Thanks Alan it clear the shadow of doubts
On Sun, Dec 21, 2014 at 8:43 PM, Alan Bromborsky <abro...@verizon.net>
wrote:
> Attached is another view of tensors where if a dot product of vectors is
> defined then given a set of basis vectors [image: $e_{i}$] we have a
> metric [image: $g_{ij}=e_{i}\cdot e_{j}$] and a set of reciprocal basis
> vectors defined by [image: $e^{i}\cdot e_{j} = \delta_{i}^{j}$]. Then [image:
> $g^{ij}=e^{i}\cdot e^{j}$].
>
>
>
>
>
> On 12/21/2014 09:12 AM, Kalevi Suominen wrote:
>
>
>
> On Friday, December 19, 2014 8:20:58 AM UTC+2, Alok Gahlot wrote:
>>
>> my question is what is the difference in contravariant and covariant
>>> tensor
>>>
>> The simple (and stupid) answer is that they are objects of different
> type.
> In more detail, their components behave differently under general
> coordinate
> transformations.
>
>
>> and how can we decide to choose tensor in any problem?
>>
> There is no freedom of choice in general. Only if an inner product
> ("metric") is given,
> contravariant, covariant, and mixed tensors can be transformed into each
> other.
> Even then there is usually one form that is best suited to the problem in
> question.
>
>
>> what is the rule by which we know that we use covariant or
>> contravariant or mixed tensor ?
>>
> This is a hard question. I cannot write down any general rule valid in all
> situations.
> Instead I try to give some examples in the case of tensors on a (single)
> vector space V
> as in http://en.wikipedia.org/wiki/Tensor_%28intrinsic_definition%29 .
>
> Loosely speaking, a tensor can be regarded as a (linear or multilinear)
> mapping,
> often in more than one way.
> A tensor is of type (p,q) if there are q vector arguments and p vector
> values.
> If the value is a scalar, then p = 0, and q = 0, if there are no
> arguments.
>
> 1. A vector x in V is a tensor of type (1,0). It may be thought of as a
> mapping with no arguments
> and the single vector x as its value.
>
> 2. A covector x* in V* is a tensor of type (0,1). It is a linear mapping
> from V to the scalars.
> So its argument is a single vector and its value is a scalar.
>
> 3. A linear mapping u: V -> V is a tensor of type (1,1). It takes a
> single vector x as its
> argument and has another vector u(x) as its value.
>
> 4. An inner product (or metric) in V is a tensor of type (0,2). It has two
> vector arguments x and y
> and their scalar product x.y as the value.
>
> 5. A bilinear mapping VxV -> V is a tensor of type (1,2). It takes two
> vector arguments
> and the value is a vector. An example is the cross product in the
> three-dimensional space.
>
> Finally, if a metric is given, each vector x in V corresponds to a
> covector x* in V*, namely the
> linear form x*: y -> x.y . Then the difference between contravariance and
> covariance disappears
> under transformations that preserve the metric.
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--
With regards,
Dr. Alok Gahlot
Department Of Mathematics
+91 9897514806
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