There is only one* model consistent with CPT Symmetry

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Re: There is only one* model consistent with CPT Symmetry

Postby WibblyWobbly » Fri Aug 16, 2013 8:33 pm UTC

Treatid wrote:
andyisagod wrote:(some) Possible states for charge of a system:

+1, +2, +3, +4 ....

QED

Fair point.

That system with a charge of +4 is composed of multiple objects who's charge sums to +4. If you consider the components of that system you get down to objects that have a charge +-1 when you drill down far enough. A bit is still a bit even when collected together with other bits to form a Byte.


Why not drill down further? An electron may be a fundamental particle, but what about a proton?

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Re: There is only one* model consistent with CPT Symmetry

Postby Schrollini » Sat Aug 17, 2013 4:03 am UTC

Treatid wrote:We know that {b, b, b} fully defines the behaviour of a single system (the argument is over whether CPT implies {b, b, b} - not the properties of {b, b, b} itself). That is, {b, b, b} is the fully defining axiom of a single system. There is no room to add additional axioms - because the system is already fully defined. There are no choices in the system for an additional axiom to address. As such, the extra axiom can only be a re-statement of an aspect of the original system (a redundant axiom).

Herr Gödel would beg to differ. Every sufficiently interesting system (i.e., capable of expressing arithmetic) that is consistent is incomplete. There is always another independent axiom you can add.

Your two outs are (1) that your system can't express arithmetic (but then it has no hope of describing the universe) or (2) that your system is inconsistent. Pick your poison.

Treatid wrote:
Do you disagree with my assessment that both equations have CPT symmetry? If so why? If not, do you consider them to be parts of the same "system" or not? You keep redefining words, so I really have no idea what you mean by a "system".

I disagree with your assumption that both system are well formed (have no contradictory axioms).


Prove it.

For the record, these are the two I'm talking about. Nothing about GeV photons.
Schrollini wrote:Let's consider two models of the universe:
  1. F = m x''(t)
  2. F = m T2 x''''(t)
F is some force (a vector), x is the position vector of a particle, m is the particle mass (a scalar), and T is a constant with units of time (the Planck time, if you insist on a specific value).


Treatid wrote:For what it is worth... the model is a dynamic directed graph. The graph itself is composed of a number of vertices each with two edges. Vertices and edges are conserved. Distance within the model has a correspondence with edges.

Now I'm disappointed. I thought you were going to derive the model from CPT symmetry. But you're just making shit up. And when I just make shit up, you dismiss it with vague worries. That's not fair!

Well, there's only one thing left to do, I guess.

"I disagree with your assumption that this system is well formed (has no contradictory axioms)."
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Re: There is only one* model consistent with CPT Symmetry

Postby Forest Goose » Sat Aug 17, 2013 4:16 am UTC

I go back to what I said before: I have no idea what you're even saying. Either I'm far stupider than I thought or your posts have a lot of nonsense in them. I do know that most of what you say seems to be vague hinting at something, I'm guessing the something is nothing. I have yet to see you present actual arguments, you just state vague sentences full of terms as if they were fact; all of your replies are, essentially, "fair point! but I was saying...blah blah blah CPT blah blah blah flux capacitor, unitary." At any rate, while it's fun to play "poke the crank" for a bit, I'm done.

As a bit of parting advice: this appears to be important to you, you don't appear to be able to communicate/develop it (whatever it is) however; if you care about/think you have an idea, spend some more time learning about the ideas of others, then return to your own stuff later. You may get better responses.
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Re: There is only one* model consistent with CPT Symmetry

Postby doogly » Sat Aug 17, 2013 12:14 pm UTC

Yeah, if CPT is something that excites your interest, you should go learn about it. Maybe read Streater and Whiteman, that's a lovely book.
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Re: There is only one* model consistent with CPT Symmetry

Postby Treatid » Sat Aug 17, 2013 1:41 pm UTC

Regarding CPT Symmetry...

I have done a woeful job of communicating on this. My language has been ambiguous and plain wrong in places. I'm sorry I put you through that.

Let me see if I can improve a little...

S = {s1, ..., sn} are possible states of the universe or part thereof (anything from an electron or photon on up).

C, P and T are each an operator that if applied to a state si maps that state to states sCi, sPi and sTi respectively. C:(si) |-> (sCi),...

C is such that C:(sCi) |-> (si)

Thus for all si there is an sCi, sPi and a sTi.(Some states may map to themselves under one or more of C, P or T).

That is, there are many possible states. For each of these states, and each operator of C, P and T there exists a corresponding state with C, P and T being the map between those states.

Each operator, C, P and T implies a pair of states. There are many such pairs.

Each si, sCi (and the operator C) is a binary object... Two states with a mapping between them. Likewise for P and T.

An individual state si is one state in each of the three binary systems, {si, sCi}, {si, sPi}, {si, sTi}

Hopefully this makes it clear what I'm referring to as binary objects. Again, I apologise that I haven't made this clear earlier... I have no good excuse for not doing so.

It is this set of binary objects {si, sCi}, {si, sPi}, {si, sTi} that I am equating to the set {b, b, b}.

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Re: There is only one* model consistent with CPT Symmetry

Postby krogoth » Sat Aug 17, 2013 2:00 pm UTC

This sounds like it would have issues with the double-slit experiment/quantum superposition principle?
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Re: There is only one* model consistent with CPT Symmetry

Postby vbkid » Sat Aug 17, 2013 2:56 pm UTC

Treatid wrote:Regarding CPT Symmetry...

I have done a woeful job of communicating on this. My language has been ambiguous and plain wrong in places. I'm sorry I put you through that.

Let me see if I can improve a little...

S = {s1, ..., sn} are possible states of the universe or part thereof (anything from an electron or photon on up).

C, P and T are each an operator that if applied to a state si maps that state to states sCi, sPi and sTi respectively. C:(si) |-> (sCi),...

C is such that C:(sCi) |-> (si)

Thus for all si there is an sCi, sPi and a sTi.(Some states may map to themselves under one or more of C, P or T).

That is, there are many possible states. For each of these states, and each operator of C, P and T there exists a corresponding state with C, P and T being the map between those states.

Each operator, C, P and T implies a pair of states. There are many such pairs.

Each si, sCi (and the operator C) is a binary object... Two states with a mapping between them. Likewise for P and T.

An individual state si is one state in each of the three binary systems, {si, sCi}, {si, sPi}, {si, sTi}

Hopefully this makes it clear what I'm referring to as binary objects. Again, I apologise that I haven't made this clear earlier... I have no good excuse for not doing so.

It is this set of binary objects {si, sCi}, {si, sPi}, {si, sTi} that I am equating to the set {b, b, b}.


And a little foreshadowing, treatid. If you manage to formulate something that isn't trivial, it will be fairly expected of you to derive something from it. Like if I proposed a new equation for gravity, I should be able to, starting with that equation, derive the orbit of earth around the sun.

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Re: There is only one* model consistent with CPT Symmetry

Postby skeptical scientist » Sat Aug 17, 2013 3:20 pm UTC

Treatid wrote:Each operator, C, P and T implies a pair of states. There are many such pairs.

No, each operator is a function mapping states of the universe to states. For example C is the function s -> sC. The only thing "binary" about it is the fact that C(C(s)) = s for every state s. In other words, C○C is the identity, as are P○P and T○T. CPT symmetry is the fact that none of C, P, and T are individually isomorphisms of the universe (or rather of our model of the universe), but the composition C○P○T is.
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Re: There is only one* model consistent with CPT Symmetry

Postby Treatid » Sat Aug 17, 2013 5:25 pm UTC

A change of statements:

Given any system that contains three operators X, Y and Z such that X(X(s)) = s, Y(Y(s)) = s, Z(Z(s)) = s and X○Y○Z is also invariant for all states s then that system may be expressed by the set {b, b, b} where b is a binary object such that f(f(s)) = s.

{b, b, b} fully describes the behaviour (rules) of a system (the trivial system excluded). All such systems are equivalent (have the same rules/physics).

This is provable (the statement resides entirely within mathematics).

Observation:
CPT Symmetry is such that C(C(s)) = s, P(P(s)) = s, T(T(s)) = s and C○P○T is also invariant for all states s.

Either CPT Symmetry constrains the rules of the universe to those implied by {b, b, b}, or the observation of CPT Symmetry is in error.

A hearty, Thank You, to all those who have helped me reach this much better statement.

...

skeptical scientist wrote:
Treatid wrote:Each operator, C, P and T implies a pair of states. There are many such pairs.

No, each operator is a function mapping states of the universe to states. For example C is the function s -> sC.

You say "No" but...

The only thing "binary" about it is the fact that C(C(s)) = s for every state s.

Yes. Precisely. Exactly. An excellent definition of 'binary'. A much better definition than my "object with 2 states". Might need to expand it to exclude null functions/operations/maps.

That is the 'binary' to which I am referring.

krogoth wrote:This sounds like it would have issues with the double-slit experiment/quantum superposition principle?

On the contrary, trying to understand/explain double-slit was a significant motivator. An assumption of Quantum Mechanics is that particles travel between interactions. We never see this travelling - we simply infer it from the sequence of interactions ('distance' is also inferred).

Since particle interactions are the only thing we know happens, all the things between those interactions are assumptions or constructions. If Quantum Mechanics is making assumptions (such as distance, dimensions, velocity, ...) that aren't an accurate reflection of reality then one might expect Quantum Mechanics to be over-complicated in that it is trying to describe behaviours within the wrong context. It turns out that once all the assumptions are removed the universe is incredibly simple in its underlying behaviour (simple behaviour can lead to complex behaviour - Chaos/Complexity Theory).

Schrollini wrote:Herr Gödel would beg to differ. Every sufficiently interesting system (i.e., capable of expressing arithmetic) that is consistent is incomplete. There is always another independent axiom you can add.

Apples and Oranges. We can readily fully define the rules (behaviour) of a state machine. Feed that state machine an arbitrary tape and life becomes interesting. We can build a computer with a known instruction set. The programs/data we put on that computer are beholden (ahem) to Herr Gödel.

However - your point does highlight that I have been loose with terms (yet again - what a surprise). We are specifically talking about the instruction set of the universe. That instruction set could be fed any possible universe - An instance of a universe is constrained by the limits on knowledge that Gödel outlines... but the instruction set (physics) can be known/fully defined.

Treatid wrote:I disagree with your assumption that both system are well formed (have no contradictory axioms).


Prove it.


I have (see above).

For the record, these are the two I'm talking about. Nothing about GeV photons.
Schrollini wrote:Let's consider two models of the universe:
  1. F = m x''(t)
  2. F = m T2 x''''(t)
F is some force (a vector), x is the position vector of a particle, m is the particle mass (a scalar), and T is a constant with units of time (the Planck time, if you insist on a specific value).


Exchanging "photon" for "particle" doesn't help. Force, Mass, Time, Particle are all observations of physics - not mathematically derived objects. 1 apple is not "1". That Force is associated with a vector does not mean that Force is the mathematical object 'vector'.

This is obvious when you remember that the whole purpose of physics is to understand what these various observations are. How does Force apply to a particle? What is mass? What is space-time? These things are manipulated with mathematics... and often those manipulations correspond or predict actual observation very well... but the objects manipulated are not themselves mathematics.

A mathematically rigorous system needs to be constructed from axioms so that all the properties of the system are known from the start. Only then can you build mathematically rigorous proofs.

No theory of physics can be proven right because the final test of the theory is whether it matches up with observation. Observation is not amenable to proof. As such, anything which incorporates physical observation cannot be used as a proof.

Now I'm disappointed. I thought you were going to derive the model from CPT symmetry.

Given there is only one such model - whether a model is derived from the proof - or a model is constructed that matches the proof is the same either way.
Last edited by Treatid on Sat Aug 17, 2013 6:21 pm UTC, edited 1 time in total.

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Re: There is only one* model consistent with CPT Symmetry

Postby doogly » Sat Aug 17, 2013 5:56 pm UTC

Yeah it seems the core of the problem is that we point out "C^2 = identity" and you get from there through some belaboured logics to some binary set. And there is no way to do this.

And also, C P and T individually are all quite boring. What's special is that "CPT = identity" (order irrelevant.)

Schrollini's examples are really perfect for this, and you can't just dismiss them. You write

F = m x''
and
F = m x''''

It is irrefutable that, as he showed, CPT F = F, CPT m x'' = m x'', and CPT m x'''' = m x''''.

These two dynamical laws are not equivalent to each other, but they are both CPT symmetric.
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Re: There is only one* model consistent with CPT Symmetry

Postby Schrollini » Sat Aug 17, 2013 6:44 pm UTC

Treatid wrote:
The only thing "binary" about it is the fact that C(C(s)) = s for every state s.

Yes. Precisely. Exactly. An excellent definition of 'binary'. A much better definition than my "object with 2 states". Might need to expand it to exclude null functions/operations/maps.

That is the 'binary' to which I am referring.

But this doesn't mean that there are two states -- it means there are two operators. Specifically, C can either be the identity or "negation". But there is no limit on the number of states s on which C can act.

Edit: This is somewhere between misleading and just plain wrong. For each element s of a set S, there's "two" things C can do: map it to itself or to an other element s' that gets mapped back to s. But this certainly doesn't mean that there are only two such functions C. The number of functions with C(C(s)) will depend somehow on the cardinality of S. If S = {a,b}, there are only two functions, I and {a <-> b}. But if S = {a,b,c}, there are four functions: I, {a <-> b, c <-> c}, {a <-> c, b <-> b}, and {a <-> a, b <-> c}. And if S is infinite, you'll have an infinite number of functions.

But as this discussion shows, you can have any cardinality of S you want, not just 2.

And this is why everyone's been insisting that C, P, and T are operators, not states.

Treatid wrote:
Schrollini wrote:Prove it.


I have (see below).

[...]

Exchanging "photon" for "particle" doesn't help. Force, Mass, Time, Particle are all observations of physics - not mathematically derived objects. 1 apple is not "1". That Force is associated with a vector does not mean that Force is the mathematical object 'vector'.

This is obvious when you remember that the whole purpose of physics is to understand what these various observations are. How does Force apply to a particle? What is mass? What is space-time? These things are manipulated with mathematics... and often those manipulations correspond or predict actual observation very well... but the objects manipulated are not themselves mathematics.

A mathematically rigorous system needs to be constructed from axioms so that all the properties of the system are known from the start. Only then can you build mathematically rigorous proofs.

No theory of physics can be proven right because the final test of the theory is whether it matches up with observation. Observation is not amenable to proof. As such, anything which incorporates physical observation cannot be used as a proof.

I don't know what this is (the ramblings of a sophomore philosophy major after toking?), but it sure ain't a proof that either of my proposed systems is inconsistent.

Treatid wrote:
Now I'm disappointed. I thought you were going to derive the model from CPT symmetry.

Given there is only one such model - whether a model is derived from the proof - or a model is constructed that matches the proof is the same either way.

That's sort of begging the question. If your proof were constructive, then you wouldn't have to show other models were wrong. But it isn't, so you better be able to show that your model has CPT symmetry and no other model does. You've done neither.

doogly wrote:It is irrefutable that, as he showed, CPT F = F, CPT m x'' = m x'', and CPT m x'''' = m x''''.

A minor nitpick, but all of those terms actually change sign under the combined action of C, P, and T. Specifically, both F and x are vectors and change sign under P. But since both sides change sign, the model is CPT symmetric.
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Re: There is only one* model consistent with CPT Symmetry

Postby doogly » Sat Aug 17, 2013 6:46 pm UTC

"The mark of a good theoretical physicist is making an even number of sign errors."
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Re: There is only one* model consistent with CPT Symmetry

Postby Schrollini » Sat Aug 17, 2013 9:06 pm UTC

Schrollini wrote:The number of functions with C(C(s)) will depend somehow on the cardinality of S. If S = {a,b}, there are only two functions, I and {a <-> b}. But if S = {a,b,c}, there are four functions: I, {a <-> b, c <-> c}, {a <-> c, b <-> b}, and {a <-> a, b <-> c}. And if S is infinite, you'll have an infinite number of functions.

And because I nerd-sniped myself on this, I'll note that the number of such functions on a set with n elements is given by n!/m! * (1 - 2-m) + 1, where m = floor(n/2). The first few elements of this sequence are 1, 2, 4, 10, 46, 106, 736, 1576, 14176, 29296, ... . Maybe someone else can figure out why all but the first four end in 6. :)

If you look at the number of functions other than the identity (subtract that one), these numbers can be factored to give 0, 1, 3, 3·3, 3·3·5, 3·5·7, 3·5·7·7, 3·3·5·5·7, 3·3·3·3·5·5·7, and 3·3·3·5·7·31. Well, there was almost a nice pattern there.
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Re: There is only one* model consistent with CPT Symmetry

Postby Treatid » Mon Aug 19, 2013 5:10 pm UTC

doogly wrote:Schrollini's examples are really perfect for this, and you can't just dismiss them. You write

F = m x''
and
F = m x''''

You don't know what 'mass' is.

We have some equations that approximate to what we observe - but that is all. Those equations aren't derived from a set of axioms.

This isn't some weird philosophy (Schrollini) or even anything that should be controversial. This is not an attack on existing physics... it is simply an awareness of the limitations of the tools available.

By using 'mass' in an equation you are wrapping mathematics around a great big unknown. The physical conception of mass is particularly shaky - but the same can be said for any physical quantity - we don't know what any physical artefact (Force, photon, charge, spin) actually is. No matter how close Quantum mechanics comes to accurately describing the behaviour of something... we still don't know to what degree that something corresponds to the mathematics of Quantum Mechanics. And Quantum Mechanics is not a formal axiom based system.

Simply put, the current models of physics match and predict observation. They are not 'proven' in any way, shape or form. Sticking a bunch of unknowns in an equation proves nothing. Even when that equation approximately describes an observed behaviour, that is just physics - not proof. (Again - not attacking or criticising current physics... just describing what it is).

Distance, velocity, acceleration, .... can be achieved from Euclidean Geometry... Force and mass are incompletely understood. In providing an equation that is supposed to be consistent with CPT Symmetry you are setting up a "Solve for x" problem. You are not even defining a single system.

doogly wrote:Yeah it seems the core of the problem is that we point out "C^2 = identity" and you get from there through some belaboured logics to some binary set. And there is no way to do this.

Perhaps you could go into detail about why there is no way to do this... Given that I have, in fact, done this...

Can you point to a specific error in the logic? Or are you rejecting it because you don't like the conclusion?

It really is a trivial piece of logic. it isn't "axiom of choice" stuff. One could even say that it is obvious... You have previously made points that addressed what I wrote. Here you are simply stating that I must be wrong. Can you be more constructive in helping me understand why I am wrong?

If it helps... you can take X(X(s)) = s, excluding X = null, as the definition of a binary object (axiom).

Schrollini wrote:
Treatid wrote:
The only thing "binary" about it is the fact that C(C(s)) = s for every state s.

Yes. Precisely. Exactly. An excellent definition of 'binary'. A much better definition than my "object with 2 states". Might need to expand it to exclude null functions/operations/maps.

That is the 'binary' to which I am referring.

Edit: This is somewhere between misleading and just plain wrong. For each element s of a set S, there's "two" things C can do: map it to itself or to an other element s' that gets mapped back to s. But this certainly doesn't mean that there are only two such functions C. The number of functions with C(C(s)) will depend somehow on the cardinality of S. If S = {a,b}, there are only two functions, I and {a <-> b}. But if S = {a,b,c}, there are four functions: I, {a <-> b, c <-> c}, {a <-> c, b <-> b}, and {a <-> a, b <-> c}. And if S is infinite, you'll have an infinite number of functions.

You are arguing that the operator changes for each state.

Thus f(a, b) |-> f(b, a) is a different function for each of {(0,0), (0,1), (1, 0), (1,1)}?

You are trying to tell me that there is no function that applies to more than one state... if the state changes, the function changes?

So - C, P and T are each undefined - or have to be redefined for every possible state that they could be applied to?

However, even if C, P and T are different for every state or pair of states... that doesn't actually contradict X(X(s)) = s being a binary object. We still have pairs of states that are related. You would have been better off arguing that the states are different in each binary object so that binary objects aren't equivalent (please brush up on a little set theory before arguing this).

And this is why everyone's been insisting that C, P, and T are operators, not states.

Yes. And I am not disagreeing with them.

I am saying that X(X(s)) = s specifies a binary object. If you look closely you will see that I am not saying that X = binary object.


I don't know what this is (the ramblings of a sophomore philosophy major after toking?), but it sure ain't a proof that either of my proposed systems is inconsistent.

No - it isn't proof. It is showing that you don't know what your systems are. Because you don't know what you are describing you cannot know whether they are consistent or not.

This is standard physics. You should not be finding this controversial. Physics is tested by observation - not proof. An equation containing physics contains objects that are not fully understood. You cannot guarantee that something you don't understand has or hasn't certain properties (such as being consistent with CPT). If you construct an equation from mathematical primitives that is consistent with CPT Symmetry then you have a reasonable argument... but you are constructing equations around incompletely understood objects. You don't know what mass is. Nobody does. So what are you actually demonstrating when you put mass into an equation?

This isn't a flaw in physics. It is simply what physics is.

That's sort of begging the question. If your proof were constructive, then you wouldn't have to show other models were wrong.

I am not trying to show that other models are wrong. Your counter evidence contains unknown qualities (mass, particles, force, ...) that prevent conclusions being drawn from that evidence.

On the other hand, I have a proof that something with XYZ Symmetry can only be one possible model.

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Re: There is only one* model consistent with CPT Symmetry

Postby LaserGuy » Mon Aug 19, 2013 5:30 pm UTC

Treatid wrote:
doogly wrote:Schrollini's examples are really perfect for this, and you can't just dismiss them. You write

F = m x''
and
F = m x''''


You don't know what 'mass' is.

We have some equations that approximate to what we observe - but that is all. Those equations aren't derived from a set of axioms.


It doesn't matter. Those equations still have CPT symmetry. That's kind of the point: CPT symmetry alone is insufficient to distinguish between models of real things and models of false ones.

Simply put, the current models of physics match and predict observation. They are not 'proven' in any way, shape or form. Sticking a bunch of unknowns in an equation proves nothing. Even when that equation approximately describes an observed behaviour, that is just physics - not proof. (Again - not attacking or criticising current physics... just describing what it is).


Uh, sure, but the same is true of CPT symmetry. It's just something we observe. It's not something that has been proved in any rigorous way. Certain models will necessarily have CPT symmetry as a consequence of their axioms, but that doesn't imply that CPT symmetry is a thing, because we don't know which models correspond to reality. Talking about "proof" in science isn't really meaningful. Science doesn't work in terms of rigorous proof; it works in terms of induction and Bayes' Theorem. We gather evidence to support a particular claim, and the more/stronger evidence that we have, the more we are inclined to believe that claim is true.

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Re: There is only one* model consistent with CPT Symmetry

Postby eSOANEM » Mon Aug 19, 2013 6:11 pm UTC

Treatid wrote:
doogly wrote:Schrollini's examples are really perfect for this, and you can't just dismiss them. You write

F = m x''
and
F = m x''''

You don't know what 'mass' is.

We have some equations that approximate to what we observe - but that is all. Those equations aren't derived from a set of axioms.

This isn't some weird philosophy (Schrollini) or even anything that should be controversial. This is not an attack on existing physics... it is simply an awareness of the limitations of the tools available.

By using 'mass' in an equation you are wrapping mathematics around a great big unknown. The physical conception of mass is particularly shaky - but the same can be said for any physical quantity - we don't know what any physical artefact (Force, photon, charge, spin) actually is. No matter how close Quantum mechanics comes to accurately describing the behaviour of something... we still don't know to what degree that something corresponds to the mathematics of Quantum Mechanics. And Quantum Mechanics is not a formal axiom based system.

Simply put, the current models of physics match and predict observation. They are not 'proven' in any way, shape or form. Sticking a bunch of unknowns in an equation proves nothing. Even when that equation approximately describes an observed behaviour, that is just physics - not proof. (Again - not attacking or criticising current physics... just describing what it is).

Distance, velocity, acceleration, .... can be achieved from Euclidean Geometry... Force and mass are incompletely understood. In providing an equation that is supposed to be consistent with CPT Symmetry you are setting up a "Solve for x" problem. You are not even defining a single system.


Mass is observed to be unaffected by C, P and T independently. We could take that as an axiom if we chose. We could also take either of those equations as axiomatic and, along with some other rules about forces adding linearly and something like Newton's 1st and 3rd laws, we could deduce a significant amount of physics mathematically. We could prove stuff (in this system).

One of these systems would correspond to classical mechanics (which is properly axiomatised so no getting around that one for you I'm afraid) whilst the other does not match observed physics at all (and you'd probably need to modify newton's 1st law to say objects maintain constant x''' in order to get a nicely formulated system); regardless, we can prove (as schrollini did) that both systems (which are both axiomatic systems) are CPT invariant. They are both models and they are both distinct (they give vastly different predictions). It follows then that there are at least 2 CPT invariant models. We also know that SR, GR, Schrodinger QM, QFT, the standard model and many quantum gravity theories are CPT invariant.

If your theory is true, there are three possibilities: these are not models (in which case you have an idiotic definition of model), they are not distinct (in which case your definition of distinct has nothing to do with observables) and/or they are not CPT symmetric (which is mathematically false).

Pick one. Please. Just say which. Not a big waffley wall of philosophy. I just want you to say which you think is the explanation as to why these theories do not disprove your claim.
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Re: There is only one* model consistent with CPT Symmetry

Postby doogly » Mon Aug 19, 2013 6:27 pm UTC

This is actually just sad, I think.

Look, all sets have only their cardinality to define them, so in that sense, yes, everything that is (b,b,b) is the same.

But this has nothing at all to do with CPT symmetry. You have at no point actually dealt with them. You could take an axiom like,
" X(X(s)) = s means X is a member of a two element set," or something like this,
but then this X would have nothing to do at all with an operator that exists in quantum field theory.

Your argument reads to me like this.
1 Rigorous constructive axiomatic quantum field theory AND observations of physics suggest that CPT symmetry is a true symmetry of nature.
2 Let's completely ignore quantum field theory and regard with suspicion all observations of how things like force and mass work, and any math that sounds complicated, like functional analysis or C* algebras or even really linear algebra
3 Now, completely ripped from context, let's invent a new context for C P and T, such that now they correspond to 3 choices of a bit
4 3 choices of a bit can model the universe

So yeah you can construct a model from the ground up, but then it is not at all based on our observation of CPT symmetry.

So you should say, "I have a wildly new and exciting theory that the universe can be described by a set with 8 elements" and then if you can predict some observable quantities... more than 8 of them maybe? like, more than 8 things ever? then this would be fantastic and very exciting to other people too.
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Re: There is only one* model consistent with CPT Symmetry

Postby Treatid » Mon Aug 19, 2013 7:54 pm UTC

LaserGuy wrote:It doesn't matter. Those equations still have CPT symmetry. That's kind of the point: CPT symmetry alone is insufficient to distinguish between models of real things and models of false ones.


The issue is whether such an equation is well formed (contains no paradoxes).

You can put together any equation you like. Here: http://www.math.hmc.edu/funfacts/ffiles/10001.1-8.shtml shows that 1 = 0.

In an equation such as F = m x" you don't know what F, or m are. They are not just variables... if I express the equation as a = b x" and tell you that this equation is consistent with CPT Symmetry, what does that mean? If a and b are just numbers then how do C, P and T operate on those numbers?

Uh, sure, but the same is true of CPT symmetry. It's just something we observe. It's not something that has been proved in any rigorous way. Certain models will necessarily have CPT symmetry as a consequence of their axioms, but that doesn't imply that CPT symmetry is a thing, because we don't know which models correspond to reality. Talking about "proof" in science isn't really meaningful. Science doesn't work in terms of rigorous proof; it works in terms of induction and Bayes' Theorem. We gather evidence to support a particular claim, and the more/stronger evidence that we have, the more we are inclined to believe that claim is true.

I mostly agree.

Mathematics is able to prove certain results based on axioms. I have thus shown a provable result for XYZ Symmetry. This result is true within the scope of the axioms.

What I can't do (nobody can do) is to show that CPT Symmetry is an instance of XYZ Symmetry. There are striking similarities... but CPT Symmetry is, as you say, an observation and not a proof.

I can say that if CPT Symmetry is an instance of XYZ Symmetry then there is only one possible model of the universe.

So - there is doubt within the observation of CPT Symmetry. But there is no doubt that where X(X(s)) = s, Y(Y(s)) = s, Z(Z(s)) = s and X○Y○Z are invariant for all s then there is only 1 possible model (trivial X, Y and Z excluded).

That CPT Symmetry is an instance of XYZ Symmetry can never be proven (it can be dis-proven). On the other hand, the similarity is plain enough, I think, that the possibility of one being an instance of the other cannot be ruled out.

eSOANEM wrote:Pick one. Please. Just say which. Not a big waffley wall of philosophy. I just want you to say which you think is the explanation as to why these theories do not disprove your claim.

You are not fully defining the systems. You don't know their properties. You haven't specified a formal set of unambiguous axioms.

We could also take either of those equations as axiomatic and, along with some other rules about forces adding linearly and something like Newton's 1st and 3rd laws, we could deduce a significant amount of physics mathematically. We could prove stuff (in this system).

You can make up any axioms you like - yes. But those axioms may not be consistent. It is the consistency of axioms we are arguing over. You are stating that such and such system is consistent... but you aren't demonstrating that.

Take a moment to specify whatever axioms you choose... then build a proof that shows those axioms are consistent... that they do not lead to a contradiction or paradox. Do this for two different system that both contain CPT Symmetry and you have demonstrated that I am wrong.

At the moment you are just stating that this can be done. You have yet to provide any evidence that this is the case.

doogly wrote:Look, all sets have only their cardinality to define them, so in that sense, yes, everything that is (b,b,b) is the same.

Good.

" X(X(s)) = s means X is a member of a two element set," or something like this,

X is the relationship between two states. X is not, itself, a state. So X is such that {s1, s2} are related by X. s1 and s2 may in some circumstances be the same. X may not be null.

but then this X would have nothing to do at all with an operator that exists in quantum field theory.

Given any system that contains three operators X, Y and Z such that X(X(s)) = s, Y(Y(s)) = s, Z(Z(s)) = s and X○Y○Z is also invariant for all states s then that system may be expressed by the set {b, b, b} where b is a binary object such that f(f(s)) = s.

{b, b, b} fully describes the behaviour (rules) of a system (the trivial system excluded). All such systems are equivalent (have the same rules/physics).

This is a statement that is purely mathematical. Really easy set theory with a plain definition thrown in.

Presumably you are not comfortable with the idea that all X(X(s)) = s (excluding X = null) are equivalent?

but then this X would have nothing to do at all with an operator that exists in quantum field theory.

That is a statement. It isn't an argument. You aren't giving me any reason for that statement.

C(C(s)) = s where C is the operator in (say) QFT. You tell me that X(X(s)) = s and C(C(s)) = s are in no way connected, related or have any similarity of any kind... Why not? What exactly distinguishes them?

Your argument reads to me like this.
1 Rigorous constructive axiomatic quantum field theory AND observations of physics suggest that CPT symmetry is a true symmetry of nature.
2 Let's completely ignore quantum field theory and regard with suspicion all observations of how things like force and mass work, and any math that sounds complicated, like functional analysis or C* algebras or even really linear algebra
3 Now, completely ripped from context, let's invent a new context for C P and T, such that now they correspond to 3 choices of a bit
4 3 choices of a bit can model the universe

That is surprisingly accurate. I am pleased that some communication has occurred.

I would put a slightly different emphasis on 2 & 3, perhaps. I am taking a single facet of of physics... C(C(s)) = s, P(P(s)) = s, T(T(s)) = s and CPT Invariant for all s.

Based on this structure I have constructed a proof that there is only one possible model.

What would including the rest of physics do? Adding axioms/constraints tends to reduce rather than increase the degrees of freedom of a system. Since the proof has already constrained the system to one possibility, adding extra constraints is going to... reduce the possibilities from 1?

The "new context" that you disparage is based on set theory. It is certainly worth questioning whether the axioms are broad enough for any result to count as a general result... Is that an argument you would like to make?

You may notice that the proof and CPT Symmetry are distinct. The proof above says nothing about CPT Symmetry. No doubt I constructed the statement such that any similarity between CPT Symmetry and XYZ Symmetry is clear... but the proof itself is stand-alone.

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Re: There is only one* model consistent with CPT Symmetry

Postby LaserGuy » Mon Aug 19, 2013 8:27 pm UTC

Treatid wrote:You can put together any equation you like. Here: http://www.math.hmc.edu/funfacts/ffiles/10001.1-8.shtml shows that 1 = 0.


Please do not insult my intelligence with grade school problems.

In an equation such as F = m x" you don't know what F, or m are. They are not just variables... if I express the equation as a = b x" and tell you that this equation is consistent with CPT Symmetry, what does that mean? If a and b are just numbers then how do C, P and T operate on those numbers?


The symmetries of F, m, and x with respect to CPT are well-defined. Show me the paradox.

So - there is doubt within the observation of CPT Symmetry. But there is no doubt that where X(X(s)) = s, Y(Y(s)) = s, Z(Z(s)) = s and X○Y○Z are invariant for all s then there is only 1 possible model (trivial X, Y and Z excluded).


But those aren't the objects that you constructed the proof with. The proof that you posted at the very beginning is not defined in terms of operators at all (you didn't know what operators were until a page or two ago), let alone operators with these particular definitions. Given that the details of the proof don't change at all regardless of the type of mathematical objects you put into the proof, I think that there are two possibilities that are likely: 1) Your proof is showing a completely trivial result. 2) Your proof is self-contradictory. You can't claim that there is only one model consistent with CPT symmetry, then change the inputs of the model entirely and claim that the new model is also the only one consistent with CPT symmetry.

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Re: There is only one* model consistent with CPT Symmetry

Postby andyisagod » Mon Aug 19, 2013 9:07 pm UTC

The way I see it there are two distinct problems with your idea. The first is that there is no observational evidence supporting the claim that universe is described by 3 binary values. The second is the weird way in which you build a model from the ground up but fail to even acknowledge the possibility of other models built up in the same way.

The first problem is a big problem but does not necessarily preclude your model since you could have just happened to guess the correct model for the universe without any evidence. If we want to take this model seriously however there needs to be some sort of evidence for this claim. The arguments you have made so far do not do this. Mainly this is because the evidence for CPT symmetry is evidence for a type of symmetry in a QFT and does not lead to binary values as you suggest. This has been explained several times but you have still not understood.
Treatid wrote: But there is no doubt that where X(X(s)) = s, Y(Y(s)) = s, Z(Z(s)) = s and X○Y○Z are invariant for all s then there is only 1 possible model (trivial X, Y and Z excluded).


Is untrue, all this says is that s and X(s) are pairs of states related by the operator X. In terms of CPT that they are (for example) charge conjugates of each other. This does not mean that there are only two such states for example it is also true that X(X(s'))= s' for s' =/= s. This is essentially the same point that people have tried to make again and again that there are different consistent models that have CPT symmetry with different predictions.

Another way of posing the problem of evidence would be to ask what the 3 binary quantities in your model are. As we have said they clearly are not C P and T since they are not binary quantities. So what are those quantities and where is the experimental evidence for them?

The second problem is more to do with how you have built up your model. For this we can ignore the lack of evidence for the axioms of your model and instead concentrate on the model itself. As a model for the universe you have the following problems.

1)Your model is not predictive. You have said that your model is predictive in that your prediction is that your model is the fundamental model of the universe but this is not really a prediction. At this point we would settle for correctly describing stuff we already know but again there has been nothing. You have to have some real prediction i.e something that can actually be compared to experiment otherwise your model is useless.

Assume for a minute that you were actually right about the motivation in terms of CPT in that case making a prediction for anything becomes a way to test CPT, if your model doesn't reproduce compton scattering correctly then we would have evidence that CPT is in fact broken. This lack of predictive power is really important for two reasons firstly that it is required so that we can compare your model to other theories, I can after all come up with a QFT that conserves CPT and my model might have supersymmetry does your model have supersymmetry? that could be an important question if the LHC starts turning up sparticles and we would want to know who's model we should use.

The other problem with not having predictions is just practical, we know we can use QFT to make astoundingly precise predictions about the physical world and there are lots of extensions to the standard model that people are working on that could potentially solve some of the outstanding problems in particle physics. We can do calculations in those theories and make predictions, if we can't do calculations and make predictions in your theory then it doesn't matter if your theory is in fact correct nobody will have any use for it.

2)From the ground up your model seems to have some rather large implications. If I have understood correctly your position is that your model is the only model that contains the CPT symmetry (where for CPT here I mean your binary definition), that is there is no possible extension of this symmetry in a yet unobserved sector that would approximate CPT symmetry in the standard model and that any models that contain this symmetry + some large symmetry either continuous or discrete are either Inconsistent systems or in fact determined by the original CPT model. This to me is very interesting it reminds me of the Coleman–Mandula theorem of symmetries of the s-matrix. If you could prove this then that would be interesting. It might be easier to prove this for a single example, could you provide as examples an extended set of axioms that is consistent and derived from your model vs a set that is inconsistent? It seems that in the absence of other predictions predicting the symmetries that are consistent with your model would be a good start since these could be compared to standard model etc.

Aside from this it is certainly true that we can build models like yours using different symmetry groups that are self consistent but break CPT at some level. The problem then becomes that you don't know how that breaking of CPT manifests so you can't say whether it is inconsistent with the current data. Plainly put without some predictions then there are a large number of similar models to yours that are also consistent (because they make no predictions!) with the current data.

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Re: There is only one* model consistent with CPT Symmetry

Postby Schrollini » Mon Aug 19, 2013 9:11 pm UTC

Treatid wrote:In an equation such as F = m x" you don't know what F, or m are. They are not just variables... if I express the equation as a = b x" and tell you that this equation is consistent with CPT Symmetry, what does that mean? If a and b are just numbers then how do C, P and T operate on those numbers?

We've told you. F and x are vectors, changing sign under P (but not C or T). m and T are scalars, remaining unchanged under C, P, and T. And of course time derivatives change sign under T (but not C or P).

Treatid wrote:You are arguing that the operator changes for each state.

Thus f(a, b) |-> f(b, a) is a different function for each of {(0,0), (0,1), (1, 0), (1,1)}?

What is this two argument function you've just introduced?

Treatid wrote:You are trying to tell me that there is no function that applies to more than one state... if the state changes, the function changes?

No, I'm trying to tell you that the number of functions with f(f(s)) = s depends on the number of states s.

Treatid wrote:So - C, P and T are each undefined - or have to be redefined for every possible state that they could be applied to?

No, they're perfectly well defined. But we have to say a lot more than just C(C(s)) = s to define C.

Treatid wrote:However, even if C, P and T are different for every state or pair of states... that doesn't actually contradict X(X(s)) = s being a binary object.

Which is binary, X or s?

Not that it matters -- we've shown that X(X(s)) = s alone is not enough to restrict either X or s to one of two possibilities. But if you limit s to two elements, then there are only two functions X, the identity and inversion. Perhaps this is what you're arguing?

Treatid wrote:I am saying that X(X(s)) = s specifies a binary object. If you look closely you will see that I am not saying that X = binary object.

Or are you saying that this creates a pairing between s and X(s)? Which is true, but says nothing about either X nor the set S which contains s.

Treatid wrote:
Schrollini wrote:Prove it.


I have (see below).


Below, Treatid wrote:(|Proof> + |Not Proof>)/√2


About this, Treatid wrote:No - it isn't proof. It is showing that you don't know what your systems are. Because you don't know what you are describing you cannot know whether they are consistent or not.


I see....
For your convenience: a LaTeX to BBCode converter

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Re: There is only one* model consistent with CPT Symmetry

Postby doogly » Mon Aug 19, 2013 9:19 pm UTC

Treatid wrote: But there is no doubt that where X(X(s)) = s, Y(Y(s)) = s, Z(Z(s)) = s and X○Y○Z are invariant for all s then there is only 1 possible model (trivial X, Y and Z excluded).

...

Given any system that contains three operators X, Y and Z such that X(X(s)) = s, Y(Y(s)) = s, Z(Z(s)) = s and X○Y○Z is also invariant for all states s then that system may be expressed by the set {b, b, b} where b is a binary object such that f(f(s)) = s.

{b, b, b} fully describes the behaviour (rules) of a system (the trivial system excluded). All such systems are equivalent (have the same rules/physics).

This is a statement that is purely mathematical. Really easy set theory with a plain definition thrown in.


Yeah, here is the problem. These don't work, as I've said before, but this is very much isolating it. This just isn't how basic set theory and function theory work. If X(X(s))=s, it does *not* follow that s \in (s_1, s_2).
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Re: There is only one* model consistent with CPT Symmetry

Postby Treatid » Tue Aug 20, 2013 4:03 am UTC

Please do not insult my intelligence with grade school problems.

The link was an illustration. If you have taken offence, it has nothing to do with me.

The symmetries of F, m, and x with respect to CPT are well-defined. Show me the paradox.

It is you who is arguing that there is no paradox. I have shown that there is only one system consistent with XYZ Symmetry. You seem to be arguing that your equations prove otherwise... but you haven't demonstrated that... you have simply thrown some equations on the page and stated that they are consistent with CPT Symmetry. You have provided me with no way to check your assertion. I cannot build the system(s) myself - because you haven't fully specified the system(s).

But those aren't the objects that you constructed the proof with.

Yes they are. All I've done is removed some ambiguity and taken on language that you understand. For extra points I have explicitly divided my statement into a pure proof component and an observation component.

Believe it or not, adapting my language so that you can better understand what I am saying is not a weakness.

Given that the details of the proof don't change at all regardless of the type of mathematical objects you put into the proof, I think that there are two possibilities that are likely:

Since your premise here is mistaken... I have not changed the structure of the proof... I have simply described 'binary objects' in a more mathematically rigorous way so that there is no confusion over what I am referring to.

andyisagod wrote:The way I see it there are two distinct problems with your idea. The first is that there is no observational evidence supporting the claim that universe is described by 3 binary values.

Yes there is. It is CPT Symmetry.

CPT Symmetry alone, all by itself, with no other supporting evidence is extremely strong evidence that the universe is described by 3 binary objects. (There are, of course, many many instances of these 3 binary objects... but the behaviour of the universe at any point is described by 3 binary objects).

The second is the weird way in which you build a model from the ground up but fail to even acknowledge the possibility of other models built up in the same way.

I'm not aware of anyone else suggesting the possibility of building other models in the same way... as such I'm not aware of anything to be acknowledged.

If you look back earlier in the thread, I did discuss other models of a similar order. So... I'm not quite sure exactly what it is you find weird. Do you have a specific construction in mind?

The first problem is a big problem but does not necessarily preclude your model since you could have just happened to guess the correct model for the universe without any evidence.

:D I'm not guessing. What I'm presenting here is the simplest, easiest to grasp proof that I am right.

Mainly this is because the evidence for CPT symmetry is evidence for a type of symmetry in a QFT and does not lead to binary values as you suggest.

Why not? (we are talking binary objects rather than binary values... is this confusing you?)

This has been explained several times but you have still not understood.

No - it has not been explained. There have been statements supported by no evidence. There have been explanations that are irrelevant to the subject at hand. What there has not been is an explanation of why the proof I have presented is flawed.

I do accept that others have a different view of this... I have no doubt that there are persons who feel they have explained until they are blue in the face and that I have wriggled and squirmed out from under their explanations in a most frustrating fashion.

I still think that I am right... and everyone else knows that I am wrong.

Bear in mind that I am using X(X(s)) = s for all s (excluding x = null) as a definition of a binary object. As such, there really isn't much to argue with... You can argue with what I then do with these objects - but the definition (axiom) is... well... an axiom. As such, I can be pretty confident that X(X(s)) = s for all s (excluding x = null) does lead to binary objects. So... when you say "This has been explained several times but you have still not understood."... maybe it isn't me who isn't understanding?

Treatid wrote: But there is no doubt that where X(X(s)) = s, Y(Y(s)) = s, Z(Z(s)) = s and X○Y○Z are invariant for all s then there is only 1 possible model (trivial X, Y and Z excluded).


Is untrue, all this says is that s and X(s) are pairs of states related by the operator X. In terms of CPT that they are (for example) charge conjugates of each other. This does not mean that there are only two such states for example it is also true that X(X(s'))= s' for s' =/= s.

You are making the wrong point. Of course there are many (many, many) instances of s. You will see in my definition "for all s".

I am not arguing that the universe is composed of just 3 binary objects. That is absurd. I am arguing that the behaviour of the universe is determined by 3 binary objects. Each interaction in the universe is governed/described by {b, b, b}. There are a very large number of such interactions. This isn't that surprising... CPT Symmetry says this directly... For every interaction... change each of C, P and T and the interaction is invariant. CPT Symmetry directly tells us that there is an aspect of the universe that is very simple... that 3 simple transformations together create an invariance. The proof I'm presenting simply formalises this to show exactly how simple... and that there is only 1 structure with this simplicity.

This is essentially the same point that people have tried to make again and again that there are different consistent models that have CPT symmetry with different predictions.

I think you are conflating two different issues. People have been trying to show that CPT Symmetry is consistent with various systems... but without formally specifying those systems. As such, there is no way to draw any conclusion from those systems. When someone builds a formal mathematical system that contains CPT Symmetry then we can move ahead with that argument. So far, all that has been done is to state that some systems are consistent with no evidence to support those statements.

Another way of posing the problem of evidence would be to ask what the 3 binary quantities in your model are.

They are binary objects. All binary objects are equivalent. They only gain meaning/relevance/distinction when they are in a context. The set {b, b, b} provides a context.

More specifically, as I've described earlier... One of those binary objects is the arrow of time, one may be visualised as a directed edge within a graph... the direction of the edge may be inverted; the third is a rotation... the direction of rotation may also be reversed.

As we have said they clearly are not C P and T since they are not binary quantities. So what are those quantities and where is the experimental evidence for them?

No C, P and T are operators such that C(C(s)) = s, P(P(s)) = s and T(T(s)) = s for all s. And X(X(s)) = s (x = null excluded) specifies a binary object for each s.

1)Your model is not predictive.

Yes it is. I simply haven't made any predictions. There is little point until the strongest evidence available is examined.

You have said that your model is predictive in that your prediction is that your model is the fundamental model of the universe but this is not really a prediction.

No - I have a proof which shows the model is the only possible model. The proof is not the model... it simply points to the model. Mathematically, {b, b, b} describes a single model... but in the abstract it isn't very informative. I have a specific instance of the model such that it can be run as a computer program... It is that specific model that has predictive power.

At this point we would settle for correctly describing stuff we already know but again there has been nothing.

I give you a feast that you turn your nose up at... You insist that you will only accept what you are familiar with even though what I offer is more delicious and healthy than what you are familiar with. :)

The proof I'm offering is tremendously strong evidence. It is far more persuasive then any single piece of evidence you think that you want.

Assume for a minute that you were actually right about the motivation in terms of CPT in that case making a prediction for anything becomes a way to test CPT, if your model doesn't reproduce compton scattering correctly then we would have evidence that CPT is in fact broken.

This is true... but we won't get there until the initial premise is accepted.

Before anyone is going to explore the actual model and test its predictions against observation; they need to be persuaded that the model could be a model of the universe. The model looks nothing like Quantum Mechanics or General Relativity. To understand a new model where none of the existing mathematical tools apply (until such time as a bridge is constructed to QM and GR) requires a commitment that won't be made lightly.

Hence I have provided a small, easy proof that shows there is a significant probability that this model is the one to look at.

2)From the ground up your model seems to have some rather large implications. If I have understood correctly your position is that your model is the only model that contains the CPT symmetry (where for CPT here I mean your binary definition), that is there is no possible extension of this symmetry in a yet unobserved sector that would approximate CPT symmetry in the standard model and that any models that contain this symmetry + some large symmetry either continuous or discrete are either Inconsistent systems or in fact determined by the original CPT model. This to me is very interesting it reminds me of the Coleman–Mandula theorem of symmetries of the s-matrix. If you could prove this then that would be interesting.

:D

I believe I have demonstrated this. Clearly the proof is not yet understood... Perhaps I'm making an assumption about the proof which I haven't made explicit... or people are bringing more assumptions into the proof than are stated.

The proof seems to me both trivial (a very simple result - none of the steps should be controversial) and, in hindsight, obvious.

The issue of X(X(s) = s representing a binary object seems to be a sticking point. As does the idea that {b, b, b} represents the behaviour of the system... obviously an instance of the system must contain a vast number of primitives...

Schrollini wrote:We've told you. F and x are vectors, changing sign under P (but not C or T). m and T are scalars, remaining unchanged under C, P, and T. And of course time derivatives change sign under T (but not C or P).

I think that m has an aspect that is treated as a scalar rather than m actually being a scalar... however... this is your demonstrations...

So... tell me what your axioms are... and then show me how you are going to prove that those axioms are consistent with one another...

Treatid wrote:So - C, P and T are each undefined - or have to be redefined for every possible state that they could be applied to?

No, they're perfectly well defined. But we have to say a lot more than just C(C(s)) = s to define C.

I simply don't understand what you are saying. I cannot see any consistency. You say that there is a different operator for each state... and then that C, P and T are defined - so they aren't different operators...

Are you saying that there are many C(C(s)) = s? This much I agree with there is an instance for every possible state (or possibly pair of states).

Are you further arguing that because there are many C(C(s)) = s, P(P(s)) = s and T(T(s)) = s that they can't be expressed as {b, b, b}? That, at best, we are looking at a large set of many binary objects - one that is in no way limited to 3?

Or something else?

Treatid wrote:However, even if C, P and T are different for every state or pair of states... that doesn't actually contradict X(X(s)) = s being a binary object.

Which is binary, X or s?

Neither. X(X(s)) = s constructs the binary object. There is an implicit s1 in there too.

Not that it matters -- we've shown that X(X(s)) = s alone is not enough to restrict either X or s to one of two possibilities. But if you limit s to two elements, then there are only two functions X, the identity and inversion. Perhaps this is what you're arguing?

No - There are indeed many such binary objects. You will note in my definition the phrase "for all s".

So, you are arguing that {b, b, b} should in fact be {b, b, b, ...}?

Treatid wrote:I am saying that X(X(s)) = s specifies a binary object. If you look closely you will see that I am not saying that X = binary object.

Or are you saying that this creates a pairing between s and X(s)? Which is true, but says nothing about either X nor the set S which contains s.

I am saying that a pairing is created between s and X(s). Since X creates the pair... I would say it does say something about X... but I assume you mean that it doesn't construct the specific X that operates on the specific state s... the implementation of the operators C, P and T is not specified beyond the constraint that C(C(s)) = s, etc... which is itself a significant constraint.

doogly wrote:Yeah, here is the problem. These don't work, as I've said before, but this is very much isolating it. This just isn't how basic set theory and function theory work. If X(X(s))=s, it does *not* follow that s \in (s_1, s_2).

Why is s not a member of {s1, s2}? How did you construct {s1, s2}? Is there a reason for me to expect s to be in this set? What relevance has this to the proof I have provided?

You have made a statement with no justification or reasoning.

I guess you are trying to say something about X(X(s)) = s not defining a binary object? In the case that X is null, then X(X(s)) = s is trivial and only relates a single state. There are other instances where s might be its own image under the function X such that X(s) = s. So are you arguing that X(X(s)) = s sometimes specifies a singleton {s} rather than a binary object {s1, s2}?

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Re: There is only one* model consistent with CPT Symmetry

Postby gmalivuk » Tue Aug 20, 2013 4:19 am UTC

X(X(s))=s could also apply to any number of larger sets S, including infinite ones.

You have yet to prove *anything* about your alleged model that could actually be used together with observations to test it. You have 24 hours to do so, after which point I'm going to lock this thread if there are no objections from any of the people who are trying to help you understand your mistakes.
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Re: There is only one* model consistent with CPT Symmetry

Postby doogly » Tue Aug 20, 2013 5:13 am UTC

Please rephrase without using the terms "binary object." This is not something that actually makes sense mathematically.

"A set with two elements" is well defined, basic set theory. Three sets like this give you a set of size eight, when order doesn't matter. Great.

"involution" is well defined, it is any function f such that f(f(s))=s for all s in the domain.

And if you'd like axiomatization, the Wightman axioms are great for QFT. Classical mechanics is also completely rigourous though.
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Re: There is only one* model consistent with CPT Symmetry

Postby Sizik » Tue Aug 20, 2013 5:21 am UTC

If CP symmetry (along with T symmetry) was true along with CPT symmetry, how would this affect your {b, b, b} "model"?
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Re: There is only one* model consistent with CPT Symmetry

Postby Schrollini » Tue Aug 20, 2013 5:37 am UTC

Treatid wrote:So... tell me what your axioms are...

We've told you what all the symbols are mathematically (scalar, pseudo-scalar, vector, pseudo-vector, etc.) and how they transform under each C, P, and T. What more do you need? And don't worry about whether mass is a scalar or "has an aspect that is treated as a scalar". This is a mathematical model and the model says it is a scalar. Experiments may prove that wrong, but that's no concern of ours.

If your worried about my axioms of calculus, I really don't care. Choose ZFC + whatever else you need.

Treatid wrote:and then show me how you are going to prove that those axioms are consistent with one another...

And how, Herr Gödel and I wonder, could we possibly do that?

The best argument I can give is that this is classical mechanics, which has been around for four hundred years without any inconsistency being found.

Treatid wrote:
No, they're perfectly well defined. But we have to say a lot more than just C(C(s)) = s to define C.

I simply don't understand what you are saying. I cannot see any consistency. You say that there is a different operator for each state... and then that C, P and T are defined - so they aren't different operators...

C(electron) = positron, C(positron) = electron, and C(electron neutrino) = electron neutrino. Thus, C(C(s)) = s for all three states {electron, positron, neutrino}. But just saying C(C(s)) = s doesn't define this. I could also define the dumb symmetry D such that D(electron) = neutrino, D(positron) = positron, and D(neutrino) = electron. This satisfies D(D(s)) = s. So to specify C, and distinguish it from D, you need to say what it does to each of these states.

(Note: Wikipedia says C takes particles to antiparticles while CP takes matter to antimatter. It's been too long since I took QFT to remember the difference, so when I say C in the above paragraph, I might actually mean CP.)

Treatid wrote:
Which is binary, X or s?

Neither. X(X(s)) = s constructs the binary object. There is an implicit s1 in there too.

Not that it matters -- we've shown that X(X(s)) = s alone is not enough to restrict either X or s to one of two possibilities. But if you limit s to two elements, then there are only two functions X, the identity and inversion. Perhaps this is what you're arguing?

No - There are indeed many such binary objects. You will note in my definition the phrase "for all s".

So, you are arguing that {b, b, b} should in fact be {b, b, b, ...}?

I think part of the problem is that we have no idea what this binary object b is. At first, we thought you were referring to states, but apparently not. Then we thought you were referring to functions, but apparently not. Now, my best guess is:

Treatid wrote:I am saying that a pairing is created between s and X(s).

These pairs are the binary objects? If so, I'd like to suggest this is a terrible nomenclature, since there are at least ceil(n/2) pairs if there are n states, and all of them are important for defining X.

And if this isn't it, you need to explain slowly and in small words, because what's obvious to you isn't getting at all across to the rest of us.

Treatid wrote:Since X creates the pair... I would say it does say something about X... but I assume you mean that it doesn't construct the specific X that operates on the specific state s... the implementation of the operators C, P and T is not specified beyond the constraint that C(C(s)) = s, etc... which is itself a significant constraint.

Yes, you've sussed out my meaning correctly. That constraint does tell you something about C, but it doesn't tell you everything.
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Re: There is only one* model consistent with CPT Symmetry

Postby andyisagod » Tue Aug 20, 2013 7:18 am UTC

Treatid wrote:
andyisagod wrote:The way I see it there are two distinct problems with your idea. The first is that there is no observational evidence supporting the claim that universe is described by 3 binary values.

Yes there is. It is CPT Symmetry.


I think that until you actually understand the mathematics and stop making weird leaps to things being binary that this isn't going to go any further. You aren't going to be taken seriously while you continue to get this bit wrong.


Treatid wrote:I think you are conflating two different issues. People have been trying to show that CPT Symmetry is consistent with various systems... but without formally specifying those systems. As such, there is no way to draw any conclusion from those systems. When someone builds a formal mathematical system that contains CPT Symmetry then we can move ahead with that argument. So far, all that has been done is to state that some systems are consistent with no evidence to support those statements.


They have specified those systems fully and you haven't shown they are inconsistent. You haven't actually proved that your system leads to consistent description of nature so I wouldn't start complaining.

Treatid wrote:I still think that I am right... and everyone else knows that I am wrong.


Because everyone else is being rational.

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Re: There is only one* model consistent with CPT Symmetry

Postby doogly » Tue Aug 20, 2013 12:57 pm UTC

Wikipedia is lying to you about what C does, CP is the matter / antimatter transform.
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Re: There is only one* model consistent with CPT Symmetry

Postby Schrollini » Tue Aug 20, 2013 3:01 pm UTC

doogly wrote:Wikipedia is lying to you about what C does, CP is the matter / antimatter transform.

Thanks. That's what a little voice in the back of my head was saying. Something about neutrinos being left-handed. But I didn't trust him.

Maybe someone who actually knows this stuff can fix up this page: http://en.wikipedia.org/wiki/CP_violation, which says:
CP-symmetry states that the laws of physics should be the same if a particle is interchanged with its antiparticle (C symmetry), and then its spatial coordinates are inverted ("mirror" or P symmetry).

and
CP-symmetry, often called just CP, is the product of two symmetries: C for charge conjugation, which transforms a particle into its antiparticle, and P for parity, which creates the mirror image of a physical system.
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Re: There is only one* model consistent with CPT Symmetry

Postby doogly » Tue Aug 20, 2013 3:28 pm UTC

I did it!

This my third wikipedia edit. It is a good one. My others were correcting that Queen Ida was the first Louisiana Creole to win a Grammy, *not* Clifton Chenier, and that Sotomayor is from Bronx, NY, not New York, NY. That one got reverted by a hater though.
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Re: There is only one* model consistent with CPT Symmetry

Postby Zamfir » Tue Aug 20, 2013 4:13 pm UTC

In the future, there will be only one wikipedia page, and all other terms will redirect to it. That single page will say {b,b,b}

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Re: There is only one* model consistent with CPT Symmetry

Postby andyisagod » Tue Aug 20, 2013 5:12 pm UTC

doogly wrote:I did it!

This my third wikipedia edit. It is a good one. My others were correcting that Queen Ida was the first Louisiana Creole to win a Grammy, *not* Clifton Chenier, and that Sotomayor is from Bronx, NY, not New York, NY. That one got reverted by a hater though.


Schrollini wrote:
doogly wrote:Wikipedia is lying to you about what C does, CP is the matter / antimatter transform.

Thanks. That's what a little voice in the back of my head was saying. Something about neutrinos being left-handed. But I didn't trust him.

Maybe someone who actually knows this stuff can fix up this page: http://en.wikipedia.org/wiki/CP_violation, which says:
CP-symmetry states that the laws of physics should be the same if a particle is interchanged with its antiparticle (C symmetry), and then its spatial coordinates are inverted ("mirror" or P symmetry).

and
CP-symmetry, often called just CP, is the product of two symmetries: C for charge conjugation, which transforms a particle into its antiparticle, and P for parity, which creates the mirror image of a physical system.


I'm not sure if this is right, Peskin and Schroeder definitely describes C as the matter anti-matter transformation. C does swap fermions with anti-fermions but the weak sector is violates C and P separately which leads to the antimatter being a mirror image of matter thing since you have to do CP to leave the weak sector invariant. I'm not sure if maybe there is confusion between the C operation which swaps matter and anti-matter and the CP operation which then makes the matter anti-matter swap indistinguishable from matter (ignoring CP violation).

Sorry that explanation is a mess...

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Re: There is only one* model consistent with CPT Symmetry

Postby PM 2Ring » Tue Aug 20, 2013 6:23 pm UTC

doogly wrote:Wikipedia is lying to you about what C does, CP is the matter / antimatter transform.

So if we assume CPT invariance, that's equivalent to T, right?

IIRC, young Feynman said that a positron was indistinguishable from an electron going backwards in time. And so if you time reverse a Feynman diagram, all the electrons become positrons & vice versa. Etc.
Last edited by PM 2Ring on Tue Aug 20, 2013 7:13 pm UTC, edited 1 time in total.

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Re: There is only one* model consistent with CPT Symmetry

Postby doogly » Tue Aug 20, 2013 6:25 pm UTC

Yes indeed. I think if P&S are saying otherwise, maybe they are exclusively in the regime of QED? Which seems weird. Or maybe they were just being sloppy. But matter/antimatter is most definitely CP.
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Re: There is only one* model consistent with CPT Symmetry

Postby andyisagod » Tue Aug 20, 2013 9:06 pm UTC

doogly wrote:Yes indeed. I think if P&S are saying otherwise, maybe they are exclusively in the regime of QED? Which seems weird. Or maybe they were just being sloppy. But matter/antimatter is most definitely CP.


But C clearly takes particles to anti-particles, it does this in QED but also in the weak interactions. P&S does refer to C as taking particles to anti-particles and this can be seen from the definition of an anti-particle, that is an anti-particle has the same mass and spin as the particle but opposite charge (where charge includes the other quantum numbers such as lepton number).

I think that the review of CP violation might clear this up http://pdg.lbl.gov/2013/reviews/rpp2012-rev-cp-violation.pdf. As this points out, C takes you from particle to anti-particle and then P reverses the handedness. I.e CP takes you from a left handed electron to a right handed positron. Which is not the same as taking the charge conjugate and going from a left handed electron to a right handed positron.

I don't think we disagree on the actual definition of C and P here just what transformation you label as the transformation of matter to anti-matter. I think that since C does take you from e- to e+ that it should be described as a transformation between anti-matter and matter. If you want to argue that the anti-particle of the left handed e- is the right handed e+ then indeed CP would be the correct transformation but I don't think that is the right convention since a lot of the literature disagrees.

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Re: There is only one* model consistent with CPT Symmetry

Postby doogly » Tue Aug 20, 2013 9:11 pm UTC

Maybe the problem is that particle physicists don't have their shit together. I'm in the cosmology camp, and we know that the matter/antimatter asymmetry is necessarily a CP violation. That's the notion of "antimatter" most often on my lips. But yeah you're right, this does just come down to labeling.

Perhaps this edit won't last any longer on wikipedia than the Sotomayor Bronx shoutout. Fuckin haters.
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Re: There is only one* model consistent with CPT Symmetry

Postby andyisagod » Tue Aug 20, 2013 9:35 pm UTC

doogly wrote:Maybe the problem is that particle physicists don't have their shit together. I'm in the cosmology camp, and we know that the matter/antimatter asymmetry is necessarily a CP violation. That's the notion of "antimatter" most often on my lips. But yeah you're right, this does just come down to labeling.

Perhaps this edit won't last any longer on wikipedia than the Sotomayor Bronx shoutout. Fuckin haters.


The matter anti-matter asymmetry does require CP violation but that isn't the same as saying that CP transforms matter to anti-matter. Sakharov's conditions require C, P and CP to be violated. Of these I think C violation is required because C transforms matter and anti-matte while CP violation ensures that you don't end up with an equal number of left handed particles and right handed anti-particles. Its probably worth bearing in mind that the conditions for the matter anti-matter asymmetry is actually CPT violation which is achieved by being out of equilibrium in the case of baryogensis rather than CPT violation at the lagrangian level.

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Re: There is only one* model consistent with CPT Symmetry

Postby doogly » Tue Aug 20, 2013 10:14 pm UTC

Yeah, this sounds righter.
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Re: There is only one* model consistent with CPT Symmetry

Postby Treatid » Wed Aug 21, 2013 3:48 am UTC

gmalivuk wrote:X(X(s))=s could also apply to any number of larger sets S, including infinite ones.

X(s1) |-> (s2). X(s2) |-> (s1). The maximum size of the set is {s1, s2}. There are many such pairs. But X(X(s)) = s constructs a set of maximum size 2.

I'd be grateful if you could explain how you obtain an infinite set.

You have yet to prove *anything* about your alleged model that could actually be used together with observations to test it. You have 24 hours to do so, after which point I'm going to lock this thread if there are no objections from any of the people who are trying to help you understand your mistakes.

I disagree with your premise. However, I accept that it is your intention to lock this thread.

doogly wrote:Please rephrase without using the terms "binary object." This is not something that actually makes sense mathematically.

I accept this. The construction provided by X(X(s)) = s where X is not null is much less ambiguous than "binary object".

"A set with two elements" is well defined, basic set theory. Three sets like this give you a set of size eight, when order doesn't matter. Great.

Yes-ish. The set {b, b, b} has cardinality 3. Given that b is a binary digit then there are 8 possible instances of this set. The general case versus a specific instance.

"involution" is well defined, it is any function f such that f(f(s))=s for all s in the domain.

Fine.

And if you'd like axiomatization, the Wightman axioms are great for QFT. Classical mechanics is also completely rigourous though.

Okay - and has anyone proven that the Wightman axioms are consistent? That they are in no way contradictory?

My argument on this point has always been that the systems that people are claiming are consistent with CPT Symmetry have not been proven to be consistent with CPT Symmetry. Creating a set of axioms for a system is a step in the right direction - but those axioms still need to be shown to be consistent.

Sizik wrote:If CP symmetry (along with T symmetry) was true along with CPT symmetry, how would this affect your {b, b, b} "model"?

A good and relevant question. Thank you.

T Symmetry means that if just the 'arrow of time' is inverted then all observations continue unchanged. Currently we observe that entropy increases with our familiar 'arrow of time'. If T is invariant then the system that T relates to cannot have increasing entropy.

Entropy is a measure of information loss. A function may operate on a state to map it to another state. Where multiple states map to a single state (many-to-one); the reverse must either map a single state to many states or choose which state to map to...

T Symmetry requires that the function only ever maps one state to one other state (one-to-one). Such a system can only consist of a chain or loop of states (all states are at equal entropy).

As such... the CP Symmetry is redundant information. The T Symmetry has entirely defined the behaviour of the system and the system would be represented as {b}.

It follows then that within {b, b, b} the whole system forms a symmetry but no subset of the system may form a symmetry... Which I haven't explicitly stated and is probably relevant information.

Treatid wrote:and then show me how you are going to prove that those axioms are consistent with one another...

And how, Herr Gödel and I wonder, could we possibly do that?

The best argument I can give is that this is classical mechanics, which has been around for four hundred years without any inconsistency being found.

CPT Symmetry is classical mechanics? CPT Symmetry has been around for four hundred years?
Your claim is that you know that these system are consistent with CPT Symmetry. That it is obvious. That you can dismiss my ideas because there is no possibility that there is an inconsistency between CPT Symmetry and whatever system you are connecting it with. But you don't have proof to back up that claim. Absent proof - there is the possibility that your claim is not true.

Even while you are pretty sure that your assertions are correct... that CPT Symmetry is consistent with a variety of distinct models... there is doubt until you provide a rigorous mathematical proof.

Perhaps you feel that you are sufficiently certain that a proof is a mere formality...

Treatid wrote:
No, they're perfectly well defined. But we have to say a lot more than just C(C(s)) = s to define C.

I simply don't understand what you are saying. I cannot see any consistency. You say that there is a different operator for each state... and then that C, P and T are defined - so they aren't different operators...

C(electron) = positron, C(positron) = electron, and C(electron neutrino) = electron neutrino. Thus, C(C(s)) = s for all three states {electron, positron, neutrino}. But just saying C(C(s)) = s doesn't define this. I could also define the dumb symmetry D such that D(electron) = neutrino, D(positron) = positron, and D(neutrino) = electron. This satisfies D(D(s)) = s. So to specify C, and distinguish it from D, you need to say what it does to each of these states.

I don't disagree with any of what you are saying.

"Given any system that contains three operators X, Y and Z such that X(X(s)) = s, Y(Y(s)) = s, Z(Z(s)) = s and X○Y○Z is also invariant for all states s..."

I am not trying to construct CPT Symmetry. I am not trying to define what a state is or how the operators are specified.

If an operator has a particular property such that X(X(s)) = s... the state can be anything... the operator can be anything... All that matters is that X(X(s)) = s.

Yes, the actual states are many. There are many mappings. There might even be an infinite number of possible states... and each of these states is mapped by each of the operators to (usually) another state. So there are an infinite number of states and their mapped pairs. And you can argue that the mappings are unique to each pair... there are thus an infinite number of mappings.

None of that impinges on the original statement: X(X(s)) = s for all s. Is C(C(s)) = s, P(P(s)) = s and T(T(s)) = s for all s true of the operators C, P and T in CPT Symmetry. If it isn't then I don't have a leg to stand on. If it is then the weakness must be found elsewhere in the proof.

Your argument isn't wrong... it just doesn't impact on on the proof I have presented.


I think part of the problem is that we have no idea what this binary object b is. At first, we thought you were referring to states, but apparently not. Then we thought you were referring to functions, but apparently not. Now, my best guess is:

Treatid wrote:I am saying that a pairing is created between s and X(s).

These pairs are the binary objects? If so, I'd like to suggest this is a terrible nomenclature, since there are at least ceil(n/2) pairs if there are n states, and all of them are important for defining X.

I do agree that I started out with an ambiguous definition of a binary object - this has led to confusions galore...

"X(X(s)) = s" is what I am talking about. As you have been saying - there are many instances of this corresponding to all the possible states s. For any single state s there is one X(X(s)) = s, one Y(Y(s)) = s, one Z(Z(s)) = s and one XYZ invariance.

As such {b, b, b} is shorthand for {{b, b, b}, {b, b, b}, ....}

Treatid wrote:Since X creates the pair... I would say it does say something about X... but I assume you mean that it doesn't construct the specific X that operates on the specific state s... the implementation of the operators C, P and T is not specified beyond the constraint that C(C(s)) = s, etc... which is itself a significant constraint.

Yes, you've sussed out my meaning correctly. That constraint does tell you something about C, but it doesn't tell you everything.

As noted by doogly... I am stripping away all the information that (I consider) redundant. I am taking a very abstracted view. I am taking a single set of observations related to CPT Symmetry and stripping away all the details.

And yet, with that stripped down information, I can show that there is only one possible model. This is remarkable. That such a small nugget of information separated from the vast wealth of physics should alone provide such specific information is really quite surprising. I am sorry that I haven't been able to convince you of this result and show you its beauty...

They have specified those systems fully and you haven't shown they are inconsistent. You haven't actually proved that your system leads to consistent description of nature so I wouldn't start complaining.

Repeating something does not make it true (at least not in mathematics).

To fully specify a system in mathematics is hard. To know that a set of axioms is consistent is hard. It takes a great deal more than writing down an equation to prove anything in mathematics. Nobody has proven that any particular system is consistent with CPT Symmetry.

Because everyone else is being rational.

Because I disagree with the majority I must be insane?

Most people have had the courtesy to address my comments... and to argue against what I have written.

Zamfir wrote:In the future, there will be only one wikipedia page, and all other terms will redirect to it. That single page will say {b,b,b}

:D

However, while physics and chemistry determine biology... it is still useful to have a separate discipline. Building everything from first principles gets old pretty fast.

...

Thank you to all those who have taken the time to respond to me. I feel this has been productive in showing me some of my hidden assumptions, improving my language and further refining my ideas.

It has been a little frustrating for me at times, as I suspect it has been for those who have made efforts to enlighten me regarding my errors. Even so, I have felt that progress was being made by both sides in explaining themselves to the other.

Thus it is with some regret that I bid you farewell...


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