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Correct Definition? 
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#1
Apr1507, 04:38 PM

P: 4,777

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. 


#2
Apr1507, 04:41 PM

P: 529

They look like the same thing to me.



#4
Apr1507, 05:03 PM

Mentor
P: 8,325

Correct Definition?
What's the context here? Your expression looks very like grad(f).



#5
Apr1507, 05:09 PM

P: 529

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) 


#6
Apr1507, 06:15 PM

Mentor
P: 15,205

The notation [itex]df/d\vec x[/itex] is just a nonstandard 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] 


#7
Apr1507, 06:34 PM

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P: 8,325




#8
Apr1507, 06:49 PM

P: 4,777

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? 


#9
Apr1507, 07:02 PM

PF Gold
P: 4,087

According to post #6 by DH



#10
Apr1507, 07:57 PM

P: 2,954

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


#11
Apr1507, 08:06 PM

P: 4,777




#12
Apr1507, 08:09 PM

Emeritus
Sci Advisor
PF Gold
P: 16,091

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) 


#13
Apr1507, 08:12 PM

P: 4,777

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. 


#14
Apr1507, 08:27 PM

Emeritus
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PF Gold
P: 16,091




#15
Apr1507, 08:29 PM

P: 4,777

Well, in Dynamics it makes more sense to define things as column vectors and not row vectors.



#16
Apr1507, 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. 


#17
Apr1507, 08:53 PM

P: 4,777

I dont understand what your talking about christian. Can you explain what your saying in more detail?



#18
Apr1507, 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] 


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