Tensor transformations for change of coordinate systemby aeson25 Tags: coordinate, tensor, transformations 

#1
Aug1211, 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? 



#2
Aug1211, 01:29 PM

Sci Advisor
P: 1,563

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.




#3
Aug1211, 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.




#4
Aug1211, 02:18 PM

Sci Advisor
P: 1,563

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.




#5
Aug1511, 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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