Does Distance exist without time?

Discussion in 'Physics & Math' started by Quantum Quack, May 31, 2008.

  1. Quantum Quack Life's a tease... Valued Senior Member

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    there is no light to see and no time to see it in....and given that was the case do you still consider the question as meaningless and result nonsense?
     
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  3. AlphaNumeric Fully ionized Registered Senior Member

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    That doesn't mean that relativity gives quantum results. It means that there's (hopefully) a theory which combines the two.

    Relativity makes a set of statements A
    Quantum mechanics makes a set of statements B

    A and B can be put together but that doesn't mean A=>B.
    No, you just don't seem to understand logic and physics.
    I'm well aware of fermionic QFT. I'm certain I know more about it than you.
    You still haven't learnt that just throwing buzzwords at me isn't going to work, have you?
    That isn't what I said. Well done on not understanding. I didn't say they were incompatible, I said relativity doesn't imply quantum statements. If it did they would be trivially compatible.
    No, ∞/∞ is undefined. And renormalisation is something different and I'm certain you don't know the details of it. And you know I do.

    Renormalisation is a way of computing physical quantities from manifestly divergent quantities. In other words quantities which are unbounded. When you do cut off renormalisation you get expansions of the form \(a\Lambda + b + cO \left( \frac{1}{\Lambda}\right)[tex] where [tex]\Lambda \to \infty\).
    As usual, you're wrong.
     
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  5. Prince_James Plutarch (Mickey's Dog) Registered Senior Member

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    Quantum Heraclitus:

    As I said in the other thread, no I do not agree. The reason being: There is THREE things there. Past PRESENT and future. This isn't two swords touching, but three things converging sequentially.

    If the moment has no duration, then it is not substantial, ergo, there is no present, therefore no past and future, which are defined in relation to the present. This would also destroy all your efforts at trying to bypass SRT by virtue of making "now" even more incoherent than in multiple reference frames of SRT.

    I told you: Infinite time.

    Then no time, as you could not see at all - nor is there anything existent.

    It does not follow: Again, Zeno's arrow. No time, yet space.

    No. Infinity/infinity = 1.

    Zero/zero is undefined. It is part of those things in Mathematics subject to Godel's incompleteness theorem.

    You can't divide by zero.

    Zero/infinity = zero. As it would with any other number. Zero is not divisible.
     
    Last edited: Jun 15, 2008
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  7. Quantum Quack Life's a tease... Valued Senior Member

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    and your response to Alpha's post?
     
  8. Reiku Banned Banned

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    ''That doesn't mean that relativity gives quantum results.''

    Wrong. If relativity can define an electrons speed to be in zig-zagged motion through spacetime, which it does, then you are TOTALLY WRONG.

    ''No, you just don't seem to understand logic and physics.''

    Or that you simply don't understand how language and math are inexorably tied.

    ''That isn't what I said. Well done on not understanding. I didn't say they were incompatible, I said relativity doesn't imply quantum statements. If it did they would be trivially compatible.''

    Not what you said? It seemed quite clear to me. But if i was wrong, fine.

    ''No, ∞/∞ is undefined. And renormalisation is something different ''

    Not true. In fact, the only way to renormalize an infinity, is by subtracting them from other infinities. Sure, you will know more about the area than i do, but if divide an infinity by another, surely it's along the same lines of logic.

    It should leave either a zero or non-zero value, even if it is normally defined as undefined. Just think about it.
     
  9. AlphaNumeric Fully ionized Registered Senior Member

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    *sigh* Still not getting it, are you?

    The equations of motion of an electron in free space in relativity don't give things like effects from virtual particles. An electron will move in a smooth line without interacting. Without quantum mechanics you don't get loop processes.

    Yes, you describe the motion of the electron as moving through space-time but it's motion is NOT a result of the relativity equations, you MUST have the quantum mechanical ones.
    Yes, you were wrong.
    If you can renormalise by subtraction then you can do it by division.
    So it's either zero or it's not? Wow, stellar logic.

    The mathematical expressions \(\frac{\infty}{\infty}\) and \(\infty-\infty\) are undefined. Renormalisation doesn't apply. This is because with renormalisation there is an expression which is generating the 'value' \(\infty\). You can alter it to be finite and then consider it's divergence. With the expression \(\frac{\infty}{\infty}\) you can't, because you're not given the function which generates it.

    If you are considering something like\(\frac{x}{x^{2}+1}\) then you can define the 'value at infinity' as a limit and in this case it's 0. But just given \(\frac{\infty}{\infty}\) is undefined.
    Wrong. It's undefined. What about \(\frac{kx}{x+1}\)? It goes to k as x->infinity.
    Evidence?
     
  10. Reiku Banned Banned

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    11,238
    *sigh* Still not getting it, are you?

    The equations of motion of an electron in free space in relativity don't give things like effects from virtual particles. An electron will move in a smooth line without interacting. Without quantum mechanics you don't get loop processes.

    Yes, you describe the motion of the electron as moving through space-time but it's motion is NOT a result of the relativity equations, you MUST have the quantum mechanical ones


    Me

    Right, you are still wrong. The electrons zig-zagged path through spacetime would never have been discovered if it wasn't for the unification of relativistic electron dynamics.

    ''Yes, you were wrong.''

    I concur

    ''So it's either zero or it's not? Wow, stellar logic.

    The mathematical expressions and are undefined. Renormalisation doesn't apply. This is because with renormalisation there is an expression which is generating the 'value' . You can alter it to be finite and then consider it's divergence. With the expression you can't, because you're not given the function which generates it.

    If you are considering something like then you can define the 'value at infinity' as a limit and in this case it's 0. But just given is undefined.''

    I still don't agree. It may make me wrong, but who really gives a fuck other than me or you?

    James

    I don't usually side with AN, however, he is right that ∞/∞ does not equal 1, unless there is an equality between the value of N=∞ with N'=∞.
     
  11. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    Precisely, you NEED quantum mechanics for that. Special relativity predated QFT by about 20~30 years. So quantum results are not special relativistic. They need quantum theory too.

    Think about it carefully. Do you get those results if you don't know about quantum mechanics? No. Then you need quantum mechanics.
    You should, being corrected when you're mistaken is part of how you learn.
     
  12. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    I agree with AN. The only way to understand this expression is to look at it as a limit, eg

    \(\lim_{x\rightarrow 0}\frac{\sin x}{x} = 1\),

    which can be seen from L'Hopital's theorem. In this case, 0/0 = 1.
     
    Last edited: Jun 16, 2008
  13. CptBork Valued Senior Member

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    Er... shouldn't that be \(\lim_{x\rightarrow 0}\frac{\sin x}{x} = 1\)?
     
  14. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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  15. CptBork Valued Senior Member

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    Hahahaha! That's what I yell out whenever I make a mistake too!
     
  16. Vkothii Banned Banned

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    If something is zero, then it certainly should be as well.
    I hate it when they do that.
     
  17. Quantum Quack Life's a tease... Valued Senior Member

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    Is it possible to define zero as (1) relative and (2) absolute in mathematics?

    relative zero is a zero that exists because there are values in existance.
    Where-as absolute zero is a zero that no-exists simply as there is no values in existance.
     
  18. Prince_James Plutarch (Mickey's Dog) Registered Senior Member

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    AlphaNumeric, BenTheMan:

    Regarding Godel's Incompleteness Theorem (GIT) and division by zero and other undefines. As you are aware, GIT postulates that every formal system with a complexity that can encompass mathematics will either be complete and inconsistent, or consistent and complete. THat is, if it is truly powerful enough to encapsulate all theorems from the axioms necessary for mathematics, there will be statements which are improvable in the system. As such, we come upon things in mathematics where the application of the axioms cease to make sense and we create theorems which can neither be proven or disproven either which way without a sacrifice somewhere in the system, an inconsistency. 0/0 is traditionally held to be undefined. Infinity/infinity, too (although as seen I prefer the answer "1" for that). These are examples where the theorems cease to be descriptive of a coherent and complete system, and instead represent the Godelian breach that lead to the GIT. 0^0 is another one of these cases. 0 * n = zero; n^0 1. Any answer requires throwing out one of these principles, or inconsistency. As such, it is an example of GIT's hydric heads raising up and ruining logistic programmes of mathematics.

    As to the other aspects of mathematics you raise, I am frankly not prepared to debate with you on those topics.
     
  19. CptBork Valued Senior Member

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    I thought Godel's incompleteness incompleteness theorem just says that if the axioms of math are self-consistent, then there is no way to prove all of these original axioms from first principles.

    In a case like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we first have to say that these things exist, and define them, before we can prove or disprove anything regarding these objects. We know that from the basic axioms of math, these types of terms are undefinable, so in order to define them we'd have to add extra axioms. I don't see how Godel's theorem talks about this anymore than how it talks about whether math can be used to prove or disprove that Hollywood stars wear nice clothes.
     
  20. Vkothii Banned Banned

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    3,674
    What happens when the index of a number is 0?

    What if the number with an index of 0 is 0?
    \(0^0 \)
    (for some reason tex tags don't do anything for me, but that's life)
    (Edit: Ha! stoopid spaces... now it works)
     
    Last edited: Jun 16, 2008
  21. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    \(0^0 = 1\), I think.

    Prince James---

    The concept of infinity is ill-definied, I think. Someone please correct me, but if infinity was well defined, then one could make statements such as \(\infty - 1 = a\) for some \(a \neq \infty\). But this statement is clearly false, i.e. infinity is never obtainable by adding a finite number of integers together. In other words, infinity is NOT a number. If it were, then it must obey the following algebraic equation:

    \( x - 1 = x\).

    So we MUST define infinity in terms of a limit.
     
  22. CptBork Valued Senior Member

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    To clarify my point, I don't think this problem arises due to Godel's theorem. Godel's theorem talks about how any consistent logical system will always contain unprovable statements, i.e. the fundamental axioms. However, \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\) aren't statements in math- nothing in the axioms says such objects must exist. They are more aptly described as statements arising from mathematical notation- i.e. we can define \(\frac{1}{1}\), so why can't we define \(\frac{0}{0}\)? However, just because we can define an object using notation, doesn't mean we can define it as being an object that exists within a formal mathematical system.
     
  23. CptBork Valued Senior Member

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    6,464
    I know that in a lot of numerical programming problems, it's extremely convenient to define \(0^0=1\) when running various kinds of loops and such. However, it's still undefined from a mathematical perspective- i.e.

    \(0^0=0^{(2-2)}=\frac{0^2}{0^2}=\frac{0}{0}\), so it can't be defined.
     

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