The Weyl tensor equates to the Riemann tensor in a vacuum

$$ C_{\mu \nu \eta \lambda} = R_{\mu \nu \eta \lambda} $$

So it makes me wonder about the tensor

$$T_{\mu \nu \eta \lambda} = C_{\mu \nu \eta \lambda} - R_{\mu \nu \eta \lambda} $$

and how it relates to the 2nd-rank stress-energy tensor $T_{\mu \nu}$. In particular, both tensors are zero and non-zero in the same domains, so they must be related. On the other hand, general relativity says that matter affects geometry only through the 2nd-rank tensor, so in theory, no higher-rank tensor should contain more information about the matter fields than what it (the 2nd-rank tensor) already does

I'm trying to figure out if the 4th rank tensor can be interpreted or not as containing more information about the energy-matter fields or if the extra degrees of freedom are strictly geometric. The question is relevant for considering alternative formulations of the matter curvature relationship that tend to the Einstein equation in some sensible limit


2 Answers 2


This object is related to the Schouten tensor, $$S_{ab} = \frac{1}{n-2}\left(R_{ab} - \frac{R}{2(n-1)} g_{ab}\right).$$ We find $${C_{ab}}^{cd} - {R_{ab}}^{cd} = -4 S_{[a}^{[c} \delta_{b]}^{d]}.$$ As @Luboš Motl mentions, this tensor depends only on the Ricci tensor and scalar curvature. That is, it doesn't "know" any more than Ricci---on the geometry side it is nothing new.

It doesn't make sense to ask how this is related to $T_{ab}$ without the Einstein equations or some explicit "alternative formulation." If the Einstein equations can be assumed it is a straightforward exercise to work it out using $R = -\frac{2}{n-2} 8\pi T$ and $R_{ab} = 8\pi(T_{ab} - \frac{1}{n-2} g_{ab}T)$.


One may simply calculate and simplify your tensor $T_{4}$. Following the definition of the Weyl tensor, it's obvious that your $$ T_{abcd} = -\frac{2}{n-2} (g_{a[c}R_{d]b} - g_{b[c}R_{d]a}) + \frac{2}{(n-1)(n-2)} R g_{a[c} g_{d]b} $$ so even without using any equations of motion, your tensor is a function of the Ricci tensor only (and the Ricci scalar which is the trace of the Ricci tensor, anyway). So indeed, in the vacuum where the Ricci tensor vanishes by Einstein's equations without a right hand side, your tensor vanishes, too.

It's completely illogical to call this tensor a "generalization of the stress-energy tensor" because as you defined it, it depends on the metric and its derivatives only. You may also replace the Ricci tensor in the formula above by the appropriate multiple of the stress-energy tensor (with the inverted trace), using Einstein's equations. Then, you will see that $T$ with four indices may indeed be obtained from objects such as $T_{ab} g_{cd}$.

But what it would be good for? If your question is whether this object is natural, useful, or justified by something, my answer is No. Nevertheless, if you're excited by various other fields with 4 indices and symmetries of the Riemann tensor, you may want to read the following 2000 paper by Chris Hull (page 10 and so on):


  • $\begingroup$ Hi Lumo, i appreciate that it is all expressed in terms of the metric and the Ricci tensor. Thinking about it, your argument should apply to the Einstein tensor as well, since it depends on the same quantities. Does it mean that the Einstein tensor depends "on the metric and its derivatives only"? clearly, that is simply not true. $\endgroup$
    – lurscher
    Commented Jul 19, 2012 at 17:52
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    $\begingroup$ Yes, obviously, Einstein's tensor only depends on the metric and its derivatives, much like the Riemann tensor, Ricci tensor, Ricci scalar, Weyl tensor, or anything else of this kind. All these objects describe the curvature of the spacetime geometry and nothing else. WTF? Einstein's equations link Einstein's tensor with the stress-energy tensor of matter – that's how matter curves the spacetime – but these equations are not definitions of either objects and they don't hold by definition. They're dynamical laws of Nature that hold when they hold. $\endgroup$ Commented Jul 19, 2012 at 17:54
  • $\begingroup$ also, the argument that it can be expressed as non-contracted tensor products of the 2-nd rank stress-energy tensor and the metric is valid, but note that it relies on assuming the Einstein equation. I was trying to make this question more of a meta sort of question, without assuming the Einstein equation (which would lead to failure by tautology) $\endgroup$
    – lurscher
    Commented Jul 19, 2012 at 17:54
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    $\begingroup$ when you say "because they are the only way how the stress-energy tensor may be related to geometry", thats it! that is what i'm asking. If you know how to prove this, then post that as the answer. thanks! $\endgroup$
    – lurscher
    Commented Jul 19, 2012 at 18:02
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    $\begingroup$ I have already explained why your speculative part of the question you just added is based on elementary childish errors in logic. $\endgroup$ Commented Jul 19, 2012 at 18:03

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