# Can a tensor be represented as the average of vector triple products?

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Generally a tensor may be represented as a product of vectors.

A “reducing tensor” may be represented as the average of vector triple products. A reducing tensor will also “reduce” to an average of scalar products.

Acceleration may be represented as a vector. A field (gravitational, electric, or magnetic) may be represented using “reducing tensors” of acceleration.

An “equivalent EFE” may be written as reducing tensors. The equivalent EFE will then reduce to scalar products of acceleration. Suitable definitions of acceleration will give the Schwarzschild metric. Christoffel symbols are not required.

Is a “reducing tensor” mathematically valid?

Reference;         http://newstuff77.weebly.com                02 The Reducing Tensor

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A quick look reveals the following.

Cutting back on the Einstein summation convention (much as I hate it) removes the opportunity to distinguish between contravariance and covariance and the distinction between components and projections.
In other words the extra stuff that is introduced in moving from vector calculus to tensor calculus seems to be missing when you 'reduce' the tensor set to the cartesian product of two (or more) vector sets.

But thanks for the link they papers look sufficiently interesting that you might not hear form me for a while.

+1

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14 hours ago, chemguy said:

Generally a tensor may be represented as a product of vectors.

A tensor is a multilinear map that maps vectors and 1-forms into other objects. This is true even if the manifold isn’t endowed with a metric, and regardless of the number of dimensions.

14 hours ago, chemguy said:

A “reducing tensor” may be represented as the average of vector triple products.

The vector cross product which you are using here is defined only in 3 dimensions, and is not generally covariant. It also requires the presence of a metric. Furthermore, if you replace a higher rank tensor by a real number, you loose both information and a number of degrees of freedom.

14 hours ago, chemguy said:

A field (gravitational, electric, or magnetic) may be represented using “reducing tensors” of acceleration.

Again, you cannot simply replace replace a rank-n tensor field by a scalar field; they are not equivalent objects. Physically speaking, you need to be able to capture all relevant symmetries of the field (in the case of GR, diffeomorphism invariance), which is possible only if it is a rank-2 tensor field. Remember also that, in terms of QFT, the graviton needs to be a massless spin-2 boson, which again means you need a rank-2 tensor field.

14 hours ago, chemguy said:

Suitable definitions of acceleration will give the Schwarzschild metric.

It appears that you chose your “characteristics” in order to obtain the solution you wanted, not because those choices are motivated mathematically or physically; there is no mention of any boundary conditions.

14 hours ago, chemguy said:

Christoffel symbols are not required.

Vanishing Christoffel symbols imply a flat region. Clearly, this would not lead you to the Schwarzschild metric, or any spacetime other than the trivial Minkowski one.

14 hours ago, chemguy said:

Is a “reducing tensor” mathematically valid?

I’m afraid it is not.
Also, the fact that the EFE is a rank-2 tensor equation has deeper reasons; you can’t just replace it with a scalar relation.

14 hours ago, chemguy said:

An “equivalent EFE” may be written as reducing tensors.

No, mostly because these are not generally covariant objects (no diffeomorphism invariance), and do not capture the underlying physics.

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Before tensors were properly developed there was a half way house between vectors and tensors called dyadics, whose elements were called dyads.

This appears to be a similar half way house, a simplification answering some questions but not others.

Quote
A dyad is a tensor of order two and rank one, and is the result of the dyadic product of two vectors (complex vectors in general), whereas a dyadic is a general tensor of order two (which may be full rank or not).

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