Yang–Mills and Mass Gap

Discussion in 'Physics & Math' started by Thales, Nov 29, 2017.

  1. arfa brane call me arf Valued Senior Member

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    Suppose you construct a set of parallel lines in \( \mathbb {R}^2 \). Now you mark regularly spaced points along the lines and map the points in adjacent lines to each other. Does it matter if different lines have different spacing? (yes it does, in \( \mathbb {R}^2 \) because integer arithmetic is based on the integers being separated by unit lengths).

    Ok, so suppose you start with a square lattice of points in \( \mathbb {R}^2 \); How many ways can you construct sets of parallel lines, so you can map points to points between adjacent parallel lines?
     
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  3. hansda Valued Senior Member

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    Is \(60 minutes > 1 hour ? \)
     
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  5. NotEinstein Valued Senior Member

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    You indeed seem to think so.
     
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  7. hansda Valued Senior Member

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    So, maximum \(T \) is \(1 \) unit of time.
     
  8. NotEinstein Valued Senior Member

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    That is indeed your claim. But your claim is inconsistent: if \(T\) can at maximum be \(1\) hour, then it's maximum is also \(60\) minutes, and thus larger than \(1\). In other words, if \(T\) has a maximum of \(1\), it can have a value of \(60\), which is a logical contradiction, and thus your claim is proven false. (Tip: Look up "proof by contradiction".)
     
  9. Thales Registered Member

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    Got lots of reading still to do, before I reply.
     
  10. NotEinstein Valued Senior Member

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    Hansda, have you finally realized the error you have made?
     
  11. hansda Valued Senior Member

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    Consider any time duration/interval as \(t \)units of time. Consider time-period of oscillation as \(T \)units of time. Here to get atleast one complete oscillation within the time interval \(t \) units, \(T\leq t \) . Here frequency \(f=\frac{t}{T} \) or \( fT=t\) .
     
  12. NotEinstein Valued Senior Member

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    And now substitute \(t\) with \(2t\). Note that this means that \(T\) has to be substituted with \(\frac{T}{2}\) to keep the physics the same. So if \(T=1\) is the minimum, so is \(T=2\), etc. Your claim of a minimum \(T\) of \(1\) is thus incoherent.
     
  13. hansda Valued Senior Member

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    Here \( t\) is any unit of time. Why do you want to substitute \( t\) with \(2t \). Seems you are trying to twist my statement and make it complicated.

    Just your imagination. There is no need of it.

    Again your imagination. I am not claiming \(T=1 \) is the minimum.
     
  14. NotEinstein Valued Senior Member

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    Because for that formula to mean anything physical, it must be true independent of the unit of choice. That's basic physics.

    No, I'm trying to make you understand why you are wrong.

    There is only no need, if you think there's no need to make sense.

    Sorry, I keep mixing it up. You think \(T=1\) is a maximum. Just throw in some inverses in the right places.
     
  15. hansda Valued Senior Member

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    Here \(t \) is having, any unit of time. That means it is independent of unit of choice.

    I am also not claiming this. This is your another imagination. You can re-read my post #148.
     
  16. NotEinstein Valued Senior Member

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    But its numerical value isn't; that's my point. Look: 1 hour = 60 minutes. By changing the unit of time, I have to change the numerical value too. So when you say \(f\geq 1\) without specifying the units, it's meaningless. Do you mean: \(f\geq 1 s^{-1}\)? Or \(f\geq 1 hour^{-1}\)? Your original claim is nonsensical.

    Post #54: You claim that the minimum of frequency is \(1\).
    Post #97: You say that \(fT=1\), and thus that \(f\geq\frac{1}{T}\)
    Combine the two:
    \(f=\frac{1}{T}\geq 1\)
    Invert it:
    \(\frac{1}{f}=T\leq 1\)

    QED

    Please go back and read/understand what you yourself have posted.
     
  17. hansda Valued Senior Member

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    Which post you are replying? I told you to read my post #148 and reply.
     
  18. NotEinstein Valued Senior Member

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    Ah, I see you fail to understand the quoting system:
    - Look very carefully at those pieces of texts that are displayed in boxes: they were originally written by you! You can even tell they came from post #152. So that's the post I'm replying to.
    - I see I didn't quote your post #148 in my reply #149. Typically, if no post is quoted it's either a stand-alone post (which clearly isn't the case here), or a reply to the post directly above. So my reply to your post #148 is post #149.
     
  19. hansda Valued Senior Member

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    Here time duration \(t \) can have any unit of time. \( t\) can be any amount of time. So substituting \(t \) with \(2t \) is unnecessary.
     
  20. NotEinstein Valued Senior Member

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    No, you are misunderstanding. I'm not changing the physical time, I'm changing both the numerical value and the unit. 1 hour = 60 minutes. Also, something being unnecessary doesn't mean it can't be done.
     
  21. hansda Valued Senior Member

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    This way, what you are going to achieve? What you think about my post #148. It is right or wrong?
     
  22. NotEinstein Valued Senior Member

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    I'm trying to make you understand that the statement \(f\geq 1\) you made is incoherent.

    Read my reply #149.
     
  23. hansda Valued Senior Member

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    Where I have made this statement in post #148.

    Your post #149 does not say whether my post #148 is right or wrong. Anyway, you can give a fresh answer also.
     

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