Foreword: I'm a video game developer and writer. This question is about a hypothetical set of physics pertaining to a fictional setting. However, I feel that neither the Game Development StackExchange nor the Worldbuilding StackExchange would be appropriate for this, because it's essentially pure physics (an extension of real physics that happens to not be real). Hopefully this is appropriate here, and if not, I'll try the Math StackExchange.

Anisotropic Gravity

With regards to Newtonian physics, Newton's law of gravitational attraction uses a gravitational constant $G. G$ is isotropic in our universe. For the purposes of a video game I am making, I want to construct a set of physics where Newton's law is a special form of a more general equation, wherein $G$ may have different values along three absolute perpendicular axes of 3-space (i.e. the game's three-dimensional coordinate system), so $G$ is anisotropic in this hypothetical space. Newton's law would thus be a special form where all three values are the same.

So I want to expand Newton's law so that it can calculate the attractive force between two massive objects (I'm currently just working with rigid-body spheres in Unity) on the axes $X$, $Y$, and $Z$, where each axis has a gravitational constant, respectively $G_x,$ $G_y$, and $G_z$. I have tried to do this by applying the formula for the diagonal of a rectangular prism to calculate the distance along each axis and then multiply the square of each by its gravitational constant. Here's one attempt:

$$F = \sqrt{\frac {G_x(m_1m_2)}{r_x^2} + \frac {G_y(m_1m_2)}{r_y^2} + \frac {G_z(m_1m_2)}{r_z^2}}$$

(I apologize for not being familiar with MathJax).

I have run into a problem, however, because I cannot find a way to express the more general equation so that it behaves identically to Newton's law when $G_x$ = $G_y$ = $G_z$. The above expansion does not work. But I feel that an expansion can exist. I recognize that in real space there are no absolute axes, so I think in that case you'd have to use a sort of value-gradient. But for my purposes, three axes are enough.

So, how can I expand Newton's law to account for anisotropic gravity?

  • $\begingroup$ Hi, I think I have got the expresions sorted properly, will you please check them also mathjax is covered here if you need to alter your post, thanks meta.math.stackexchange.com/q/5020 $\endgroup$
    – user108787
    Nov 9, 2016 at 12:25
  • $\begingroup$ @CountTo10 Thank you, that's how I had it. $\endgroup$ Nov 9, 2016 at 16:05
  • $\begingroup$ You didn't apply the expression for the diagonal of a prism. Each quantity has to be squared. (Like pythagorean theorem). $\endgroup$
    – garyp
    Nov 11, 2016 at 16:26
  • $\begingroup$ I cannot understand the way of thinking of people who vote for closing as "non-mainstream physics" every request of the kind "what if..." as the present question. This seems to me a perfectly in-topic question about what mainstream physics would say about a non conventional situation. I would say that it is a clear case of "specific questions evaluating new theories in the context of established science", then an allowed question. $\endgroup$ Dec 22, 2019 at 15:17

2 Answers 2


For an anisotropic gravity, Newton's gravitational law $$\vec{F}=\dfrac{G.m_1.m_2}{r^2}.\vec{e_r}=\dfrac{G.m_1.m_2}{r^3}.\begin{bmatrix}x \\y\\z \end{bmatrix}$$ will still hold, though $G$ wouldn't be a scalar, but rather a rank-2 tensor (a matrix).

The new formula would be :$$\vec{F}=\dfrac{m_1.m_2}{r^3}\begin{bmatrix}G_x & 0 & 0\\0 & G_y & 0\\ 0 & 0 & G_z\end{bmatrix}\begin{bmatrix}x \\y\\z \end{bmatrix}$$

Which can be expanded to $$\vec{F_x}=\dfrac{G_x.m_1.m_2}{r^3}.x\vec{e_x}$$ with similar expressions for $\vec{F_y}$ and $\vec{F_z}$.

Notice that for $G_x=G_y=G_z=G$, the equation reduces back to its familiar form as stated by Newton with $G$ a scalar.

However, it is important to note that the total force in an anisitropic field will not be a head to head attraction, $\vec{F}$ and $\vec{e_r}$ will not be collinear if $G_x$, $G_y$ and $G_z$ don't have the same value.


I'm not sure that this is a physics question, but is there any reason why you should not just use

$$ \vec{F} = -(\alpha r_x, \beta r_y, \gamma r_z) \frac{Gm_1m_2}{|\vec{r}|^3} $$

Where $\vec{r}$ is the separation, and $(\alpha r_x, \beta r_y, \gamma r_z)$ is a scaled separation expressed in cartesian coordinates (so $\vec{r} = (r_x, r_y, r_z)$)? Setting $\alpha = \beta = \gamma = 1$ yields the normal law:

$$\vec{F} = -\vec{r}\frac{Gm_1m_2}{|\vec{r}|^3}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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