun Jan 08, 2012 2:08 pm UTC

### Re: RELATIVITY QUESTIONS! (and other common queries)

2D surfaces have 4 possibilities:

3D surfaces have 8 possibilities:

When you get to 4D surfaces, things get strange:

- ++ (positive curvature)
- +- (negative curvature)
- -+ (negative curvature)
- -- (positive curvature)

3D surfaces have 8 possibilities:

- +++ (positive curvature)
- ++- (negative curvature)
- +-+ (negative curvature)
- +-- (negative curvature)
- -++ (negative curvature)
- -+- (negative curvature)
- --+ (negative curvature)
- --- (positive curvature)

When you get to 4D surfaces, things get strange:

- ++++ (positive curvature)
- +++- (negative curvature)
- ++-+ (negative curvature)
- ++-- (???)
- +-++ (negative curvature)
- +-+- (???)
- +--+ (???)
- +--- (negative curvature)
- -+++ (negative curvature)
- -++- (???)
- -+-+ (???)
- -+-- (negative curvature)
- --++ (???)
- --+- (negative curvature)
- ---+ (negative curvature)
- ---- (positive curvature)

- doogly
- Dr. The Juggernaut of Touching Himself
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### Re: RELATIVITY QUESTIONS! (and other common queries)

And zeros exist as options as well.

LE4dGOLEM: What's a Doug?

Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.

Keep waggling your butt brows Brothers.

Or; Is that your eye butthairs?

Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.

Keep waggling your butt brows Brothers.

Or; Is that your eye butthairs?

- Carlington
**Posts:**1574**Joined:**Sun Mar 22, 2009 8:46 am UTC**Location:**Sydney, Australia.

### Re: RELATIVITY QUESTIONS! (and other common queries)

In my head, I'm working with two dimensions, because that's how I usually come to terms with these sorts of things before I extend them. I'm doing that by envisaging a flat plane, say a piece of paper. If we give it positive curvature, we're pulling the edges up as much as we can, and eventually they're going to touch each other, because they just keep on curving.

For negative curvature, I'm doing the same thing, only pushing all the edges down, like a mirror image. Is this where I'm coming unstuck?

For negative curvature, I'm doing the same thing, only pushing all the edges down, like a mirror image. Is this where I'm coming unstuck?

Kewangji: Posdy zwei tosdy osdy oady. Bork bork bork, hoppity syphilis bork.

Eebster the Great: What specifically is moving faster than light in these examples?

doogly: Hands waving furiously.

Eebster the Great: What specifically is moving faster than light in these examples?

doogly: Hands waving furiously.

Please use he/him/his pronouns when referring to me.

### Re: RELATIVITY QUESTIONS! (and other common queries)

Yes it is.

Negative curvature (for a two dimensional surface), is like pushing one set of opposite corners of the square and pulling the other two (which is why you get the saddle shape in the diagrams people have shown).

Negative curvature (for a two dimensional surface), is like pushing one set of opposite corners of the square and pulling the other two (which is why you get the saddle shape in the diagrams people have shown).

my pronouns are they

Magnanimous wrote:(fuck the macrons)

- thoughtfully
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### Re: RELATIVITY QUESTIONS! (and other common queries)

The intersection of the saddle with the planes that bisect the saddle ("planes of principle curvatures" in the diagram linked), and any plane parallel to these, are hyperbolas and parabolas. If you visualize these, you should be able to see how they don't "wrap around". Remember your conic sections if you need a little help.

Last edited by thoughtfully on Wed Jan 11, 2012 8:22 pm UTC, edited 1 time in total.

- shawnhcorey
**Posts:**42**Joined:**Sun Jan 08, 2012 2:08 pm UTC

### Re: RELATIVITY QUESTIONS! (and other common queries)

thoughtfully wrote:The intersection of the saddle with the planes that bisect the saddle ("planes of principle curvatures" in the diagram linked) are a hyperbola and a parabola. If you visualize these, you should be able to see how they don't "wrap around". Remember your conic sections if you need a little help.

In the diagram shown, they are conic sections. If 3D space is a "surface" in 4D (or higher), it need not be a conic section and it may wrap around in ways that are hard to imagine but mathematically possible.

- thoughtfully
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Ok, but intuitions aren't going to be useful there. Out of curiousity, is there an analogue to conic sections for quartic surfaces? Can they be classified as intersections of 3-space with a "hypercone"? Hypercones aren't all that much harder to visualize than hyperspheres or hypercubes or simplices. Then it's one more step up in dimensions to the shape of space-time, yeah?

- doogly
- Dr. The Juggernaut of Touching Himself
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### Re: RELATIVITY QUESTIONS! (and other common queries)

oooooh let's be careful.

So, what does the word "hyperbolic" mean. Who are you talking to?

Maybe you are big on conic sections, and when you go to generalize to higher dimensions, you want to talk about intersections and things, and you are an algebraic geometer.

But in the sense we have been using it, it is a differential geometry phrase. Hyperbolas are the "trope namer," but a given manifold of negative curvature isn't a hyperbola necessarily.

And if you go to 3 manifolds, you can have a manifold which is hyperbolic but also compact. That is not at all like what you have in 2D. Compact hyperbolic manifolds are very exciting though.

So, what does the word "hyperbolic" mean. Who are you talking to?

Maybe you are big on conic sections, and when you go to generalize to higher dimensions, you want to talk about intersections and things, and you are an algebraic geometer.

But in the sense we have been using it, it is a differential geometry phrase. Hyperbolas are the "trope namer," but a given manifold of negative curvature isn't a hyperbola necessarily.

And if you go to 3 manifolds, you can have a manifold which is hyperbolic but also compact. That is not at all like what you have in 2D. Compact hyperbolic manifolds are very exciting though.

LE4dGOLEM: What's a Doug?

Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.

Keep waggling your butt brows Brothers.

Or; Is that your eye butthairs?

Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.

Keep waggling your butt brows Brothers.

Or; Is that your eye butthairs?

### Re: RELATIVITY QUESTIONS! (and other common queries)

doogly wrote:OK, 2 dimensions.

You're on the surface of a sphere, like the earth. The curvature is +R^2. So you say to yourself, hmmm, what if I just changed my perspective, and declare that I am inside this sphere. What happens to the curvature now? Ah, well the principal curvature along this here latitude direction is now -R! And along this longitude is -R! Negative waaaaait..... The curvature is still +R^2!

In terms of arithmetic, this fact is just due to the product of negative numbers being a positive number.

In terms of geometry, what happens is that you haven't actually *done* anything. The sphere doesn't care where you're standing.

Higher dimensions remain a bit trickier, naturally.

So at any point on a negatively curved surface the curvatures on your latitude and longitude directions have opposite signs, ie the curves in one direction are concave, but the orthogonal curves are convex. Here's an image of a pseudosphere, which is a surface of constant negative curvature (apart from the singularity at the equator).

OTOH, thinking about curved spaces by considering their properties when embedded in a higher dimensional flat space isn't necessarily that helpful. In some ways, it's better to avoid invoking an embedding space, since it may have no actual physical existence. The alternative approach is to look at the metric properties of the curved space. A simple way to do that is to examine the properties of triangles. In a flat space, the sum of the angles of any triangle always equals pi. In a positively curved space, the angle sum of a triangle is greater than pi, and in a negatively curved space, the angle sum of a triangle is less than pi. In both curved cases, if the space is of uniform curvature then the area of a triangle in that space is proportional to the absolute value of the difference between the angle sum and pi.

See

http://en.wikipedia.org/wiki/Spherical_trigonometry#Lines_and_angles_on_a_sphere

http://en.wikipedia.org/wiki/Hyperbolic_triangle

http://en.wikipedia.org/wiki/Gauss-Bonnet_theorem

### Re: RELATIVITY QUESTIONS! (and other common queries)

This comes from this game. http://images.4channel.org/f/src/gravity.swf

See that sine and curl going off to the lower left? Could that really happen? Happen in space?

ED:

Oh wait, of course it can. The moon makes the Earth wobble too.

See that sine and curl going off to the lower left? Could that really happen? Happen in space?

ED:

Oh wait, of course it can. The moon makes the Earth wobble too.

Last edited by Nosforit on Mon Jan 16, 2012 8:25 pm UTC, edited 1 time in total.

Submit.

- Yakk
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### Re: RELATIVITY QUESTIONS! (and other common queries)

That looks like two orbiting bodies heading off into the void together. As a guess, the sin wave is heavier than the curl-wave. The curl-wave sometimes goes in a retrograde direction, before it is pulled forward by the heavier sin-wave body.

But that is just a guess.

But that is just a guess.

One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

### Re: RELATIVITY QUESTIONS! (and other common queries)

I was able to replicate it with just a few dozen tries, so unless I have a magic touch it should be pretty common in physical space. Unless tidal forces shred them?

IIRC retrograde meant that a celestial object appered to move backwards, and did not physically do so...

IIRC retrograde meant that a celestial object appered to move backwards, and did not physically do so...

Submit.

- gmalivuk
- GNU Terry Pratchett
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### Re: RELATIVITY QUESTIONS! (and other common queries)

It can mean either. If you're talking to astrologers for some reason, something is retrograde if it looks like it's moving backwards relative to other stuff. If you're talking to astronomers, it means it's actually moving backward relative to other stuff.

- Eebster the Great
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### Re: RELATIVITY QUESTIONS! (and other common queries)

"Retrograde" is also sometimes used in historical contexts to describe the apparent backward motion of planets. Of course in Ptolemaic cosmology it was true retrograde motion, so I guess the line is blurred there.

- Robert'); DROP TABLE *;
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Since Newton's gravitation is supposed to be an approximation of Relativity, can anyone explain to me where does the value of the gravitational constant comes from? Is it still a free variable, or does it pop out elegantly from GR?

...And that is how we know the Earth to be banana-shaped.

- doogly
- Dr. The Juggernaut of Touching Himself
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Still a free parameter. That and the cosmological constant are the only two in GR though, which is nice.

Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.

Keep waggling your butt brows Brothers.

Or; Is that your eye butthairs?

- Eebster the Great
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### Re: RELATIVITY QUESTIONS! (and other common queries)

doogly wrote:Still a free parameter. That and the cosmological constant are the only two in GR though, which is nice.

It was nicer when they thought the cosmological constant was zero.

- doogly
- Dr. The Juggernaut of Touching Himself
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### Re: RELATIVITY QUESTIONS! (and other common queries)

It has no more reason to be zero than any other number.

Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.

Keep waggling your butt brows Brothers.

Or; Is that your eye butthairs?

- Eebster the Great
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### Re: RELATIVITY QUESTIONS! (and other common queries)

doogly wrote:It has no more reason to be zero than any other number.

That's only true in a theory that incorporates the constant in the first place. The constant doesn't come out of the original theory, it was Einstein's reaction to its prediction of an unstable universe.

From a quantum perspective it was MUCH nicer, since it is hard to figure out why the value is so small yet nonzero.

- doogly
- Dr. The Juggernaut of Touching Himself
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### Re: RELATIVITY QUESTIONS! (and other common queries)

You have to have the cosmological constant. You are writing down the Lagrangian, you include all terms up to a certain order. Einstein didn't include it, but there was no reason not to. Essentially he just figured it should be zero; it wasn't *necessary.* But so what?

Making things 0 in quantum mechanics is just as hard as small things. Unless of course you are willing to just wave your hands and mumble something about symmetry, then 0 is easy to justify. But if you want these symmetries to be in some sense natural and not entirely ad hoc, you still have some work to do to get 0 cosmological constant. I haven't seen any compelling arguments why that should be.

Making things 0 in quantum mechanics is just as hard as small things. Unless of course you are willing to just wave your hands and mumble something about symmetry, then 0 is easy to justify. But if you want these symmetries to be in some sense natural and not entirely ad hoc, you still have some work to do to get 0 cosmological constant. I haven't seen any compelling arguments why that should be.

Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.

Keep waggling your butt brows Brothers.

Or; Is that your eye butthairs?

- Christophoros
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Hi. I have a slightly-higher-but-not-much-higher-than-A-Level physics understanding. I understand that when you are in motion, you experience time at a slower rate than someone who is not in motion, or who is travelling more slowly than you. However, I have been thinking about this recently, and I run into a problem when I consider all motion as relative (Since we don't have any absolute reference point to use).

My question is this - Suppose a spaceship travels away from a planet at, say, 0.2 the speed of light, and then returns. Since it's been going fast, time has passed more slowly on board, so an atomic clock on the ship will have counted less time than an atomic clock on the planet. That's taking the planet as our reference point.

Why can't we take the ship as our reference point? From that perspective, the planet is travelling away at 0.2 the speed of light, and then returning. Surely that means the planet (having been going fast) will have experienced less time than the ship, and the atomic clock on the planet will have counted less time than the clock on the ship.

Am I missing something fundamental here?

I've tried to be concise with my question, but if it's not clear I can try rephrasing.

My question is this - Suppose a spaceship travels away from a planet at, say, 0.2 the speed of light, and then returns. Since it's been going fast, time has passed more slowly on board, so an atomic clock on the ship will have counted less time than an atomic clock on the planet. That's taking the planet as our reference point.

Why can't we take the ship as our reference point? From that perspective, the planet is travelling away at 0.2 the speed of light, and then returning. Surely that means the planet (having been going fast) will have experienced less time than the ship, and the atomic clock on the planet will have counted less time than the clock on the ship.

Am I missing something fundamental here?

I've tried to be concise with my question, but if it's not clear I can try rephrasing.

"Analogies in writing are like feathers on a snake."

"Exaggeration is a billion times worse than understatement."

Never Forget

"Exaggeration is a billion times worse than understatement."

Never Forget

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### Re: RELATIVITY QUESTIONS! (and other common queries)

Yes, you are correct in the framework of special relativity; both of them experience less time than the other. However, this is fixed by general relativity, where the ship's acceleration is not relative, and has the effect of speeding up external clocks. This means that the ship objectively experiences less time than the planet.

...And that is how we know the Earth to be banana-shaped.

- Christophoros
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Ah, thankyou.

"Analogies in writing are like feathers on a snake."

"Exaggeration is a billion times worse than understatement."

Never Forget

"Exaggeration is a billion times worse than understatement."

Never Forget

### Re: RELATIVITY QUESTIONS! (and other common queries)

An interesting case which seems paradoxical which is closely related is if you have two space ships which leave a planet going in different directions. Both of these ships will see the other ship's clock moving slower than there own.

For an intuitive justification for why the spaceship experiences less time rather than the earth doing so (in the original situation with one ship leaving a planet and returning), the spaceship ends up in the planet's reference frame so it makes sense that that reference frame should be more correct than the ship's one when talking about the time experienced (also, the earth is in an inertial frame (meaning SR is valid for it) whereas the ship is not (meaning SR is not valid for it)).

For an intuitive justification for why the spaceship experiences less time rather than the earth doing so (in the original situation with one ship leaving a planet and returning), the spaceship ends up in the planet's reference frame so it makes sense that that reference frame should be more correct than the ship's one when talking about the time experienced (also, the earth is in an inertial frame (meaning SR is valid for it) whereas the ship is not (meaning SR is not valid for it)).

my pronouns are they

Magnanimous wrote:(fuck the macrons)

- doogly
- Dr. The Juggernaut of Touching Himself
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### Re: RELATIVITY QUESTIONS! (and other common queries)

SR can handle acceleration just fine. It is the flat space limit of GR. Non inertial frames are just different coordinate systems. They're just as easy in SR as, say, a rotating one is in Newtonian mechanics. Which is to say, not entirely easy, you have to do a little work, but it's not a theory or system of mechanics. It's just new coordinates.

Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.

Keep waggling your butt brows Brothers.

Or; Is that your eye butthairs?

### Re: RELATIVITY QUESTIONS! (and other common queries)

I was under the impression that, under GR, mach's principle holds so gravitating frames are identical to accelerating frames. If this is the case, surely any predictions of SR with accelerating frames would be on shaky ground should the accelerations be sufficiently large or for the effects to be measured across a sufficiently large length scale?

my pronouns are they

Magnanimous wrote:(fuck the macrons)

### Re: RELATIVITY QUESTIONS! (and other common queries)

You can easily observe the effect by following the motion of the ship in special relativity, assuming instantaneous acceleration.

The ships starts in the Earth's Frame at (0,0,).

Boost this to the moving ship's frame, travelling at velocity v (still at (0,0))

The ship travels for some time t in its own reference frame, so it's located at (t,x)

Now perform the Lorentz transform to get it travelling back towards the Earth, instead of away, and you'll notice something incredibly interesting — the act of turning suddenly adds a huge amount of time to the clocks on Earth. This is because acceleration is not relative in relativity; you can always tell when you're accelerating, independent of any other object in the Universe.

I do suggest doing the maths, though, starting at the Earth, boosting to the frame v, travelling some (t,x), boosting back to the initial frame, boost again to -v (turning around) and coming back, and you realise the solution to this problem (the Twins Paradox) simply falls out of the maths

The ships starts in the Earth's Frame at (0,0,).

Boost this to the moving ship's frame, travelling at velocity v (still at (0,0))

The ship travels for some time t in its own reference frame, so it's located at (t,x)

Now perform the Lorentz transform to get it travelling back towards the Earth, instead of away, and you'll notice something incredibly interesting — the act of turning suddenly adds a huge amount of time to the clocks on Earth. This is because acceleration is not relative in relativity; you can always tell when you're accelerating, independent of any other object in the Universe.

I do suggest doing the maths, though, starting at the Earth, boosting to the frame v, travelling some (t,x), boosting back to the initial frame, boost again to -v (turning around) and coming back, and you realise the solution to this problem (the Twins Paradox) simply falls out of the maths

cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

Pronouns: Feminine pronouns please!

- Eebster the Great
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### Re: RELATIVITY QUESTIONS! (and other common queries)

doogly wrote:You have to have the cosmological constant. You are writing down the Lagrangian, you include all terms up to a certain order. Einstein didn't include it, but there was no reason not to. Essentially he just figured it should be zero; it wasn't *necessary.* But so what?

I don't see why that isn't sufficient justification. The cosmological constant has no more inherent right to exist than the divergence of the magnetic field. At the time, there was no reason to believe vacuum energy existed. There still isn't a great reason to believe magnetic monopoles exist. You make your theory no more complicated than it needs to be.

Making things 0 in quantum mechanics is just as hard as small things. Unless of course you are willing to just wave your hands and mumble something about symmetry, then 0 is easy to justify. But if you want these symmetries to be in some sense natural and not entirely ad hoc, you still have some work to do to get 0 cosmological constant. I haven't seen any compelling arguments why that should be.

Quantum field theory naively predicts a constant over 100 orders of magnitude too large. In such cases, of course it is easier to search for symmetries that make it zero than to tune the parameters to make it extremely small but nonzero. I don't understand how it could be otherwise.

- doogly
- Dr. The Juggernaut of Touching Himself
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### Re: RELATIVITY QUESTIONS! (and other common queries)

The equivalence of gravity and acceleration is pointlike. So special relativity would be all you need if you did everything where the acceleration was just 9.8 m/s^2, and you never left that regime, looked for tidal effects, or what have you.

That calculation, "100 orders of magnitude off," is with a naive cutoff at the planck scale. The actual answer is infinite, and there's no reason to put in a cutoff. But if you do, at the planck scale, then you get... not a right answer. So, maybe that was the wrong thing to do? Cutoffs are just awful ways to do QFT.

That calculation, "100 orders of magnitude off," is with a naive cutoff at the planck scale. The actual answer is infinite, and there's no reason to put in a cutoff. But if you do, at the planck scale, then you get... not a right answer. So, maybe that was the wrong thing to do? Cutoffs are just awful ways to do QFT.

Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.

Keep waggling your butt brows Brothers.

Or; Is that your eye butthairs?

### Re: RELATIVITY QUESTIONS! (and other common queries)

That calculation, "100 orders of magnitude off," is with a naive cutoff at the planck scale.

Its also from using rough estimates for the vacuum expectation of the Higg's field/potential energy of the Higgs field (and the chiral condensate of QCD). Even if you figure out a better method to deal with vacuum fluctuations, you still run into problems from the non-zero vacuum expectation values.

Thats why SUSY forces exactly 0- the constraints on the super-potential guarantee the potential energy is 0, even with a non-zero vev.

- doogly
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### Re: RELATIVITY QUESTIONS! (and other common queries)

So since we observe a nonzero cosmological constant, I'm putting that in the "con" column for SUSY ; )

Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.

Keep waggling your butt brows Brothers.

Or; Is that your eye butthairs?

### Re: RELATIVITY QUESTIONS! (and other common queries)

Indeed. Its also a clear problem with the higgs mechanism that we don't talk about very often.

Of course, someone who likes SUSY would suggest that we obviously know it is broken- perhaps leading to a small cosmological constant.

Of course, someone who likes SUSY would suggest that we obviously know it is broken- perhaps leading to a small cosmological constant.

- Robert'); DROP TABLE *;
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### Re: RELATIVITY QUESTIONS! (and other common queries)

(Not entirely sure this goes here, but it's kinda on-topic)

I've got a couple of questions, since I was recently trying to calculate (out of interest) various things about pulsar periods.

1. Does anyone have any estimate of how many pulsars are in the Milky Way?

2. I tried assuming that all pulsars have a random period between 10 and 1000ms, (I couldn't find real data) and then calculating the lowest common multiple of the lot of them, so I could see the "period" of the entire galaxy. However, I always got an answer between 10

I've got a couple of questions, since I was recently trying to calculate (out of interest) various things about pulsar periods.

1. Does anyone have any estimate of how many pulsars are in the Milky Way?

2. I tried assuming that all pulsars have a random period between 10 and 1000ms, (I couldn't find real data) and then calculating the lowest common multiple of the lot of them, so I could see the "period" of the entire galaxy. However, I always got an answer between 10

^{28}and 10^{32}ms, even with a "galaxy" consisting of only 100 pulsars. (At minimum, about 50k times longer than the universe has existed) Am I doing something wrong here, or is that actually how periods work?...And that is how we know the Earth to be banana-shaped.

- phlip
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### Re: RELATIVITY QUESTIONS! (and other common queries)

I don't know about part 1, but for part 2... two periods will only even have a LCM if their ratio is rational. If one is, like 1ms, and another is sqrt(2) ms, then they simply don't have an LCM. And even if they are co-rational, the LCM can be huge even if the two numbers are small... if the ratio of the two numbers is p/q, in least terms, then the LCM is pq times their GCD (or, equivalently, p times the first number, or q times the second number)... So numbers that don't form a nice simple ratio means a huge LCM.

Picking random numbers "properly" will almost certainly give you numbers which aren't co-rational... picking random numbers with a computer will always give you rational numbers, for all the usual systems, but liable to give you numbers where pq is huge.

Picking random numbers "properly" will almost certainly give you numbers which aren't co-rational... picking random numbers with a computer will always give you rational numbers, for all the usual systems, but liable to give you numbers where pq is huge.

Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

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### Re: RELATIVITY QUESTIONS! (and other common queries)

Oops, I meant to specify that the periods were an integer number of milliseconds. The calculation went something along the lines of:

And I suppose on an abstract level I knew that the lcm would generally come out quite large, but I generally wasn't expecting it to be quite as large as it was. Am I doing the right calculation, even if the answer isn't what I was expecting?

Code: Select all

`l = [random.randint(10, 1000) for i in range(100)]`

reduce(lcm, l)

And I suppose on an abstract level I knew that the lcm would generally come out quite large, but I generally wasn't expecting it to be quite as large as it was. Am I doing the right calculation, even if the answer isn't what I was expecting?

...And that is how we know the Earth to be banana-shaped.

- Yakk
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### Re: RELATIVITY QUESTIONS! (and other common queries)

There are ~145 primes < 1000. (very approx)

If you only sampled the primes, after 100 you'd have about half of them.

Now, because you are also sampling composite numbers, you'll over-select small primes and under-select big primes. And while a given number can add more than one prime to your list, this will happen more often with small primes.

The other thing that LCM cares about is prime powers.

So these are all complications. But, because I'm lazy, I'll just use the "sampling primes" case to get an initial spitball estimate.

The product of the first 145 primes is known as 145#. We only have about half of them, so we'll take the sqrt. Now, n# is bounded below by e^(n ln n) I believe -- so this is on the order of e^361 =~ 10^156. That's big.

So I'm vaguely surprised the result is that small. I guess the dominance of the low-order values ends up having a large impact, not surprisingly.

If we ignore prime powers, the odds of getting a new prime end up being the product of (p-1)/p, which sadly is not a function which has been researched much to my knowledge.[1]

---

You'll note that this value is a function of your temporal resolution, or how close you want it to get. It is sort of fractal that way -- the more you zoom in, the bigger it gets.

[1] I kid.

If you only sampled the primes, after 100 you'd have about half of them.

Now, because you are also sampling composite numbers, you'll over-select small primes and under-select big primes. And while a given number can add more than one prime to your list, this will happen more often with small primes.

The other thing that LCM cares about is prime powers.

So these are all complications. But, because I'm lazy, I'll just use the "sampling primes" case to get an initial spitball estimate.

The product of the first 145 primes is known as 145#. We only have about half of them, so we'll take the sqrt. Now, n# is bounded below by e^(n ln n) I believe -- so this is on the order of e^361 =~ 10^156. That's big.

So I'm vaguely surprised the result is that small. I guess the dominance of the low-order values ends up having a large impact, not surprisingly.

If we ignore prime powers, the odds of getting a new prime end up being the product of (p-1)/p, which sadly is not a function which has been researched much to my knowledge.[1]

---

You'll note that this value is a function of your temporal resolution, or how close you want it to get. It is sort of fractal that way -- the more you zoom in, the bigger it gets.

[1] I kid.

Last edited by Yakk on Fri Apr 13, 2012 2:47 pm UTC, edited 2 times in total.

One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

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