In this segment, we will talk about transforming normals. So far, all of our activity has been about transforming geometric objects and points on the surface. However, the surface orientation of the surface normal is also critical for computer graphics. One important use is in lighting. The intensity you see from the surface depends on the angle between the light source and the surface normal.

For highlights on the surface it depends on the angle between the viewer, the surface normal and the light source.

And the normals do not transform the same way as surface positions. To see why, we can look at this diagram. Here, I've just shown the effect in 2D, and that's all you need to really understand it. I have a rectangle, and of course, the surface normals are pointing straight upwards.

I now apply a shear to the rectangle, and so it becomes a parallelogram like this. But of course, the surface is normal on the top is still pointing upwards.

However, if I were to apply the same transformation as I apply to the geometry, imagining the normals attached to the geometry in this part of the geometry of the object. Then when I shear it, what is going to happen, is that it is going to move, because I'm moving points here more than I'm moving points here. And so, while this point will move a little bit in the direction of the shear, this point will move more. And so, the normal will appear tilted, when that's not really the case. So therefore, we need to derive the correct transformation for the surface normals.

And we will go through some algebra, which is the easiest way to see how they deform. If you want to think about this, also think about the transformations in a circle when it's stretched into an ellipse. The normals will not transform the same way as the geometry. In fact, they'll transform in the inverse way in a certain sense.

So, to find the normal transformation, we consider a point on the surface, and we consider the geometric location around it. So, we consider a plane around the point, here is the point, here is the normal, here are some points on the plane, which we consider the tangents. Tangents are actual geometric locations on the surface and they transform just as the same transformation matrix on the surface, so M times t.

Normals transform with another matrix which we'll call Q. And the question is, what is Q equal to?

One of the important things we know is that the normal has to be at 90 degrees to the tangents, so its dot product with respect to the tangents has to be equal to 0. This is the basic definition of a plane or the definition of normals. So n dot t has to be equal to 0.

Now, the question is, we want to solve these equations, and we want to find what the matrix Q is. But, this is the condition, n transpose T is = to 0 (n^T t = 0). So n is equal to QN. And so let's transpose that. So let's look at QN transpose ((Qn)^T).

Now when you transpose, you transpose the individual elements. And you invert the order in which they apply. So this will be equal to n transpose times Q transpose ( (Qn)^T = n^T * Q^T ).

And t will be equal to Mt. So we will have this times Mt is equal to 0.

So I've just reproduced that here. So I have n transposed, Q transposed Mt is equal to 0 ( n^T * Q^T * Mt = 0 ). And this implies that Q transposed M must be equal to the identity ( Q^T * M = I ). This is because this must hold for any n and t in order to get the original condition that n transposed t is equal to 0. So of course, if Q transposed M is equal to the identity, it's very easy to solve for the matrix.

And we have that Q is equal to M inverse, because I multiplied by M inverse on both sides and then I have to get rid of the transpose and so I take M inverse transpose ( Q = (M^-1)^T ). One point I want to make very clear, since we've just had the discussion of homogeneous coordinates, is that this inverse transposition is applied to the upper 3x3 matrix only. So it's applied to the upper 3x3 matrix only.

It doesn't really apply to the homogeneous coordinates, and in particular the normal, being a vector, isn't really acted on by translations. And so, what you do to the translational coordinates doesn't matter. But, in order to change this vector you do have to apply this inverse transpose to the upper left 3x3, which includes things like rotations and shears, and so I'll also say there is no effect of translation, right. So no T translation applied.

Finally, we have the formula here. Q is equal to M inverse transpose ( Q = (M^-1)^T ). And this formula must be applied to all surface normals.