Why is del a Vector? Answers & Explanations

[misogynisticfeminist]

I have been thinking about this for a while, but why is the del operator a vector?? The book i have states no reason why and i was thinking if you guys could tell me why.

Thanks...

 

[robphy]

Del is not an ordinary vector. It's an example of a vector differential operator.

 

[HallsofIvy]

Strictly speaking, it isn't. "grad f", at a given point, is a vector (by definition). Del is a notation for the operator that take a real valued function into that vector. It is, more correctly, a differential operator, as robphy said, that takes real valued functions to vector valued functions.

 

[misogynisticfeminist]

hmmm, but from what I know, the definition of del is,

$$\nabla = \frac {\partial}{\partial {x}} \hat {x} +\frac {\partial}{\partial {y}} \hat {y}+\frac {\partial}{\partial {z}} \hat {z}$$

Why the need for the unit vectors? And how does del differ from the total differential of a function?

 

[quantumdude]

Well, if the unit vectors weren't there, $\nabla$ wouldn't be a vector differential operator. It would just be a plain vanilla differential operator (sans "vector").

And it should be clear how $\nabla$ differs from the differential of a function: One is an operator, and one is a differential. On the other hand, you can express a differential of a function as $df=\nabla f \cdot d \mathbf{r}$.

 

[jcsd]

yep, it operates on a scalar field to give a vector field, infact really it gives a covector field.

 

[dextercioby]

The del is a one-form.It acts on an (n,m) tensor field -------->(n+1,m) tensor field.It is called the "covariant derivative operator"...Quite useful in GTR...

Daniel.

 

[misogynisticfeminist]

Hmmm, actually one thing which puzzles me is the gradient operator,

if the gradient operator operates only on scalars and the del is a scalar but a vector differential operator. Why is the del being a vector differential operator allowed to work on a scalar function? And why is the gradient using the del of a scalar function a vector?

Is my understanding wrong? I can't understand the terminology of tensors yet though, sorry, so i was wondering if this can be explained in a way without tensors.

 

[Hurkyl]

Riddle me this: if I define $\vec{f}(s) = (s, s^2, s^3)$, why is $\vec{f}$ a vector? Doesn't the same answer apply to del?

 

[misogynisticfeminist]

Riddle me this: if I define $\vec{f}(s) = (s, s^2, s^3)$, why is $\vec{f}$ a vector? Doesn't the same answer apply to del?

That vector f has a direction? Or because of the arrowhead? I'm really not too sure...

 

[robphy]

Note that $$\vec \nabla$$ satisfies the Leibniz rule:
$$\vec\nabla(\psi\phi)=\psi\vec\nabla(\phi)+\phi\vec\nabla(\psi)$$
whereas a vector $$\vec v$$ does not
$$\vec v(\psi\phi)=\vec v \psi\phi$$

 

[cepheid]

That vector f has a direction? Or because of the arrowhead? I'm really not too sure...

I'm pretty sure the example Hurkyl gave is a vector because he defined it that way: it is a vector-valued function that has three components. Each is a different function of s.

 

[Crosson]

misogynisticfeminist, vectors are not actually "quantities with magnitude and direction". It is more true that (some of) such quantities can be expressed as vectors.

An n dimensional vector is an n-tuple (i.e. a set of n ordered elements) which obeys certain transformations.

Formally, Del is a vector, and (to your satisfaction) you will in your studies see a distinction made between these vectors and the ones you are familiar with now.

 

[Data]

And note that vectors don't even necessarily have to be representable as n-tuples. Calling something a vector just means that it's an element of some vector space.

 

[misogynisticfeminist]

Note that $$\vec \nabla$$ satisfies the Leibniz rule:
$$\vec\nabla(\psi\phi)=\psi\vec\nabla(\phi)+\phi\vec\nabla(\psi)$$
whereas a vector $$\vec v$$ does not
$$\vec v(\psi\phi)=\vec v \psi\phi$$

hmmm that kinda helped. And thanks for the help also everyone.

 

[matt grime]

hmmm, but from what I know, the definition of del is,

$$\nabla = \frac {\partial}{\partial {x}} \hat {x} +\frac {\partial}{\partial {y}} \hat {y}+\frac {\partial}{\partial {z}} \hat {z}$$

Why the need for the unit vectors? And how does del differ from the total differential of a function?

But what are the coefficients by the unit vectors? are they real numbers?

 

[HallsofIvy]

Hmmm, actually one thing which puzzles me is the gradient operator,

if the gradient operator operates only on scalars and the del is a scalar but a vector differential operator. Why is the del being a vector differential operator allowed to work on a scalar function? And why is the gradient using the del of a scalar function a vector?

Is my understanding wrong? I can't understand the terminology of tensors yet though, sorry, so i was wondering if this can be explained in a way without tensors.

Yes, it appears your understanding is very wrong! No one said "del is a scalar". "del" is just a symbol-that is used for a variety of similar things. "gradient" is a differential operator whose domain is "scalar valued functions" and range is "vector valued functions". Does that make sense?

Why the need for the unit vectors? And how does del differ from the total differential of a function?

We don't need the unit vectors- that just makes it easier to write in terms of that specific coordinate system. The important thing about vectors- and why vector "operators" are useful- is that they don't depend on anyone coordinate system (or "basis").

Again, I think you should be saying "gradient" rather than "del"- "del" is just a symbol- it is the gradient that is the operator. If f(x,y,z) is a scalar valued function of three variables, then its gradient is $$\nabla f= \frac{\partial f}{\partial x}\vec i + \frac{\partial f}{\partial y}\vec j+ \frac{\partial f}{\partial z}\vec j$$

The "total differential" is the dot product of that with the "differential position vector"
$$\frac{\partial f}{\partial x}\vec i + \frac{\partial f}{\partial y}\vec j+ \frac{\partial f}{\partial z}\vec j \bullet dx\vec i+ dy\vec j+ dz\vec k$$.

 

[Crosson]

And note that vectors don't even necessarily have to be representable as n-tuples. Calling something a vector just means that it's an element of some vector space.

This statement is contradictory. All vector spaces have basis, and any element of a n-dimensional vector space can be written as an n-tuple.

 

[jcsd]

This statement is contradictory. All vector spaces have basis, and any element of a n-dimensional vector space can be written as an n-tuple.

Though the dimension of a vector space though needn't be finite or even countable, but as you say under the axiom of choice all vector spaces have a basis.

 

[mathwonk]

Well I know I am old and stupid, but I can't understand any of these answers to the opriginal question, at elast not taken together. i recommend the questioner to just read a good book like spivaks' calculus on amanifolds.

no offense meant here to the man well intended attempts to answer, some correct, but it is very confusing. the language is hard to get straight, for one thing,k i.e. the main thing. people are calling del something different from d, when apparebntly thet are intended to be essentially the same thing.

people seem to be confusinf at least in their language if not in their minds, d with df, with dp with dfp. Even if the posters are not confused i would not blame the questioner if he/she were so by this discussion. I am, and I have taught this stuff for decades, and feel entirely comfortable with using it.

some of the confusion my stem from learning these ideas from inadequate sources such as bachman's book, a few of whose many imprecisions and errors I have pointed out elsewhere.

but so what, learning is a process, and apparently some of this helped misogynistf. i am just suggesting she/he not feel alone if she is still puzzled.

 

[quantumdude]

no offense meant here to the man well intended attempts to answer, some correct, but it is very confusing. the language is hard to get straight, for one thing,k i.e. the main thing. people are calling del something different from d, when apparebntly thet are intended to be essentially the same thing.

Now that would be confusing. I don't know if you are aware of this, but every calculus book that I have ever seen refers to $d$ and $\nabla$ as two different things. Assuming that misogynisticfeminist is just learning $\nabla$ for the first time, her book no doubt does the same.

 

[mathwonk]

instead if just criticizing, let me add my two cents. then you can all see and point out my errors as well. (curly d's printed below as question marks for some reason.)

This is purely my own vulgar view of the topic. I hope it helps someone.

A manifold M is a space of points, on which it makes sense to define a real valued smooth function say f:M-->R.

Then we want to differentiate this function, but a derivative is a linear function at each point p of M which approximates M locally near p.

So how to do this when M is not a linear space?

first one approximates M itself locally near p by a linear space called the tangent space to M at p, i.e. Tp(M).

this construction takes work, unless M is already embedded in R^n, and then it just consists of the set of velocity vectors at p of all smooth curves in M, through p. [Even if M is not embedded, one can embed it and check that the equivalence relation "same velocity vector at p" is independent of the embedding. then one defines the tangent space as the set of equivalence classes of curves having the same velocity vector in all embeddings, instead of as the set of those velocity vectors.]

Once it is done, we want the derivative dfp, or f'(p), of f at p, to be a linear map from Tp(M) to R, i.e. f'(p) = dpf:Tp(M)-->R must be defined and linear.

well one way would be, for each vector v in Tp(M), to choose a smooth curve s through p, in M, with velocity vector at p equal to v, [i.e. s(0) = p, and s'(0) = v.] Then the composition fos is a smooth map from an interval to R, and has itself a derivative at 0. this is the value of dfp(v).


Of course invariance under choice of s must be checked (chain rule).

Anyway, we have at last defined a procedure for sending a smooth function f:M-->R, a point p of M, and a tangent vector v of Tp(M), to a number dfp(v).

If we fix any of these objects, say f, and p, and let v vary, we have a linear function dfp, from Tp(M) to R. this is called a "cotangent vector" at p, or an element of "T*p(M)".


If we fix only f say, and let both p and v vary, we have a "covector field" df, that assigns to each point p of M, a linear function or covector on Tp(M).

If we fix only M, we have an operator d, which assigns to functions f and points p,... etc etc...


Now if we fix nothing at all, not even M, we still have a construction d, which associates to manifolds M and functions f, a covector field df on M.

We can call d a total differential, or exterior derivative or whatever. the terminology is less important than the behavior.


Now if we do not like to learn abstract definitions, or even if we do, but we want to make some calculations, we can begin to introduce coordinates to represent these objects.

for instance we can put in a coordinate system on M, and hence a basis of each space

Tp(M), and call these basis elements ?/?x, ?/?y, ?/?z, or whatever. [recall ? means "curly d".]

[If M is already sitting in R^n, then one can simply restrict the basis elements for the ambient tangent space TpR^n, but then they are of course no longer independent on Tp(M) usually.]

This of course immediately chooses also a dual basis for the dual linear space T*p(M) of linear functions on Tp(M). This basis is called dxp, dyp, dzp, i.e. to get a basis for the local derivatives of all functions f at p, one takes the differentials at p of the basic coordinate functions x,y,z.


Then since dfp belongs to T*p(M), one can of course express dfp as a linear combination of dxp, dyp, and dzp. the coordinates in such an expression are called, essentially by definition, ?f/?x(p), ?f/?y(p), ?f/?z(p).

so one has the operation del?" that assigns to each f and each p, the triple (i.e."vector" to some people) of coordinates for dfp, in terms of the basis dx,dy,dz.


Thus the operation assigning f -->(?f/?x(p), ?f/?y(p),?f/?z(p)) might be called a differential operator of some kind. Then some people might argue over whether to call it a scalar operator or a vector operator, to distinguish the two operators

f -->(?f/?x(p), ?f/?y(p),?f/?z(p))

and f --> ?f/?x(p) dxp + ?f/?y(p) dyp + ?f/?z(p) dzp,

from each other.

this is a linguisitic discussion. the point is to understand what is going on. one is approximating a real valued smooth f, by a family of real valued linear functions.

note too that the operator del above depends on the unnatural choice of coordinates x,y,z, hence is a computational aid, and not a concept.

 

[quantumdude]

Great post, and no one is saying you're wrong. But the issue you've raised is that of being confusing, and I can guarantee you that your responses are going to be more confusing than most of ours to anyone who is taking a Calculus III course. Granted, I am only assuming that misogynisticfeminist is taking that course because she seems to be new to $\nabla$.

 

[mathwonk]

thanks tom. my point is always just the same. the geometric concept is less subject to confusion than the notation. but i could be wrong. i often am.

by the way, what does mis fem think of my post? [by the way the answer to a question like that in the original post is always the same: read the definition of del, then read the definition of a vector, and see if it applies.]

 

[Data]

All vector spaces have basis, and any element of a n-dimensional vector space can be written as an n-tuple.

Agreed. As jcsd noted I was only referring to infinite-dimensional vector spaces with that comment. An infinite sequence doesn't fall under the common definition of an n-tuple~