# I Geometrical interpretation of gradient

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1. May 26, 2017

### binei

In 'Introduction to Electrodynamics' by Griffiths, in the section of explaining the Gradient operator, it is stated a theorem of partial derivatives is:
$$dT = (\delta T / \delta x) \delta x + (\delta T / \delta y) \delta y + (\delta T / \delta z) \delta z$$
Further he goes onto say:
$$dT = (\dfrac{\delta T} {\delta x} {\bf x} + \dfrac{\delta T}{\delta y} {\bf y} +\dfrac{\delta T}{\delta z} {\bf z} ) . (dx {\bf x} + dy {\bf y} + dz{\bf z} )$$
$$= \triangledown T . d{\bf l}$$

Further, in the geometrical interpretation of the gradient it is said that:
$$dT =\triangledown T . d{\bf l} = |\triangledown T||d {\bf l}|\cos \theta$$

My question is:
1. The magnitude $dT$ is greatest when $\theta = 0$ , i.e. when $\bf l$ is in same direction of $\triangledown T$ . Since now $d{\bf l} = (dx {\bf x} + dy {\bf y} + dz{\bf z} )$ , to vary the direction of $d{\bf l}$ , the relative magnitudes of $dx, dy, dz$ need to be different. Am I correct?

2. Does the magnitude of the vector $\triangledown T$ have any physical significance, given that it gives the length of the vector at some point (x,y,z)?

2. May 26, 2017

### Orodruin

Staff Emeritus
1. Yes. The direction of $\nabla T$ is the direction in which $T$ grows the fastest for a fixed $|d\vec \ell|$.

2. It is the rate at which the quantity increases when you go in the direction that it is pointing in.

Last edited: May 26, 2017
3. May 26, 2017

### binei

Thank you for the answer. I however found it strange that the magnitudes of these dx, dy, dz are relatively different, when they themselves are all infinitesimally small quantities......

4. May 26, 2017

### Staff: Mentor

5. May 26, 2017

### Staff: Mentor

Another physical interpretation is the derivative of T in the direction normal to the contours of constant T.

6. May 26, 2017

### binei

For example in a 2-dimensional case, we have the vector $d{\bf l} = dx {\bf i} + dy {\bf j}$. The angle or direction of this vector can be said to be $tan\theta = dy/dx$ radians w.r.t x-axis. But both $dx, dy$ are infinitesimally small quantities. Perhaps introducing limits, we can say $\theta = 1$, as both $dy, dx \rightarrow 0$. But how do we get other angles?

7. May 26, 2017

### Staff: Mentor

$$dy=\sin{\theta} dl$$
$$dx=\cos{\theta} dl$$

In 3D,
$$dy=\sin{\theta} \sin{\phi}dl$$
$$dx=\cos{\theta}\sin{\phi} dl$$
$$dz=\cos{\phi}dl$$

8. May 26, 2017

### Stephen Tashi

You can't reason with infinitesimal quantities in the same way that you reason with finite quantities. In fact, you can't reason with infinitesimal quantities in a logically consistent manner at all unless you use some very complicated definitions and axioms for them (e.g. https://en.wikipedia.org/wiki/Non-standard_analysis ), which are quite different than the approach taken in physics texts.

Infinitesimals in physic texts are treated in an intuitive manner. To help your intuition, consider that the infinitesimal formulation of the derivative of a real valued function of one real variable is "dy/dx". So there you have an example where a ratio between two infinitesimal quantities can be different than 1. Reasoning with infinitesimals is an attempt to deduce results that logically require reasoning about limits without actually doing the labor of thinking about limits.