2
$\begingroup$

I am wondering if my line of thought is correct - and thus the resulting answer to the problem above would be correct.

As we know the gravitational force (of two point masses) is given by $$F = G\frac{m_1m_2}{r^2}.$$

So the gravitational force/vector field reduces with the distance squared. Now this is the formula in 3 spatial dimensions - and I always picture it as a point with gravitational field lines moving outward. Then the "strength" of the field would be the density of the lines. And hence the density drops with the distance squared (as it is inversely proportional to the area of the sphere at that distance).

Now taking this line of thought to other situations we can think of course about a hypothetical 2 dimensional world. Here gravity would also be. And here we can also see the density of the "gravitational field lines". However as they propagate only in 2 spatial dimensions the density would be inversely proportional to the circumference of the circle at a distance $r$. And hence the formula would lose the square and become like:

$$F = G\frac{m_1m_2}{r}$$

(With change $G$, and obviously we can't talk about mass in 2d).

Is this line of thought correct?

$\endgroup$
5
$\begingroup$

Yes, this is correct. Formally, you could write for circles around the point mass $C_1$ and $C_2$: $$\int_{C_1}\vec F\left(r_1\right)\cdot \mbox{d}\vec s=\int_{C_2}\vec F\left(r_2\right)\cdot\mbox{d}\vec s$$

By rotational symmetry, this can be written as:

$$2\pi r_1 F_1=2\pi r_2 F_2 \Rightarrow F_2r_2=F_1r_1$$

See also.

$\endgroup$
  • $\begingroup$ I was wondering, isn't there something weird happening in 2 dimensions given $\int_1^\infty \frac{1}{r} dr = \infty$ ? $\endgroup$ – reuns Jul 27 '18 at 22:40
-2
$\begingroup$

You can never know whether your line of thought is correct or not because you have asked an unscientific question.In science the test of all ideas is experiment and you can never conduct an experiment in your hypothetical 2 dimensional world.

$\endgroup$
  • 1
    $\begingroup$ A circle in 2 dimensional space, would exhibit the same behaviour as a infinitely long cylinder in 3 spatial dimensions. Or almost similar when staying close to a "very long" cylinder - which is something we CAN test. -- Funny: when I actually saw the "gauss law" I knew I was asking a stupid question and was correct. $\endgroup$ – paul23 Sep 29 '13 at 13:32
  • $\begingroup$ There are hundreds of peer-reviewed papers on 2d and 1d gravity (not speaking about of thousands of papers on string theory and such, most of which also cannot be experimentally verified in the foreseeable future). I think your definition of science needs to be considerably broadened. $\endgroup$ – user23660 Sep 29 '13 at 13:55
  • $\begingroup$ @paul23 Well we have a mass problem. Do you take the limit with the length $\to \infty$ and the density of mass being proportional to $1/L$ to obtain that 2d gravity is a limiting case of 3d gravity ? What would it mean in term of compatibility of 3d and 2d physical laws ? $\endgroup$ – reuns Aug 6 '18 at 23:18

Not the answer you're looking for? Browse other questions tagged or ask your own question.