Gradient vs. Directional Derivative

1MileCrash
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On a quiz, a true/false statement was given along the lines of:

"The gradient is a specific example of a directional derivative."


I marked "true" and got it wrong. I see why, I think, since the gradient is an actual "guide," a vector, towards the max rate of change, while the directional derivative is a scalar value given some direction.

However, would it be correct to say that the magnitude of the gradient IS a specific example of a directional derivative?
 
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You don't take the gradient of a function f in a particular direction, you simply take its gradient, df. On the other hand, given a vector v, you can find the rate a which f changes along that vector: df(v).
 
The object df contains information about the rate of change of f in all directions.
 
1MileCrash said:
On a quiz, a true/false statement was given along the lines of:

"The gradient is a specific example of a directional derivative."I marked "true" and got it wrong. I see why, I think, since the gradient is an actual "guide," a vector, towards the max rate of change, while the directional derivative is a scalar value given some direction.

However, would it be correct to say that the magnitude of the gradient IS a specific example of a directional derivative?

yes. it is is the directional derivative in the direction of the vector that is gradient divided by its length, i.e. in the direction of the unit vector that points in the same direction as the gradient vector.

The definition of the gradient is df(X) = <gradf,X>

The unit vector is gradf/|gradf| so the directional derivative is <gradf,gradf/|gradf|>

= (1/|gradf|)<gradf,gradf> = |gradf|.BTW: df, the differential of f, is not the gradient of f. It is not even a vector. It is a 1 form.

When you have a metric, i.e. an inner product, then there is always a vector,Y, that satisfies the equation df(X) = <Y,X> for all X. Y is the gradient of f with respect to this metric.

If you think about it, it will be clear that you do not need the idea of an inner product to take a directional derivative.
 
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The gradient is not a vector. In three dimensional vector calculus, the gradient of a function is represented by a vector perpendicular to the level surfaces of the function at that point, but this is not possible on a general manifold, where there is no notion of perpendicularity. The gradient is in fact a 1-form.
 
dx said:
The gradient is not a vector. In three dimensional vector calculus, the gradient of a function is represented by a vector perpendicular to the level surfaces of the function at that point, but this is not possible on a general manifold, where there is no notion of perpendicularity. The gradient is in fact a 1-form.

This is really a semantic point but in all of the books I have ever read the gradient requires an inner product on the tangent space as do other operators such as the Laplacian of a function.

The differential df of a function does not require an inner product. So for each df there is a different gradient for each different inner product. Otherwise put, the graident is the dual of the differential with respect to the inner product.
 
A function defined on a smooth manifold has a gradient, even though the tangent spaces of the manifold have no inner product. The gradient of f in coordinate representation is

df = f,i dXi

where dXi are the gradients of the coordinate functions p → Xi(p). This object contains information about how f varies in all directions. Given a vector v, the directional derivative of f in the direction v is given by the action of the 1-form df on the vector v.
 
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dx said:
A function defined on a smooth manifold has a gradient, even though the tangent spaces of the manifold have no inner product. The gradient of f in coordinate representation is

df = f,i dXi

where dXi are the gradients of the coordinate functions p → Xi(p). This object contains information about how f varies in all directions. Given a vector v, the directional derivative of f in the direction v is given by the action of the 1-form df on the vector v.

In all of the books I have read,df ,what you are calling the gradient, they call the differential.

The gradient is the vector which is dual to the differential with respect to a metric.

I will be happy to send you some differential geometry books that go over this.BTW: what do you call the dual of the differential if you do not call it the gradient?
 
Yes it is called the differential too, also exterior derivative. But the fact remains, it is the generalization of the gradient operator of vector calculus to general spaces which do not necessarily have an inner product. The idea of a gradient of a function makes sense in spaces that have no metric. It only requires the linear/affine structure.
 
  • #10
dx said:
Yes it is called the differential too, also exterior derivative. But the fact remains, it is the generalization of the gradient operator of vector calculus to general spaces which do not necessarily have an inner product. The idea of a gradient of a function makes sense in spaces that have no metric. It only requires the linear/affine structure.

http://en.wikipedia.org/wiki/Gradient

here is the first sentence of the link.

"In vector calculus, the gradient of a scalar field is a vector field that points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is that rate of increase."

Note that the gradient is a vector field and has a magnitude.
 
  • #11
Yes, in vector calculus. My point is simply that vector calculus is a formalism that only works in three dimensional euclidean space. The notion of gradient is more general.
 
  • #12
The concepts are the same on any manifold. In Euclidean space, the domain of vector analysis, the gradient is the dual of the differential of a function with respect to the standard metric.
 
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  • #13
"The gradient is a specific example of a directional derivative."

I'd be more inclined to say that the directional derivative is a specific example of the gradient, grad . n
 

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