Space time is reality Pseudo

Discussion in 'Pseudoscience' started by chinglu, Oct 19, 2013.

1. chingluValued Senior Member

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1,637
Uh, this is the 6th time.

d(x,y) = 0 under a metric space iff x = y.

I am sure you can google.

Yet, under space-time d(x,0) = 0 and x != 0.

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3. paddoboyValued Senior Member

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27,543

No as usual you are wrong....Just as you were wrong with your mathematics at the other forum, and which you were banned for being a moron.

http://mathworld.wolfram.com/MinkowskiSpace.html

Minkowski space is a four-dimensional space possessing a Minkowski metric

dtau^2=-(dx^0)^2+(dx^1)^2+(dx^2)^2+(dx^3)^2.
Alternatively (but less desirably), it can be considered to have a Euclidean metric but with imaginary time coordinate x^0=ict, where c is the speed of light, by convention c=1 is normally used, and i is the imaginary number sqrt(-1). Minkowski space unifies Euclidean three-space plus time (the "fourth dimension") in Einstein's theory of special relativity.

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5. paddoboyValued Senior Member

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http://ncatlab.org/nlab/show/Minkowski space

Definition

For d−1∈N, d-dimensional Minkowski space is the Lorentzian manifold whose underlying smooth manifold is the Cartesian space Rd and whose pseudo-Riemannian metric is at each point the Minkowski metric.

This is naturally a spacetime.

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7. paddoboyValued Senior Member

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27,543
ENERGY-DEPENDENT MINKOWSKI METRIC
IN SPACE-TIME

http://www.mathem.pub.ro/proc/bsgp-10/0MIRON01.PDF

Highly mathematical this one chinglu, way above my head, and your maths has been shown to be rather faulty at the best of times....anyway, have a look.
Or are you going to refuse to do that?

8. arfa branecall me arfValued Senior Member

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7,832
Wow! That proves that nobody, even you, can measure any distance.
But doesn't it really say that if t = 0, then an object doesn't travel any distance, in which case it doesn't seem to prove much after all.

9. chingluValued Senior Member

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1,637
Uh, use the Minkowski metric and measure the distance between any point of the light sphere and the origin. Let me know what you conclude.

10. chingluValued Senior Member

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1,637
I am not seeing this prove that the Minkowski metric distance between a point on the light sphere and the origin is not zero. That is your task to prove your case.

11. paddoboyValued Senior Member

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27,543
You show your ignorance once again.
You are the one disputing mainstream beliefs and models.
It is up to you to show the scientific community is in error...
You cannot do that either here or with the gravity SR/GR thread.
That's why you are in pseudoscience.

12. chingluValued Senior Member

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1,637
Talk is cheap.

Prove the Minkowski metric creates a metric space.

13. paddoboyValued Senior Member

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27,543
The onus of proof is on you and no matter how much you try and twist to get out of that dilemma, that is always the case.
Yes, talk is cheap, and you have shown that to the nth degree.

14. UndefinedBannedBanned

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1,695
Let's calm down a mo, guys.

Has anyone yet provided a proof to support the (as interpreted by chinglu) current 'mainstreamer' claim: "Minkowski metric DOES creates a metric space"?

If not, then chinglu appears to be within his science method/debating rights to FIRST ASK for proof of same from those who HAVE made that 'mainstreamer' claim (as interpreted by chinglu).

That's my impartial observation of where the onus of proof stands at this juncture, guys (unless someone can point to and explain what/where that proof has already been put here?).

Good luck and good discussing, guys!

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15. paddoboyValued Senior Member

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27,543

Disagree entirely Undefined.......
but anyway......

http://mathworld.wolfram.com/MinkowskiSpace.html

Minkowski space is a four-dimensional space possessing a Minkowski metric

dtau^2=-(dx^0)^2+(dx^1)^2+(dx^2)^2+(dx^3)^2.
Alternatively (but less desirably), it can be considered to have a Euclidean metric but with imaginary time coordinate x^0=ict, where c is the speed of light, by convention c=1 is normally used, and i is the imaginary number sqrt(-1). Minkowski space unifies Euclidean three-space plus time (the "fourth dimension") in Einstein's theory of special relativity.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

http://ncatlab.org/nlab/show/Minkowski space

Definition

For d−1∈N, d-dimensional Minkowski space is the Lorentzian manifold whose underlying smooth manifold is the Cartesian space Rd and whose pseudo-Riemannian metric is at each point the Minkowski metric.

This is naturally a spacetime.

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

ENERGY-DEPENDENT MINKOWSKI METRIC
IN SPACE-TIME

http://www.mathem.pub.ro/proc/bsgp-10/0MIRON01.PDF

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

16. chingluValued Senior Member

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1,637
Prove the Minkowski metric creates a metric space.

Messages:
27,543

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19. paddoboyValued Senior Member

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27,543
The onus is on you, to [1] learn to read, [2] learn to interprete it properly, and [3] admit you are wrong as far as SR/GR goes.

I mean if you were correct, you would not be here would you?? You would be getting your papers peer reviewed, wouldn't you?

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20. chingluValued Senior Member

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I proved it six times now.

21. paddoboyValued Senior Member

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Great work!!!!
Now all you need to do is get it peer reviewed!

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22. chingluValued Senior Member

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prove it wrong yes or no.

23. paddoboyValued Senior Member

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Get it peer reviewed, yes or no.
I follow accepted mainstream views in this instant.
The onus is on you to show where the proof I gave you in the three links is wrong.
You are unable to do that.
As a consequence, you will continue to wallow in your own dung, as you are doing in at least two other threads.

Last edited: Nov 9, 2013