Understanding Triply Periodic Surfaces: Translations and Invariance?

  • Context: Graduate 
  • Thread starter Thread starter Demon117
  • Start date Start date
  • Tags Tags
    Periodic Surfaces
Click For Summary
SUMMARY

A surface is classified as triply periodic when it remains invariant under three translations in R³. The discussion centers around the primitive defined by the equation cos(x) + cos(y) + cos(z) = 0. The translation is represented as (x, y, z) → (x + Δx, y + Δy, z + Δz). It is established that the gradient of the level set, given by ∇(cos(x) + cos(y) + cos(z)) ≠ 0, indicates that the surface is invariant under these translations, as long as x, y, z are not equal to zero.

PREREQUISITES
  • Understanding of triply periodic surfaces in R³
  • Knowledge of gradient and level sets in multivariable calculus
  • Familiarity with trigonometric functions and their properties
  • Basic concepts of invariance and translations in mathematical contexts
NEXT STEPS
  • Research the mathematical properties of triply periodic surfaces
  • Study the implications of gradient non-vanishing in level sets
  • Explore the concept of invariance under transformations in R³
  • Learn about applications of triply periodic surfaces in materials science and crystallography
USEFUL FOR

Mathematicians, physicists, and researchers in materials science who are exploring the properties of triply periodic surfaces and their applications in various fields.

Demon117
Messages
162
Reaction score
1
So I understand that a surface is triply periodic when the surface is invariant under three tanslations in R^{3}. When looking at the primitive for example, how is that translation defined? Say that the primitive is a set defined by the equation

cos(x)+cos(y)+cos(z)=0

My guess is that the translation would take (x,y,z)\rightarrow(x+\Delta x, y+\Delta y, z +\Delta z). Is that incorrect thinking?
 
Physics news on Phys.org
Readjustment

So I have new information now. Apparently showing that the gradient of a level set does not vanish somehow also shows that a set defined as above is invariant under three translations. How is that the case?

With that in mind, the gradient of the above is

\nabla(cos(x)+cos(y)+cos(z)) \ne 0 \rightarrow (-sin(x),-sin(y),-sin(z)) \ne 0 \rightarrow x,y,z \ne 0

I don't see how this guarantees the surface is invariant under three translations in R^{3}. Any suggestions?
 

Similar threads

  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 12 ·
Replies
12
Views
4K
Replies
19
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
3K
Replies
9
Views
895
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K