Register to reply

Correct Definition?

by Cyrus
Tags: correct, definition
Share this thread:
Cyrus
#1
Apr15-07, 04:38 PM
Cyrus's Avatar
P: 4,780
Which one is the correct definition? I have seen both being used in texts and it makes a world of difference which one is used when doing proofs.


[tex] \frac {\partial f}{\partial \bar{x} } = [ \frac {\partial f}{\partial x_1 }, \frac {\partial f}{\partial x_2 }, \frac {\partial f}{\partial x_n }] [/tex]

OR:

[tex] \frac {\partial f}{\partial \bar{x} } = \left[\begin{array}{c} \frac {\partial f}{\partial x_1 } \\ \\ \frac {\partial f}{\partial x_2 }\\ \\ \frac {\partial f}{\partial x_n }\end{array}\right] [/tex]

My book and wiki have the first definition, but I have seen some people use the second.
Phys.Org News Partner Physics news on Phys.org
Physicists discuss quantum pigeonhole principle
First in-situ images of void collapse in explosives
The first supercomputer simulations of 'spin?orbit' forces between neutrons and protons in an atomic nucleus
christianjb
#2
Apr15-07, 04:41 PM
P: 529
They look like the same thing to me.
Cyrus
#3
Apr15-07, 04:44 PM
Cyrus's Avatar
P: 4,780
One is a column vector and one is a row vector.

cristo
#4
Apr15-07, 05:03 PM
Mentor
cristo's Avatar
P: 8,309
Correct Definition?

What's the context here? Your expression looks very like grad(f).
christianjb
#5
Apr15-07, 05:09 PM
P: 529
Quote Quote by cristo View Post
What's the context here? Your expression looks very like grad(f).
It's very nearly the same thing as writing grad.

f(r+dr)=grad[f(r)].dr
=dr df/dr
=dr. I df/dr

So grad[f(r)]=I df/dr

(I'm too lazy to latex the above, but all the r, dr are vectors)
D H
#6
Apr15-07, 06:15 PM
Mentor
P: 15,065
The notation [itex]df/d\vec x[/itex] is just a non-standard way to write [itex]\nabla f[/itex].

The standard representation of a displacement vector is as a column vector. With this representation, a displacement vector transforms from frame A to frame B as

[tex]\vec x_B = T_{A\to B} \, \vec x_A[/tex]

The gradient of a scalar field transforms by a different set of rules. The gradient lives in the dual space of the displacement vector field. The gradient should thus be represented as a row vector. With the gradient expressed as a row vector, the gradient of a function transforms according to

[tex]\nabla f_B(x) = (\nabla f_A(x)) \, T_{A\to B}[/tex]
cristo
#7
Apr15-07, 06:34 PM
Mentor
cristo's Avatar
P: 8,309
Quote Quote by D H View Post
The notation [itex]df/d\vec x[/itex] is just a non-standard way to write [itex]\nabla f[/itex].
That's what I thought; although the bar in the original post confused me!

The standard representation of a displacement vector is as a column vector. With this representation, a displacement vector transforms from frame A to frame B as

[tex]\vec x_B = T_{A\to B} \, \vec x_A[/tex]

The gradient of a scalar field transforms by a different set of rules. The gradient lives in the dual space of the displacement vector field. The gradient should thus be represented as a row vector. With the gradient expressed as a row vector, the gradient of a function transforms according to

[tex]\nabla f_B(x) = (\nabla f_A(x)) \, T_{A\to B}[/tex]
Of course, some people gloss over this and write all "vectors" as row vectors!
Cyrus
#8
Apr15-07, 06:49 PM
Cyrus's Avatar
P: 4,780
The bar means its a scalar differentiated W.R.T a vector.

Also, It never dawned on me to think of it as the gradient, but in fact thats exactly what it's doing (only if x is spatial coordinates). In that case, the correct definition is definition 2.

But Im asking which one is the correct definition? Why would you need context to answer my question?
Mentz114
#9
Apr15-07, 07:02 PM
PF Gold
P: 4,087
According to post #6 by DH

The gradient should thus be represented as a row vector.
pmb_phy
#10
Apr15-07, 07:57 PM
P: 2,954
Quote Quote by cyrusabdollahi View Post
Which one is the correct definition? I have seen both being used in texts and it makes a world of difference which one is used when doing proofs.


[tex] \frac {\partial f}{\partial \bar{x} } = [ \frac {\partial f}{\partial x_1 }, \frac {\partial f}{\partial x_2 }, \frac {\partial f}{\partial x_n }] [/tex]

OR:

[tex] \frac {\partial f}{\partial \bar{x} } = \left[\begin{array}{c} \frac {\partial f}{\partial x_1 } \\ \\ \frac {\partial f}{\partial x_2 }\\ \\ \frac {\partial f}{\partial x_n }\end{array}\right] [/tex]

My book and wiki have the first definition, but I have seen some people use the second.
The components of the first is a vector and as such is is layed out using a row matrix, which is the correct thing to do here. The second has components of a vector but is a column matrix of a 1-form which is the wrong thing to do. Only 1-forms are represented with the variables in the denuminator using subscripts and a change of sign for the spatial components.

Well .... it sounds good anyway. Does anyone agree?

Pete
Cyrus
#11
Apr15-07, 08:06 PM
Cyrus's Avatar
P: 4,780
Quote Quote by Mentz114 View Post
According to post #6 by DH
Incorrect, the gradient is a column vector because my spatial coordinates is a column vector in R^n. So the gradient also has to be a column vector in R^n.
Hurkyl
#12
Apr15-07, 08:09 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,098
Quote Quote by cyrusabdollahi View Post
Which one is the correct definition?
For better or worse -- the right answer to this question is "the one in the book." (Or "the one the instructor gave", or whatever is appropriate)

Also, for better or worse (worse, IMHO), it is common to formalize the mathematics so that no distinction is made between the two.


But if you are going to make a distinction (which I certainly advise) -- then [itex]\partial f / \partial \vec{x}[/itex] is going to be the opposite of what you use to denote [itex]\vec{x}[/itex]. The most common convention is that [itex]\vec{x}[/itex] is a column, thus [itex]\partial f / \partial \vec{x}[/itex] is a row.



And, for maximum confusion, the notation [itex]\nabla f[/itex] is often used both to denote the gradient and to denote [itex]\partial f / \partial \vec{x}[/itex]. (e.g. see Spivak's Differential Geometry)
Cyrus
#13
Apr15-07, 08:12 PM
Cyrus's Avatar
P: 4,780
Thats my problem. The book uses one way, my notes from class, which is from a different book, uses another way.

Here is a paper that basically says the same thing I've been saying in the footnote. But I wanted a second opinion from you guys.

http://www.colorado.edu/engineering/.../IFEM.AppD.pdf

Also note, the Jacobian is defined as the Transpose in my book of whats in the link above.
Hurkyl
#14
Apr15-07, 08:27 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,098
Quote Quote by cyrusabdollahi View Post
Thats my problem. The book uses one way, my notes from class, which is from a different book, uses another way.

Here is a paper that basically says the same thing I've been saying in the footnote. But I wanted a second opinion from you guys.

http://www.colorado.edu/engineering/.../IFEM.AppD.pdf

Also note, the Jacobian is defined as the Transpose in my book of whats in the link above.
Ye gads! That's (IMHO) a very poor conventional choice -- so much that my first instinct would be to call it wrong! I would be a thousand times happier if x and y were row vectors, rather than the column vectors they were defined to be. (Or keep them as columns, but transpose everything else in that appendix)
Cyrus
#15
Apr15-07, 08:29 PM
Cyrus's Avatar
P: 4,780
Well, in Dynamics it makes more sense to define things as column vectors and not row vectors.
christianjb
#16
Apr15-07, 08:48 PM
P: 529
I still maintain that

[itex]\nabla f(\mathbf{r})=\mathbf{I}\frac{\partial f(\mathbf{r})}{\partial \mathbf{r}}[/itex]

Grad and df/dr are not completely equivalent.
Cyrus
#17
Apr15-07, 08:53 PM
Cyrus's Avatar
P: 4,780
I dont understand what your talking about christian. Can you explain what your saying in more detail?
christianjb
#18
Apr15-07, 09:04 PM
P: 529
[itex]f(\mathbf{r}+d\mathbf{r})=f(\mathbf{r})+\nabla f(\mathbf{r}).d\mathbf{r} [/itex]

Also

[itex]f(\mathbf{r}+d\mathbf{r})=f(\mathbf{r})+\frac{\partial}{\partial \mathbf{r}} f(\mathbf{r})d\mathbf{r}
= f(\mathbf{r})+\frac{\partial}{\partial \mathbf{r}} f(\mathbf{r})\mathbf{I}.d\mathbf{r}[/itex]

So

[itex]\nabla f(\mathbf{r})=\mathbf{I}\frac{\partial f(\mathbf{r})}{\partial \mathbf{r}}[/itex]


Register to reply

Related Discussions
Correct definition of planetary precession Astronomy & Astrophysics 0
Is this correct? Introductory Physics Homework 0
Cant get it correct! Calculus & Beyond Homework 3
Time dilation(correct/not correct) Introductory Physics Homework 1
Is this correct? Introductory Physics Homework 8