What would be the physical consequence of a non-metric connection?

[pellman]

Suppose in whatever physical theory we are using, our derivative does not satisfy \nabla_X g=0 for tangent fields X. Of course, this means that parallel-transported vectors do not preserve their length. But what would this look like physically?

Perhaps there are various consequences which depend on the details of the connection coefficients, but I just want to get a general idea so that the signficance of why we want \nabla_X g=0 will sink in. Right now the most it means to me is that it simplifies the calculations :-)

 

[Bill_K]

This is like asking, "What would be the physical significance of a rank three tensor?" The answer of course is, whatever meaning your theory has assigned to it. After Einstein's success in explaining gravity in terms of geometry, people tried to extend the idea to somehow incorporate electromagnetism as an aspect of geometry. Unsymmetrical metric tensors, for example, and of course the Kaluza-Klein theories. None of them led anywhere.

 

[pellman]

Thanks, Bill.

Maybe I should have asked the inverse question. Why do we assume that \nabla_X g=0 in GR?

 

[member 11137]

Thanks, Bill.

Maybe I should have asked the inverse question. Why do we assume that \nabla_X g=0 in GR?

I presume that you gave your self the answer in the initial intervention: to preserve the length. Why do we preserve the length? For a part because of the Morley-Michelson experiment and the analysis we have made of it.

 

[pellman]

I don't get it though. If lengths changed as we passed from one point to another, the lengths of our measuring rods would change as well, so how would you notice?

Edit: unless the change in length was path dependent. ??

 

[member 11137]

I don't get it though. If lengths changed as we passed from one point to another, the lengths of our measuring rods would change as well, so how would you notice?

In observing the proportions.

L -> L' = 2. L
S = L² -> S' = (2.L). (2.L) = 4. S for a square
V = L³ -> V' = (2.L).(2.L).(2.L) = 8. V for a cube

So if your rod a initial length r, it becomes 2.r and effectively L/r = L'/r'.

But now consider V/r (or V/L) and compare it with V'/r' (or resp. V'/L') and state that V'/r' = 8.V/2.r = 4. V/r which is obviously not V/r. Your eyes will tell you the difference immediately!

 

[HallsofIvy]

I don't get it though. If lengths changed as we passed from one point to another, the lengths of our measuring rods would change as well, so how would you notice?

Edit: unless the change in length was path dependent. ??

Essentially, yes, the problem is path dependence- that, as you might remember from Calculus III, implies there are closed paths over which length changes.

Use your measuring rod to measure and object, then carry your measuring rod around a closed path that changed lengths, to remeasure the same object (which has not moved).

 

[pellman]

Thanks, all. Much appreciated