Second postulate of SR quiz question

martinbn

Let us be less abstract and try to do something more concrete in differential geometry. Consider a two-dimensional sphere with unit radius, immersed in the 3-dimensional space with Euclid metric. Can you prove that the area of the sphere is $4\pi$ without using any coordinates? (I can't.)
Probably not, unless you do something clever with limits. But that's beside the point, there will be questions, for which you need coordinates, that's for sure. But my wondering came from your post, where you said that for defining a manifold somewhere along the definition and axioms you will need them. So, is that the case? With charts and atlases you can get away without coordinates, but will they be needed at some point (before you want to do a calculation that requires them).

vanhees71

Gold Member
Hm, I'm only a naive theoretical physicist, but a chart is a mapping of an open part of the manifold to an open subset of $\mathbb{R}^d$, and these $d$-tupels of real mumbers are usually called (local) coordinates, right? Is there any possibility to define a differentiable manifold without first introducing such charts and atlasses and then forget about the specific choice of these charts and atlasses by taking them modulo diffeomorphisms between open subsets of $\mathbb{R}^d \rightarrow \mathbb{R}^d$, describing "changes of coordinates" between different choices of charts and atlasses? If so, is there a textbook explaining this?

For physics it's of course very abstract, because there you deal with measurements in the real world, and these measurements are provided by some (usually real) numbers anyway. As far as spacetime measures are concerned these are in some sense (local) coordinates in the theory describing this experiment, but I think now we really drift apart to metaphysics ;-)).

martinbn

A map to an open subset of an Euclidean space. No coordinates! If you choose coordinates to identify the Euclidean space with $\mathbb R^d$, then you'll have coordinates.

• atyy

stevendaryl

Staff Emeritus
A map to an open subset of an Euclidean space. No coordinates! If you choose coordinates to identify the Euclidean space with $\mathbb R^d$, then you'll have coordinates.
Isn't a coordinate system for a patch the same thing as a smooth map between that neighborhood and $R^d$?

Demystifier

2018 Award
With charts and atlases you can get away without coordinates, but will they be needed at some point (before you want to do a calculation that requires them).
I think it is crucial to answer the following question. Without the coordinates, are the calculations merely difficult and impractical but possible in principle, or are they strictly impossible even in principle? I think that only in the former case we can say that geometry, in principle, can be formulated without coordinates.

Dale

Mentor
A clock doesn't do an abstract integral; it is a real-world device
Why not? Because you put the qualifier "abstract" there? If so, then the same can be said of a reference frame, which is an abstract set of orthonormal basis vector fields. What makes one abstraction impossible for a real-world device and the other abstraction something intrinsic to the device?

loislane

That is fine. I didn't clarify that i was using units where c=1.
I was actually referring to the distinction between proper time defined as the local readout of a clock and the interval that is what includes the integral in its computation. You don't seem to be clear about this distinction that has also been remarked by vanhees71.
Why not? Because you put the qualifier "abstract" there? If so, then the same can be said of a reference frame, which is an abstract set of orthonormal basis vector fields. What makes one abstraction impossible for a real-world device and the other abstraction something intrinsic to the device?
I don't think it is the word abstract, it is the simple fact that the integral implies a computation, a proper time local measurement doesn't require any operation to define a local frame, it is just a clock readout associated to a specific triad for the position in wich the readout is made, the integral operation is needed in the analysis to connect the local frame(the orthonormal basis tetrad that includes a specific local proper time) with a general frame and its orthonormal tetrad. The reason is that there is no global time in relativity.

Dale

Mentor
I don't think it is the word abstract, it is the simple fact that the integral implies a computation,
That doesn't seem right to me. Integrals show up all the time in the laws of physics.

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