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

1. Jun 3, 2010

### JustinLevy

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:
$$T^{ab} T_{ab} \ge 0$$
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:
$$T^{abc} T_{abc} \ge 0$$
and
$$T^{abcd} T_{abcd} \ge 0$$

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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
$$2(N-1) \le N(N-1)/2$$
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
$$T^{ab} T_{ab} \ge 0$$.

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 $R_{abcd}R^{abcd} \ge 0[/tex]?) Last edited: Jun 3, 2010 2. Jun 3, 2010 ### haushofer 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. Jun 3, 2010 ### JustinLevy 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. Alright, so using the notation from http://en.wikipedia.org/wiki/Curvature_invariant_(general_relativity) $$K_1 = R_{abcd} R^{abcd}$$ $$K_1 + K_3 = 4 R_{ab} R^{ab} - R^2$$ 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,...$ we get
$$R_{ab}R^{ab} = \sum_i (\lambda_i)^2$$
and
$$R^2 = [- \lambda_0 +\sum_{i>0} \lambda_i ]^2$$

So R^2 can be arbitrarily larger than R_ab R^ab. So it looks like K_1 can have either sign.

Last edited: Jun 3, 2010