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Tensor transformations for change of coordinate system

by aeson25
Tags: coordinate, tensor, transformations
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aeson25
#1
Aug12-11, 11:44 AM
P: 7
In school I've always learned that tensor transformations took the form of:

[tex]\mathbf{Q'}=\mathbf{M} \times \mathbf{Q} \times \mathbf{M}^T [/tex]

However, in all the recent papers I've been reading. They've been doing the transformation as:

[tex]\mathbf{Q'}= \frac {\mathbf{M} \times \mathbf{Q} \times \mathbf{M}^T}{det(\mathbf{M})}[/tex]

Where Q is the tensor in question and M is the transformation matrix and M^T is the transpose of M.

Does anyone know why the difference?
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Ben Niehoff
#2
Aug12-11, 01:29 PM
Sci Advisor
PF Gold
P: 1,594
Such on object is properly called a "tensor density". Specifically, it is a "tensor density of weight -1", since the determinant appears with the power -1.
aeson25
#3
Aug12-11, 02:14 PM
P: 7
I can't seem to find a clear definition of "tensor density" online. How does this differ (or provide an advantage) (practically, not mathematically) from a regular coordinate transformation? FYI, I'm trying to follow the transformation of a anisotropic density (actual matter density) in a paper.

Ben Niehoff
#4
Aug12-11, 02:18 PM
Sci Advisor
PF Gold
P: 1,594
Tensor transformations for change of coordinate system

Did you even try Googling it? For your claim that you "can't find it", I have serious doubts.
aeson25
#5
Aug15-11, 11:40 AM
P: 7
Of course I tried googling it. I didn't say I didn't find ANY definitions, I said I couldn't find a CLEAR definition which included information pertaining to the entirety of the second sentence of my last post with an emphasis on the parenthetical parts. The online definitions basically say its a coordinate transformation with a weight to it based on a Jacobian determinant. Big whoop, those definitions tell me nothing about why it's used over a normal transformations. Why use a weight of -1 rather than 1000?


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