How much Avogadrian mass is involved in this "you can measure"? Isn't there a problem with Newtonian G and nuclear I?
You're embarrassing yourself again. Momentum is not a ensemble variable. Individual particles have momentum. Even massless photons have momentum.
G is an ensemble variable, or can you (please) tell me how many protons are in 10^-23 moles? I'd like to know within one unit of Planck mass, thanks.
Oh, come on! You're not even trying anymore. "G is an ensemble variable," - Eh? Last I checked, G was a constant, and works fine in non-statistical systems... "how many protons are in 10^-23 moles?" - You need define a specific substance that you have "10^-23 moles" of, for the question to make sense. \(10^{-23}\) moles of kittens will have rather more protons than \(10^{-23}\) moles of \(C_{12}\). "I'd like to know within one unit of Planck mass, thanks." - You'd like to know a dimensionless quantity (the number of protons) within one unit of Planck mass? Dimensonal analysis red flag, here.
What compels you spout such nonsense? Do you know the difference between variables and physical constants? Between a mechanical variable and an ensemble variable? G is a physical constant. In terms of the Planck mass, \(G=\frac{{m_P}^2}{\hbar c}\). Mole is a count. Your question did not specify moles of what. What you asked is akin to asking how many protons are in one million. Even worse, 10[sup]-23[/sup] moles is not an integer; it is about 6.02214. With that, 10[sup]-23[/sup] moles of protons is 6.02214 protons while 10[sup]-23[/sup] moles of hydrogen molecules contains 12.04428 protons, neither of which makes a lick of sense (what is 0.02214 protons?) You apparently used "Planck mass" thinking it represents a tiny unit of mass, similar to how Planck length and Planck time are very tiny units of length and time. The Planck mass, being about 21.76 micrograms, is not all that tiny. The mass of one proton is a lot, lot less than one Planck mass. It is about 7.6844×10[sup]-20[/sup] Planck mass.
You want to know a number to within an accuracy of a unit of mass? Does this exchange make sense : Q: How many fingers am I holding up? A : 3, to within 0.5kilograms of error Alternatively, if you mean 10^-23 moles, to within a Planck mass then its easy, none! The nearest Planck mass to the mass of 10^-23 moles of protons is zero. As pointed out, G is a function of the Planck mass and other common units in particle physics. It's a function of the number and size of space-time dimensions but its by no means an ensemble variable.
...when mass is sufficiently scaled as atoms, there are 10^23 of these in a N_A gas of hydrogen atoms, according to Avogadro. 1 x atom of hydrogen = 10^-23 mole of hydrogen. Planck mass in grams is based on mass at Avogadro's scale. G, when you get close to 0K, has a different value, G is constant if you are Newtonian. So, far we haven't really been able to connect mass and momentum at Newton's scale to what 1 atom sees. So what does a mole of protons weigh, or a mole of h do, or an 'inverse mole' = 1 proton or h J.s?
Is that what a proton is? I wasn't expecting an answer that made sense, so your response is entirely appropriate; you have rejected the question and started babbling.
Question, Vkothii: What does any of the above response have to do with G being a physical constant? Other than being non-responsive, this response starts with gibberish ("when mass is sufficiently scaled as atoms"). The middle ("there are 10^23 of these in a N_A gas of hydrogen atoms") would be a tautology if you had the numbers right. The correct number is 6.02214×10[sup]23[/sup], not just 10[sup]23[/sup]. The end, at least, is almost right. Avogadro's number was named after Avogadro's, but posthumously. Wrong. - Avogadro's number is 6.02214×10[sup]23[/sup], not "10^23". - The Planck mass is defined as \(m_P = \sqrt{\frac{\hbar c}G}\). Note well: N[sub]A[/sub] does not appear anywhere in that definition. Reference needed, please. Wrong. Why in the world do you think we build particle accelerators?
That's great; so it has a scale that's also a real number, apart from 23 zeros? So if I ask you to weigh out 6.02214 x 10^-23 moles of hydrogen, you know what to do? But G does, I can see it in the equation, what's Newton's constant doing in an equation that defines a smallest unit of mass? How well does this measure match up with an electrons mass? Or the mass of a quark? "G has a different value", because Newtonian G fails to measure very small amounts of matter, you need another constant but it appears to vary - the gravitational interaction between fundamental amounts of m. Therefore, G does not appear until the number n, of atoms or molecules is scaled to a significant fraction of N_A; it cannot apply to atoms scaled with small n, or, as n -> 1 G is undefined in Newtonian mechanics (i.e. at atomic scales, it doesn't fit). In fact, QM treats mass as a distributed momentum, and with nonlocality; an electron can quantum-mechanically transport its momentum through a barrier, or tunnel through it
That sentence is not coherent. Avagadro's number is defined to be the number of carbon atoms in 12 grams of carbon. It's the 'Bakers dozen' of large numbers. No, it isn't. The Planck mass relates to gravity. In particle physics you work in units such that energy and length have inverse units of one another, which means that to probe small distances you need lots of energy, hence why particle accelerators get bigger yet the things they probe get smaller. At low energies gravity is not relevant to quantum processes. However, when you get to small enough scales (the Planck scale) quantum gravity is important. The energy scale associated with the Planck length is the Planck mass. Unlike the length and time Planck units, the Planck mass is big. Very big. At no point have I had to make references to Avagadro's number in that. N_A is an arbitrary quantity, it has no 'natural' justification. We defined a metre by the size of the Earth. We define a gram by a volume of water, whose volume relates to a metre. We pick carbon for no particular reason and we define N_a by how many carbon atoms in a particular mass of carbon. If the Earth were a different size, so would N_a be. The Planck mass is defined by the fundamental properties of space and time. Before mouthing off BS on a physics forum, don't you think its a wise idea to check your claims? Come on now, you know you're either making that up or are dubiously remembering something badly, yet you try to tell people who you know do physics or maths things about their area of research. Ever heard the phrase 'Straight from the horse's mouth'? Stop telling the horse what it said.
IF you use Newton's G, to define Planck mass, it has units of Newtonian space and time. It does not have units of quantized space or time. Does it? Before you start mouthing off about particle physics and Planck mass, shouldn't you check what dimensions you're using, if you want to be sure "you haven't mentioned Avogadro"?
Firstly, I generally work in units where G, c and h-bar are 1, thus making the Planck mass 1. G appears in relativity too, as seen in the Einstein Field Equations so its not restricting analysis to Newtonian space-time at all. Further more, the relationship between the gravitational coupling, number of dimensions and signature of the metric is well known and is discussed in any book on string theory. Its units depend on the dimensionality and signature of your metric, because you need for the action to be a dimensionless quantity. It's simple power counting of units of energy which leads to the very straight forward conclusion that gravity in 3+1 dimensions is a non-renormalisable theory, yet gravity in 2+1 dimensions is renormalisable. The units of G in usual space-time formulations can be found by using \(F = G\frac{mm'}{r^{2}}\). Sure, this is a Newtonian equation but its the same G you see in \(G_{ab} = \frac{8\piG}{c^{4}}T_{ab}\) for the EFE. Also, a length is still a length even when you quantise it. Things like quantised momentum still carry units of momentum (ie [M][L]/[T]), it just means the momentum is some integer number times by some small unit of momentum. This week I've been staring at about 4 different textbooks in relation to U duality, a nonperturbative symmetry of string theory. To understand U duality you need to be able to grasp non-perturbative transformations on tori compact spaces, both in 10D Type II models and 11D M theory (and 12D F theory, if you're so inclined). U duality involves intertwining S and T dualities, which alter the size and structure of the compact dimensions which in turn alters the value of the Planck mass. I actually do research into models where the Planck mass is a dynamical quantity. And the last time I used Avagadros number was 8 years ago, in high school chemistry. So yes, I'm pretty f'ing sure \(N_{A}\) has sweet F A to do with the Planck mass. Of course, if you want to get into the deep and dark details of the symmetries various string theory formulations possess which involve the Planck scale I'm happy to do so. Something tells me that if you have trouble working out intersections of circles and curves, you're really going to struggle with Planck scale physics. Or are you going to continue telling me what my research involves?
Where in the world did you get the idea that the Planck mass is the smallest unit of mass? 21.76 micrograms is a factor of 10[sup]19[/sup] times the proton mass. Once again, citation needed.
Where did you get the idea that "a smallest unit" = "the smallest unit"?? BTW Avogadro's number "has sweet FA" to do with Planck mass, according to the resident physics expert, so please stop using Newton's G to define it. (which is found - N_A, that is - by using mass at Avogadro's scale of die-screte particles we sometimes call "atoms") Thank you.
The mass of a typical bacteria is several orders of magnitude smaller than the Planck mass. Dust particles and even some insects mass less than a Planck mass. The Planck mass is not the smallest possible mass. End of story. Where did you get the idea that it was, and why are you sticking with it?
Where did you get the idea that I got the idea that Planck mass, in terms of Newton's constant G, is the smallest possible mass? ...that means, (here's the hint) where in this thread did I post anything that suggests the conclusion you draw with: "you got the idea that", etc. Can you point this out?
I got it from this poorly written sentence: Your use of "a smallest" is grammatically incorrect. "Smallest" is a superlative, which means it must be used with a definite article rather than an indefinite article.
I think you are wrong DH, beg my pardon. After all, consider the mathematical following. As ''superlative'' as it is, Vkothi hs a good point, under the analysis of the relationship between theb Planck Mass and the Planck Time. It can be seen as a small mass, when these following relationships are considered: \(d_{pi}=\frac{c^5}{hG^2}\) where the small mass is related to the ''infinitesimal time of Planck's Constant \(t_{pi}=\sqrt{\frac{Gh}{c^5}}\)
