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If I have an arbitrarily shaped object made of a uniform material of some specified resistivity, how would I go about calculating the resistance between two measurement points with known contact geometry?

Is there a general formula for this? (other than just Maxwell's equations) If so, where would I find a derivation?

Edit: Some simulations re answer below: enter image description here enter image description here

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  • $\begingroup$ Are you talking about a single path integral? The resistivity is, from dimensional analysis, the resistance times a distance. The simple formula is just the scalar resistance times the cross sectional area, divided by a length. Looking at that, I would convert the divergence of the resistivity tensor into a dot product with the unit length. On the RHS, you would have an integral over area of some average scalar resistance or total resistance. This is only a guess, which is why I left it as a comment. $\endgroup$
    – honeste_vivere
    Commented Aug 10, 2015 at 11:21

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Well, yes you can, but it is usually very hard. Here are the steps:

  1. Solve the Laplace equation: $$ \nabla^2V = 0 \, .$$ In your case, find the general solution in spherical coordinates. Try to use every simplification you can. You might wonder why you don't solve Poisson's equation:
    $$ \epsilon\nabla^2V = \rho \, .$$ That's because a conductor is an equal number of lattice positive and moving negative, so you have a net null density of charge.

  2. Find the electric field with:
    $$ E = -\nabla V $$ You should still have a term dependent on your $V_0$, the potential difference between your two points.

  3. Find the current density with: $$ J = \sigma E \, .$$

  4. Find the total current $I$ by integrating over any closed surface containing only one of your two contact point.

  5. $$V_0 = R I \, !$$

That's it. I have used it to find the resistance for certain geometries when you could do lots of simplifications to find the solution of $V$, but I don't know how it can be applied for more general problems.

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  • $\begingroup$ yes, this is what I was looking for. I'd figured it talking with RossMillikan below, but it never made it to a proper answer. $\endgroup$
    – Lucas
    Commented Aug 10, 2015 at 12:40
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Not that I am aware of. The best I could do is to make a numerical simulation. Make a mesh of points that fill the object, compute the resistance between each neighbor pair, and do a numerical relaxation to determine the potential throughout the object.

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  • $\begingroup$ I'd decided this approach was incorrect, though perhaps you could persuade me otherwise. The reason I had decided this was this: Lets say we use a rectangular grid. Now consider a relatively thin straight object. If the object, of length $l$ is aligned with the grid, it would contain $n$ resistors and have a resistance $R$, if it lies diagonally to the grid, then it would contain $~\sqrt{2}n$ resitors and have a resistance of $\sqrt{2}R$. Perhaps you are suggesting modifying the resistances as part of the relaxation, though I'm not sure how one would go about that. $\endgroup$
    – Lucas
    Commented Jun 20, 2015 at 22:45
  • $\begingroup$ No, the resistances are fixed during the relaxation, only the potentials at each node are varied. When the relaxation converges, you compute the current flow for unit input voltage and get the resistance. The node spacing has to be small compared to all the feature sizes of the object for exactly the reason you mention. If your node spacing is small compared to the width of the rod, you will get $\sqrt 2$ more lines of points sharing the current to compensate for the added length and get the same resistance. $\endgroup$
    – Ross Millikan
    Commented Jun 20, 2015 at 22:53
  • $\begingroup$ I'm not 100% convinced, but you've persuaded me to code something up and see. Assuming this is correct though, doesn't it suggest an equation expressed in terms of the minimisation of some function? $\endgroup$
    – Lucas
    Commented Jun 20, 2015 at 23:32
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    $\begingroup$ My model above has about 10,000 nodes. It took about 100,000 iterations to converge. I hope I've done it right, some reverse engineering implies what I've been doing is solving Laplace's equation $\nabla^2 \phi = 0$ with boundary conditions $(\hat{n} \cdot \nabla) \phi = 0$ for non-contacts and $\phi = const$ for the contacts. This seems very plausible to me now. You've been very helpful, thanks. $\endgroup$
    – Lucas
    Commented Jun 21, 2015 at 13:19
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    $\begingroup$ Is this the same as finite element method? $\endgroup$
    – user122089
    Commented Feb 12, 2018 at 13:02

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