R_speir: questions about cosmology and dimensions

@PdotdQ. Comment and question

If the universe is a sphere (strange, but why not? Let’s go with it for now) and a hyper-dimension is returning flat space (I said the time dimension was split, but folks don’t like that. Ok, fine. I don’t like extra spatial dimensions so we are even. Ha.) But ok, I will concede for discussion sake. Give the system a hyper spatial dimension. Don’t you still have to hang an extra clock on the extra spatial dimension??? I say yes - at least I would certainly want to in order to get a good reading on the overall expansion. And two clocks that are clocking the expansion of a sphere should return an accelerating universe. That’s my view. What’s yours?
Just for the record, I have no problem whatever with an accelerating universe.

You don’t need to “hang a clock” on any spatial dimensions. A time dimension is just another axis in spacetime. Unless I am misunderstanding what you are saying, we never “hang a clock” on any spatial dimensions.

Hyper dimension. Does not an added spatial dimension require its own timepiece?

Nope, I can add as many spatial dimensions I want to my theory without adding any time dimension and vice versa.

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Do you think that each of the 3 dimensions of Euclidean space requires it’s own timepiece? If so, please explain what you mean by a timepiece. If not, why would you think that any further dimensions require timepieces?

I’m gonna say you’re right and I have a misunderstanding.

Now, consider this. Take the normal line metric with 3 + 1 dimensions, but split one of the spatial dimensions such that you are not really adding a spatial dimension but rather simply taking the hypotenuse of the two arms of the split dimension. Won’t the resolution of that metric return an acceleration of the system?

I don’t quite understand what operation you are referring to here. Perhaps a picture would help.

What is the “resolution” of a metric? There is a “resolution” of spaces from topology, but from the context I don’t think that is what you mean.

Simplistically shown

dr = -dt + dr(r1+r2) + dr + dr (I hope that metric is correct, of course each terms should be squared)

where r1 + r2 is understood to be the Pythagorean sum (r1^2 + r2^2)^.5

See the split spatial dimension? r1+r2. I want the measure of the hypotenuse only.

Hmm, that metric doesn’t make sense, even if each terms are squared. In any case, you cannot “split” a dimension like that without changing the number of dimensions. What happens is that either your new coordinates are not independent to each other (i.e. there are constraint equations that reduces the number of dimensions), or that you have implicitly embedded the space to a different space with a different number of dimensions.

I need to understand this more clearly. The arms of the split dimension of course are not moving independently of one another. They are moving in step with one another. Now help me along by restating what you did state and commenting on my comment.

Suppose I have a vector space with only one dimension, let’s call it r. I can perform a coordinate transformation and write, for example,

r = r_1+r_2 \; ,

It is not the case that r_1 and r_2 are two different coordinates that somehow “splits” the original coordinate r. To wit, these two coordinates are not linearly independent. What you are proposing is the same, but with the coordinate transformation being

r = \sqrt{r_1^2 + r_2^2} \; ,

instead. At least, that’s how I understand what you are trying to do.

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Correct. Better stated than mine. And correct to the comment about not being linearly independent.

Now, given your better visual, what on earth (or rather in spacetime :)) would such a transformation do to an expanding sphere (if such a construction could exist)?

I am not sure what you are trying to ask.


So before I get ahead of what is physical, can the above line element exist (in theory)?

If I may catch my own mistake. I am expecting all terms on the right to advance, say, 1 unit (hoping that r1 and r2 will also move 1 unit as well? Why would I expect that? Wouldn’t it be dx only that advanced 1 unit and not its individual components?)

For now I see an error. Do you agree? Your comments are welcome.

The above line element can of course exist mathematically, as all you do is perform the coordinate transformation:

x = \sqrt{r_1^2 + r_2^2} \;,

but you need to be careful with what is the actual mathematical meaning of such coordinate transformation! Writing it this way makes it seem that x is some sort of hypothenuse. However, this is a mistake! There is no hypothenuse of anything; the equation x = \sqrt{r_1^2 + r_2^2} is just an arbitrary coordinate transformation, there is no triangle anywhere. In fact, the equation x = \sqrt{r_1^2 + r_2^2} is not even the right equation for the hypothenuse of a triangle in arbitrary spaces (only true for flat space in cartesian coordinates).

Further, as stated before, you are not adding any new dimensions, and r_1 and r_2 are not linearly independent. As such, the above line element should instead be written as

ds^2 = -c^2 dt^2 + [A dr_1 + dy + dz ]^2\;,

where A is just a constant.

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Very good. Thank you. You know I was looking for an acceleration of the sphere thinking an added dimension came with its own timepiece by necessity. You said otherwise, and so that pretty much ends it. And I certainly won’t find it by a coordinate transformation as above.

So let me think about something out loud. In my paper I found an acceleration because as Tom Bridges told me “you constructed it that way”. Ok, maybe or maybe not. So let me ask a question.

Given a dynamical sphere, one that expands or contracts over time, how would one go about describing the dynamic of expansion without invoking a radius? Or a circle will do.

The radius is just another coordinate. You can write your expansion in terms of other coordinates (e.g., x and y in 2 dimensions) without ever writing down what a radius is.

Exactly. So why can I not find a formal write-up anywhere in the literature about a sphere, circle without the use of the radial coordinate? Does it not exist?

I don’t have the time to check right now, but this is probably in university textbooks on vector calculus. You can easily write it down just by making the usual coordinate transformations. Also, don’t mistake the radial coordinate, with 1-form basis dr and the curvature parameter r, which is just a parameter that determines the curvature of a circle. The second is just a parameter and not a “radial coordinate”.