Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Let's say that we want to measure the gravity force in 1D, 2D, 3D and higher spatial dimensions.

Will we get the same force strength in the first 3 dimensions and then it will go up? How about if we do this with Electromagnetic force?

I've include the Electromagnetic force just to see if I can find an analogy to the gravity force behavior.

share|improve this question
4  
Your questions are poorly worded. Please rephrase them to explain the context. –  Vladimir Kalitvianski Jan 6 '13 at 15:17
1  
...or if you're not sure about the maths, at least give some examples that could clarify what you mean or explain your intuition. –  Vibert Jan 6 '13 at 15:48
    
@user17530 I agree with Vibert and Vladimir. I propose the following rewording "What are the force vs distance relationships for gravity and electromagnetism for small integer numbers of dimensions?". If this is what you're asking, you can edit the question in this way, or one of us can do it. –  twistor59 Jan 6 '13 at 17:32
1  
If OP is just asking for the radial dependence of the non-relativistic generalizations of Newton's gravitational force and the Coulomb force in $d$ spatial dimensions, then the answer is that $F\propto r^{1-d}$, due to Gauss' law, because a Gauss surface is $d-1$ dimensional. –  Qmechanic Jan 6 '13 at 22:32
1  
user17530, if you are asking about how gravity falls off in different spaces, see Qmechanic's comment above: gravity force goes down faster for higher dimensions, not slower. Gravity stays constant for 1D, falls off more slowly than we are accustomed for 2D (at 1/s), and drops off faster and faster for higher $d$. The same relationship applies to light from a star, so it would be very easy to freeze to death if you were, say, nine dimensional. See this related answer for a longer informal discussion. –  Terry Bollinger Jan 30 '13 at 4:04
show 1 more comment

1 Answer

Let us begin with an example from electromagnetism, which you may be familiar with. Gauss' law is given in 3 dimensions by:

$\int\int E.ds= \frac{Q}{\epsilon}$, where $E$ is the electric field produced, $Q$ is the charge and the integral is performed over a surface that encloses the charge. In 3 dimensions, the simplest shape to enclose the charge is the sphere, so we will choose this to make the mathematics simpler.

Making the reasonable assumption that $E$ is constant along all points in the sphere, the above equation becomes:

$4\pi r^2E =\frac{Q}{\epsilon}$ and therefore, $E = \frac{Q}{4\pi r^2\epsilon}$.

It is not hard to see how to generalize this approach to 2 dimensions. Instead of considering a surface integral around a sphere, we simply need to consider a line integral around a circle. The modified 2-D Gauss' law will then become:

$\int E.dl= \frac{Q}{\epsilon}$, where the integral is performed around a circle enclosing the charge $Q$.

Again, evaluating this integral gives:

$2\pi rE = \frac{Q}{\epsilon}$, and therefore, $E = \frac{Q}{2\pi r\epsilon}$

Finally we can generalize to 1-D, where a circle in 1-D becomes a line, and line integral changes to simply adding together points. If we pick a line of length 2r, enclosing the charge, gauss' law will become:

$2E = \frac{Q}{\epsilon}$, that is, $E = \frac{Q}{2\epsilon}$


Now we must see how to generalize this approach to gravity.

In 3-Dimensions, the gravitational field produced by a mass $M$ is given by:

$g = \frac{GM}{r^2}$

If we introduce a new variable, $k$, defined as $k = \frac{1}{4\pi G}$, then we can re-write the field as:

$g = \frac{M}{4\pi r^2k}$. Compare this to the electric field in 3-D, which is $E = \frac{Q}{4\pi r^2\epsilon}$

One could thus construct a "Gauss'" law for the gravitational field and construct the 2D and 1D fields, in the same way I did for the electric fields above.

The results will be the same as that for the electric field, but with $\epsilon$ replaced with $k = \frac{1}{4G\pi}$, and $Q$ replaced with $M$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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