1. Feb 12, 2005

### Reshma

I am not completely satisfied with Griffith's explanation for the geometrical interpretation of a gradient. Can someone elaborate on the geometrical meaning of the magnitude and the direction of a gradient?

2. Feb 12, 2005

### arildno

Let $$(x_{0},y_{0},z_{0})$$ be a point, and f(x,y,z) be some function.
Define the function:
$$g(t)=f(x_{0}+tn_{x},y_{0}+tn_{y},z_{0}+tn_{z}), \vec{n}=(n_{x},n_{y},n_{z}), ||\vec{n}||=1$$
That is, g(t) measures the value of f along the direction given by $$\vec{n}$$
By differentiating g, and setting t=0, we find the rate of change of f in the direction of $$\vec{n}$$ :
$$\frac{dg}{dt}\mid_{t=0}=\nabla{f}\cdot\vec{n}\mid_{t=0}$$
But, now we have:
$$|\frac{dg}{dt}\mid_{t=0}|\leq||\nabla{f}||\mid_{t=0}||\vec{n}||=||\nabla{f}||\mid_{t=0}$$
That is, the direction in which f grows strongest, is given by the direction of its gradient.

3. Feb 12, 2005

### Reshma

Hey, this is even more complicated! Can you demonstrate using a more simpler function?

4. Feb 12, 2005

### arildno

What are you looking after?
Hand-waving?

If there are some specific points which you find difficult, focus on them one at a time when posting your questions.

5. Feb 12, 2005

### dextercioby

It's not really that easy to give a simple explanation.I think the mathematized way in which Arilldno put things is the best.

Daniel.

P.S.Why think of the gradient as having a geometrical interpretation,when the differential one is highly digestable...?

6. Feb 12, 2005

### reilly

Reshma -- First, your problem is discussed in at least a thousand books. I'm sure it's discused in many Google references. So read.

Think about the fact that df(x)/dx gives the rate of change w.r.t x. Then ponder the fact that a 2-D curve has an associated normal vector, actually given by the gradient. In fact, if a circle is entered at (0,0) then at the intersection of the circle and the x axis, the normal has magnitude 2x, and is directed along the x axis. Similarly, at the y axis intersection, the normal is of magnitude 2y directed along the y axis. The next step to an arbitrary point on the circle is yours, should you choose to accept the challenge.

Regards,
Reilly Atkinson

7. Feb 12, 2005

### Crosson

Thing of a rectangle, on the ground. At each point (x,y) I will define a function h(x,y) to be the height of a hill at the point x,y. h is continuous , so imagine this function as a surface (in the shape of a hill).

So we have, that we know the exact height of the hill at each point and so we can construct a picture of the hill.

But now suppose we would like some different information. I want to know, if I put a round ball at any point on the hill, which direction will it travel in and with what speed? Notice, I am now asking for a vector function, that is at each point (x,y) I want to associate a magnitude and direction (that the ball would role down). Let me call this function v (velocity), and exprese it in unit vector notation:

V = X(x,y) i" + Y(x,y) j" + Z(x,y) k"

Using unit vector notation, my vector function can be represented by three scalar component functions. Now here is the punchline, the way I have got this set up

Isn't that cool? h(x,y) contains all of the information that those three scalar components of V contain. We feel intuitively that this should be true, that the height of the hill does contain all of that information. Replace height with voltage, and velocity with electric field and you should be set for Grifiths.

8. Feb 12, 2005

### Biology

:zzz: :zzz: :zzz: :zzz: :zzz:

9. Feb 13, 2005

### Reshma

That's more like it! Thanks for the replies everyone :-)

10. Feb 13, 2005

### Reshma

11. Feb 13, 2005

### samoth1

lol, that was inappropriate, but the timing and placement was quite humerous.

A good book describing that kinda stuff geometrically is Weinreich's "Geometrical Vectors". He describes and depicts vectors, operators, fields, and more in a relatively simple geometric manner.

12. Aug 31, 2005

### aranyakam

Gradient: Its interpretation is quite qood in Griffiths. Assume u have a scalar field. For example, we take the case of a solid body where the temperature has differnt values at different point in the solid. We would like to find the temperature difference between two close by points along some arbitrary direction. For this, just take the dot product of gardient of temperature with 'dr', the vector connecting the two points. Thus if the dot product vanishes, the temperature difference is zero. So, along a direction perpendicular to the direction ur gradient, the variation the corresponding scalar field is zero and the variation of temperature is maximum when measured in the direction of the gradient.

Last edited: Aug 31, 2005
13. Apr 21, 2010

reshma, gradient to a vector function at a point gives the direction in which that vector function having its maximum value .the magnitude of the gradient gives how fast we can reach that max value.so at max point of a vector function the gradient is zero.since there is no other direction for maxima

14. Apr 24, 2010

reshma,gradient at a point to a function gives the direction towards the max value of a function.the magnitde of gradient gives ,how fast the maximum value is attained.at the maximum value the gradient of a function is zero,since there is no other direction for maximum.

15. Apr 24, 2010

### maverick280857

I'm absolutely sure she understood it over 5 years ago.

PS -- Came to the thread looking for a new post, only to discover a 5 year (mass?) gap. That applies to me too. Couldn't resist pointing that out.

16. Apr 24, 2010

### LukeD

5 years old, and not a very good geometrical interpretation of the gradient..
Properly treated, the gradient is not a vector at all, but a covector.

There is a much better picture of the gradient in Misner, Thorne, and Wheeler.