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Can I use a higher rank tensor field as the metric field?

+ 3 like - 0 dislike
228 views

What a metric should necessarily do is it should give me a way to associate a frame invariant number with a given pair of spacetime events. Now, if I use a higher rank tensor field (say, for example, a tensor field of rank 3: $g_{\mu \nu \rho}$) then also I can certainly produce a frame invariant scalar out of a given displacement $\vec{A}$ in this trivial way: $ I := g_{\mu \nu \rho} A^{\mu}A^{\nu}A^{\rho}$. Since the components of the higher rank tensor and that of the displacement vector are going to transform in a covariant and a contravariant manner respectively, the cooked up quantity is certainly a scalar.

Another crucial property, which I think has more direct physical content than the previous one, is that between two _inertial_ frames, at least one such transformation should exist that leaves at least one metric invariant. i.e., There should exist at least one combination of transformation matrix $\displaystyle\frac{\partial{x^{\alpha}}}{\partial{x^{\mu '}}}$ and metric $g_{\alpha \beta \gamma}$ that satisfies the following equation:

$\displaystyle\frac{\partial{x^{\alpha}}}{\partial{x^{\mu '}}}\displaystyle\frac{\partial{x^{\beta}}}{\partial{x^{\nu '}}}\displaystyle\frac{\partial{x^{\gamma}}}{\partial{x^{\rho '}}} g_{\alpha \beta \gamma} - \delta_{\mu '}^{\alpha}\delta_{\nu '}^{\beta}\delta_{\rho '}^{\gamma} g_{\alpha \beta \gamma} =0$

The last thing I can think of that can put a restriction on the choice of a tensor as a metric is the existence of a possibility of finding a metric compatible symmetric connection field. Following the usual procedure of finding the expression for a metric compatible symmetric connection field, I reached following condition (unlike the case of the usual two rank metric where we get a full-fledged expression) for the connection in the terms of the metric:

$g_{\nu \rho k_1} \Gamma^{k_1}_{\mu \lambda} - g_{\lambda \mu k_2} \Gamma^{k_2}_{\rho \nu} = \displaystyle\frac{1}{2} (\partial_{\nu}g_{\rho \lambda \mu} + \partial_{\rho}g_{\lambda \mu \nu}  - \partial_{\lambda}g_{\mu \nu \rho}  - \partial_{\mu}g_{\nu \rho \lambda}) $

My question is that if it is possible to satisfy the two highlighted conditions then can we use such a 3 rank (or even higher rank tensors with similarly produced conditions) tensor fields as metric fields?

PS: This is **NOT** a proposal for a new home-production theory of gravity (or that of anything for that matter) but rather it is just that I am trying to understand why a two rank tensor is used in General Relativity as the metric. Thank you.

asked Nov 28, 2016 in Theoretical Physics by Dvij (40 points) [ revision history ]
edited Nov 29, 2016 by Dvij
Most voted comments show all comments

That is under Lorentz transformation. Can't there be other transformations which preserve another set of metric components?

You said 

but rather it is just that I am trying to understand why a two rank tensor is used in General Relativity as the metric.

, so I assume Minkowski space with Lorentz transformations is a very pertinent limit. So you in fact meant to ask about general manifold with general transformation laws?

I am, of course, not questioning the validity of Lorentz transformation and Minkowskian metric. But I am trying to understand whether it is a choice made to keep things simpler or it is the only possible choice. Yes, I am considering all the possible transformations but constraining myself to efforts of describing the classical physical spacetime of GR only.

Your question is still unclear to me, without Lorentz transformations you can hardly call it relativity, and if on top of Lorentz transformations you are considering adding other transformations to the picture, that can only reduce the number of invariant tensors.

If on the other hand, you are asking the question in a pure mathematical context about general manifold with some transformations, then one potential example is the $SU(N)$ group manifold with $N\geq 3$, the rank-3 symmetric tensor with adjoint indices, defined as $d^{abc} = \text{tr}(T^a\{T^b,T^c\})$. This is an invariant tensor, but I need to check if it's compatible with group-manifold connection.

I am not sure I understand well the question but Riemannian structure always has the form $ <_,_> $. The choice of a metric is not by any means restrictive as long as the measure is invariant under the diffeomorphism group. The job of the metric tensor is to calculate angle between two sections of $TM$. The metric Jia (I think want to) proposing in group manifolds is the Cartan Killing form and it again has two indices making it a rank 2-tensor. The notion of metric tensor for higher than rank 2 tensors, to my knowledge, does not exists. Lorentz invariance is independent of a metric, it is a physical requirement (that restricts even further).

@conformal_gk I think the physical requirement imposed by SR is that when the Christoffel symbols are vanishing, there should be a combination of metric and transformation that preserves the metric components. Am I mistaken?
 

@conformal_gk, no, I didn't want to use Cartan-Killing form, which would be $\text{tr}(T^aT^b)$, and which would've been rank-2 as you said. What I wrote is a valid rank-3 invariant tensor, called "d-symbols" by some physicists, which is a very prominent object appearing in many of Yang-Mills amplitudes calculations (most famous perhaps being the discussion of anomaly cancellations).

2 Answers

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Lorentz transforms are in fact fully derivable from physical postulates (you can find one procedure for the elementary boost e.g. here). The fact that it is a two-form which falls out of these transformation procedures as invariant is, in the end, the consequence of the Euclidean norm, or the fact that Pythagoras theorem hold in our space.

There is really no way to argue why instead of 

$$a^2 + b^2 = c^2$$

 for the sides of a right triangle, we should not have e.g.

$$a^3 + b^3 =c^3$$

which would correspond to your metric 3-form. The fact that the metric is quadratic has little obvious explanation in standard physics and is simply a physical postulate derived from what is observed in reality.

There is, however, a class of geometries where the metric is more than quadratic, the so-called Finsler geometries, and you can happily construct gravitational dynamics there. You can find one nice, well-discussed example of gravitational equations in Finsler geometry by Christian Pfeifer and Mattias Wohlfart here.

These theories almost always violate special relativity in the rest frame but provide frameworks for incorporating Lorentz-violations such as Very special relativity into a theory of gravitation. It is not immediately obvious but the postulates of the Finsler geometry can be understood as an instruction on how to construct a good physical clock postulate and thus a variational principle. - You can also use the Finsler geometry with a Finsler function $F=(g_{\mu \nu \kappa} u^\mu u^\nu u^\kappa)^{1/3}$ to get a connection and in general everything you need.
---

As to your original approach, I would suggest you make a Waring decomposition (local diagonalization) of your metric tensor and proceed in a tetrad formalism. There are however cases when you obtain more than four eigendirections in your "tetrad" for a symmetric 3-tensor. This means that as a result you cannot probably generally have the usual covariant derivative annihilating the metric 3-form, you will have to have a "Finsler covariant derivative" where the Christoffel symbols have one extra index contracted with a basis vector.

answered Nov 30, 2016 by Void (1,505 points) [ no revision ]
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From a physical viewpoint, I would dare to say that a metric must be a (symm. cov.) tensor field of rank 2 because of the principle of equivalence! But to understand why I said that, lets see the mathematics first. By definition, a metric on a (pseudo)-Riemannian manifold should be a symmetric covariant tensor field, and the reason is purely mathematical. Indeed, in its most elementary form, the notion of a metric arises in mathematics when we have an abstract set \(X\) together with a mapping \((p,q)\in X \times X\mapsto d(p,q)\in \mathbb R\) that satisfy the obvious conditions of what we expect to be a "distance" between the points \(p,q \in X\) (namely, \(d(p,p)=0\), symmetry \(d(x,y)=d(y,x)\) and the triangle inequality; I have not required it to be positive, so mathematicians call it a pseudo-metric). In physics, one is usually interested in spaces \(X\) that carry more structure, for example, that it be a vector space. The metric structure on such spaces should somehow be related to its linear structure. The obvious way is to introduce an inner product \((p,q)\in X \mapsto g(p,q)\), a symmetric nondegenerated bilinear form on \(X\), define a (squared) norm \(||p||^2=g(p,p)\) and declare \(d(p,q)^2=||p-q||^2\). So the metric structure derives from the linear structure.

In Euclidean spaces, of course, there is no problem in taking the square root since the inner product \(g\) is positive definite. But in relativity, we are interested in the case that the inner product is of Lorentz signature, so it is better to work with the squared norm, which is actually called the quadratic form. So the famous \(ds^2\) that appears in physics textbooks is in reality a quadratic form sending a vector \(p \in X \mapsto ds^2(p)=g(p,p)\). See for example O'Neill, Semi-Riemannian Geometry.

In special relativity, we can identify the spacetime manifold with its own tangent space, so Minkowski space is just the vector space \(M=\mathbb R ^4\) with the Lorentz inner product \(g(x,y)=\eta_{ab}x^ay^b\forall x,y\in M\). Now, in a general spacetime manifold, the principle of equivalence, which says that locally the spacetime should look like Minkowski, translates into the framework of differential geometry as saying that the tangent space at each spacetime point (or event) is isomorphic to Minkowski space. (Recall that in DG, infinitesimal objects goes to tangent space!) So, a metric on a general spacetime \(\mathcal M\) should be an assignment to each event \(p \in \mathcal M\) of an inner product on the tangent vector space \(T_p \mathcal M\), or: \(p\in \mathcal M \mapsto^g g_p:T_p(\mathcal M)\times T_p(\mathcal M) \longrightarrow \mathbb R\), and this precisely symm. cov. tensor field \(g \in \mathrm{sec} T^{(0,2)} \mathcal M\).

All this having being said, I think that there is simply no sense in saying that a higher order tensor (namely, higher than 2) could represent a metric. However, if what one is interested in is in constructing the metric tensor from a higher order one, than the question and therefore the answer is completely different. For example, you naturally construct the metric from the curvature tensor if you assume the curvature to be that derived from the Levi-Civita connection of the metric. But observe that I can give you a curvature without giving you a metric at all. It could be the curvature derived from a teleparallel connection, for example.

answered Dec 3, 2016 by Igor Mol (420 points) [ no revision ]

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