In video 89 of Tensor Calculus, we'll examine how vectors are decomposed into projections. Understanding how this process differs between curves and surfaces is an important precursor to completing our review of those aspects that are common. I want to start with a brief review of what we did back in video number 68. There we started with this definition for the covariant derivative as it would apply to a plane.
that is a simple flat two-dimensional euclidean manifold well this was based on our original definition of the christoffel symbol we said the christoffel symbol is a set of scalar functions that can be used in conjunction with our basis vectors to form linear combinations to represent this partial derivative well we then ask ourselves the question do these things apply to a curved surface and we soon discover that the answer to that is no so to remind ourselves of why that is the case let's look at this diagram here i've drawn a curved surface outlined in yellow and i've included the tangent plane in red along with the basis vectors and the normal now what we concluded is that even though these objects lie in the tangent plane this operation may well introduce a normal component in other words this partial derivative does not necessarily lie in the tangent plane but may result in a vector that points in any direction in space this means that this particular term by itself cannot be a correct expression for this partial derivative because these linear combinations are limited to vectors in the tangent plane well at this point we modified our definition for the Christoffel symbol to this form. And by doing so, we were able to show that what this term really represents is the projection of our partial derivative onto the tangent plane. So with this definition for the Christoffel symbol, this term represents this vector right here and not the partial derivative itself.
Note that this enhanced definition does not invalidate anything we've already done for a plane. in that case the partial derivative does lie on the plane and the surface projection of any such vector is the vector itself so this definition continues to work for a plane as a special case now to fix this expression we need an additional term and we get that first of all by finding the projection of our partial derivative onto the normal that gives us this vector we then assign a numerical value to the magnitude of this to a new object we call the curvature tensor and we then added that linear combination to our expression here and this gives us the form that we need what we've got here is the surface projection that's this vector this is the normal projector that's this vector when we add them together we get the result that we're looking for right here all right with that review in mind Let's perform a similar analysis with respect to curves to see where that takes us. Here we have the definition for the Christoffel symbol and the covariant derivative, both expressed in curve coordinates. And we'll refer to this diagram over here. Now what we're going to do is to ask the same question we ask for a surface, and that is, does this relationship adequately describe the results for this partial derivative?
Does this one term by itself cover all the possibilities for the outcome of this partial derivative? And the answer to that question is no. And the reason is, as before, that this partial derivative could result in a vector that points in any direction in space, such as this, while this term over here can only describe vectors that lie, in this case, along a line tangent to the curve. The use of this basis vector means that whatever vector this is must lie parallel to our basis vector. So if we're going to have a general expression here, we need additional terms to cover the other possible components of this vector.
Now, we said that this new definition is one that will give us the projection of the vector onto the manifold. With the surface, it gave us a projection onto the... tangent plane, but here because we're dealing in only one dimension, this projection is going to be a projection onto the the line tangent to the curve.
It has to be limited to the one dimension parallel to our basis vector right here. So this one term is really this vector right there. So we need other terms to add the additional components properly describe the general expression for that result.
Now we solve this problem for a surface by finding the projection of our resultant onto the normal. And that projection, the normal projection, plus the surface projection, gave us the result that we wanted. Add the two together and we have the general result.
Well, that approach won't work here because we don't have a tangent plane and therefore we don't have a well-defined normal that we could use to take a projection onto. So we need a different approach. And here's how we go about it. We construct a plane that is perpendicular to our curve at the point of interest.
So our basis vector here, in a sense, becomes the normal of the newly defined plane that we're talking about. So what we'll do now is to find the projection of our resulting vector onto this newly defined plane. And it looks like this. We've dropped a perpendicular from this point back to the newly defined plane.
And since both of these are perpendicular, they're parallel, that means all three of these vectors are in a plane together. And that means this vector plus this vector. will yield that vector.
So what we're going to do is to define this new vector as something that we call the vector curvature normal. Now let's pause here for a minute to see that what we've done is very similar to what we did for a surface manifold, in that to represent a general expression here, we had to take two projections. We took a projection here, onto the manifold.
We took a projection here away from the manifold. The difference is that for a surface this projection was a projection onto a plane and therefore covered two of the possible three degrees of freedom of our result. While this projection was a projection onto a normal which covered only one degree of freedom.
So together they covered the three degrees of freedom necessary for the general result. Well, the difference for a curve is that this projection covers only one degree of freedom because it has to be along this line. But this projection covers two degrees of freedom. And the reason that it covers two degrees of freedom is because we can rotate this vector on the plane in any direction we want to.
And we orient it in the way that makes this vector in the same plane as these two vectors together. And so... The two degrees of freedom are covered by the orientation of this vector on the plane. Together, in either the case of a surface or a curve, we're covering three degrees of freedom to represent a general result here.
Now before we move on, let me reemphasize the fact that although these indexes can only have a value of one, and therefore there's only one object here, we want to retain these indexes as part of our notation. And we do that because we're going to be consistent with the indexes in each of these terms. We've got two free indexes in these terms, so we need to maintain them here.
Among other things, it tells us, first of all, that what we've defined here is not an invariant object, because we cannot form a contraction with two indexes in the lower position like this. But we can see right away, however, that the relationship is a symmetric relationship. We know that these indexes are symmetric, these are symmetric, so these have to be as well.
So we can swap the two indexes, but we don't have an invariant object. Alright, so let's move on and explore a little bit about what we have with regards to this vector curvature normal. The first thing I'll do is define the dot product of both sides of this expression with respect to our contravariant basis vector. So all I've done here is to dot each of these terms with this factor throughout the expression. All right, well, if you look at it, this first term right here, according to our definition up there, is just gamma, sigma, phi, psi.
That's according to our definition right there. Well, here, these two together will form a delta factor of sigma theta, and that will absorb the theta index and turn it into sigma. So this term is gamma sigma phi psi. And then, of course, our last term out here.
Well, these... two gamma terms of course will cancel out they're the same so they drop out and that means that this by itself has to be equal to zero what does that tell us well it simply tells us what we already know and that is that the new object here this vector curvature normal is orthogonal to the contravariant basis vector which lies along this line so that tells us that wherever this is oriented It is perpendicular to our curve at this point. Next, I want to find the covariant derivative with respect to psi of our covariant basis vector u phi. As always, we start with a partial derivative.
So the partial of u phi with respect to u psi. And because we have a lower index, we'll need a negative. term and that's gamma free indexes are psi and phi and will form a contraction with you and we'll need a dummy index of theta here okay next thing is I want to replace this term right here with its equivalent in this form so I'm going to put these two terms in place of this one So you can see that this is the equivalent of this as per this relationship right there. Okay, and immediately, of course, you see that these two terms cancel out, and that means that this term is equal to this. And this gives us two things.
First of all, it tells us that our covariant basis vector here is not metronilic. This covariant derivative does not give us zero, but it's equal to our new term here. which of course should not be surprising because it's exactly what happens with a surface manifold.
And the second thing it does, it gives us an explicit relationship for this vector curvature normal. So let's go over and explore this in more detail. This is what we just derived. Of course we can express it this way because these indexes are symmetric. So what we want to do now, we said this was not an invariant expression, So let's fix that.
What we'll do is to raise this index, raise this index with it, and then form a contraction. So what we'll have is this. We've got b, phi, phi, and it's a vector. And we're raising this first index, so it'll be the contravariant derivative, phi and u phi, like this.
Okay, so that's all we've done is to raise the index and form a contraction. All right, well, let's expand this out using the definition of our basis vector. So we've got B with indexes in the upper and lower position being equal to the contravariant derivative of our basis vector. But the definition of this is just the partial of our position vector with respect to.
u phi like this. And you'll remember from previous discussions that whenever we have the partial derivative of an object that is, say, an invariant object, such as a vector or scalar, then that's the same as its covariant derivative. The covariant derivative of an invariant object is equal to its partial. So we can replace this with the covariant derivative.
So we'd have the contravariant derivative of the covariant derivative. of our position vector r and of course that's equal to the laplacian so we can represent it this way the laplacian of our position vector like this now we've run across something like this before back in video number 79 we were working with the curvature tensor in a surface and uh we had said that this was the kind of thing we got with b alpha alpha times n hat We said that this was the curvature normal when it relates to a surface. Well, you can see the relationship here. It's just that we can't use this here because we don't have a normal when we're talking about a curve.
So we leave it in the form of this vector over here. But it's the same idea. Well, let's start over with this and expand it out a different way.
We'd have our new object here, this vector curvature normal. And it's going to be equal to this side. But this time I'm going to replace this with a pair of factors.
Delta, phi, psi times the contravariant derivative with respect. to psi. So these two factors together would equal this because this index would be absorbed. And then of course we have this last factor like this.
Okay, now next I'm going to replace the delta factor here with a pair of Levy-Savita symbol factors. And this is something I can only do in one dimension because we've only got one index here. So this relationship, this delta factor equaling this pair of these two is only true in one dimension. Okay, but we can apply it here nonetheless. And we next note that the Levy-Savita symbol is metric-Nilich, which means we can take this factor and move it inside this derivative operation and associate it with our basis factor here.
At the same time, I can raise this index and lower this one. So the next expression would look like this. I'm raising psi in the upper position here, lowering it on this operation, which makes it a covariant derivative. And then this factor is being associated with the vector. So this Levy-Savita symbol associated with our vector like this.
And of course, I need parentheses. because i'm applying this covariant derivative to this product well first thing to see is that this operand here is an invariant expression because it has a full contraction here and anytime we're taking the covariant derivative of an object that's invariant we can replace that with its partial derivative or in this case just its derivative because there's only one variable here Also, you remember in the last video that our Levi-Savita symbols can be replaced, in this case, in the upper position, that's equal to 1 over the square root of u. You can look back in the last video to see that that's what this equates to. So we're replacing this with 1 over the square root of u.
This we're going to replace with the ordinary derivative with respect to u1, and we can do that because this is an invariant expression. and then inside the parentheses we again have our Levy-Savita symbol, which can be replaced by 1 over the square root of u, and then of course our covariant basis vector of just u1 here. So what I've got is an expression that is very useful in terms of actually evaluating the object over here. So I can come up with all the values of these objects, and work through this operation here to find out exactly what this looks like. Well, there's more, so let's keep going.
We've got our object here, and in the last video, you may recall that we said that ds is equal to the square root of u times du1, and that in turn means that 1 over the square root of u is equal to du1 over ds. So what we're going to do now is to replace the 1 over square root of u term with this. So we'll have this with this factor of du1 with respect to ds times the derivative with respect to du1.
And then inside the parentheses, We'll do the same thing. This is du1 with respect to ds, and our definition for this covariant basis vector is just dr, our position vector, with respect to du1. And I think you can see what's going on.
This combination of terms is really the chain rule that just equates simply to dr ds. This is also the chain rule. which would result in this operation just being the derivative with respect to ds.
So this resolves down to dds of dr ds, which of course is the second derivative of r with respect to s, and this should bring back a memory of video 76. That's back where we started the discussion of the curvature of a line in the first place. We said that that's equal to the curvature times the principal normal, That's all back in video 76, and if you look at these two things together, we realize now that we can express this as a magnitude b, p, phi, times p-hat. So, all that having been said, what we're pointing out here is that where a curve is concerned, p-hat plays the role of the normal, because we had this kind of expression for for curvature on a surface, only this factor was n-hat. So for a curve, all we have to do is to replace n-hat with p-hat, and everything we've said about the equations in surface manifolds continues to have a parallel or an analog with the curve, as long as we use p-hat instead of n-hat going forward.
So let's review real quick what we've got here. I've thrown a lot at you. And what we're saying is that we have several relationships with this new object. One right here that tells us that the covariant basis vector is not metronilic.
Then we have this relationship that shows that the object is equal to the Laplacian of the position vector. Now we can express it this way, which allows us to fully evaluate in real terms what this looks like. And finally, we have this relationship. And what that tells us is that we can express our vector curvature normal as this linear combination. Now, you'll remember throughout our analysis for surface that we ran across expressions like this, b alpha alpha times n hat.
Well, this is the analogy for curve that equates or parallels with this expression for surface. So. If we'll replace this and our surface equations with this form, we can continue on looking at the surface equations and see how they apply to a curve.
And that's exactly where we're going to go in the next video.