Physics Of N Dimensions

[petrushkagoogol]

How does the physics of n dimensions differ from that of (n + 1) dimensions in general ?  :zip:

[CraigD]

How does the physics of n dimensions differ from that of (n + 1) dimensions in general ? :zip:

I can’t think of a general description a difference between the physical laws of an universe with n and n+1 Euclidean spatial dimension, but I can think of an obvious specific difference: their “inverse square laws” are different, one or both’s universes not being “square”, but some other exponent.

 

For example, the power of radiation at distance X in 3-D space is [math]P(x) = \frac{P(1)}{x^2}[/math].

In 1-D space it’s [math]P(x) = P(1)[/math], 2-D space [math]P(x) = \frac{P(1)}{x}[/math], 4-D space [math]P(x) = \frac{P(1)}{x^3}[/math], n-D space [math]P(x) = \frac{P(1)}{x^{n-1}}[/math].

 

So in a 1-dimensional universe, light doesn’t dim as you move away from its source. In a high-dimensional space, it dims very rapidly with distance.

 

Since universal Newtonian gravity is an inverse square law, all the laws derived from it might change for [math]n \ne 3[/math]. I wonder what <a data-ipb="nomediaparse" data-cke-saved-href="https://en.wikipedia.org/wiki/Kepler" href="https://en.wikipedia.org/wiki/Kepler" s_laws_of_planetary_motion"="">Kepler’s laws look like in 2 D, 4 D, 5 D, etc?

[xyz]

I can’t think of a general description a difference between the physical laws of an universe with n and n+1 Euclidean spatial dimension, but I can think of an obvious specific difference: their “inverse square laws” are different, one or both’s universes not being “square”, but some other exponent.

 

For example, the power of radiation at distance X in 3-D space is [math]P(x) = \frac{P(1)}{x^2}[/math].

In 1-D space it’s [math]P(x) = P(1)[/math], 2-D space [math]P(x) = \frac{P(1)}{x}[/math], 4-D space [math]P(x) = \frac{P(1)}{3^2}[/math], n-D space [math]P(x) = \frac{P(1)}{^{n-1}}[/math].

 

So in a 1-dimensional universe, light doesn’t dim as you move away from its source. In a high-dimensional space, it dims very rapidly with distance.

 

Since universal Newtonian gravity is an inverse square law, all the laws derived from it might change for [math]n \ne 3[/math]. I wonder what <a data-ipb="nomediaparse" data-cke-saved-href="https://en.wikipedia.org/wiki/Kepler" href="https://en.wikipedia.org/wiki/Kepler" s_laws_of_planetary_motion"="">Kepler’s laws look like in 2 D, 4 D, 5 D, etc?

There would be a need to also consider the possible space between universes where it may be void like, that  I think you are meaning by the difference in the inverse square law. It does not need to be different in my opinion, I try to  imagine to equal watt light bulbs ''floating'' around in a void, but neither light bulb can observe each other because the length of space exceeds both lengths of light leaving a void in the space between them. There is also a need to consider volume contraction relatively to distance to the observer and the  volume of the  object contracting while moving away  to a zero point source where the point does not exist visually any more.  

 

Just my opinion, hope this helps you in some way. 

[A-wal]

Just to restate what CraigD said:

Gravity and electro-magnetism are proportional to the volume they fill. In zero dimensions any amount of gravity or light would be infinitely strong because it's being divided by zero volume. In a one dimensional straight line, there would be no loss of strength over distance because there's no spreading out of the energy. In two dimensions, doubling the distance would halve the strength. In three dimensions it's an inverse square (doubling the distance divides the strength by four). In four spatial dimensions, it would be an inverse cube (doubling the distance divides the strength by eight). In five dimensions it would be an inverse tesseract (doubling the distance divides the strength by sixteen) and so on. Because it halves with each new dimension, it would take that many more spatial dimensions for the sun to become invisible to the naked eye.

 

There would be a need to also consider the possible space between universes where it may be void like, that  I think you are meaning by the difference in the inverse square law. It does not need to be different in my opinion, I try to  imagine to equal watt light bulbs ''floating'' around in a void, but neither light bulb can observe each other because the length of space exceeds both lengths of light leaving a void in the space between them. There is also a need to consider volume contraction relatively to distance to the observer and the  volume of the  object contracting while moving away  to a zero point source where the point does not exist visually any more.  

 

Just my opinion, hope this helps you in some way. 

Huh? :huh: Nothing you say makes any sense.

[CraigD]

Because it [brightness] halves with each new dimension, it would take that many more spatial dimensions for the sun to become invisible to the naked eye.

It’s fun to calculate how many more spatial dimensions it would take to make the Sun invisible from Earth.

 

Assuming atoms get re-jiggered to work the same in whatever number of dimensions, to the surface of the sun has the same brightness, the brightness formula is

[math]B(n) = \left(\frac{r_1}{r_0}\right)^{-(n-1)} \dot= 215^{-(n-1)} [/math]

where [math]n[/math] is the number of dimensions, [math]r_0[/math] the Sun’s radius, and [math]r_1[/math] Earth’s orbit.

 

The dimmest magnitude you can see with the naked eye is 3 to 6. The Sun’s is about -27. Magnitude is [math]-2.5 log_{10}( B )[/math], so the Sun becomes invisible for [math]n \ge 9 > \frac{33}{2.5 log_{10}(215)} +3[/math].

 

In even 4 D space, stars in their present positions, size, and surface brightness wouldn’t be visible. The largest stars have radii less than [math]r_0=[/math]2000 time the Sun’s, while the closest stars are over [math]r_1=[/math]250000 times the Earth’s distance from the Sun, which adds more than 60 to their magnitudes. The most sensitive telescopes built or likely to be built by a civilization of our technological level can detect stars can detect perhaps magnitude 40 (the Hubble can detect up to about 31.5)

 

The atom re-jiggering assumption I’m making for all this is pretty big, though – but not as big as imagining more than 3 non-compact spatial dimensions. The smart, imaginative, and well-educated Rudy Rucker wrote a very cool novel exploring the biochemistry, ecology, and politics of 4 D space, Spaceland (2003).

Edited February 26, 2016 by CraigD
Added missing - signs to equation

[A-wal]

In even 4 D space, stars in their present positions, size, and surface brightness wouldn’t be visible in 4 D space .

Huh? In four spatial dimensions the brightness of the stars would be halved. Lots would still be visible.

Edited February 26, 2016 by A-wal

[CraigD]

Huh? In four spatial dimensions the brightness of the stars would be halved. Lots would still be visible.

That would be true if our distance to the surface of every star ([math]r_1-r_0[/math]) were exactly the same as the distance from the center of the star to its surface ([math]r_0[/math]), but we’re actually many times further than [math]r_0[/math] from every star, including the Sun.

 

Because [math]\frac{r_1}{r_0}[/math] for the Sun is astronomically (no pun) smaller – only about 215 – than any other star – eg: For Alpha Centuri, it’s about 2.35 x 10²⁵ – it takes a lot of additional dimensions for it to be dimmed to invisibility - 9 D space. But for all the other stars, 1 additional dimension is all that’s needed – 4 D space.

 

The key concept here is that the distance that you use as a unit in your doubling of distance is important. It has to correspond to an absolute brightness (which is power, energy/time) associate with an actual physical process, the glowing of the star’s photosphere. Because ordinary astronomy doesn’t get into questions of how things would look in other than 3 spatial dimensions, we don’t usually care, or in most cases know with much precision, [math]r_0[/math], which is why absolute magnitude is not a measurement of the brightness per unit area of a star’s photosphere, but just its apparent magnitude at a standard distance of 10 parsecs.

 

When we start playing with the number of spatial dimensions, we’ve got to take into account the actual size of the surfaces radiating at a given power (which varies so little compared to its reduction due to distance if we have more than 1 dimension that we can consider it nearly constant for all lit stars). If, like I’ve done, you do the arithmetic for various naked eye visible stars, greater and smaller, nearer and further, you’ll find that for 4 D space, they all have apparent magnitudes of 65 to 62. To see them, you’d need a telescope with about 10¹² the light-gathering of the best existing ones. So for a space telescope like the Hubble, you’d need to increase it’s 2.4 m mirror to about 2400 km.

 

My habit of talking about areas and lengths shows how rooted in 3 D space I am. In 4 D space, telescopes we’d talk about the volume of a telescope’s mirror rather than its area – that’s what would be proportional to its light-gathering ability. Our retinae wouldn’t be 2 D surfaces inside little gooey 3 D spheroids, but 3 D volumes inside 4 D hyperspheroids, or whatever shape 4 D biology evolved.

 

I think it’s safe to say 4 D biology would be 3 D imagination-strainingly weird. I find it nice that folk like Rudy Rucker spend time imagining it for us, saving us some imagination strain. :)

[A-wal]

:huh: I don't get it. What does the surface area of the source have to do with the brightness of the light from the perspective of a distant observer? The brightness of the stars would be halved in four dimensional space because of the four dimensional volume it fills as it spreads out in every direction.

[CraigD]

:huh: I don't get it. What does the surface area of the source have to do with the brightness of the light from the perspective of a distant observer? The brightness of the stars would be halved in four dimensional space because of the four dimensional volume it fills as it spreads out in every direction.

Consider the geometry underlying an inverse square law (which is a special case of d^-(n-1) law where n=3). We have a surface of dimension n-1 (in 3D space, a 2D surface) that has a given number of quanta of something (in the case of a star, photons of EM radiation) per unit of its dimension and time (eg: x photons / m² / s) – let’s just call it “area-time” for convenience, understanding that for cases other than n=3, it’s not ordinary 2 D area. Each quantum is traveling in a direction originating in a single point (the center of the star). So to know the quanta per unit area-time, we just have to find the ratio of the areas of 2 spheres (again, if n isn’t 3, they’re not ordinary 3 D spheres), one the source of the light (the star’s photosphere), the other a virtual sphere the observer is on the surface of.

 

We start with the assumption that the star has the same power – that is, with the simplifying assumption that it emits photons of only one frequency, which isn’t too unrealistic, that it emits the same number of photons per second – regardless of our universe’s number of dimensions.

 

The brightness (just another term for power per unit area) of a star at a distance of 2 lightyears from its center is 2^-(n-1) what it is at a distance of 1 ly. But the brightness of the star is not the same at a distance of 1 lr in n=2 or n=4 dimensions as it is in n=3 dimensions. To calculate what it is at 1 ly, we must have a starting point where we know its brightness. That starting point is the surface of its photosphere.

 

This gives us the brightness formula I gave (fancied up a bit to include all its arguments and constants):

[math]B(r_1,n) = B(r_0,n) \left(\frac{r_1}{r_0}\right)^{-(n-1)}[/math]

 

[math]B(r_0,n)[/math] is the constant we get from our starting assumption, which we can easily calculate from real observation in real 3 D space.

 

Put another way, we can say “every star is 1/2 as bright from 2 times the ([math]r_1[/math]) distance in 4 D ([math]n=[/math]4) space as it is from 2 times the distance in 3 D space”, but we can’t say “every star is 1/2 as bright in 4 D space as in 3 D space”. This is because what we’re changing is not [math]r_1[/math], but [math]n[/math].

 

We can also see that “every star is 1/2 as bright in 4 D space as in 3 D space” can’t be true by noting that, per our starting assumption that stars have the same total power regardless of [math]n[/math] at a distance of [math]r_1=r_0[/math], stars have the same brightness for any [math]n[/math].

[A-wal]

You mean that not only is the light being spread out in an extra dimension, the brightness of the source would be decreased as well because it's surface area is being spread out over an extra dimension.

 

How is that relevant? The power per unit area as you put it, halves with each additional spatial dimension, so the only way the brightness at any distance could be less than half is if the total output of light from the star is decreased. It shouldn't matter that its surface isn't as bright due to an increased surface area, or surface volume to be precise.

[CraigD]

You mean that not only is the light being spread out in an extra dimension, the brightness of the source would be decreased as well because it's surface area is being spread out over an extra dimension.

No, I assume the brightness of the source is the same for all number of dimensions.

 

How is that relevant? The power per unit area as you put it, halves with each additional spatial dimension...

[math]B(r_1,n) = B(r_0,n) \left(\frac{r_1}{r_0}\right)^{-(n-1)}[/math]

“halves with each additional spatial dimension” only when [math]\frac{r_1}{r_0}=2[/math]. That is, the [math]\frac{1}{2}[/math] factor is an artifact of choosing an easy-to-say example phrase like “in 3 dimensions, doubling the distance decreases brightness by a factor of 4. In 4 D, doubling the distance decreases brightness by a factor of 8”, or “when doubling distance, brightness halves with each additional dimension”. A different choice of example like “in 3 D, increasing distance by 13 times decrease brightness by 169 time. In 3 D, increasing distance by 13 times decreases brightness by 2197 times,” or “when increasing distance by 13, brightness decreases by 13 with each additional dimension.”

 

That is, there’s nothing special about “2” or “13” – they’re just arbitrarily choses numbers to make easy-to-say sentences.

 

The distance from the center of the star to the surface of its photosphere, and to the observer, on the other hand, are not arbitrarily chosen examples. Since those [math]\frac{r_1}{r_0}[/math] factors are like 10²⁵, our spoken example would be more sensible if we said something like “for a typical bright star in the night sky, brightness decreases by something like 10,000,000,000,000,000,000,000,000 times for each additional dimension.”

[A-wal]

:confused: I'm really not getting this. I don't see what the source has to do with anything other than how much light it's putting out, that's all that should matter.

 

The intensity of the light depends how many dimensions it's being spread through as it moves away from the source. It's just an increase in volume, which doubles with each additional dimension.

 

How does the photosphere play any role in determining the amount of light that reaches an observer if we assume that the brightness of the start remains constant? It's just the amount of light that's traveling through space that matters. Start some distance away from the source, it's the same thing.

[CraigD]

:confused: I'm really not getting this. I don't see what the source has to do with anything other than how much light it's putting out, that's all that should matter.

It is all that matters. The confusion isn’t from that, I think, but from jumping to the conclusion that “the brightness of [all] stars would be halved in four dimensional space [vs. 3 D space]”, and trying to work backwards from that.

 

To minimize confusion, let’s not use my “power per unit area(oid) is the same at the photosphere(oid) in any number of dimensions” assumption, and imagine a super-simple toy universe with only 2 identical stars and an observer at “Earth”, star A r_A = 2 units from Earth, star B r_B = 3 units from Earth.

 

It’s easy to calculate the brightness of A and B as seen from Earth in a given number of dimensions, giving this table

           Dimensions

  Distance 2   3   4

A 2        1/2 1/4 1/8

B 3        1/3 1/9 1/27

 

Notice that the assertion that increasing the number of dimensions by one halves the brightness of all stars isn’t true even in this simple toy example.

 

Often, the best way to understand is to just do the arithmetic until you get an “ah-ha” moment when they make intuitive sense. Trying to work from even simple, sensible seeming natural language sentences is, in my experience, more confusing than sticking with numbers.

[A-wal]

Star A's brightness is divided by two with each additional dimension and star B's brightness is divided by three with each additional dimension because it's further away? Why?

 

Put them in a line. Star B - Star A - Earth. The light from star B passes star A. Star B is brighter than start A. The light intensity from star B is equal to star B after one unit. Now the two stars are the same brightness from Earth. They'll continue to be the same brightness in any number of dimensions.

[CraigD]

Star A's brightness is divided by two with each additional dimension and star B's brightness is divided by three with each additional dimension because it's further away? Why?

Because brightness follows an inverse square law – better described here as an “inverse power of distance” law. Adding a dimension doesn’t reduce the brightness an observer sees for every light source by half. Adding a dimension changes the exponent in the brightness function which gives the brightness of light sources at any distance.

 

Put them in a line. Star B - Star A - Earth. The light from star B passes star A. Star B is brighter than start A. The light intensity from star B is equal to star B after one unit. Now the two stars are the same brightness from Earth. They'll continue to be the same brightness in any number of dimensions.

No, they won’t.

 

Let’s add your conditions to the table, by adding a brightness column to allow the two stars to have different absolute brightnesses. You’re condition is Apparent brightness of B at distance 2 equals apparent brightness of A at distance 1. So, in 3 D, we calculate the absolute brightness of B from [math]\frac{b_B}{2^2} = \frac{b_A}{1^2}[/math], giving [math]b_B = 4 b_A[/math], and the table is now

                        Dimensions

  Brightness  Distance  2   3   4

A 1           2         1/2 1/4 1/8

B 4           3         4/3 4/9 4/27

 

Again, I think you’re getting into trouble by attempting to reason in words rather than numbers, A-wal The realization that numbers can be used to understand reality is a key part of the scientific revolution. Though an oversimplification of its history, the rise of numbers over words mirrors the emergence of science from sophistry. Unfortunately it also give us the split of people who like to think into what C P Snow famously termed “The Two Cultures”.

[A-wal]

Again, I think you’re getting into trouble by attempting to reason in words rather than numbers, A-wal

It's not that. I understand the maths of what you're saying. Maths can't help to understand something, only to accurately define something. You need to understand it or the maths could be correct but not applicable.

 

I think this is one of those occasions where I'm missing something obvious. I'll try not to think about and then come back to it, that usually does the trick. I'm stuck in the same neural pathways, my mind needs a reboot. :)

 

Cheers for explaining, hopefully the penny will drop soon.