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I am having problem to admit why in a laminar flow of a fluid we assume a small area and take a direction normal to it.Similarly in stress tensor analysis introductory diagram area of each face of the cube is treated as a plane with direction.Please explain me why area turns from a directionless quantity to a quantity with direction and also in that context explain me what is a tensor in a simple language.

gateprep
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Tensors are a broad class which includes the scalars and vectors. The transformation of a quantity under coordinate transformation defines whether the quantity is a scalar etc. For instance we know that the magnitude of a vector is a scalar since it does not change under a transformation but the componenet of a vector does. As regards area the direction is usually associated with a direction perpendicular to the infinitisimal area. It is useful, for instance in obtaining various quantities like electric flux, flow of current etc.

SAKhan
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Since I cannot delete an accepted answer, I will edit, because I think the comments are correct, that I was misrepresenting what actually a tensor is.

We live in a three dimensional space and it is easy to understand that three numbers in an (x,y,z) coordinate system on each of the axis define a point in space and the line joining with the origin is a vector, i.e. it has a direction. Vectors are useful because they model forces. There are zero dimensional vectors, i.e. points, there are two dimensional vectors defined in a plane also.

Continue with this link for a correct formulation.

Notice that the effect of multiplying the unit vector by the scalar is to change the magnitude from unity to something else, but to leave the direction unchanged. Suppose we wished to alter both the magnitude and the direction of a given vector. Multiplication by a scalar is no longer sufficient. Forming the cross product with another vector is also not sufficient, unless we wish to limit the change in direction to right angles. We must find and use another kind of mathematical ‘entity.’

For detailed exposition also go to the duplicate indicated above.

To model physical systems one has to consider the independent variables defining the system, whether a Cartesian coordinate system of n dimensions can be defined with the proper coordinate transformations. Tensors are used in modeling physics problems wherever there are variations in quantities that depend on two vectors and one can fit the mathematical demands of tensor algebra.

anna v
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    Actually, by increasing the number of dimensions one simply increases the dimension of a vector. The vector stays a vector. The point is that a vector (having one index) is a tensor of rank 1; a matrix (having two indices) is a tensor of rank 2, etc. In general tensors can have arbitrary numbers of indices. A better answer is provided in: https://physics.stackexchange.com/questions/32011/what-is-a-tensor. – flippiefanus Aug 26 '17 at 06:37
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    @flippiefanus sure they are better answers mathematically, but are they responding to the level of this question? note "simple language" – anna v Aug 26 '17 at 07:16
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    The issue is that the notion of a tensor has nothing to do with the number of dimensions, but rather with the rank of the object. This can be explained in terms of mathematics or in terms of "simple lamguage." – flippiefanus Aug 26 '17 at 07:18
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    @flippiefanus please give a link where the rank of a tensor is not the same as a dimension in an n dimensional mathematical space. I find this :"The rank (or order) of a tensor is defined by the number of directions (and hence the dimensionality of the array) required to describe it." https://www.doitpoms.ac.uk/tlplib/tensors/what_is_tensor.php – anna v Aug 26 '17 at 07:21
  • The Faraday tensor has rank 2 and the dimensions of the indices are 4. – flippiefanus Aug 26 '17 at 07:23
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    @flippiefanus I find a confusion in the use of "rank" here http://mathworld.wolfram.com/TensorRank.html where symmetries of the system are taken into account and in effect reduce the dimensions. I think my explanation above is simple enough for somebody interested , to then continue on the more confusing terminology. The main answer in the duplicate uses dimensions and ranks equivalently – anna v Aug 26 '17 at 07:32
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    The problem is that in the context of tensors the terms dimension and rank means different things. These terms are (mostly) used correctly in physics.stackexchange.com/questions/32011/what-is-a-tensor. One shouldn't get fixated on the use of one particular person or source, but rather determine what is the generally accepted use of these terms. – flippiefanus Aug 26 '17 at 07:47
  • @flippiefanus: there's more than one meaning of rank in linear algebra, I wouldn't get hung up on it. – Mozibur Ullah Aug 26 '17 at 08:00
  • @anna v.I saw this link doitpoms.ac.uk/tlplib/tensors/what_is_tensor.php. I am having confusion on why they took 3X1 for 1st order tensor and not 1X3 for the same. – gateprep Aug 26 '17 at 14:09
  • 1X3 would be a row vector, and is isomorphic and is usually considered when dot products are discussed. – anna v Aug 26 '17 at 16:26
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    @MoziburUllah: linear algebra deals with vectors and matrices, not tensors in general. The rank of a matrix is not the same thing as the rank of a tensor. – flippiefanus Aug 30 '17 at 17:48
  • @flippiefanus: yeah, thats elementary linear algebra; 'higher' linear algebra does deal with tensors - have a look at |Northcotts Multi-linear Algebra – Mozibur Ullah Sep 05 '17 at 10:43