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Offline JRowe

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Re: On Stuff
« Reply #40 on: March 22, 2019, 11:01:49 AM »
Your altitude boundary condition, Mz=-g(h-z)^2, has units of s^(-2)m^(-1). So this means that it is not time invariant. That is, it will not permit the time interval to be the same between adjacent spacetime coordinates.
It's g the mathematical constant, not the acceleration.

Quote
Try tinkering with this a while. The next stage is to discover transformation equations so that the proper velocity appears different to observers depending on their coordinates in that frame.
Will try to unpack it in a sec.

Do you mean Newton’s constant? That’s G=6.7*10^(-11) m^3 kg^(-1) s^(-2).

In this case, Mz has units of m*kg^(-1)*s^(-2) and is still not time invariant.

Cool, let me know how it strikes you.
The mathematical constant, no units. It's just there to give the rate of variation, which is connected to the cause of gravity but isn't directly it.
My DE model explained here.
Open to questions, but if you're curious start there rather than expecting me to explain it all from scratch every time.

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Offline QED

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Re: On Stuff
« Reply #41 on: March 22, 2019, 11:58:49 AM »
Your altitude boundary condition, Mz=-g(h-z)^2, has units of s^(-2)m^(-1). So this means that it is not time invariant. That is, it will not permit the time interval to be the same between adjacent spacetime coordinates.
It's g the mathematical constant, not the acceleration.

Quote
Try tinkering with this a while. The next stage is to discover transformation equations so that the proper velocity appears different to observers depending on their coordinates in that frame.
Will try to unpack it in a sec.

Do you mean Newton’s constant? That’s G=6.7*10^(-11) m^3 kg^(-1) s^(-2).

In this case, Mz has units of m*kg^(-1)*s^(-2) and is still not time invariant.

Cool, let me know how it strikes you.
The mathematical constant, no units. It's just there to give the rate of variation, which is connected to the cause of gravity but isn't directly it.

You mean it is just the number in front? If it gives a rate then it must have units. That’s what a rate means.

The difference between math and physics is that math is pure, it uses no units because it has not yet been applied to something. When you apply it - and do applied mathematics, then all the equations, functions, graphs, numbers, acquire a dimension that represents the application. It is erroneous to develop physical models without units. It is an oxymoron. We do not have the option.

If it is pure mathematics that you wish to do instead, then I can try to help you with that. Just understand that it will not relate to anything physical, and will instead reference previously substantiated mathematics as its assumption set.
The fact.that it's an old equation without good.demonstration of the underlying mechamism behind it makes.it more invalid, not more valid!

- Tom Bishop

We try to represent FET in a model-agnostic way

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Offline JRowe

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Re: On Stuff
« Reply #42 on: March 22, 2019, 06:08:19 PM »
You mean it is just the number in front? If it gives a rate then it must have units. That’s what a rate means.

The difference between math and physics is that math is pure, it uses no units because it has not yet been applied to something. When you apply it - and do applied mathematics, then all the equations, functions, graphs, numbers, acquire a dimension that represents the application. It is erroneous to develop physical models without units. It is an oxymoron. We do not have the option.

If it is pure mathematics that you wish to do instead, then I can try to help you with that. Just understand that it will not relate to anything physical, and will instead reference previously substantiated mathematics as its assumption set.
It's always pure until it's applied, getting it working in the abstract is far more important to me because if that can't be done then there's no point making all the tweaks necessary to apply it physically. At the end of the day all that's going to be required to bring units in line are a couple of constants as multiples, and that's as good as trivial. The same goes for Mx and My. Working without them is just a whole lot simpler, and I've seen far too many instances of math being used to bamboozle rather than aid understanding and I'm well aware that making this any more complicated is just going to have people brush it off as pseudoscientific evasion.
Mz is the rate of change. g is a constant to give quantity, and in retrospect it probably wouldn't be the 9.8... value anyway as that's be Mt over the direction z, while this is just connected to that. It's not different to any constant, it's just g for gravity because it's defining the value of that quantity, but alone it means nothing.
My DE model explained here.
Open to questions, but if you're curious start there rather than expecting me to explain it all from scratch every time.

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Offline QED

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Re: On Stuff
« Reply #43 on: March 23, 2019, 01:42:03 AM »
You mean it is just the number in front? If it gives a rate then it must have units. That’s what a rate means.

The difference between math and physics is that math is pure, it uses no units because it has not yet been applied to something. When you apply it - and do applied mathematics, then all the equations, functions, graphs, numbers, acquire a dimension that represents the application. It is erroneous to develop physical models without units. It is an oxymoron. We do not have the option.

If it is pure mathematics that you wish to do instead, then I can try to help you with that. Just understand that it will not relate to anything physical, and will instead reference previously substantiated mathematics as its assumption set.
It's always pure until it's applied, getting it working in the abstract is far more important to me because if that can't be done then there's no point making all the tweaks necessary to apply it physically. At the end of the day all that's going to be required to bring units in line are a couple of constants as multiples, and that's as good as trivial. The same goes for Mx and My. Working without them is just a whole lot simpler, and I've seen far too many instances of math being used to bamboozle rather than aid understanding and I'm well aware that making this any more complicated is just going to have people brush it off as pseudoscientific evasion.
Mz is the rate of change. g is a constant to give quantity, and in retrospect it probably wouldn't be the 9.8... value anyway as that's be Mt over the direction z, while this is just connected to that. It's not different to any constant, it's just g for gravity because it's defining the value of that quantity, but alone it means nothing.

Okay, so this an entirely different ball game then. Developing the pure math is much, much harder.

So the place to begin is by stating your assumptions. What mathematics are you building off of? Are you assuming an arithmetic? If so, is it binary? If you’re assuming the modern fundamental theorem of arithmetic than this has consequences that you need to learn. Otherwise it is very easy to contradict one of them without knowing. Are you using standard algebraic structures? Rings, fields, group theory? How about vector spaces and complex analysis?

If you want to use topology, which I bet you’ll end up needing, then understanding how you develop manifolds and perform algrebra on them is essential.

In summary, I applaud and encourage your pursuit into pure math, but do so with your eyes open - it is a much more abstract and technically demanding task than building a physical model. The development would proceed with a series of lemmas leading to the fundamental proof of the work. Then this would serve as the stepping stone for corollaries, and further lemmas to build a network of consistent proofs that detail your theory.
The fact.that it's an old equation without good.demonstration of the underlying mechamism behind it makes.it more invalid, not more valid!

- Tom Bishop

We try to represent FET in a model-agnostic way

- Pete Svarrior

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Offline Rama Set

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Re: On Stuff
« Reply #44 on: April 14, 2019, 01:51:12 PM »
Did this progress any further?
You don't get races of anything ... accept people.