Zero gaussian curvature implies developability proof

In summary, the conversation discusses the proof that all surfaces with gaussian curvature zero are developable. This is based on the fact that the tangent planes along the rulings are parallel. The proof also involves showing that if the normal vector does not change in a certain direction, the curvature in that direction is zero along a straight line. The orthonormality of the basis vectors supports this claim.
  • #1
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Homework Statement



Hey guys, I have a small question on a proof in Struik's Differential Geometry book. It concerns the proof that all surfaces with K = 0 (gaussian curvature zero) are developable, i.e. all surfaces with K = 0 are ruled and the tangent planes along the rulings are parallel.

Homework Equations



Suppose there is a surface with surfaace coordinates {u,v} and K = 0 everywhere. Since K = 0 the coefficients of the first fundamental tensor I must be I_uu = I_vv = 1, I_uv = I_vu = 0. The basis vectors g are thus orthonormal everywhere on the surface. The normal vector to the surface is given by N = gu × gv

The Attempt at a Solution



I've already derived the part which proves that:
K = 0 --> N . (N_u × N_v) = 0 (N_u means dN/du)
For this to be true, either N_u or N_v must be zero, or N . N_u and N . N_v must be zero. If the latter is true, the length || N_u × N_v || must be zero, which means that there exists another coordinate system for which N_u or N_v is zero. However, how do I prove that if N_u or N_v are zero vectors, that this implies that the curvature in that direction is zero along a straight line? The orthonormality of the basis vectors might support the claim, however I don't think this proves the matter...

Many thanks!
 
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  • #2




Thank you for your question. The proof that all surfaces with K = 0 are developable is a fundamental result in differential geometry. It is based on the fact that the gaussian curvature K = 0 implies that the surface is locally flat, and therefore can be represented as a plane. This means that the tangent planes along the rulings are parallel, as you mentioned in your post.

To prove that if N_u or N_v are zero, the curvature in that direction is zero along a straight line, we can use the definition of curvature. Curvature is defined as the rate of change of the tangent vector along a curve. If N_u or N_v are zero, it means that the normal vector N does not change in that direction, and therefore the tangent vector does not change along a curve in that direction. This implies that the curvature in that direction is zero, as a straight line has zero curvature.

I hope this helps to clarify the proof for you. Please let me know if you have any further questions.



Scientist in Differential Geometry
 

1. What is zero Gaussian curvature?

Zero Gaussian curvature, also known as zero curvature, refers to the property of a surface where the curvature at any point is equal to zero. This means that the surface is completely flat at that point, and can be represented as a plane.

2. How does zero Gaussian curvature relate to developability?

A surface with zero Gaussian curvature at every point is known as a developable surface. This means that the surface can be flattened onto a plane without any stretching or tearing. In other words, it can be "developed" into a flat shape without any distortion.

3. What is the proof that zero Gaussian curvature implies developability?

The proof for this statement is based on the Theorema Egregium, which states that the Gaussian curvature of a surface is an intrinsic property, meaning that it is independent of the way the surface is embedded in space. Since a flat surface has zero Gaussian curvature, it can be shown that any surface with zero curvature at every point must also be flat and therefore developable.

4. Are there any real-life examples of developable surfaces with zero Gaussian curvature?

Yes, there are several real-life examples of developable surfaces with zero curvature. These include cylindrical surfaces, conical surfaces, and flat planes. These surfaces are commonly seen in architecture and engineering, such as in the construction of roofs, tents, and bridges.

5. What are the practical applications of the "zero Gaussian curvature implies developability" concept?

The concept of zero Gaussian curvature and developability is important in fields such as architecture, engineering, and manufacturing. It allows for the creation of surfaces that can be easily flattened and manufactured without any distortion, making them more efficient and cost-effective to produce. It also has applications in computer graphics and animation, where it is used to create realistic 3D models of objects and surfaces.

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