Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

I'm plotting the equipotential of the potential function: $$\phi(x,y,z):=\frac{1+x^{2}+y^{2}+z^{2}}{[\left(1+x^{2}+y^{2}+z^{2}\right)^{2}-4z^{2}]^{1/2}}$$ Mathematica code:

Φ[x_, y_, z_] := (1 + x^2 + y^2 + z^2)/((1 + x^2 + y^2 + z^2)^2 - 4 z^2)^(1/2)

So I write, for example

ContourPlot3D[Φ[x, y, z] == 1, {x, -100, 100}, {y, -100, 100}, {z, -100, 100}]

But, I get this image: enter image description here

And, sometimes if I change the plot region, I get absolutely nothing.

My question is, what did I wrong? I think I should obtain smooth 3D curves, but I don't.

share|improve this question
2  
The problem is that $\Phi(x,y,z)\ge1$ everywhere... –  Rahul Narain Jun 8 '13 at 19:46
2  
I think you should manipulate it mathematically a bit further. –  Spawn1701D Jun 8 '13 at 19:49
3  
Try p[x_, z_] := (x x + z^2)/((x x + z^2)^2 - 4 z^2)^(1/2) a = 5; Plot3D[p[x, z], {x, 1, a}, {z, -a, a}] –  belisarius Jun 8 '13 at 20:10
1  
Or a = 2; ContourPlot3D[{\[CapitalPhi][x, y, z] == 1 + 1/100, \[CapitalPhi][x, y, z] == 1 + 1/10}, {x, -a, a}, {y, -a, a}, {z, -a, a}] –  belisarius Jun 8 '13 at 20:53
2  
As Rahul already pointed out, your contour is never crossed, only touched. ContourPlot3D fails with such things which I already answered here. For a quick solution you should follow belisarius' last comment. –  halirutan Jun 8 '13 at 21:46
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.