Magnitude of the gradient of a constant scalar field

[Gamdschiee]

Hey!

Short definition: A gradient always shows to the highest value of the scalar field. That's why a gradient field is a vector field.

But let's assume a constant scalar field f(\vec r) The gradient of f is perpendicular to this given scalar field f.

My Questions:
1. Why does the gradient points away? I mean yes, it is clear that there isn't any other higher value, so it just points away?

2. Does the magnitude of the gradient represent the alteration of the scalar field f, although the field itself is constant?

 

[phyzguy]

I don't understand your question. If the scalar field f is constant, the gradient is zero everywhere.

 

[Gamdschiee]

I was confused, because I read this thread: https://www.physicsforums.com/threads/constant-scalar-field.811084/

Isn't that the case I've described?

 

[phyzguy]

I wondered, because I read this thread: https://www.physicsforums.com/threads/constant-scalar-field.811084/
Isn't that the case that I've described?

That thread talks about a scalar field that is constant on a surface S, meaning that the field has the same value everywhere on the surface S, but varies elsewhere. I understood your OP to say the scalar field was constant everywhere.

 

[Dale]

. I understood your OP to say the scalar field was constant everywhere.

That is also how I understood it.
@Gamdschiee can you clarify, do you intend to ask about a scalar function which is constant everywhere or constant on some surface?

 

[Gamdschiee]

Ahh! Thank you, I see what I did wrong there. Yes, I meant that the scalar field is only constant on certain surface/line etc. - Sorry!

Let's view page 2 and fig. 1 of this link please: http://www.phys.ufl.edu/~pjh/teaching/phz3113/notes/week5.pdf

So you can say that those closed lines in this figure are the whole scalar field. And it makes sense when the gradient points away, because there is the higher value. Is that right?

 

[Dale]

Ahh! Thank you, I see what I did wrong there. Yes, I meant that the scalar field is only constant on certain surface/line etc. - Sorry!

Let's view page 2 and fig. 1 of this link please: http://www.phys.ufl.edu/~pjh/teaching/phz3113/notes/week5.pdf

So you can say that those closed lines in this figure are the whole scalar field. And it makes sense when the gradient points away, because there is the higher value. Is that right?

Yes, that is correct. You can think of it as a topographical map.

 

[Gamdschiee]

I see thanks.
Now it's more clear to me. But how can you in general describe the magnitude of such graditude?

 

[Dale]

The magnitude is equal to the rate of change of f in the steepest direction.

 

[Gamdschiee]

So basically you can say that a gradient from any location in the gradient field always points to the maximum value of the scalar field. And the magnitude from a certain picked gradient is just the slope at this certain location? The higher the slope the nearer you come to a inflection point.

Is that right?

And what do you exactly mean by "steepest direction"? Hasn't any direction (i.e. gradient) a different slope?

 

[Dale]

So basically you can say that a gradient from any location in the gradient field always points to the maximum value of the scalar field.

Unfortunately, no. It points in the locally steepest direction, but the maximum is a global feature not a local one. If you follow the gradient around it will eventually bring you to the maximum, but not usually in a straight line.

 

[Gamdschiee]

Thank you, I think I got that now.

Here an example: http://d2vlcm61l7u1fs.cloudfront.net/media/dd4/dd4ee8d1-733b-4f93-a05f-b6178210dac1/phpxPlrNF.png

1. The maximum should be at C. It looks like a local maximum around the C-spot. Because you don't know how high the value is outside F and A e.g.
2. The longer the vectors the steeper is f at that location, so the f is steepest at F.
3. "The gradients are pointing in the locally steepest direction" - What does that exactly mean to point in the locally steepest direction? Does that mean, that the gradient always points to the nearest higher "slope-value" than its current "slope-value"?

 

[Dale]

"The gradients are pointing in the locally steepest direction" - What does that exactly mean to point in the locally steepest direction?

Sorry, this is difficult to express in words. Imagine that f is a 2D scalar function f(x,y). One way that you could plot f is to draw a surface which is curved in 3D in such a way that the height is equal to the potential, ie z=f(x,y).

If you were a mountain climber on such a surface then at any point you could go uphill, downhill, or along the hill. The gradient tells you which direction is uphill and how steep the hill is at that point. If you walk in any other direction you won’t be climbing as steeply, thus the direction of the gradient is the “locally steepest direction”

 

[jbriggs444]

Unfortunately, no. It points in the locally steepest direction, but the maximum is a global feature not a local one. If you follow the gradient around it will eventually bring you to the maximum, but not usually in a straight line.

There is no guarantee that a local maximum found in this manner will be a global maximum.

 

[Dale]

There is no guarantee that a local maximum found in this manner will be a global maximum.

Yes, that is correct

 

[ChuckH]

I don't understand your question. If the scalar field f is constant, the gradient is zero everywhere.

then why isn't the gradient of the dot product of two vectors always zero? Del (A.B) =0??

 

[Dale]

then why isn't the gradient of the dot product of two vectors always zero? Del (A.B) =0??

Because the dot product is not always constant. Indeed, why would it be?