
#1
Mar912, 02:18 AM

P: 734

One of the definitions of the tensors says that they are multidimensional arrays of numbers which transform in a certain form under coordinate transformations.No restriction is considered on the coordinate systems involved.So I thought they should transform as such not only under rotations but also under transformation from cartesian to plane polar coordinates,so I tried it on the contravariant tensor below:
[itex] \left(\begin{array}{cc}xy&y^{2}\\x^{2}&xy\end{array}\right) [/itex] But I got zero for all four elements.I got confused then I thought maybe curvilinear coordinates are somehow different from cartesian.Is it correct?If not,what's the reason? thanks 



#2
Mar1212, 11:50 AM

Sci Advisor
HW Helper
PF Gold
P: 4,768

Maybe you can show us your computations.




#3
Mar1212, 12:46 PM

P: 734

[itex] r= \sqrt {x^2+y^2}[/itex]
[itex] \theta=tan^{1}{\frac{y}{x}}[/itex] [itex] \frac {\partial r} {\partial x} = \frac {x}{\sqrt {x^2+y^2}} [/itex] [itex] \frac {\partial r} {\partial y} = \frac {y}{\sqrt {x^2+y^2}} [/itex] [itex] \frac{\partial \theta}{\partial x}=\frac {y}{x^2+y^2}[/itex] [itex] \frac{\partial \theta}{\partial y}=\frac {x}{x^2+y^2}[/itex] And then I used the transformation rule below and the partial derivatives above: [itex] A^{' kl} = \frac {\partial x^{' k}} {\partial x^{i}} \frac {\partial x^{' l}} {\partial x^{j}} A^{ij} [/itex] I calculate one of them here: [itex] A^{' 11}=( \frac {x}{\sqrt {x^2+y^2}} )^2 \times (xy) + \frac {x}{\sqrt {x^2+y^2}} \times \frac {y}{\sqrt {x^2+y^2}} \times (y^2) + \frac {x}{\sqrt {x^2+y^2}} \times \frac {y}{\sqrt {x^2+y^2}} \times x^2 + ( \frac {y}{\sqrt {x^2+y^2}} )^2 \times xy =0 [/itex] I did the calculations several times,I'm sure there was nothing wrong with them. 



#4
Mar1612, 11:03 PM

P: 1

An ambiguity in the definition of tensors
There are different types of tensors. If you're only worried about how your objects transform between cartesian coordinate systems, then you'll define what you mean by a tensor in terms of orthogonal transformations (and you'll get Cartesian tensors). On the other hand, if you're interested in considering how objects transform under general coordinate transformations, you'll define a tensors (or general tensor) by the transformations of the components as you did above.
Now, not all collections of components will be tensors. Maybe the matrix you defined above doesn't represent the components of a general tensor? 



#6
Mar1712, 03:44 AM

P: 615

Tensors are defined as things that transform like [itex]T_{a,b}=\frac{ \partial x^n}{\partial x^a}\frac{ \partial x^m}{\partial x^b}T'_{n,m}[/itex] or the other way if it's mixed or contravariant. If it doesn't transform like this then it isn't a tensor. There's nothing to really 'understand' about it, it's just a definition, things that don't transform like tensors aren't tensors! 



#7
Mar1712, 05:33 AM

P: 734

Ok.That's right.
But the matrix I've given in my first posts,transforms as a contravariant tensor under rotations but gives zero under transformation from cartesian to polar coordinates. That's what I wanna know the reason. 



#8
Mar1712, 09:52 AM

P: 134

I went through your calculation as well, your transformation rule seems correct...
Maybe the problem is that your coordinate matrix has determinant 0, but I don't see why it's wrong. 



#9
Mar1712, 12:39 PM

P: 734

What you mean by coordinate matrix?
I just know one matrix relating this and that's the Jacobian which does not have zero determinant here. I think if we analyze another tensor which works well here,we can find the wrong thing. 



#10
Mar1712, 05:04 PM

P: 134

If we took [itex] \left(\begin{array}{cc}x&y\\y&x\end{array}\right) [/itex], it would be [itex] A^{' 11}=( \frac {x}{\sqrt {x^2+y^2}} )^2 \times x + \frac {x}{\sqrt {x^2+y^2}} \times \frac {y}{\sqrt {x^2+y^2}} \times y + \frac {x}{\sqrt {x^2+y^2}} \times \frac {y}{\sqrt {x^2+y^2}} \times y + ( \frac {y}{\sqrt {x^2+y^2}} )^2 \times x [/itex] which obviously is not zero. So it seems to be as unfortunate choice of tensor and coordinate system. But still, it is wierd. If we tried coordinate transformation to polar coordinates and back to cartesian, result would be zero. Which is wrong. 



#11
Mar1712, 10:52 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,890

You have yet to give any reason why you think that is a tensor!




#12
Mar1812, 12:51 AM

P: 734

Very good point HallsofIvy.
Just the professor at university told that it is a tensor and has done the transformation for rotations and it proved to be a tensor under rotations. Maybe its not a tensor because it does not work well under this kind of transformation. So very unfortunate choice. 


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