Gravitational Potential Energy in a 2 dimensional Universe

[Vashta Nerada]

Hello everyone! I was trying to imagine how some properties of physics (like energy and forces) would be in a 2D Universe. But I found some irregularities...

Thinking in 3D terms and considering that gravity exists, we can easily deduce the equation for the Gravitational Force (like Newton did):

1. It must be proportional to both two-body masses;
2. Since we live in a 3D universe we can imagine that the gravity extends by a distance in 1 dimension, remaining the other 2 dimensions for it to spread, forming a spherical propagation. So it must be inversely proportional to the distance squared;
3. We need a constant to fix the units and values.

So, the equation must be:   F_G = G.M.m/d²

 

Now, to derive the Gravitational Potential Energy formula we just need to integrate the force with respect to the distance that it was applied (Energy = force x distance).

We also need to think that this force was applied since an infinite distance all the way through a distance "d" (the distance where the two bodies are now appart), these will be our integration limits.

So:   E_PG = Integral[∞->d](G.M.m.dx/x²) = -G.M.m.Integral[d->∞](dx/x²) = -G.M.m.(-1/x)[d->∞] = G.M.m/x[d->∞] = G.M.m/∞ - G.M.m/d = 0 - G.M.m/d = -G.M.m/d

 

Ok, all good. But now, if we think in a 2D universe, we can deduce that the Gravitational Force would be inversely proportional to the distance, since the gravity would spread like a circle.

So, the equation must be:   F_G = G.M.m/d   (of course, the G constant will have different units)

 

Now, the irregularity that I found was in trying to derive the Gravitational Potential Energy, because following the same idea, we have:

E_PG = Integral[∞->d](G.M.m.dx/x) = -G.M.m.Integral[d->∞](dx/x) = -G.M.m.ln(x)[d->∞] = -G.M.m(ln(∞) - ln(d)) = G.M.m.ln(d) - ∞     :(

 

That was the irregularity... this energy makes no sense... So it must exist some wrong definition that I made... I don't know if it was in the 2D Gravitational Force or in the concept of Energy in a 2D universe... But if someone know, please share with me the knowlage. Thanks!

[Genady]

You don't have to integrate to infinity. You can choose anything convenient for your 0 energy and calculate from it. The only meaningful value (in Newtonian world) is energy difference, not its absolute value.

[Vashta Nerada]

4 hours ago, Genady said:

You don't have to integrate to infinity. You can choose anything convenient for your 0 energy and calculate from it. The only meaningful value (in Newtonian world) is energy difference, not its absolute value.

Okay, but. Why in 3D we can get a general equation that gives us the absolute value for any distance but in 2D we can't?

Edited February 16, 2022 by Vashta Nerada

[Genady]

1 hour ago, Vashta Nerada said:

Okay, but. Why in 3D we can get a general equation that gives us the absolute value for any distance but in 2D we can't?

The value calculated from infinity is as absolute as values calculated from any other choice of distance. You can choose the energy being 0 at some finite distance L. Then in 3D, the general equation for absolute value for any distance is GMm(1/L-1/d). In 2D, it would be GMm*ln(d/L). They all are absolute values. You can choose L to be 1 km, 1 parsec, diameter of the Milky Way galaxy, radius of the observable universe, whatever. It doesn't matter because in all physical calculations the constant L will cancel out.

The difference between the formulas for 3D and for 2D is that the former allows you to choose L being infinity and to get a nicer formula without L to start with, since it will cancel anyway, -GMm/d, while in 2D you can't do this, although it will cancel there as well.

[joigus]

13 hours ago, Vashta Nerada said:

Ok, all good. But now, if we think in a 2D universe, we can deduce that the Gravitational Force would be inversely proportional to the distance, since the gravity would spread like a circle.

What's the argument? How can you deduce the gravitational law?

[Sensei]

2D world is equivalent to 3D world but with one axis constant zero.. e.g. z=0

(which is used all the time by computer graphics cards/modern operating systems, i.e. even when you look at the screen it's actually an accelerated 3D gfx)

 

Edited February 16, 2022 by Sensei

[Genady]

10 hours ago, Vashta Nerada said:

Okay, but. Why in 3D we can get a general equation that gives us the absolute value for any distance but in 2D we can't?

 

8 hours ago, Genady said:

The value calculated from infinity is as absolute as values calculated from any other choice of distance. You can choose the energy being 0 at some finite distance L. Then in 3D, the general equation for absolute value for any distance is GMm(1/L-1/d). In 2D, it would be GMm*ln(d/L). They all are absolute values. You can choose L to be 1 km, 1 parsec, diameter of the Milky Way galaxy, radius of the observable universe, whatever. It doesn't matter because in all physical calculations the constant L will cancel out.

The difference between the formulas for 3D and for 2D is that the former allows you to choose L being infinity and to get a nicer formula without L to start with, since it will cancel anyway, -GMm/d, while in 2D you can't do this, although it will cancel there as well.

In fact, you can get a nice formula in your 2D case as well: since adding a constant to energy doesn't change anything, you can simply remove the  ln(L) part and get the general equation for absolute value for any distance, GMm*ln(d). 

1 hour ago, Sensei said:

2D world is equivalent to 3D world but with one axis constant zero.. e.g. z=0

(which is used all the time by computer graphics cards/modern operating systems, i.e. even when you look at the screen it's actually an accelerated 3D gfx)

 

Is there a magnetic field in 2D?

Edited February 16, 2022 by Genady

[Vashta Nerada]

8 hours ago, Genady said:

The value calculated from infinity is as absolute as values calculated from any other choice of distance. You can choose the energy being 0 at some finite distance L.

Ok. But I thought that we consider the force "starting" at the infinity because at that imaginary distance the two bodies wouldn't experience any force or energy at all, It's like there's never been an interaction between them, because they have never even been close. At the infinity, the body A doesn't exerce Gravitational Force in body B (consequently they have no E_PG) and vice versa. If the bodies are from a finite distance apart, that mean that they already had a Gravitational Force applied, so they already have E_PG associated (not zero).

So, for me it feels very wrong to define the E_PG being 0 at a finite distance L because that would imply that for distances greater then L the energy would be inverted in signal, that's madness.

Edited February 16, 2022 by Vashta Nerada

[mistermack]

I know this is going to sound a bit thick, but what the hell. How can you have gravitational potential energy greater than zero, with only two spatial dimensions? 

If you take any 3d object, and reduce one of it's dimensions to zero, then you get zero mass. So how can you have gravity, without mass? 

[Genady]

16 minutes ago, Vashta Nerada said:

So, for me it feels very wrong to define the E_PG being 0 at a finite distance L because that would imply that for distances greater then L the energy would be inverted in signal, that's madness.

When calculating gravitational potential energy near the Earth surface, where the gravitational acceleration g is almost constant, we choose it to be 0 on the floor and the energy is mgh where h is height above the floor. It changes sign, too. And it works just fine.

[Vashta Nerada]

2 hours ago, joigus said:

What's the argument? How can you deduce the gravitational law?

Just like I wrote at the beginning... Actually, I think Newton derived using Kepler 3th law (T²/r³ = constant). But anyway...

If we think, let's say, the light intensity of a point source spreading in the space (in our 3D universe), the formula would be simply: B(x) = P₀/4.π.x² given in [W/m²].
It's obvious to see that the power is diluted in the surface of a sphere of radius x. So it is inversely proportional to the distance squared because it is spreading in an area.
So it's reasonable that we imagine that gravity would kind of do the same. It's intensity would spread in the universe inversely proportional to the distance squared.

Now, for two dimensions, the power will expand in a circle shape, being diluted in the linear circumference of radius x. So the formula is: B(x) = P₀/2.π.x given in [W/m]

The same idea for gravity... For two dimensions it would be inversely proportional to the distance, not the distance squared.

[Genady]

2 hours ago, joigus said:

What's the argument? How can you deduce the gravitational law?

 

14 minutes ago, mistermack said:

I know this is going to sound a bit thick, but what the hell. How can you have gravitational potential energy greater than zero, with only two spatial dimensions? 

If you take any 3d object, and reduce one of it's dimensions to zero, then you get zero mass. So how can you have gravity, without mass? 

Because physical 2D world is a pure fantasy, I guess one can make up any 2D physics they like. However, they still cannot make up mathematics.

[joigus]

16 minutes ago, Vashta Nerada said:

Just like I wrote at the beginning... Actually, I think Newton derived using Kepler 3th law (T²/r³ = constant). But anyway...

If we think, let's say, the light intensity of a point source spreading in the space (in our 3D universe), the formula would be simply: B(x) = P₀/4.π.x² given in [W/m²].
It's obvious to see that the power is diluted in the surface of a sphere of radius x. So it is inversely proportional to the distance squared because it is spreading in an area.
So it's reasonable that we imagine that gravity would kind of do the same. It's intensity would spread in the universe inversely proportional to the distance squared.

Now, for two dimensions, the power will expand in a circle shape, being diluted in the linear circumference of radius x. So the formula is: B(x) = P₀/2.π.x given in [W/m]

The same idea for gravity... For two dimensions it would be inversely proportional to the distance, not the distance squared.

You're right. That's what Newton did. If Kepler's laws are satisfied, what kind of distance-dependent power law do I need to reproduce them? And the answer is indeed an inverse-square law.

But how do we know Kepler's laws would be satisfied in a 2D world?

I don't think the gravitational law can be deduced.

Your argument here is more akin to a deduction. But it still rests on a guess. If it's true that the overarching principle of gravity is that it reproduces the way field lines escape from a point in the same way that the lines representing a substance that once starts flowing away from a point, is conserved; then that would be what gravity would do in 2D.

I agree. But that's a guess.

15 minutes ago, Genady said:

Because physical 2D world is a pure fantasy, I guess one can make up any 2D physics they like. However, they still cannot make up mathematics.

Agreed.

Nevertheless, one can still manage to find examples in which thinking in 2D is useful. Graphene being one example.

Edited February 16, 2022 by joigus
minor correction

[Genady]

22 minutes ago, Vashta Nerada said:

Just like I wrote at the beginning... Actually, I think Newton derived using Kepler 3th law (T²/r³ = constant). But anyway...

If we think, let's say, the light intensity of a point source spreading in the space (in our 3D universe), the formula would be simply: B(x) = P₀/4.π.x² given in [W/m²].
It's obvious to see that the power is diluted in the surface of a sphere of radius x. So it is inversely proportional to the distance squared because it is spreading in an area.
So it's reasonable that we imagine that gravity would kind of do the same. It's intensity would spread in the universe inversely proportional to the distance squared.

Now, for two dimensions, the power will expand in a circle shape, being diluted in the linear circumference of radius x. So the formula is: B(x) = P₀/2.π.x given in [W/m]

The same idea for gravity... For two dimensions it would be inversely proportional to the distance, not the distance squared.

Not necessarily. Imagine a 3D sphere with the gravitating mass in the center and the gravitational field lines going from there. The force goes inversely proportional to the distance squared. Now cut the sphere in the middle and look at the cross section. This gives you the 2D image of the field. The force still goes inversely proportional to the distance squared.

[Vashta Nerada]

21 minutes ago, mistermack said:

I know this is going to sound a bit thick, but what the hell. How can you have gravitational potential energy greater than zero, with only two spatial dimensions? 

If you take any 3d object, and reduce one of it's dimensions to zero, then you get zero mass. So how can you have gravity, without mass? 

Yeah... Just like @Genady said, 2D World is (yet) only a fantasy that we create. We don't know if it exists or if it's even possible to exist. However, it makes sense.

An analogy that I like to make is to just think in a Universe with 4 dimensions, the intelligent forms of life at that universe would be so more complex than we are that from their perspective, a 3D universe would seem easily impossible or unlikely to exists (just like 2D are from our 3D world), but here are us . We exists.

So maybe it's very possible to exist a 2D universe, but all the things would be total different, particles would be total flat, without thickness, mass is now something attached to the Area of things, and not the Volume. Remember... there's no Volume in 2D. You can't imagine a functional 2D world without letting go of our models of the universe.

But maybe some concepts of physics is similar to ours, who knows... That's why I started this topic.

[Genady]

4 hours ago, Sensei said:

2D world is equivalent to 3D world but with one axis constant zero.. e.g. z=0

(which is used all the time by computer graphics cards/modern operating systems, i.e. even when you look at the screen it's actually an accelerated 3D gfx)

 

Not necessarily. The 2D world doesn't have to be flat.

[mistermack]

40 minutes ago, Vashta Nerada said:

So maybe it's very possible to exist a 2D universe, but all the things would be total different, particles would be total flat, without thickness, mass is now something attached to the Area of things, and not the Volume.

But if that were the case, we would detect that effect now. Just because we have 3 dimensions, that doesn't make us blind to 2D effects, if they exist. We might not be able to concieve of 4D phenomena, but that doesn't mean that a 4D creature would be unable to detect and measure 3D things. 

Just as volume can't exist in 2D, I don't think gravity can exist either.  

[studiot]

On 2/15/2022 at 8:26 PM, Vashta Nerada said:

Hello everyone! I was trying to imagine how some properties of physics (like energy and forces) would be in a 2D Universe. But I found some irregularities...

Thinking in 3D terms and considering that gravity exists, we can easily deduce the equation for the Gravitational Force (like Newton did):

1. It must be proportional to both two-body masses;
2. Since we live in a 3D universe we can imagine that the gravity extends by a distance in 1 dimension, remaining the other 2 dimensions for it to spread, forming a spherical propagation. So it must be inversely proportional to the distance squared;
3. We need a constant to fix the units and values.

So, the equation must be:   F_G = G.M.m/d²

 

Now, to derive the Gravitational Potential Energy formula we just need to integrate the force with respect to the distance that it was applied (Energy = force x distance).

We also need to think that this force was applied since an infinite distance all the way through a distance "d" (the distance where the two bodies are now appart), these will be our integration limits.

So:   E_PG = Integral[∞->d](G.M.m.dx/x²) = -G.M.m.Integral[d->∞](dx/x²) = -G.M.m.(-1/x)[d->∞] = G.M.m/x[d->∞] = G.M.m/∞ - G.M.m/d = 0 - G.M.m/d = -G.M.m/d

 

Ok, all good. But now, if we think in a 2D universe, we can deduce that the Gravitational Force would be inversely proportional to the distance, since the gravity would spread like a circle.

So, the equation must be:   F_G = G.M.m/d   (of course, the G constant will have different units)

 

Now, the irregularity that I found was in trying to derive the Gravitational Potential Energy, because following the same idea, we have:

E_PG = Integral[∞->d](G.M.m.dx/x) = -G.M.m.Integral[d->∞](dx/x) = -G.M.m.ln(x)[d->∞] = -G.M.m(ln(∞) - ln(d)) = G.M.m.ln(d) - ∞     

 

That was the irregularity... this energy makes no sense... So it must exist some wrong definition that I made... I don't know if it was in the 2D Gravitational Force or in the concept of Energy in a 2D universe... But if someone know, please share with me the knowlage. Thanks!

Hello Vashta,

This question of yours is about upper highschool / first year college level. Your presentation shows you are thinking hard about the subject +1 for that.

However you seem you be having some trouble matching the  classical Physics (where you have placed the question) with the Maths.

This is not suprising as you will likely not yet have gone far enough in integration theory.

First let me state quite plainly that gravity works well in one, two, three or more dimensions.

It is probably easier to explain in one dimension what you need.

The definition of the gravitational potential is an integral yes.

But it is a definite integral.

Using the definite integral avoids two problems.

The constant of integration 'cancels out', so enabling the fact that you can choose any position as your bas point (as others have mentioned).

It allows the use of the limiting process in evaluating an integral if one of the limits of integration (don't get these two different uses of the maths term 'limits' mixed up) is infinity.

I assume you are aware of the difference between a definite and an indefinite integral ?

But you may not have come across what comes next.

If we integrate  the work done F.dx from -∞  to ∞ or 0 to _∞ we have a problem.

The curve is infinite, but we want the area under it to be finite as it represents the energy in the gravitational field.

This type of integral is known as an improper intergral.

Techniques for dealing with these are usually left until first year maths in college.

 

OK in order to integrat this we need an expression for F in terms of x

This is where Newton's Law comes in.

At this point it is worth comparing Newtons Law for gravity, Coulombs Law for Electric charge and Michells Law for Magnetic Monoploles, which all have the same format.

However there are also some not so suble differences.

Gravity is the simplest because no other information is required
Gravity has one direction (attraction) or sign (positive) only, in addition to magnitude.

Electric fields have both direction or sign ( positive or negative) and magnitude.

Magnetic fields have direction, magnitude and rotation or vorticity or vector curl.

This last is most important because the vector curl represents a pointer or way into an extra dimension to the ones in use.

So for a 2D surface carrying the mag field the curl points into the 3rd dimension and so on.

There are no corresponding pointer vectors in the cases of charge or mass.

Other differences are that it is possible to 'shield' agains an electric or magnetic field or introduce a physical device to set that field to zero.

This is not possible in the case of gravitational fields.
 

 

[Vashta Nerada]

2 hours ago, studiot said:

Hello Vashta,

This question of yours is about upper highschool / first year college level. Your presentation shows you are thinking hard about the subject +1 for that.

However you seem you be having some trouble matching the  classical Physics (where you have placed the question) with the Maths.

This is not suprising as you will likely not yet have gone far enough in integration theory.

First let me state quite plainly that gravity works well in one, two, three or more dimensions.

It is probably easier to explain in one dimension what you need.

The definition of the gravitational potential is an integral yes.

But it is a definite integral.

Using the definite integral avoids two problems.

The constant of integration 'cancels out', so enabling the fact that you can choose any position as your bas point (as others have mentioned).

It allows the use of the limiting process in evaluating an integral if one of the limits of integration (don't get these two different uses of the maths term 'limits' mixed up) is infinity.

I assume you are aware of the difference between a definite and an indefinite integral ?

But you may not have come across what comes next.

If we integrate  the work done F.dx from -∞  to ∞ or 0 to _∞ we have a problem.

The curve is infinite, but we want the area under it to be finite as it represents the energy in the gravitational field.

This type of integral is known as an improper intergral.

Techniques for dealing with these are usually left until first year maths in college.

 

OK in order to integrat this we need an expression for F in terms of x

This is where Newton's Law comes in.

At this point it is worth comparing Newtons Law for gravity, Coulombs Law for Electric charge and Michells Law for Magnetic Monoploles, which all have the same format.

However there are also some not so suble differences.

Gravity is the simplest because no other information is required
Gravity has one direction (attraction) or sign (positive) only, in addition to magnitude.

Electric fields have both direction or sign ( positive or negative) and magnitude.

Magnetic fields have direction, magnitude and rotation or vorticity or vector curl.

This last is most important because the vector curl represents a pointer or way into an extra dimension to the ones in use.

So for a 2D surface carrying the mag field the curl points into the 3rd dimension and so on.

There are no corresponding pointer vectors in the cases of charge or mass.

Other differences are that it is possible to 'shield' agains an electric or magnetic field or introduce a physical device to set that field to zero.

This is not possible in the case of gravitational fields.
 

 

Hi, Studiot,

Yeah, I actually already done calculus and I'm well aware of the difference between a definite and an indefinite integral.
But the thought in question was actually in combine three things: Classical Physics, Calculus and The fantasy of 2D universes.

 

3 hours ago, studiot said:

The definition of the gravitational potential is an integral yes.

But it is a definite integral.

Here I didn't understand why you reinforced that "for energy we must use a definite integral" because it was exactly what I did... I made an integral going from Infinity to "d"... It is in the first post.
And actually, We can technically use a Indefinite Integral, but for that we must use a boundary condition to found the value of C, like that:

E_PG(x) = Integral(G.M.m.dx/x²) = -G.M.m/x + C
Boundary Condition: E_PG(∞) = 0, so:      0 = -G.M.m/∞ + C      ->      0 = 0 + C      ->      C = 0
Finally: E_PG(x) = -G.M.m/x

So the question was all the time in how would it be the E_PG for a 2D universe since the Energy (Area below the Force graph) tends to infinity in that case, and for that I got no answer.

[Genady]

12 minutes ago, Vashta Nerada said:

So the question was all the time in how would it be the E_PG for a 2D universe since the Energy (Area below the Force graph) tends to infinity in that case, and for that I got no answer.

I thought you got the answer, even more than one. Maybe you have missed this one, for example:

13 hours ago, Genady said:

In fact, you can get a nice formula in your 2D case as well: since adding a constant to energy doesn't change anything, you can simply remove the  ln(L) part and get the general equation for absolute value for any distance, GMm*ln(d). 

 

[Vashta Nerada]

3 hours ago, Genady said:

I thought you got the answer, even more than one. Maybe you have missed this one, for example:

 

Hi, Genady!
Sorry, I actually saw your comment, but I'm still trying to convince myself that this equation may be right, but the truth is I kind of don't think so. It just don't feel right.

Because if E_PG(x) = G.M.m.ln(x) , then the Energy at the infinity would be: E_PG(∞) = G.M.m.ln(∞) = ∞

And I don't kow... That makes no sense for me...

How could the Energy of two bodies that don't interact with each other be Infinite?

[studiot]

9 hours ago, Vashta Nerada said:

Here I didn't understand why you reinforced that "for energy we must use a definite integral" because it was exactly what I did... I made an integral going from Infinity to "d"... It is in the first post.

Gravitational potential is not 'energy'

The PE results from a potential difference

A potential is not the same as a potential difference, although they will have the same units.

 

Take an isolated single body, possessing mass.

There exists a gravitational potential around it.

But there is no energy, potential or otherwise involved.

 

Now introduce a second massive body.

Potential energy now arises from the configurational interaction of the two bodies and perhaps also kinetic energy.

This is the same for electric and magnetic fields.

It is also worth noting that the potential energy is conventionally set to negative infinity at the source, reducing to zero at infinite distance.

I repeat it is easiest to try these questions out in one dimension first.

Edited February 17, 2022 by studiot

[Genady]

6 hours ago, Vashta Nerada said:

How could the Energy of two bodies that don't interact with each other be Infinite?

It will take an infinite amount of energy to remove a body from any given position to infinity.

[studiot]

23 minutes ago, Genady said:

It will take an infinite amount of energy to remove a body from any given position to infinity.

If that were the case then the gravitational field would contain an infinite amount of energy.

The point that I already made is that whilst the length of the bounding curve is infinite, the area under it remains finite.

https://en.wikipedia.org/wiki/Improper_integral

Edited February 17, 2022 by studiot

[Genady]

1 minute ago, studiot said:

If that were the case then the gravitational field would contain an infinite amount of energy.

The point that I already made is that whilst the length of the bounding curve is infinite, the area under it remains finite

In my response above I refer to the OP "fantasized" gravitational force that is inversely proportional to distance rather than to distance squared. Not to the Newtonian gravity.