# Is the gradient really just a first-order approximation?

## Main Question or Discussion Point

In physics texts, its customary to write (and even to define the gradient as) the following:

$$dT = (\nabla T) \cdot dl$$

Working in Cartesian coordinates, we can expand this as follows:

$$dT = \frac{\partial T}{\partial x} dx + \frac{\partial T}{\partial y} dy + \frac{\partial T}{\partial z}dz$$

But the above equation is really just a "first-order" approximation for the differential dT, is it not? Working just in the differential dx, a more "complete" way to write this expression would be:

$$dT \approx \frac{\partial T}{\partial x} dx + \frac{1}{2} \frac{\partial ^2T}{\partial x^2} (dx)^2 + \frac{1}{6} \frac{\partial ^3T}{\partial x^3} (dx)^3 + ...$$

along with the additional, commensurate expansions in terms of the y and z variables as well.

So while we can define equations in terms of gradient, and work with gradients, etc, is it fair to say that all these gradient expressions are really just giving us answers which are first-order approximations? And that if we desired more precise answers, we would have to delve deeper into the higher-order terms?

Hurkyl
Staff Emeritus
Gold Member
Nope; $dT = (\nabla T) \cdot d\ell$ is an exactly correct equation of differential forms.

You're thinking about differential approximation, which says $\Delta T \approx (\nabla T) \cdot \Delta \ell$, or more explicitly,

$$T(\ell + v) - T(\ell) = \nabla_v T(\ell) + \frac{1}{2} \nabla_v \nabla_v T(\ell) + \cdots$$

Hurkyl
Staff Emeritus
Gold Member
You have to remember that differential forms are not numbers (or number-valued fields); they're differential forms. Their very definition is arranged so that these "first-order" equations are exactly correct. If you want to pass back into the world of "ordinary" mathematical objects, you have to use some sort of integral.

That said, people do often use expressions like dT when they really mean $\Delta T$.

dx
Homework Helper
Gold Member
They are first order approximations of ∆T, not of dT.

dT is exactly equal to $\partial_{x}T dx + \partial_{y}T dy + \partial_{z}T dz$

They are first order approximations of ∆T, not of dT.

dT is exactly equal to $\partial_{x}T dx + \partial_{y}T dy + \partial_{z}T dz$
Where can I read up on the difference between dT and $$\Delta T$$?

I don't think I've ever (formally) encountered that distinction before..

So in my first equation above, what I should have written, should have been:

$$\Delta T \approx \frac{\partial T}{\partial x} \Delta x +\frac{1}{2}\frac{\partial^2 T}{\partial x^2}\Delta x^2 + \frac{1}{6} \frac{\partial^3 T}{\partial x^3} \Delta x^3 + ...$$

Yes?

dx
Homework Helper
Gold Member
So in my first equation above, what I should have written, should have been:

$$\Delta T \approx \frac{\partial T}{\partial x} \Delta x +\frac{1}{2}\frac{\partial^2 T}{\partial x^2}\Delta x^2 + \frac{1}{6} \frac{\partial^3 T}{\partial x^3} \Delta x^3 + ...$$

Yes?
Yes.

∆T is an actual difference of T between two points. dT on the other hand is what is called a 1-form. You can read about 1-forms in books about differentiable manifolds and differential geometry.