Why Torsion = 0 => Planar Curve in this Proof

The discussion revolves around the relationship between torsion and the planarity of curves in three-dimensional space, specifically addressing the claim that zero torsion implies the curve is planar. Participants explore the implications of the mathematical argument presented in a lecture regarding the dot product of a curve with a constant vector and its relation to defining a plane.

Discussion Character

Technical explanation
Conceptual clarification
Debate/contested

Main Points Raised

Some participants explain that if torsion is zero, then the binormal vector is constant, leading to the conclusion that the curve lies in a plane defined by the dot product with a constant vector.
Others question the clarity of the argument that a constant dot product with a vector represents a plane, expressing confusion about the geometric interpretation of this relationship.
One participant provides an example of a circle parameterized in three-dimensional space and discusses how the dot product with a constant vector yields a constant value, suggesting the curve lies in a specific plane.
Another participant challenges the choice of the constant vector and discusses the implications of different vectors on the dot product and the resulting geometric interpretation.
Some participants acknowledge the difficulty in understanding the proof and the necessity of the professor's explanation to clarify the relationship between the curve and the plane.

Areas of Agreement / Disagreement

Participants express a mix of agreement and confusion regarding the implications of the proof. While some understand that the zero torsion condition leads to planarity, others remain uncertain about the geometric interpretation and the specific examples discussed.

Contextual Notes

There are limitations in understanding the geometric implications of the dot product and its relation to defining a plane, as well as the specific conditions under which the examples are valid. The discussion highlights the need for clarity in the definitions and assumptions used in the proof.

Rippling Hysteresis Messages 24 Reaction score 0

I was watching a lecture that made the conclusion about the torsion being equal to zero necessitated that the path was planar. The argument went as follows: -Torsion = 0 => B=v, which is a constant -(α⋅v)'=(T⋅v)'= 0 => α⋅v= a, which is a constant (where α is a function describing the path and T is the tangent vector) -If α⋅v = constant, then the curve is planar.

Everything up to the end made sense, but I wasn't sure why the professor said that α⋅v= constant is the equation of the plane when the velocity was constant. Can you help me fill in the gaps?

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Science Advisor Gold Member Messages 1,084 Reaction score 658

For a fixed vector ##v\in\mathbb^3## and constant ##C\in\mathbb##, the set ##\## is a plane.

Your professor's argument shows that ##\alpha(t)\cdot v=C## for all ##t##, and hence the curve ##\alpha## lies in this plane.

Could you say which part of this argument isn't clear?

Likes Reactions: Rippling Hysteresis Rippling Hysteresis Messages 24 Reaction score 0 Infrared said:

For a fixed vector ##v\in\mathbb^3## and constant ##C\in\mathbb##, the set ##\## is a plane.

Your professor's argument shows that ##\alpha(t)\cdot v=C## for all ##t##, and hence the curve ##\alpha## lies in this plane.

Could you say which part of this argument isn't clear?

The lecturer has used α(s) in the past to be any curve in R^3. I'm not sure why dotting that curve with a constant value represents a plane. I can see how the dot product will equal a scalar constant, but think of a plane as something as ax+by+cz=d, but am having trouble envisioning that from α⋅v= a.

Further, while I seemed to follow the very long build-up in which he derived the Frenet basis (with T, N, and B) and worked to the steps I've written, I'm also missing something deeper that's conceptual. If α(s) is a curve, one example is a circle parameterized as r(cos (t/r), sin (t/r), 0). How would taking the dot product of this with a constant vector, v, equal a constant when the curve is not time-independent?

I'm showing my supreme ignorance here I'm sure (this is a lecture series I stumbled onto on Youtube and am enjoying. I have take vector calc before, but this is my introduction to differential geometry).

Science Advisor Gold Member Messages 1,084 Reaction score 658 Rippling Hysteresis said:

f α(s) is a curve, one example is a circle parameterized as r(cos (t/r), sin (t/r), 0). How would taking the dot product of this with a constant vector, v, equal a constant when the curve is not time-independent?

Let's just take ##r=1##, so in your example ##\alpha(t)=(\cos(t),\sin(t),0)##. Then ##\alpha(t)\cdot\begin0\\0\\1\end=0## for all ##t##, so ##\alpha(t)## lies in the plane ##\left\,## which is just the plane ##z=0##. And indeed you can see this to be the case.

Edit: maybe the word "planar" is causing you confusion. It just means that the curve lies in a plane, not that it is a (whole) plane.

Last edited: Aug 26, 2020 Likes Reactions: Rippling Hysteresis Rippling Hysteresis Messages 24 Reaction score 0 Infrared said:

Edit: maybe the word "planar" is causing you confusion. It just means that the curve lies in a plane, not that it is a (whole) plane.

Couldn't the vector also be v= (1/sqrt(2), 1/sqrt(2), 0), which then wouldn't equal a constant? So the dot product would be 1/sqrt(2)(cos(t) + sin(t)). Or perhaps that choice of v violates a rule?

Science Advisor Gold Member Messages 1,084 Reaction score 658

The vector ##v## is the binormal (which is constant, so this makes sense). The argument you gave in your OP shows that ##\alpha(t)\cdot v## is a constant.

Rippling Hysteresis Messages 24 Reaction score 0 Infrared said:

The vector ##v## is the binormal (which is constant, so this makes sense). The argument you gave in your OP shows that ##\alpha(t)\cdot v## is a constant.

So would the α(s) (a circle) I've chosen be wrong or would it be the vector (1/sqrt(2)(1,1,0). It seems like the vector is constant and the circle has zero torsion which is what I'm hung up about. The proof makes sense, but there's a disconnect with a specific example.

And further, I can see why this example with the circle dotted with my vector gives a plane, I don't think it's obvious to me that it HAS to be that way.

Science Advisor Gold Member Messages 1,084 Reaction score 658 Rippling Hysteresis said: So would the α(s) (a circle) I've chosen be wrong or would it be the vector (1/sqrt(2)(1,1,0).

Your vector ##v## is wrong. The binormal vector for ##\alpha## is ##(0,0,1)##, as you should compute.

Rippling Hysteresis said:

And further, I can see why this example with the circle dotted with my vector gives a plane, I don't think it's obvious to me that it HAS to be that way.

You're right, it's not obvious. That's why the professor had to supply a proof. Likes Reactions: Rippling Hysteresis Rippling Hysteresis Messages 24 Reaction score 0 Infrared said:

Your vector ##v## is wrong. The binormal vector for ##\alpha## is ##(0,0,1)##, as you should compute.You're right, it's not obvious. That's why the professor had to supply a proof.

Thanks for your patience. It's still a challenge for me to make sense of proofs in concrete terms-- work in progress. I went back earlier into the lecture and see now how (after computing the cross product) the binormal vector must be (0,0,1) in this particular case. The trouble all started when he used the statement "α⋅v=a, therefore it's a plane." I suppose I still don't recognize that form as something I've seen as a plane in the past. But the example at least is in consonance with the conclusions.

So to re-state and make sure I get this, let's say you had a circle drawn in the plane of x+y+z=1. In this case, once you parameterize, the binormal vector would work out to v= 1/sqrt(3)(1,1,1) and when dotted with alpha you know you'd get a constant. If so, I think I'm starting to wrap my head around this and just need to keep at it.

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