On Monday, October 19, 2020, Bruce Kellett <[email protected]> wrote:
> On Mon, Oct 19, 2020 at 11:08 PM Jason Resch <[email protected]> wrote: > >> On Sun, Oct 18, 2020, 11:55 PM Bruce Kellett <[email protected]> >> wrote: >> >>> On Mon, Oct 19, 2020 at 3:31 PM Jason Resch <[email protected]> >>> wrote: >>> >>>> On Sun, Oct 18, 2020, 10:33 PM Bruce Kellett <[email protected]> >>>> wrote: >>>> >>>>> On Mon, Oct 19, 2020 at 1:09 PM Jason Resch <[email protected]> >>>>> wrote: >>>>> >>>>>> On Sun, Oct 18, 2020, 8:47 PM Bruce Kellett <[email protected]> >>>>>> wrote: >>>>>> >>>>>>> >>>>>>> Remember that entropy is basically related to the volume of phase >>>>>>> space, not of ordinary space. And phase space relates to the number of >>>>>>> particles (hence mass-energy). Spatial volume is essentially irrelevant >>>>>>> for >>>>>>> volumes greater than that of the corresponding black hole. >>>>>>> >>>>>> >>>>>> No. Consider an infinite length. With a single atom you can encode >>>>>> infinite information through placement of the atom along that length. >>>>>> This >>>>>> is with finite mass energy, but unrestricted spatial volume. >>>>>> >>>>> >>>>> That does not encode infinite information. There is, after all, only >>>>> one particle, and it can have only one position. If you want to encode >>>>> more >>>>> information, you need more particles. You might need an infinite number of >>>>> bits to encode the position of one particle as a real number, but the >>>>> single particle cannot encode this. >>>>> >>>> >>>> This is plainly false. Every 1 mile distance that particle is placed >>>> along the line encodes a unique number. Travel up to 2^N miles and you can >>>> encode N bits. With infinite range there's no upper bound. >>>> >>> >>> A single particle can be in only one place, and encode on ly one bit. >>> >> >> If I were building a hard drive and could only write a fixed number of >> crosses, let's say 5 crossed, on a LxL grid, the number of possible >> combinations would be (L^2 choose 5). >> >> So for L = 3 that is (9 choose 5) = 126. >> Which means I can encode Log2(126) = 6.97 bits >> >> For L = 4 that is (16 choose 5) = 4368. >> Which means I can encode Log2(4368) = 12.09 bits >> >> My total number of crosses (let's say I use a single atom to represent >> each, is 5 in both cases). It is constant. But if I have more space, I have >> more ways of arranging them, and can use them to encode more bits, or >> conversely it takes more information to describe the system. >> >> Why does this analogy not extend to a quantum system of particles in >> larger or smaller regions of space? >> > > > I think you are forgetting the physical nature of your atoms and your > grid. Because information is physical, it requires mass-energy to encode. > Look again at the Bekenstein bound you have used: > With enough cells in a grid, you could store an entire hard drives worth of data by placing a single atom placed in the right cell. > > S <= 2pi RE > > That does imply that if you increase the volume, you can fit in a greater > entropy. But it does not mean that increasing the volume for fixed > mass-energy automatically increases the entropy. > I agree it's not automatic. Is it correct to say it increases the ceiling of distinguishable quantum states the system might be in? > > In order to increase the entropy to that allowed in the larger volume, you > have to also increase the mass-energy. > Doesn't this contradict what you just said? Do you agree that the bound is the same for 1 Kg in a 2 meter sphere as for 2 Kg in a 1 meter sphere? > > Or think of a grid of naughts and crosses, with a larger grid but fixed >>>> number of crosses, the number of possible combinations for drawing a fixed >>>> number of crosses still increases with more spaces to place them. >>>> >>> >>> Each combination encodes only one combination. >>> >> >> >> More unique combinations mean more states a system can possibly be in, >> meaning it takes more information to uniquely define the state the system >> can be in, or alternatively the more information the system may encode. >> >> >>> An arbitrary volume can only hold a limited amount of energy, or >>>>> entropy, as given by the Bekenstein bound. >>>>> >>>> >>>> Energy isn't the same thing as entropy. >>>> >>> >>> Bekenstein relates them. >>> >>>> But the maximum entropy for a particular mass is given when that mass >>>>> forms a black hole -- which saturates the Bekenstein bound. >>>>> >>>> >>>> The bound is always satisfied. Black holes just reach the maximum of >>>> the bound at a given VOLUME. >>>> >>> >>> I said saturated, not 'satisfied'. The bound gives the maximum possible >>> enclosed mass for a given volume, or the volume is that for which entropy >>> is maximum for a given mass which saturates the bound. >>> >> >> I think you're thinking of the black hole entropy equation, which is >> related to, but distinct from, the Bekenstein bound. >> > > > The Bekenstein bound as you have used it merely means that the amount of > entropy in a given volume is limited. > > Agree. > > Increasing the volume will allow for greater entropy, > Agree. > > > but the entropy at the bound increases only if the mass is also increased. > > > When you say "at the bound" you are talking about black holes, which is the point of maximum mass and maximum entropy for a given volume. But I have to disagree with your conclusion that adding volume doesn't increase the total entropy of the system. Use my grid example: with placement of the block hole in the grid instead of the atom. If the volume is big enough, you can encode more bits through the placement of the hole somewhere in the grid than the Bekenstein bound gives for the black hole's total entropy. > Entropy (information) is a physical thing, and coding or storing > information requires energy. > It takes energy to encode information, yes, but that is irrelevant to the bound. It's the difference between powering up a hard drive or CD burner to write the information once, and then later considering the hard drive or CD with the data on it as an isolated system. You can ignore the input energy used to write when considering the bound for the powered down storage media. > > I realize that it is difficult to say this clearly and precisely, because > in general statistical physics, the entropy is so far below the maximum > possible in the considered volume, that the Bekenstein bound is largely > irrelevant. It becomes an issue only if you look at situations, such as > black holes, where the bound is in fact saturated, and you consider > increasing either the mass or the volume. It is then that the fact that > the bound depends on their interdependence becomes important. > > > >>>> Increasing the volume does not increase the actual entropy unless you >>>>> simultaneously increase the mass. >>>>> >>>> >>>> You keep saying this but don't provide any justification or sources. I >>>> implore you to read the wikipedia article and if it is wrong, please point >>>> me to a source with the right/corrected equation. >>>> >>> >>> The justification is that it is impossible to increase the mass of a >>> black hole without at the same time increasing its radius (volume). For a >>> black hole, the radius is 2M, in natural units. So the mass and radius are >>> directly related. Any greater volume for the same mass does not saturate >>> the bound. >>> >> >> Forget about saturating the bound, that's not the point. Saturating the >> bound requires maximizing entropy for a given volume. On that we agree. >> >> My point is that the bound implies that a larger amount of volume, for >> fixed energy, allows for higher entropy. >> > > That is correct, provided you realize that increasing the entropy beyond a > saturated bound requires the input of more mass-energy. > I don't see how your previous sentence can be interpreted consistently. It seems like the seconds clause contradicts the first. > > Put your black hole in a larger volume and now the black hole has a very >> well defined position in that volume, which is more information than you >> had before. >> >> It's a generally accepted in computer science that a turing machine >> allowed to use infinite space could store infinite information, even with >> fixed total mass/energy. >> > > > You do not have massless tapes on which to store your infinite > information. So this would appear to be nonsensical. A Turing machine in a > physical object, and it is subject to the laws of physics. > The tape can be empty space, while the information can be represented by placement of one particle in a definite location within that infinite space. Imagine an infinite grid. By marking a single X in a single cell of that grid you can encode any number. Therefore you can encode any data of any length in a system of finite energy. The grid can be empty space. > > As explained on that page, the bound is not limited to black holes, it >>>> says something more general which relates entropy bounds to the product of >>>> spherical radius and mass. >>>> >>> >>> The entropy bound you are talking about is >>> >>> S <= 2pi RE. >>> >>> This is saturated when the radius and energy are related as for a black >>> hole: >>> >>> R = 2M, for which S = 4pi M^2. >>> >>> Nothing mysterious here. I was talking about maximum possible entropy, >>> which occurs when the bound is saturated, as for a black hole. >>> >>> That is really all that the Bekenstein bound says. It is a bound, after >>> all, and has information about the entropy only when that bound is >>> saturated. >>> >> >> If the bound strictly depends on energy, why is R included in the >> formulation? >> > > > That specifies the volume within which the energy is enclosed. But > increasing the volume does not, of itself, increase the entropy. The > maximum entropy for a fixed mass-energy is fixed by the surface area of a > black hole of radius R = 2M. > > That's false. The maximum entropy is NOT fixed unless both the mass-energy AND the volume are fixed. If the bound were as you say, determined solely by mass-energy, then R would not appear in the equation as it does. > > > >> So for a fixed amount of mass, the entropy is maximized when that mass is >>> in the form of a black hole. Increasing the volume surrounding the BH makes >>> no difference to the entropy maximum for that mass. >>> >> >> For the system as a whole it does. Now the black hole has coordinates in >> a larger volume which did not exist before, and must be included in any >> description of that system. >> >> A non-collapsed relativistic gas sits right on the edge of becoming a >> black hole and satisfying the bound. Consider such a relativistic gas >> confined to a 1 meter volume. Now considering that gas is given more space >> to occupy, it is placed in a sphere of 1 light-year. >> >> Are there not now many more degrees of freedom possible for that same >> mass energy in a 1 light-year space than when it was confined to 1 meter? >> Are not more bits and precision required to specify the coordinates of each >> particle? >> > > > Putting a black hole in a bigger volume does not increase the entropy of > that black hole. Specifying coordinates for the constituents of the BH is > either irrelevant, or requires additional mass. > See my grid example. No additional mass is needed for the black hole to occupy a certain position in the grid. If the grid has is 10^10^100 cells, then the location of the hole provides at least 10^100 bits of information. This is more information/entropy than in even a galactic mass black hole. > > The upshot of all of this is that the expansion of space in a > cosmology does not increase the maximum possible entropy. The maximum > entropy is set by the amount of mass-energy in the cosmology, and that does > not increase with the expansion. > Jason > > Bruce > > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > To view this discussion on the web visit https://groups.google.com/d/ > msgid/everything-list/CAFxXSLTUaqsbXwn%3D52hWEK_Z__ > Zw-VFKm7znHvS0_oWMTbAYjw%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAFxXSLTUaqsbXwn%3D52hWEK_Z__Zw-VFKm7znHvS0_oWMTbAYjw%40mail.gmail.com?utm_medium=email&utm_source=footer> > . > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. 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