The absence of Power of Gravity at fundamental level

Yes.
But of course I have not explained WHY the metric field is not constant over all spacetime in the presence of a gravitational source.
Later for that, as I have not the time to explain.
Stay tuned if you are interested.

 

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Yes, moderately. I appreciate your willingness to explain.

But first I think I need to understand what a metric field is. I thought a field was something disposed in space (or possibly in spacetime I suppose) with different (scalar or vector) values at different points.

Whereas I envisage a metric as a sort of 3D (or 4D?) grid or scale system determining how distance is measured, rather than a set of values.

 

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Yes I understand the idea of a metric OK. My question, though, was how it can be a field.

 

Hmmm..... I must apologize. When I say "a metric" I really mean "metric tensor" Sorry for that sloppiness

So the following is quick and very dirty.

To every vector space \(V\) there corresponds a dual space \(V^*\) such that for any \(v \in V\) and any \(v^* \in V^*\) then \(v^*(v)\) is a real number.

Now suppose a manifold. But let's not get excited - the simple 2-plane \(R^2\) is a manifold. Assume that this is the case here. Then for any point \((a,b) \in R^2\)we are entitled to a vector space, from which we can select a vector and call it a vector field. Therefore, at any point \((a,b)\) we are entitled to it's dual space, also with a selection at each point. Let's call this a dual vector (or co-vector) field

Define the tensor product of 2 co-vector spaces as the mapping \(V^* \otimes V^*:V \times V \to \mathbb{R}\). And as above we can make this tensor product into a field.

And if it is the case that for some Cartesian product say \((v,w) \in V \times V\) that \(w^*\otimes v^*(v,w)=w^*(v)v^*(w)= (v\cdot w)\) (the inner product of 2 vectors) then we may call the tensor \(w^* \otimes v^*\) as the metric tensor.

As for vectors and co-vectors, we may also define a metric tensor at every point and (for now) leave open the question as to why in GR this field my not be constant.

Await with baited breath my equally boring (and even more hand-waving) explanation. Or not

 

This is going to take me a while, as it uses a number of mathematical concept I am not familiar, with, starting with tensors. I am reading about these at the moment.

But thanks for clarifying that it is the metric tensor that is a field, rather than the metric itself. That at least makes a little more sense to me, as I can see that the tensor can have different values at different points in spacetime and thus meets the criteria for being a field.

 

From a copy of some textbook I have:

And, a vector field is a tensor field . . . tensors can be considered as compositions of vectors since then, tensors are compositions of tensors in a nicely recursive way. It's similar to how you can define a string object, or a list object, recursively.

What you want to have in a physical manifold, according to Bishop & Goldberg, is distance and length (what the difference is, I guess you have to read about), angle, and of course, energy.

.

 

arfa, you do this forum no service if you simply quote texts you do not understand yourself

This is true only in the restricted case that a vector (an element of the vector field) can be thought of as a type (1,0) tensor, which in turn is only true in general when our vector is tangent to some manifold.

If we are supposed to read it for ourselves (which I have), what is the point of your post?

Why "of course?
Bear in mind Bishop & Goldberg use non-standard notation and terminology

 

Well I guess that's why I'm here then.

. . . which manifold is locally Cartesian; if so then there are vectors tangent to a local region. You say these tangent vectors are tensors because of this?

What's the point of anything? What's the point of you asking me what the point of my post is?

Of course a physical manifold has energy defined on it. Physics without energy doesn't make a lot of sense.

I'm going to strengthen that condition--"energy exists in a physical manifold"-- by asserting that information cannot exist without energy. A corollary of this is that physics and information, are the same thing in some context.

 

Here's an interesting link:

--https://www.quantumcalculus.org/tensor-products-everywhere/

I've been able to work out, by myself, the following:

If in the Cartesian product \( R^m \times R^n \) the dimension is m + n, and in the tensor product \( R^m \otimes R^n \) the dimension is mn, then these are the same when m = n = 2.