Unravelling the Mystery of the (0,3) Symmetric Tensor

[christodouloum]

I am a bit confused by this observation.
Every tensor is it's symmetric plus antisymmetric part.

Thus for the components of a (0,3) tensor

$$F_{\lambda\mu\nu}=F_{[\lambda\mu\nu]}+F_{\{\lambda\mu\nu\}}$$

and if I write this down explicitly I end up that for the components of ANY (0,3) tensor

$$F_{\lambda\mu\nu}=(1/3)(F_{\lambda\mu\nu} +F_{\mu\nu\lambda}+F_{\nu\lambda\mu} )$$

Huh? Does this indeed hold?

 

[christodouloum]

I am replying to my self since I searched around a bit and the statement

Every tensor is it's symmetric plus antisymmetric part

holds for every pair of indices not generally. So now I know that the last equality I wrote does not hold but still, is there any way to generalize this idea? I mean, please correct me if I am wrong but we have

$$F_{\lambda\mu}=F_{[\lambda\mu]}+F_{\{\lambda\mu\}}$$
$$F_{\lambda\mu\nu}=F_{[\lambda\mu]\nu}+F_{\{\lambda\mu\}\nu}$$

how about a relation between $$F_{\lambda\mu\nu}$$ ,$$F_{\{\lambda\mu\nu\}}$$ and $$F_{[\lambda\mu\nu]}$$??

 

[AlephZero]

It is a useful idea, because symmetric and antisymmetric tensors each have useful properties, but the decomposition is basically trivial arithmetic, i.e.

a = (a+b)/2 + (a-b)/2
b = (a+b)/2 - (a-b)/2

I don't think there is much to generalize about that.

 

[christodouloum]

I know how to work it out . It is surely not a trivial idea, the symmetric and antisymmetric parts of the exponential function in the reals are the hyberbolic functions cosh and sinh which make the splitting idea quite important. Also by the property that a totally symmetric tensor contracted with a totally antisymmetric one gives nul I am just wondering, is there any way to split an arbitrary tensor in two additive parts the symmetric and antisymmetric one? I

 

[blue_raver22]

I can understand your confusion about this observation. However, it is true that every tensor can be decomposed into its symmetric and antisymmetric parts. In the case of a (0,3) tensor, the symmetric part will have the form you mentioned, where each component is equal to the average of all possible permutations of the indices. This is known as the Young symmetrizer and it ensures that the resulting tensor is symmetric under exchange of any two indices.

Now, you may be wondering why this is important or why we should care about the (0,3) symmetric tensor. Well, this type of tensor appears in many areas of physics, including fluid dynamics, relativity, and electromagnetism. It also plays a crucial role in the study of elasticity, where it describes the stress and strain of a material. By understanding the properties and behavior of this tensor, we can gain a deeper understanding of these physical phenomena.

In addition, the (0,3) symmetric tensor has many interesting mathematical properties, such as being invariant under coordinate transformations and satisfying certain symmetry relations. This makes it a useful tool for solving problems in physics and mathematics.

In conclusion, the (0,3) symmetric tensor may seem confusing at first, but it is a fundamental concept in physics and mathematics. By unraveling its mysteries and understanding its properties, we can gain a better understanding of the world around us.