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In the third chapter of the book The Large Scale Structure of Space-Time, the authors say regarding the matter fields in general relativity:

These fields will obey equations which can be expressed as relations between tensors on [the manifold] in which all derivatives with respect to position are covariant derivatives with respect to the symmetric connection defined by [the metric]. This is so because the only relations defined by a manifold structure are tensor relations, and the only connection defined so far is that given by the metric.

So I was wondering if there was some sort of mathematical "naturalness" theorem in differential geometry that this statement is referring to. I've read that world tensor fields are a subset of spinor fields and that also spinor fields can be written as tensor fields.

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  • $\begingroup$ I don't fully understand the question. It's just the statement that since you have not got anything but the tensors and the connection, any relation they fulfill can also only contain these. $\endgroup$
    – ACuriousMind
    Jan 24, 2015 at 20:19
  • $\begingroup$ Usually if someone says "you can't have anything but tensors or connections" it's implicitly followed by some sort of condition "if you want [insert] something to happen." Is this a statement about mathematics...is there something that says any object defined on a manifold satisfying X properties must be a tensor or connection? Or is it a physical conjecture positing that any physical object must be a tensor or connection? $\endgroup$
    – I Dunno
    Jan 24, 2015 at 20:25
  • $\begingroup$ It's not "you can't have anything else", but rather, "you can't have anything else without defining additional structure". On a mere manifold with metric, there simply are no objects besides the tensors and the connection. (The notion of "object X defined on the manifold" is a bit fuzzy, though.) $\endgroup$
    – ACuriousMind
    Jan 24, 2015 at 20:30
  • $\begingroup$ ...perhaps the statement follows from the fact that any two space-times are physically equivalent if isometric. $\endgroup$
    – I Dunno
    Jan 24, 2015 at 20:30
  • $\begingroup$ Moreover, tensor fields can be defined without the metric. And so can connections (although the canonical one used in GR is related to one). $\endgroup$
    – I Dunno
    Jan 24, 2015 at 20:33

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It's chapter three, section two:

These fields will obey equations which can be expressed as relations between tensors on [the manifold] in which all derivatives with respect to position are covariant derivatives with respect to the symmetric connection defined by the metric $g$. This is so because the only relations defined by a manifold structure are tensor relations, and the only connection defined so far is that given by the metric. If there were another connection on [the manifold], the difference between the two connections would be a tensor and could be regarded as another physical field. Similarly another metric on [the manifold] could be regarded as a further physical field.

So the next part is the no-harm-no-foul clause. You have a metric and its corresponding connection, and as long as you are open to other fields (as long as they are tensorial fields) then there isn't really a loss. After that they mention that tensors can also do the work of spinor fields, so there isn't really any loss from only considering tensor fields.

So they are trying to say that they will use the metric and its connection, but that this isn't a big deal at this point since the door is still open to simulating the effects of an arbitrary spinor or tensor field and the derivatives with respect to arbitrary connections, by just postulating new fields.

But it doesn't motivate why they'd do this, or why that is enough. It's literally just an excuse that no harm was done by using that metric and that connection. One way to see the naturalness of using tensors and using that metric and that connection is from the comma-to-semicolon rule.

You start out imagining some physics in SR. Then you put it in tensorial form. Then you can translate that result into GR by replacing every SR derivative with a GR covariant derivative and now you have a tensor equation for GR. There is tons of meaning there, and some assumptions, and some dangers there.

First, the assumptions. It's not just the obvious, that you assume whatever you wanted to do in SR can be written as a tensor. It's the fact that the GR manifold has local freely falling frames, and those frame aren't just locally topologically $\mathbb{R}^n$, they also have the metric in SR form at a point in the neighborhood, and the derivatives of the metric be zero there too, so that you can get SR up to small controllable corrections for a small enough freely falling neighborhood. You basically say that you want to have tensor equations in that super small freely falling neighborhood, and regular derivatives in that super small freely falling neighborhood, and that you can write them as nice tensor equations for large neighborhoods in GR. You can then really take this requirement both ways and say that if you wrote it as a tensor and covariant derivatives, then you can restrict to a small freely falling neighborhood and everything will be written entirely in terms of physical fields (the tensors), coordinate derivatives (the covariant derivatives), and the SR metric (the GR metric). So the idea is that everything in GR should have a SR analog. If you think of GR as little pieced together versions of local SR, then everything in your theory should have as its local SR part, regular coordinate derivatives, the SR metric, and physical fields. You can think of it as a promise not to have any extra magic. And that anything beyond the metric and the covariant derivative (which become the SR metric and the regular SR derivative) will be introduced explicitly as a physical field, and so translate into SR as a physical field.

The meaning is the principle of equivalence. There simply isn't a gravitational field, there are simply freely falling frames and the physics of switching from one locally freely falling frame to another will be 100% the explanation for all alleged effects of "gravity", and it too you can think of as promise.

The danger is that whenever you have two theories you have to deal with the fact that two somethings that are equivalent in one theory might not translate to two somethings that are equivalent in the other theory. So for instance there might be tensor fields that are SR equivalent, but the GR translates are not the same. That's life. And thus you don't really just get GR by starting with SR and then replacing every comma with a semicolon.

But Hawking and Ellis are saying that you can do the reverse since they say that any other metric or connection will have corresponding tensor fields that will be physical fields, and so translate to SR as a physical field.

To sumamrize

In GR you can use the metric based connection to take covariant derivatives and then in the SR limit that will be a regular derivative in a small locally freely falling frame. In GR you can also explicitly introduce any other metric as a physical tensor field and any other connection as a physical tensor field, and then in the SR limit there will be physical fields so the operations will be defined relative to that in a small locally freely falling frame. Finally any spinor field could be handled with tensor fields (and hard work) and so any physical field in GR (whether tensor or spinor) will be represented as a physical tensor field, and so in the SR limit there will be a physical tensor field in a small locally freely falling frame.

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  • $\begingroup$ Does this mean everything in SR must be a tensor or connection? $\endgroup$
    – I Dunno
    Jan 24, 2015 at 22:08
  • $\begingroup$ @IDunno In SR you don't use the connection, you use the regular coordinate derivative. But the regular coordinate derivative of a super small freely falling frame equals the covariant derivative. $\endgroup$
    – Timaeus
    Jan 24, 2015 at 22:10
  • $\begingroup$ You still use a connection in SR, particularly when working in non-inertial frames. $\endgroup$ Jan 25, 2015 at 14:50
  • $\begingroup$ @FenderLesPaul You can, you can do anything. In fact, in linearized GR you actually approximate GR by treating (approximations to) the real metric as a physical tensor field. The point is that a covariant derivative (from the real metric connection) becomes a regular derivative when you switch to a local freely falling frame. The real metric switches to the inertial SR metric when you switch to a local freely falling frame. Finally, all the physical tensor fields switch to physical tensor fields when you switch to a local freely falling frame. And the local freely falling frame is inertial. $\endgroup$
    – Timaeus
    Jan 25, 2015 at 18:28

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