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.

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.

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 s

^{1} 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 {s

_{1}, s

_{2}}? How did you construct {s

_{1}, s

_{2}}? 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 {s

_{1}, s

_{2}}?