joigus

Wrong again. It's not about re-scaling sin(x). It's about re-scaling x. First scale right, then warp. Then rinse and repeat.

I think you mistake me for somebody else. But I see more clearly what you're doing here.

Not subtle, just wrong. It shouldn't be there. Right now, what's the speed of your nose with respect to something I'm thinking of, but I'm not gonna tell you? See? That's your v. And this is a serious no, no. I hope you understand. If you don't, I can't help you, and I'd venture nobody else here can.

Ok. Riddle me this: S2 recedes with respect to what --or whom--? Not the detector --or its associate observer--, since those see no motion. Only with respect to a fiducial --but, mind you, unmentioned by you-- observer sitting at the port's pier, let's say. See the flaw? If you go over Einstein's classic papers, you will notice that all the c-v or c+v terms come from observers calculating distances or delay times from their "rest" frames! According to them, the wall (mirror, or whatever) has moved. Not according to the co-moving observer! x-posted with @Bufofrog

Thank you. But it was both you and Swanson who set me on the right track when you said, I've credited accordingly. You see, it sometimes takes some time to see how deep in shit* an OP's argument is. You have to, in a way, accept an unacceptable logic. * Pardon my French.

There is no v in the problem you're setting up. The whole point of relativity is that S2 is not receding. No observer attached to the ship can measure any such v. There is no dragging of the speed of light. The measured speed for light inside the ship is c, not c-v, as you claim. You got everything wrong. This v is only in your mind. So the very first time you wrote an equation and you said you were going to hold it against relativity. Well... You wrote the wrong equation. S2 is not receding from the POV of anybody anchored to the ship --at rest relative to it. Nobody, repeat, nobody stuck to the ship has any right to even start talking about such v. What v? What are you talking about? What you are doing is using a quantity that only makes sense in the frame attached to a certain "rest observer" you're telling us nothing about in your calculations about another frame. And yes, according to your main line of reasoning, space within a ship is non-isotropic, which is a ludicrous claim, of course. (My emphasis.) What?! This is false. In the ship's inertial system such time is t=D/c, not what you say. The whole thing is ridiculous.

Swansont, of course, didn't say such silly thing. "Simultaneous" is a relative concept, as you should know by now. In what frame? Both @studiot and @swansont have pointed out or implied... In what frame? at several points. Almost every other claim of inconsistency of SR has a flaw of this kind. Yours has it too. I confess I'm taking all this very lightly, trying not to stray too deep into your personal rabbit's hole, choosing to fix on one particular inalienable reason why your argument cannot be true. But I must be doing fine, as you chose not to address any of my concerns. Clear indication that you have no answer to them. Simultaneous? In what frame? Non-isotropic space within a Galilean ship? How could that be?

It seems that you're groping your way towards a theory of absolute orientation... Mmmm... I'm looking forward to your Earth-shattering predictions. Otherwise, you've just misinterpreted the results of experiments, which sounds like what's really happened.

BTW, sine of a degree is not that. I mean, sine of 1 degree is not 1-1³/3!+1⁵/5!-... That's sine of 1 radian. Careful, you might end up warping things too much.

I think so. What other use do you have in mind? In mathematics you have so-called functionals, which are functions of functions. They're referred to with square brackets, e.g.: G[f]

All gauge bosons are their own anti-particles. Think about it, if you will, as if they are themselves the product of annihilation. They cannot annihilate any further. Think about it as all particle-antiparticle pairs that mutually annihilate do so by offering each other opposite quantum numbers "to annihilate with". Gauge bosons have no non-zero quantum numbers to "annihilate with" in a manner of speaking. Gravitons are gauge bosons.

Oh it doesn't work on so many levels... What about starting with animals generally died in the strata where they lived. Plus T-Rex is one of the most abundant fossils out there. Swamps and marshes leave abundant organic residues easy to recognise. Not the case. Unlikely... unless they all went en masse to Utah to embrace Mormonism in a retroactive mass conversion.

The unbalance cannot come from random fluctuations. Electric charge is exactly conserved --as every other gauge charge. You really need a mechanism to nudge things out of balance. Look up Sakharov conditions for baryogenesis. Oh, look. I thought I'd said it, and indeed I did... This is no random fluctuation. Either that or everything started out unbalanced for some mysterious reason --which is always possible.

That is so passé. History always almost repeats itself, but not quite.

Mind you, the site will be down (for the same reason) in about 2 years time now. A genie told me. So brace yourselves again for the temporary disappearance of your science-minded online persona in 6.2x107 seconds.

Not me. I had completely forgotten about this conversation! This feels like a dream... Did I really say that?

Liver I think symmetry and simple patterns are part of the deal. That would be my guess anyway. I understand we're just guessing...

Reported. I'm done with this.

I have read it --well... skimmed through it, obviously. I haven't seen any calculations of the properties of the vacuum. Not even back-of-the-envelop calculations. So I'm actually not engaging in any proper discussion of your theory because I don't see anything worth discussing. I'm just pointing out to you what I see as obvious deficiencies in your reasoning, so it's more of a rebuttal than a discussion. https://www.scienceforums.net/guidelines/ Point 7 of the Guidelines at SFN: So no, I do not have to read your document in any degree of thoroughness. Copy and paste anything from your text if you think it addresses any of my concerns. That'll do. Matter goes through the horizon, and we have strong reasons to believe thermal radiation slowly leaks out of it. I don't think your claim is consistent with Liouville's theorem, and probably many other principles of physics. It would only work in two isolated universes: One only with positive energies and time going in one direction; and another with the situation reversed. Nothing from one "universe" touches anything from the other, so there's no conflict in your mind. In your mind there is no conflict because it's all taking place in your mind. In the real world, No, you haven't deployed any convincing argument that your picture is consistent, and you have shown basic misunderstanding of the connection between symmetries and conservation laws, misunderstanding of the difference between passive transformations (sheer re-labelings of coordinates) and active ones (actual changes on the system) etc. And, as I told you, you've been here before with what appears to be the same crazy idea: https://www.scienceforums.net/search/?q=muruep&quick=1&type=core_members

No, that's no what I'm asking. Again: Which was in answer to: (My emphasis.) The vacuum being stable or not is one thing. Signals, electromagnetic or otherwise (meaning radiation, not the interactions themselves) leaving the horizon of a BH is a completely different matter. (I see now though that you have chosen to completely ignore Hawking radiation as a possibility.) You haven't proven that the vacuum under your scenario is stable or, I don't know, could lead to a runaway process near the horizon, or an explosion into gamma rays, or rotating cups of tea, or... You see? You haven't proven anything (or shown a robust argument using reliable principles of physics) why what you say should be right. You hope the vacuum would be stable under your scenario. That much is clear to me. But, as they say, hope is not a strategy.

Ok. The only thing I'd say is that it's not unique. As to the sequence of 0s & 1s actually representing the expansion of the ordering in base 2, I had never noticed. I suppose that's why people in quantum computing chose that convention for the product. It's nice to know. And... wow! You're on a QC binge!

I understand they're asking for a projector that embodies this measurement. You should not use a difference of projectors, as a difference of projectors is not a projector, even though the addition of two mutuallly orthogonal projectors is: \[ \left(P+P^{\perp}\right)^{2}=P^{2}+\left(P^{\perp}\right)^{2}+PP^{\perp}+P^{\perp}P=P+P^{\perp} \] But, \[ \left(P-P^{\perp}\right)^{2}=P^{2}+\left(P^{\perp}\right)^{2}+PP^{\perp}+P^{\perp}P=P+P^{\perp} \] But the operator you've written does commute with both \( P \) and \( P^{\perp} \), so it could implement such measurement, IMO. People who work in QC are very old-school though, so I assume, for them, nothing but a projector will do. Also, I don't remember the ordering of the computational basis. Clearly, \[ \left|0\right\rangle :=\left|000\right\rangle \] and, \[ \left|7\right\rangle :=\left|111\right\rangle \] But the rest I forget. Sorry, there should be a couple of minus signs in the last equality, but it doesn't matter, as they are mutually orthogonal. That's why I was sloppy.

Exactly. As you see, Dirac's notation can handle it by using extra round braces. But it kind of becomes more awkward --or perhaps just uglier-- for those cases. And it's a huge inconvenience when you want to talk about time inversion operation, because it's not even linear, but antilinear, so it has a complex conjugation in its guts that "interferes" with the complex conjugation involved in passing from ket to bra.

Yes. On the one hand stating more clearly that what you're saying is that every \( \left| q \right\rangle) \) in the space can be written as \( \left| v \right\rangle + \left| v_{\perp} \right\rangle \) (what I've written as \( v=v_{\parallel}+v_{\perp} \) ), and on the other hand, noticing that taking the action of \( P \) from the second to the first factor in the scalar product is not simply "looking at it as acting on its left", but also complex conjugating. In matrix notation: q⁺Pr=(P⁺q)⁺r where "+" as a superindex means complex conjugate and transpose. It's only when P⁺=P that you can write, q⁺P=(P⁺q)⁺=(Pq)⁺ Otherwise, you can just say that q⁺P=(P⁺q)⁺, which is always true, because it's a definition. This you cannot say clearly in Dirac notation, as Dirac notation automatically assumes that any operator "sandwiched" between both factors is Hermitian. Otherwise the notation is ambiguous. That's why someone as careful as Weinberg sometimes drops it in his proofs. Weinberg proved most everything he said, so he was very careful about these questions. The point of the exercise being that the eigenvalues of any projector are always real. Something that is not true for any linear operator. Another way of seeing it is by investigating the eigenvalue equation. As P²=P, the eigenvalues must satisfy p²=p, so either p=0 or 1.

You're most welcome. The rest of the expansions seem right to me. As to proof 4.3, I see what you're doing there, and it's correct too, AFAICS. The only glitch is for these kind of proofs is that it's perhaps best to drop Dirac's notation, because it kind of stands in the way of distinguishing the vector, the operator, and the action of the inner product more clearly, so you wouldn't have to use the --somewhat awkward, IMO-- double parenthesis on your last line. So, for example, I would write something like, for every \( w \), \( v \) in \( \mathscr{H} \) (the Hilbert space of states), \[ \left( w, P\right) = \left( P^{\dagger}w,v \right) \] (That is just a definition of \( P^{\dagger} \), of course) I would also write, \[ v=v_{perp}+v_{parallel} \] etc, with, \[ Pv_{perp}=0 \] \[ Pv_{parallel}=v_{parallel} \] for an arbitrary vector \( v \). And then, as you say, \[ \left( w, Pv \right) = \left( w,w_{\parallel} \right) = \left( w_{\parallel}, v_{\parallel} \right) = \left( w_{\parallel}, v \right) = \left( Pw, v \right) \] So indeed \( P^{\dagger}=P. To me, it's a bit more transparent with this notation, but I understood what you meant, and if you think about it we're saying the same thing. The devil is in the details, as they say. In infinite dimension one would have to be much more careful than this, but I don't have the chops for it. 😊 Ok. Something got messed up in the LaTeX rendering, I'm afraid... I hope you can see what it is. I always have to be very careful that my text is not interpreted as rich text at some point (eg, when the editor refreshes) so the rendering is messed up. It might have to do with the software at my end. I dunno.