What happens to gravity in one dimension of space?

AI Thread Summary
In a discussion on gravity in one-dimensional space, it was noted that while gravity behaves as F=x/r^2 in three dimensions and F=x/r^3 in four dimensions, the behavior in one dimension is less straightforward. Initial assumptions suggested gravity would strengthen with distance, represented as F=xr, implying a repulsive force. However, further analysis revealed that the gravitational potential follows Laplace's equation, leading to a constant gravitational force, independent of distance. Ultimately, the consensus is that gravity does not change with distance in one-dimensional space, contradicting earlier interpretations. This clarification highlights the complexities of gravitational behavior across different dimensions.
KeplerJunior
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I recently heard that it was found how gravity would behave in different dimensions of space. Apparently, while in our 3 dimensions of space gravity is F= x/r^2, in 4 dimensions it would be F=x/r^3, so it would be weaker, and in 2 dimensions it would be F=x/r so it would be stronger.What it didn't say was what would happen in 1 dimension, it follows from the above that it would be F=x/r^-1, which simplifies to F=xr, which would indicate that gravity would get stronger as you went further away from the object. Is this what would happen? And if so does that mean gravity would become a repulsive force?
 
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You skipped over a key integer. Hint: What integer is between -1 and +1?
 
Oh of course, so would it just be x/r^0 so x/1, so x. So then distance wouldn't change anything in 1 dimension
 
KeplerJunior said:
I recently heard that it was found how gravity would behave in different dimensions of space. Apparently, while in our 3 dimensions of space gravity is F= x/r^2, in 4 dimensions it would be F=x/r^3, so it would be weaker, and in 2 dimensions it would be F=x/r so it would be stronger.What it didn't say was what would happen in 1 dimension, it follows from the above that it would be F=x/r^-1, which simplifies to F=xr, which would indicate that gravity would get stronger as you went further away from the object. Is this what would happen? And if so does that mean gravity would become a repulsive force?

Um, actually, the integer x/r dr is xln(r). Which means the equation of gravity in one dimension would be:

f(r) = gravitational force = xln(r). So gravity would according to this increase with increased distance, something I find hard to believe. It means the further you are away, the stronger it'd pull, well until you get to a distance of 0, then it starts to push you away again... because ln(r)<0 if r<1. This is something similar to strong force, it's repulsive at a certain closeness, but if they stray too for from each other it becomes a very strong attractive force.
 
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I thought that gravity did not change with distance in 1 dimensional space?
 
It doesn't. Laplace's equation in one dimension for the gravitational potential exterior to a point mass is simply ##\frac{\mathrm{d} ^2 \varphi}{\mathrm{d} x^2} = 0## so ##\varphi = Ax + B## and ##g = - \frac{\mathrm{d} \varphi}{\mathrm{d} x} = -A##.
 
WannabeNewton said:
It doesn't. Laplace's equation in one dimension for the gravitational potential exterior to a point mass is simply ##\frac{\mathrm{d} ^2 \varphi}{\mathrm{d} x^2} = 0## so ##\varphi = Ax + B## and ##g = - \frac{\mathrm{d} \varphi}{\mathrm{d} x} = -A##.

So I was wrong?
 

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