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Isotropic surfaces questions...by Dr Bwts
Tags: isotropic surfaces 
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#1
Jan2512, 08:48 AM

P: 18

Hi,
I have the following definition for an isotropic surface: the normals to the measured surface are randomly distributed. Am I right in thinking the following: 1) The surface of a sphere is isotropic? 2) The surface of a cube is anisotropic? I think (2) is correct but not sure about (1) well to be honest I not that sure about (2) either! Thanks to anybody who can put me straight. Dr B 


#2
Jan2512, 09:52 AM

P: 4,573

When you say randomly distributed are you talking about a uniform distribution? 


#3
Jan2512, 10:08 AM

P: 18

Hi Chiro,
I dont rightly know. I am reading some papers on morphology and it was one of the easier to understand definitions of an isotropic surface. Would the angular distribution of exterior surface normals be uniform for a sphere? 


#4
Jan2512, 10:13 AM

P: 4,573

Isotropic surfaces questions...
For the sphere, I would imagine that it is indeed isotropic. The way you could start to prove this is by showing that all normal vectors have the same rate of occurence as every other normal vector (in fact this is the general method you would use). By showing this you would have provide isotropy if the distribution is indeed uniform. 


#5
Jan2512, 10:19 AM

P: 18

Thanks for the reply.
So just so I'm clear, in the case of the cube I would get a distribution that would be orthotropic? 


#6
Jan2512, 10:24 AM

P: 4,573




#7
Jan2512, 10:29 AM

P: 18

Orthotropic  multually perpendicular planes of symmetry (a particular case of anisotropy)



#8
Jan2512, 11:36 PM

P: 4,573

For a cube if the normals for the edges of the cube are the same as one of the other normals in the set, then that seems to be ok. Also I think you should clarify the specific nature of a 'random' set. My interpretation is that it includes every possible normal possible within a specific geometry: in this case 3D space corresponds to every unit vector in cartesian 3D space. For the cube you should also get a uniform distribution but the number of unit normal vectors is countable unlike your sphere example. If we allow that we have any uniform distribution then this would not be anisotropic but if we specified that is to include all possibilities within some certain space, then it would still be anistropic even though each possible unit normal in the cube case is still as likely as the others. 


#9
Jan2612, 04:24 AM

P: 18

OK thanks chiron that has been very helpful



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