uncool

You have not explained why this is necessary. Again, relativity does not violate causality. Relativity states that if event A causes event B, then event B must be in the future light-cone of event A. If that is true, then no change of frame can cause event B to happen before event A, so causality is not violated. =Uncool-

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And why is ambiguity a problem? As long as cause always precedes effect - which is true in relativity - there is no problem with ambiguity. That is the reason it is called "The relativity of simultaneity". =Uncool-

You haven't exactly trisected the angle, which is the point. Math shows that you cannot exactly trisect the angle with a straightedge and compass, although you can approximate it as closely as you want. =Uncool-

You have said that if one person sees the other person slow down, then the 2nd person must see the first speed up. You have not given a specific reason why. =Uncool-

You can accomodate one at a time: =Uncool-

=Uncool-

Alex, the reason I asked you to express it in matrices is the following: The only reasoning that leads to requiring that slowdowns for one person requires speedups for the other is that you think inverse matrices in general must have the property about the upper left element. You can't explain any reasoning as to why there should be anything special about the matrices that would require speedups. You are simply asserting, and it is false. =Uncool-

This means nothing. What physically do you mean by "visual effects"? That doesn't say anything about why transformation matrices must have the property mentioned. =Uncool-

What is a "visual effect"? What does that mean physically? Then what is special about transformation matrices that it should be true for them? =Uncool-

But that simply isn't true under Galilean transformations. The speed of light changes under Galilean transformations. Which means that it is observer-dependent - it is not the same for different observers. Which makes it non-absolute. That result is there in Galilean relativity and special relativity already. It is a trivial statement. I'm asking you to make a specific statement about your theory. I'm asking you to make a specific answer, yes or no. If you have a matrix A - any matrix A - and you know that A0,0 > 1, does that mean that A-10,0 must be less than 1? =Uncool-

But you've already agreed that under your theory, the speed of light is observer-dependent. That is, you've already agreed that in different frames, the speed of light is different. I'm trying to figure out what exactly you think. So is this your definition of the speed of light being absolute? That doesn't answer my question. Is it true for any matrix that if the upper-left element is greater than 1, then the upper-left element of the inverse must be less than 1? =Uncool-

What such properties? So again, please explain precisely what you think it means for the speed of light to be absolute. Do you think that there is one preferred frame, relative to which light will expand directly from a single point, and that in all other frames, the center of the light-sphere will move with time? So do you think it is true for any matrix? If not, then why can't it be true for this transformation matrix? =Uncool-

Then what does it mean to say that they are absolute? You still have not answered that question. And am I correct that you think that that is true for any matrix - that is, for any matrix A, A0,0 > 1 implies that A-10,0 < 1? =Uncool-

Again, in that case, what precisely does it mean to say that the speed of light is absolute? You have not defined it. I have. And there is reason to agree with my definition. Furthermore, there are experiments that bear out my definition. And again: So your objection is equivalent to the statement that if the (0, 0) element of Av (that is, the upper left element (which for this transformation matrix is [math] \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/math]) is greater than 1, then the (0, 0) element of (Av)-1 must be less than 1. Do you agree? =Uncool-

Well, I don't think I am - I'm just suggesting that the word "structure" as related to particles means that the particle is not spherically symmetric. I'm guessing that you think of it as being a combination of eigenstates of the Hamiltonian, and something without structure is just an eigenstate? =Uncool-

In other words, the bare electron has no structure, but the renormalized electron has a dipole moment when chromodynamics is taken into effect, right? =Uncool-

Well, that depends on what you mean by internal structure. This would imply that it's not spherically symmetric, which to me would be the definition of having structure - it has a specific direction that it could be considered to be "pointing". =Uncool-

So these are your transformation matrices? You, too, have set the "cross-scales" to be absolute here. Furthermore, what precisely is your definition of the speed of light being absolute? So your objection is equivalent to the statement that if the (0, 0) element of Av (that is, the upper left element (which for the special relativistic transformation matrix is [math] \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/math]) is greater than 1, then the (0, 0) element of (Av)-1 must be less than 1. Do you agree? =Uncool-

Alex, you have not answered what I asked. I asked you: What is your transformation matrix? Additionally, do you agree that I have correctly summarized your objection? =Uncool-

Might want to try adjusting your quotes.

Number. That was easy. Where did this number come from? Why not just use log(pi) = .497, which is even closer? What is k here? Actually, it makes it look even more like you are doing so. Again, the only way to dispute that you are "pulling numbers out of nowhere" is to show where you are getting your numbers. And what significance does this have? Any number multiplied by its reciprocal will be 1. So what does all this "proof" have anything to do with microprocessors? =Uncool-

I would say that equipping it with the standard metric is then what restricts the automorphisms. Isotropy and homogeneity are only guarantees that those automorphisms exist. It just sounded like you were saying that any isotropic, homogeneous space would have no other symmetries. =Uncool-

I'm not sure that I see this. It seems to me that isotropy guarantees rotations, and homogeneous guarantees translations as automorphisms (in fact, that is pretty much the definition of the two) - but that other automorphisms are not out of the question. For example, say you had R^3 as your universe - that universe would still be isotropic and homogeneous, but it would also have scale invariance for distance. =Uncool-

Not what I asked. I asked you: What is your transformation matrix? Additionally, do you agree that I have correctly summarized your objection? =Uncool-