Is it true? Is the universe flat?

Discussion in 'Physics & Math' started by camilus, Dec 6, 2010.

  1. przyk squishy Valued Senior Member

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    We've already been through this with you. You conveniently cut out the part where Einstein went on to say "...compelling us to describe its state by ten functions (the gravitation potentials gμν)...". As prometheus [POST=2672735] already explained to you[/POST], symmetric tensors in space-time have ten independent components while symmetric tensors in thee-dimensional space have only six. So either you misunderstood Einstein and he was really talking about space-time, or he was contradicting himself. Which do you think is more likely?

    The stress-energy tensor is, similarly, a symmetric tensor in space-time with ten independent components. Look up a description of it anywhere and you'll find it's written in index notation as \(T^{\mu\nu}\), where the indices \(\mu\) and \(\nu\) take values in the range 0 (time index), 1, 2, and 3 (spatial indices). Furthermore, the energy density in the stress-energy tensor is the double time-like component \(T^{00}\), so if the stress-energy tensor didn't have time indices it wouldn't merit its name. It would be called the "momentum flux" or "stress-pressure" tensor or something similar instead.
     
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  3. AlphaNumeric Fully ionized Registered Senior Member

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    I only just noticed Farsight replied in this thread.

    Your first sentence is meaningless, since there is no 'universe frame', it would be another name for an absolute reference frame. This then contradicts your second sentence, as it isn't compatible with what special relativity says.

    Thanks for demonstrating you don't understand what you're talking about (again). I can't remember if I've said this to you specifically before, I definitely remember saying it to Zephir, but the CMB is not an absolute reference frame. Yes, from the Earth's point of view there is one and only one inertial motion which sets the CMB dipole to zero but it is not an absolute universal reference frame because it doesn't apply elsewhere.

    More precisely, if two people are separated by a large distance (ie a non-insignificant fraction of the observable universe's radius, else the effect would be extremely hard to measure) and both of them use rockets to give themselves motion such that they measure no CMB dipole then they will find themselves moving relative to one another. This is a consequence of the expansion of the universe, specifically involving the length scale associated to the Hubble constant.

    Besides, setting yourself to be at rest wrt to a particular huge set of photons is no different to setting yourself to be at rest wrt to anything else.

    You throw the accusation around like confetti at a gay pride parade.

    I've been asking you to provide a working testable model of any real world phenomenon for at least 2, if not 3, years now and you still can't provide one. I hardly think you're in a position to be whining about string theorists being disconnected from scientific evidence! Or shall we do this particular song and dance again where you complain about string theory, I give some an example of quantitative falsifiable prediction from string theory, ask you to provide one from your own work and then you try to change the subject?

    Where have I been insincere? I offered to help you submit to journals. I offered to engage you in discussion on GR's take on curvature where I'd get suspended if I didn't play nice. I've attempted to engage you in discussion on things directly related to your 'work', like the difference between a physical postulate and a mathematical axiom. You, on the other hand, ask people to discuss your work but when I ask you direct questions you slink off or make the excuse you just did, that I'm insincere.

    If offering to have a polite discussion where the penalty for being rude is suspension you regard as 'insincere' then I think you need to buy a dictionary.

    So take me up on my offer of an alpha rules discussion where I will be suspended if I break said rules or avoid direct relevant questions. I want such a discussion with you, where you are bound by the same rules, because I firmly believe you can't answer simple relevant questions in regards to your 'work'. But after years of trying and you making excuses its hardly surprising I don't bother sugar coating my responses to you. If you won't engage in honest discussion by answering relevant direct questions in a thread where you invite people to talk about your work its hypocritical to try to hold people to a standard you don't yourself meet.

    Besides, you're hardly innocent in terms of insults. In the past you've stated people like myself and Ben are wasting our time doing PhDs in string theory, that you 'killed' it. You tried to insult me a few months back by saying "Managed to get a job yet? In physics?", as you arrogantly thought it would be of no use to my job prospects. That blew up in your face, seeing as I got a job within a fortnight of finishing my PhD and it is in physics. So, once again, don't try to hold people to standards you don't even remotely reach yourself. Which brings me to....

    You can't use that as an excuse after turning down my offer for an alpha rules discussion. I'll happily be held to the alpha rules in a discussion with you provided you're held to them too, along with the requirement direct relevant questions asked are answered honestly by the other party. Many of the questions I'd ask are nothing you wouldn't be expected to answer or clarify when submitting a paper for review. If you really think that one day people like myself or Ben or Prom will be teaching your work then you'll have to have it accepted by the mainstream, which involves standing up to this sort of scrutiny. If your work isn't able to survive a discussion with a lowly string theorist like myself or Prom or Ben then perhaps you want to rethink how much worth it has. After all, if our PhDs are so worthless in the world of physics surely we'd be easy to respond to than more hands on experimental physicists? We should just be a warm up, so you can get used to defending your work to those in the research community. If you manage to defend your work then you can just point to the thread in the future as evidence your work can stand some level of scrutiny and you're capable of answering queries and criticism. You'd save time in the long run, you and I wouldn't be having this discussion if you'd answered me the first time long ago on PhysForums.

    The best way a crank has to get me to stop criticising their work is to defend it honestly and answer direct relevant questions. If a crank spends more time whining about how they don't have the time to defend their work than it would take to answer the questions then its a sure sign of dishonesty. If you want to be taken seriously then answering questions is a necessary step on the road.
     
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  5. chinglu Valued Senior Member

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    Setting yourself at rest to a set of any "photons" sets you are rest to c.

    So your comparison above has no value.

    Here is your correct answer.

    You cannot tell CMB is moving and so you cannot determine your absolute motion based on CMB because you cannot determine the absolute motion of CMB.
     
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  7. rpenner Fully Wired Valued Senior Member

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    4,833
    Arrogance and ignorance in one package. Must be lonely without voluntary associates.
     
  8. AlphaNumeric Fully ionized Registered Senior Member

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    Well done on failing to understand what I said. I was clearly referring to the dipole shift in the CMB, I specifically said so several times. The 'at rest' I was talking about was in regards to the observed dipole being zero. Special relativity doesn't allow a rest frame wrt a single photon and I never said otherwise.

    Engage your brain before making a post.
     
  9. chinglu Valued Senior Member

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    1,637
    Let me review your statement again.

    Besides, setting yourself to be at rest wrt to a particular huge set of photons is no different to setting yourself to be at rest wrt to anything else.

    If you consider this statement viable, then you are stupid.
     
  10. chinglu Valued Senior Member

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    1,637
    Since you have not provided any mathematical basis for your reasoning, all I can say is your icon is very pretty.
     
  11. chinglu Valued Senior Member

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    I will be more specific of your failure.

    If the lack of frequency shift is sufficient to determine the at rest state, then MMX would claim an "at rest state" with these photons on earth. On the other hand, given GPS has proven the sagnac correction, then the lack of frequency modification is insufficient to prove an "at rest state" with these photons since the GPS sagnac contradicts this "at rest" state though differential timing.
     
  12. arfa brane call me arf Valued Senior Member

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    7,832
    I bet that makes sense in an alternative universe. . .
     
  13. AlphaNumeric Fully ionized Registered Senior Member

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    Ching, if you can't actually engage in discussion and can only just parrot your own inability to understand don't bother to reply. You show enough of that kind of mindless ignorance in your thread over in pseudo, where James R has shown you have serious gaps in your understanding.
     
  14. chinglu Valued Senior Member

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    1,637
    A few more moves and he will be contradicted. So, your analysis has no mathematical basis, and neither do your comments here.

    You were claiming you can determine you are "at rest" with photons if the measured frequency is isotropic in all directions.

    Yet, we find earth based light sources and recievers at rest with ECEF based on MMX that conclude the earth is at rest with the photons by proving the measured frequency is isotropic in all directions.

    On the other hand, there are earth based applications where the light source and receiver are at rest in ECEF yet conclude the earth's rotational sagnac.

    Most minds would conclude if a saganc is not detected, you are at rest with those photons and if a sagnac is detected, you are not at rest with those photons. Your MMX methodology of proving the measured frequency is isotropic in all directions is not a reliable tool to draw any conclusions about the measured speed of light being the same in all directions.

    Therefore, your analysis has no logical basis to draw conclusions.
     
  15. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    Its not possible to literally be at rest with any photon in special relativity. I was clearly (and I stated as much) referring to the notion of the CMB and its dipole, where 'at rest' referred to there being no dipole.

    You're trying to shoe horning in your usual crap about the MMX and Sagnac, when all Farsight and I were referring to was the fact there is a frame, at each point in space, where no dipole in the CMB is measured. Whether or not other effects are measured is irrelevant to the point being raised. Farsight considered this frame 'absolute', I explained that it wasn't because it is position dependent. Everything you've come out with is irrelevant to the comments made.

    You seem incapable of engaging in discussion. You've decided you want to talk about Sagnac and the MMX and no matter what is said to you about the irrelevance of those to the point at hand you just continue on. You show the same disregard for back and forth discussion in your exchanges with James R, where you ignore direct questions and just spout your preferred talking points.

    Want to talk about Sagnac and MMX? Go find another thread. Neither of them are relevant to the question "Does the frame where the CMB has no dipole represent an absolute frame?", which is what was being discussed.
     
  16. chinglu Valued Senior Member

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    Does the frame where the CMB has no dipole represent an absolute frame

    You comment about "at rest" with the photons of course means you detect no doppler shift as I said.

    Yet, I demonstrated MMX perceives no doppler shift for earth based light sources and earth based receivers.

    You then conclude if no doppler shift, then you are at rest with the "light source". That is the part you correctly should have said but you are still learning.

    I then explained to you that ECEF light sources and ECEF light receivers detect timing differentials in terms of measuring to c.

    I then tried to help you understand with no success that you may be at rest with the light source (no doppler), but there are other measures that indicate you are not exactly at rest ie sagnac.

    I then would have hoped you would retract your silly statement because it has no scientific value as shown by the evidence. I wonder why I have to go to all this detail to explain things to you.
     
  17. arfa brane call me arf Valued Senior Member

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    It must be tough, having to explain things all the time.

    So many ignorant people, so few internet forums.
    We should be humbled by chinglu's decision to start with the sciforums ignorami--oh joy!
     
  18. chinglu Valued Senior Member

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    If you are able to refute my comment above, do it.

    It is my view that stupidity is not a merit.
     
  19. RJBeery Natural Philosopher Valued Senior Member

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    Christ, Chinglu, answer me this: do you think MMX would yield a null result under high acceleration?

    I'm all for bucking the system but if I appear the way you appear to everyone else I'm embarrassed.
     
  20. chinglu Valued Senior Member

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    I appear a certain way? I am devastated.

    And, no I do not think MMX is null under high acceleration.
    But, I was simply correcting AlphaNumeric's errors.
     
  21. RJBeery Natural Philosopher Valued Senior Member

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    OK...are GPS satellites under acceleration as they orbit the Earth? Even ones in geo-stationary orbit?
     
  22. chinglu Valued Senior Member

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    They are, but not sufficient to explain the sagnac effect.

    Here look at mathpages, the sagnac can be calculated exclusively based on linear velocity v and path length. Acceleration is not needed.

    http://www.mathpages.com/rr/s2-07/2-07.htm
     
  23. RJBeery Natural Philosopher Valued Senior Member

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    Umm, yeah bro the formula for calculating centripetal acceleration is v^2/r. This is not the same as angular acceleration. Maybe this is the source of your confusion. But centripetal acceleration IS acceleration even though it is calculated using velocity.
     

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