
#1
Oct1309, 10:40 PM

P: 103

a physical quantity which is neither vector nor scalar. is this correct definition:
pls elaborate ur ideas and suggestions. 



#2
Oct1309, 11:39 PM

Sci Advisor
P: 2,194

I would give the definition: something that transforms according to:
[tex]A'^{\alpha_1, \alpha_2, \ldots, \alpha_m}_{\beta_1, \beta_2, \ldots, \beta_n}=A^{\gamma_1,\gamma_2,\ldots,\gamma_m}_{\delta_1,\delta_2,\ldot s,\delta_n}\frac{\partial x'^{\alpha_1}}{\partial x^{\gamma_1}} \frac{\partial x^{\delta_1}}{\partial x'^{\beta_1}} \cdots \frac{\partial x'^{\alpha_m}}{\partial x^{\gamma_m}} \frac{\partial x^{\delta_n}}{\partial x'^{\beta_n}} [/tex] Note: Under this definition, a vector is simply a (1,0) tensor, and a scalar a (0,0) one. 



#3
Oct1309, 11:48 PM

P: 113

A tensor is something that transforms like a tensor, but that's not a very satisfying definition. Suppose I tell you that a car has some velocity represented by (1, 1, 0) m/s. That's a pretty useless definition to you, since you have no idea what my coordinate system is. However, if I define three orthonormal unit vectors and give you the velocity in terms of those vectors you'll have all the information you need.
A tensor works is just a generalization of the vector concept (in fact a vector is a rank 1 tensor, and even a scalar can be considered a rank 0 tensor!). Suppose I write down a matrix on a piece of paper and tell you that it represents the moment of inertia of some object. Then I tell you to transform it into a more convenient coordinate system. (Suppose the object is a torus and it would simplify things dramatically if it were represented in a toroidal coordinate system or something). You can't do it! You have no information about the coordinate system that I used when I calculated the matrix. A moment of inertia tensor is basically a matrix where each element is associated with two unit vectors in some given coordinate system. This is all the info you need to transform into other systems, and the mathematics of such transformations is extremely elegant in tensor notation. So, in short, a tensor is basically a "matrixlike" quantity that is independent of any coordinate system, and can be readily expressed in any coordinate system. 



#4
Oct1409, 07:27 AM

P: 997

What is tensor quantity?Scalar = tensor of rank 0, vector = tensor of rank 1, dyadic = tensor of rank 2, etc. A tensor is a general quantity. A scalar has magnitude with 0 direction, hence a rank 0 tensor. A vector is a magnitude acting along a line, or 1 dimension, i.e. tensor of rank 1. A dyadic has magnitude(s), acting in various planes, i.e. 2 dimensions, making it a 2nd rank tensor. These quantities are often expressed using matrices. I agree with "nnnm4" that "matrixlike" is a good way of looking at it. A scalar is a single quantity. A vector is a row or column matrix (1X2, 1X3, 2X1, etc.). A dyadic can be a matrix of 2X2, 2X3, 3X3, 3X5, etc. Does this help? Claude 



#5
Oct1409, 07:43 AM

Sci Advisor
P: 5,468

If you like, you can think of a 2ndrank tensor as a 'bivector': one vector represents the orientation of a surface (or direction of observation) and the other direction is the direction of a force acting on the surface (or field within the material). Physical properties that are (second rank) tensors include the stress tensor, strain tensor, and the index of refraction of inhomogeneous materials. There are 4thrank tensors as well: these occur in acoustooptics, mostly the acoustic field is a 2ndrank tensor and the refractive index is also a 2nd rank tensor, so relating the two requires a 4th rank tensor. 



#6
Oct1409, 01:25 PM

P: 103

the reason is: with my humble knowledge i haven't yet encountered with a physical quantity which is neither a vector nor a scalar BTW can i put it like this; any physical quantity can be classified into a vector or a scalar; am i right. i understand u all; except Nabeshins mathematical part :) .. i enountered with tensor quantity while reading about Moment of inertia (MI) as my book suggested 



#7
Oct1409, 06:30 PM

Sci Advisor
P: 5,468

The moment of inertia is a tensor because it involves two directions the axis of rotation, and the position of the center of mass (with respect to the rotation axis).
I forgot to mention the diffusion and mobility tensors, which can be measured with MRI. 


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