0

The hyper surface in my lectures is defined as a single function of coordinates which is a constant-

$f(x)= c$

and it’s stated that the normal to the hyper surface is given as

$x_\mu= \nabla_\mu f(x)$

But isn’t the above expression just 0 since $f(x)= constant$ implies

$\nabla_\mu f(x)= \nabla_\mu (constant) =0$

Won’t the above expression be just always 0?

Shashaank
  • 903
  • The gradient is not restricted to the hyper surface. That's why the normal vector "points away" from the surface. – Shil B. Feb 17 '21 at 19:43
  • Sorry I didn’t get you. I am just saying that the partial derivative of a constant is 0 – Shashaank Feb 17 '21 at 19:44

1 Answers1

2

It’s not that $f$ is a constant function, like $f(x, y, z) := 7$. Rather a hypersurface is the set of points where $f$ attains one particular value, like $\{(x,y,z) \in \mathbb R^3: f(x,y,z) = 7\}$. In particular, $f$ is changing in every direction not parallel to the hypersurface so it’s far from a constant.

Edit: Consider $f(x, y, z) = x^2 + y^2 + z^2$. The set $\{(x, y, z) \in \mathbb R^3: f(x, y, z) = 1 \}$ is a sphere with radius 1. The gradient of $f$ is $\nabla f = \langle 2x, 2y, 2z \rangle$, so you see that partial derivatives are nonzero functions. Moreover, for $(a, b, c)$ on the sphere, $(\nabla f)(a, b, c) = \langle 2a, 2b, 2c \rangle$ is a normal vector to the sphere.

Sam Freedman
  • 3,989
  • I have accepted the answer as I guess I understand what’s going on. But could you please for more clarity give an example showing the difference between the 2 things ( one where f(x) = constant and second what you are stating and how the partial derivatives don’t vanish in that case). – Shashaank Feb 17 '21 at 19:57
  • Take for example $f(x,y) = x^2 + y^2$ and consider the hypersurface defined by ${(x,y) \in \mathbb{R}^2: f(x,y) = 1}$. Then $\nabla f(x, y) = (2x, 2y)$ which is non-zero at $(x, y) = (1, 0)$ for instance. – Shil B. Feb 17 '21 at 20:14
  • @Shashaank Added in an example. Let me know if I can clarify anything. – Sam Freedman Feb 17 '21 at 22:07
  • @SamFreedman just a small follow on. Why is gradient normal to the function here – Shashaank Feb 18 '21 at 15:00
  • It's a general important theorem that the gradient is perpendicular to the level curves, see https://math.stackexchange.com/questions/1871481/why-is-the-gradient-always-perpendicular-to-level-curves. – Sam Freedman Feb 18 '21 at 15:05
  • So it's more correct to say that the gradient vector is normal to the level curve ${f = 1}$, i.e. the sphere, as opposed to the function. – Sam Freedman Feb 18 '21 at 15:06
  • @SamFreedman And like you calculated the gradient of the original $f$ and then substituted $(a, b, c) $, would it be correct if I took $g(x, y, z) = x^2+y^2+z^2-1=0$? Is taking such a $g$ correct? Because if I do this then $g$ is a constant function equal to 0, so it's derivatives will be all 0. I guess doing this is wrong but I can't spot the mistake. Could you please help me by pointing out where am I am going wrong? – Shashaank Feb 19 '21 at 05:05
  • If we have a function $g(x,y,z) = x^2 + y^2 + z^2 - 1$, we can ask where it equals zero. That zero set in this case is a sphere of radius 1. But $g$ is not a constant function on all of $\mathbb R^3$, so it’s derivatives are not all zero. Does that help? – Sam Freedman Feb 19 '21 at 05:20
  • 1
    @SamFreedman yeah thanks a lot, I got it. When we write some thing as $g(x) =0$ we are asking for the point where the function is 0, not that the function is 0 everywhere. And we could define hypersurfaces where the function attains a constant value. Would ${(x, y, z) : f(x, y, z) = g(x) }$ be also a hypersurface or not? – Shashaank Feb 19 '21 at 14:15
  • It's an interesting question. To be a hypersurface in 3D space, the set of solutions needs to be 2D. This happens with level sets due to https://en.wikipedia.org/wiki/Level_set#Level_sets_versus_the_gradient. Depending on how $f$ and $g$ relate to each other this might not be the case, but you should play around with some examples to find out! – Sam Freedman Feb 19 '21 at 15:02
  • @SamFreedman ok I see. I will try, I don’t think I am very good in this particular area. Would ${(x,y,z): f(x,y,z)=g(x,y)}$ be a hypersurface then or is there the same problem which you just said. – Shashaank Feb 19 '21 at 15:53
  • Let us continue this discussion in chat. – Sam Freedman Feb 19 '21 at 16:47