Is the tensor product T^ab T_ab non-negative?

In summary, the topic of the properties of arbitrary tensors and their relation to the inequality T^{ab} T_{ab} \ge 0 was discussed. It was suggested that the tensor must be symmetric in order for the inequality to hold, and this would result in a constraint on the fields appearing in the tensor. The Kretschmann scalar, K_1, was also mentioned as a way to check for physical singularities, and it was noted that K_1 can have either sign. Additionally, it was discussed that in the absence of wormholes, the K_3 term (related to the Euler characteristic) would be greater than 0.
  • #1
JustinLevy
895
1
I think the answer depends on the metric signature (right?), so for this topic, let's define the signature of the metric to be positive

I'm curious what properties a tensor T must have, for the following to be true:
[tex]T^{ab} T_{ab} \ge 0[/tex]
Note: I am not talking about the stress energy tensor. I'm not sure what a good symbol choice is for "arbitrary tensor".

Similarly, I'm curious for higher rank tensors, and similar relations, like:
[tex]T^{abc} T_{abc} \ge 0[/tex]
and
[tex]T^{abcd} T_{abcd} \ge 0[/tex]

-----------------
For the rank 2 tensor, I can at least start an approach for this, but can't make it rigorous.

Choose a local inertial coordinate system, so the metric is diagonal (-1,1,1,... ). the coordinate components of T^ab will differ from T_ab only by a change in sign of the ((a=0 or b=0) and (a!=b)) components. For dimension N there will be 2(N-1) of these. However we have N(N-1)/2 arbitrary spacetime rotations we can perform and still be in a local inertial frame. So as long as
[tex]2(N-1) \le N(N-1)/2[/tex]
we can naively expect to have enough "constraint" to require that those components which change sign are all zero. And therefore, for dimension N>=4, and metric signature > 0, it looks like we can always garauntee
[tex]T^{ab} T_{ab} \ge 0[/tex].

But I think there are probably cases where this fails (the naive expectation that one can constrain certain components to be zero). Consider for instance the electromagnetic tensor. If there is an electric field with no magnetic field in one coordinate system, one can't find an inertial coordinate system in which there is no electric field.

But for symmetric tensors (as discussed in another thread), it should be possible to make it diagonal in a local inertial frame. So maybe the requirement is that the tensor is symmetric? Or is that overly restrictive? (or maybe even in that case there are problems as well?)

Is there an easier way to approach this?And what conditions are necessary for the higher tank tensor cases? (For instance, does the Riemann curvature tensor have enough symmetry that [itex]R_{abcd}R^{abcd} \ge 0[/tex]?)
 
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  • #2
Well, for instance, look at which fields appear in your T, and work out the product in a simple coordinate system. This gives you constraints on the fields if you want your product to be positive.

The last one, the square of the Riemann tensor, is given by the so-called Kretschmann scalar and is often used to check for physical singularities (in many solutions to the Einstein equations the Ricci scalar is zero because the energy momentum tensor is often conformal invariant, so you have to use the Riemann tensor or Ricci tensor explicitly).
 
  • #3
haushofer said:
Well, for instance, look at which fields appear in your T, and work out the product in a simple coordinate system. This gives you constraints on the fields if you want your product to be positive.
Well yes, but the question here was the reverse. Can I know for sure that T_ab T^ab >=0 given some property of T, and therefore I could instead use this result to guarantee constraints on the fields which appear in T.

From the discussion above it looks like T_ab T^ab >= 0 if T is symmetric. This would mean we are guaranteed something like the R_ab R^ab from the Ricci curvature must be >= 0. Is this correct? Maybe I'm oversimplifying something.

haushofer said:
The last one, the square of the Riemann tensor, is given by the so-called Kretschmann scalar and is often used to check for physical singularities (in many solutions to the Einstein equations the Ricci scalar is zero because the energy momentum tensor is often conformal invariant, so you have to use the Riemann tensor or Ricci tensor explicitly).
Alright, so using the notation from
http://en.wikipedia.org/wiki/Curvature_invariant_(general_relativity)
[tex] K_1 = R_{abcd} R^{abcd}[/tex]
[tex] K_1 + K_3 = 4 R_{ab} R^{ab} - R^2 [/tex]

It says K_3 is related to the Euler characteristic. So if there are no wormholes in the spacetime under consideration, does that mean K_3 > 0 ?

If we can indeed go into a local inertial coordinate system in which R_ab is diagonal, then if I label the diagonal components [itex]\lambda_0,\lambda_1,\lambda_2,...[/itex] we get
[tex] R_{ab}R^{ab} = \sum_i (\lambda_i)^2[/tex]
and
[tex] R^2 = [- \lambda_0 +\sum_{i>0} \lambda_i ]^2[/tex]

So R^2 can be arbitrarily larger than R_ab R^ab. So it looks like K_1 can have either sign.
 
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1. What is the tensor product T^ab T_ab?

The tensor product T^ab T_ab is a mathematical operation that combines two tensors to create a new tensor. It is commonly used in physics and mathematics to represent the relationships between different quantities.

2. Why is the question of whether the tensor product T^ab T_ab is non-negative important?

This question is important because it relates to the properties of the tensor product and can give insight into the underlying physical or mathematical system. A non-negative tensor product indicates that the quantities being measured are positively correlated, while a negative tensor product indicates a negative correlation.

3. How is the non-negativity of the tensor product T^ab T_ab determined?

The non-negativity of the tensor product T^ab T_ab is determined by examining the components of the tensor product and determining if they are all non-negative. If even one component is negative, the entire tensor product will be considered negative.

4. Can the tensor product T^ab T_ab be non-negative in some cases and negative in others?

Yes, it is possible for the tensor product T^ab T_ab to be non-negative in some cases and negative in others. This depends on the specific values of the components of the tensor product. In general, the non-negativity of the tensor product can change depending on the specific system being studied.

5. Are there any real-world applications of the tensor product T^ab T_ab being non-negative?

Yes, the tensor product T^ab T_ab is commonly used in physics and engineering to represent physical quantities such as force and energy. In these applications, a non-negative tensor product can indicate a positive relationship between the quantities being measured, which can be useful for understanding and predicting physical phenomena.

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