Timeline for More (Dense) Information in Bulk than on Surface?
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24 events
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| S Jul 12, 2017 at 9:00 | history | bounty ended | CommunityBot | ||
| S Jul 12, 2017 at 9:00 | history | notice removed | user87745 | ||
| Jul 7, 2017 at 16:01 | comment | added | tparker | @valerio92 Yes, of course. But if you have an area $A$ and a volume $V$, there's no context in which the question of whether the numerical value of $A$ in square meters is greater than or less than the numerical value of $V$ in cubic meters could be physically interesting. But since the Planck length sets the natural length scale for storing one bit of entropy in a black hole, it is physically interesting to ask whether a volume $V$ contains more or fewer Planck volumes than an area $A$ contains Planck areas. At the end of the day, you're always really comparing entropies, so it makes sense. | |
| Jul 7, 2017 at 15:28 | comment | added | valerio | @tparker To be more specific, you can for sure write an expression as $V+A$ (volume plus area) but this just mean that appropriate dimensional factors are implied, like when you write $x^2+t^2$ in SR implying a $c$ factor. | |
| Jul 7, 2017 at 15:13 | comment | added | valerio | @tparker I disagree. What would be so special about the Planck units to allow us to do something that we cannot do with any other unit system? There is nothing fundamental about the Planck units: they are just one of the many ways to cook up some units by putting together fundamental constants. In atomic physics, we can take the Bohr radius as fundamental lenght scale, but this doesn't mean that we can add surfaces and volume together without appropriate dimensional prefactors. | |
| Jul 7, 2017 at 14:22 | comment | added | tparker | @valerio92 No, Rexcirus is wrong. Your analogy is invalid, because the meter is not a natural universal scale like the speed of light, so indeed there's never any natural reason to add $L / (1\text{ m}) + A / (1\text{ m}^2)$ for any physical quantities $L$ and $A$. But e.g. in SR it is very natural to add different powers of velocity - e.g. in the Lorentz factor $\gamma$ or the velocity-addition formulas. In the context of BH entropy, the Planck length is the natural length scale, so it's perfectly valid to compare areas and volumes. | |
| Jul 7, 2017 at 14:13 | comment | added | Rexcirus | You cannot add t to x. You can add c*t to x. Then if you go in a system in which c=1 you will write t+x, but your still adding quantities with the same dimension. | |
| Jul 7, 2017 at 12:13 | history | edited | user87745 | CC BY-SA 3.0 |
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| Jul 7, 2017 at 12:00 | history | edited | user87745 | CC BY-SA 3.0 |
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| Jul 7, 2017 at 11:53 | comment | added | user87745 | @valerio92 I think your analogy is not appropriate. In relativity, we can very well add $t$ with $x$ and so on. More relevantly, $v+v^2$ is a valid thing to do in relativity. And all that happens owing to setting $c=1$ - not just setting $c$ as the reference quantity for speed but setting $c$ to $1$ - a dimensionless constant. Similarly, if we set $l_P=1$, we can very well have $A/V$ dimensionless. | |
| Jul 7, 2017 at 11:44 | comment | added | valerio | @Rexcirus is right. The dimensions are still there, you just don't see them. In the international system for example the meter is the unit of length, so we have $m=1$. Even if we avoid writing the units explicitly, like it is usually done when working with natural units, we still cannot compare surfaces to volumes, because the first have units $m^2$, while the second have units $m^3$. | |
| Jul 7, 2017 at 9:46 | comment | added | user87745 | @Rexcirus I think setting $l_P=1$ doesn't mean setting just its numerical value $1$. It means making it dimensionless. Just like when we set $c=1$ it doesn't mean that only its numerical value is $1$, rather we mean that it is dimensionless and thus, space and time have the same dimensions. | |
| Jul 7, 2017 at 9:35 | comment | added | Rexcirus | The fact that you are in a unit system in which the numerical value of the planck length is 1 doesn't mean that a volume and an area have the same dimensions. And of course physical statements cannot depend on the choice of the units that you make. | |
| Jul 7, 2017 at 0:05 | answer | added | valerio | timeline score: 5 | |
| Jul 6, 2017 at 22:20 | answer | added | tparker | timeline score: -1 | |
| Jul 6, 2017 at 21:49 | comment | added | tparker | @Rexcirus That's wrong. "The units in which $S = (1/4) A$" are units in which Boltzmann's constant and the Planck length have both been set to $1$. In these units, $S/V$ is also dimensionless. | |
| Jul 5, 2017 at 11:39 | history | tweeted | twitter.com/StackPhysics/status/882564608254504960 | ||
| S Jul 5, 2017 at 11:28 | history | bounty started | CommunityBot | ||
| S Jul 5, 2017 at 11:28 | history | notice added | user87745 | Draw attention | |
| Jun 29, 2017 at 12:00 | comment | added | Rexcirus | The ratio between entropy and area is a number (using the units in which S=1/4 A), while S/V has dimensions. Unless you put some mass dimension into play (I would say the Planck Mass) the comparison in meaningless. But considering a finite constant with the dimension of mass, your question seems to apply anyway. Very interesting question, I'm wondering on this stuff from a while. | |
| Jun 29, 2017 at 5:56 | history | edited | Qmechanic♦ |
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| Jun 29, 2017 at 5:16 | history | edited | user87745 | CC BY-SA 3.0 |
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| Jun 28, 2017 at 21:23 | history | edited | valerio | CC BY-SA 3.0 |
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| Jun 28, 2017 at 21:06 | history | asked | user87745 | CC BY-SA 3.0 |