Estimating the number of atoms in the observable universe
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- Опубліковано 12 чер 2024
- This video looks at a simple method for estimating the number of atoms in the observable universe. The first step involves estimating the number of atoms in a typical star. The second step involves estimating the number of stars in a typical galaxy. The third step involves estimating the number of galaxies in the observable universe.
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I knew intuitively it would be over 500, but intuition can only take you so far.
Very under-rated comment! (at least so far)
@@chrissekely Only 22 likes at the time of writing, but intuitively I'd suggest that will probably grow over time to something in the region of 10^81 likes
Robin Williams, indeed ! At this rate we'll hardly even need time to exceed 1/10th of infinite time.
Oh, cmon, it's obvious it had to be over 9000!
My guess was around 3 tbh...
i'd like to break one myth, that is implied here: the spaces between arms of spiral galaxies are not empty. actually, there are as many stars in between the arms as in the arms themselves, and most likely the sun resides in the space between such arms. spiral arms are regions where young bright stars reside, and that makes them brighter, and more visible on the background of the whole galaxy. it's like sunspots - these spots are neither dark nor cold, they are actually very bright and very hot, but they are surrounded by even more hot surface of sun, and because of the contrast we see them dark.
There's also a lot of dust and debris in the gaps of the spiral arms that block light.
Yes, but in this over-simplified calculation, the galaxy is treated as a homogeneous disc of stars, so whether or not there are stars or dust in the space between the arms does not change the outcome.
There are a ton of myths and inaccuracies. Why is sun a typical star? It's likely not! It's pretty big. It's a single star system, binary systems are more common. Center of any galaxy is likely filled with sea of blackholes. Saying that most of galaxy mass comes from stars is an assumption. The same assumption is to say that most of observable universe's mass comes from galaxies. We simply don't know.
@Pilgrim You sound like me, ie The typical armchair astrophysicist. ie know pretty much nothing. But then there is the 000...percent chance you are right.
It's an estimation
Actually you could skip the first step altogether. The mass of the Sun is not required. Start directly with step 2 and at 11:53 use the mass of Hydrogen atom to calculate the number of atoms in galaxy. Then use that value (~2E+68) and multiply by number of galaxies from step 3 to get the desired number.
Yes, your are totally right. I included this step just as a point of interest, but it is not necessary. Thanks for the feedback
@kim burley Quality humour.
I guess the reason why he couldn't have skipped the Sun is that the relevant laws were discovered observing the Solar system.
Except, there are a lot of other elements in stars and the observable universe, not just hydrogen. Plus, the sun is unique, so using the mass of the sun and saying it's just hydrogen seems just wrong to me to estimate the number of atoms. We would need the average actual number of atoms in stars. Or the average composition and mass of stars. His estimation seem very wrong to me.
ojolafrite90 FINE ! We'll do it the old fashioned way and just COUNT ! Ahhh, ONEee... ahh, TWOoo...ah, THREEeee... ….. < > Ahhh… THREE ! It takes THREE licks to get to the center of a tootsie-pop universe ! NEXT Question ?
I like how you make what might seem inconceivable into something comprehensible using fairly simple math. I'm enjoying your videos quite a bit. Thanks
"Simple"
@@holyphainesthai286 the math is kind of simple, the logic and the methodology isnt.. not to mention the measurements used
I’m jealous of how good you can draw little diagrams!!
That's a very good estimate for rough calculation. The ball park is 10^78 to 10^82. This is perfectly in between. Excellent.
That is the same to say that the average person height was estimated between 1 mm and 10m. In other words, we have no idea.
@@geppettocollodi8945 You're right and wrong. The order of magnitude difference is nearly the same in both usecases (height vs atom count). However, the true average mangnitude of getting the value close to 10^78 vs 10^1 is a massive difference. The fact that the average on both side is order of magnitiude of 2 shows how simple and easy this calculation is to do one you have some basic data for the observable universe. I would aruge the exact opposite to you and say we have a very good idea.
what I found interesting is the amount of possible bitcoin private key addresses is about 10^77. So one could say that the odds of guessing someones private key for their bitcoin wallet close to the same as guessing a specific atom in the universe that someone picked randomly.
@@Oblivion_94 Yep, that's a good analogy.
After watching this video, I want you to tell another big number. And we are not able to believe this.
(In Chess, computer has estimated Standard, Possible, Legal 40 move games in Chess are 2x10^213 games. ) Really I enjoyed , very beautiful video.
So we could have 10^133 unique chess games on each atom in the universe.
Then there's the number of potential unique humans, using DNA as your multiplicative base. I estimate each one of every possible 40 move game of chess could be played by the earth's present population, with human's left in the queue. Old UK duffer here :)
I woke up this morning and before having my cornflakes I thought to myself , I wonder how many atoms are there in the universe, I must find out
Harry Bond : haha!! - ditto that for me.
Excellent. Such clarity and wonderful example of scientific thinking.
Glad you enjoyed it
Great ! I examined many different sources on the Internet and found that the no. of atoms in the observable Universe range between 10^78 to 10^82. And what you calculated is quite accurate... This made me so happy ! 😄
One thing missed would be the mass in the intergalactic fabric.
While it is very very thinly spread (like a lot less dense than the "atmosphere" of the moon) the sheer volume of it makes up for it - making up roughly half of all matter.
then the way of arriving at the number of galaxies in the observable universe is also rather strange - it assumes 2 things which both are not true:
1 - that the photo was able to resolve all galaxies that reside in that part of space between us and the edge of the observable universe.
2 - uniform density - the milky way is part of a local supercluster, the number of stars observable heavily depends on the direction you are looking.
So if the photo can not resolve to the edge and also looks into the direction of a large void it will show several orders of magnitude fewer stars than when it can reach to the edge and looks through superclusters.
If dark matter would consist of atoms we would have to add another order of magnitude.
@@Skandalos True that, but seems unlikely that they would form "large" objects and still have such low interaction with the rest.
@@ABaumstumpf 'Tis a weird thing. Has gravity but doesnt lump together in large objects like matter does.
@@Skandalos Try applying the idea that our entire existence within the Cosmos is a singular multiple dimensional hyper volumetric plane and that this is just one parallel plane of infinitely many other planes... I say that the Universe or Cosmos is a Perfect Symmetrical Singularity that is a Perfect Fractal: Spheres within spheres, spirals within spirals, etc. etc. etc. ...
This is just supposed to be a very rough estimate
Universe today says it's between 10^78 and 10^82. Wikipedia says it's around 10^80 and that the number is called Eddington number. So, the result is spot on.
I knew the number atoms equal to 10^80 from the Sagan's old series but he didn't tell the how he got the value in the first place. That's nice to see these calculations
10^80 thanks for the quick info
Here is a related calculation.
total dark energy in universe = (energy denisty) x (volume of observable universe) = (7 x10E30 g/cm3) x (4x10E86 cm3)
= 2.8E54 g
total blackhole mass in obs. universe = (Mgalactice center*Ms) x (ngalaxies) = (1E8 solar masses x 1.8E33g/Ms) x (1.5E12 galaxies)
= 2.7E53 g
The galactic center mass is that for Andromeda, not the Milky Way. The total BH mass could be an overstimate if Andromeda is not typcal, or (more likely) an underestimate becasue there are black holes outside of galactic centers (quasars, maybe lots of little one spread around).
Do you think it is curious that the total energy in the form of dark energy is close to the total energy in the form of BH gravity?
When you did your final calculation of atoms in a star, you went a different way than I had initially thought of. I thought you were going to express the mass in moles and then multiplying by the avagadro constant.
He used Avogadro's constant without showing it in making use of the mass of a single hydrogen atom.
I did the same as you also used atomic weight for hydrogen and heilium
Can we sit back and appreciate the guy's handwriting!
This is really high-quality stuff. I love it.
This is great
Thanks! Glad you liked it
Except, there are a lot of other elements in stars and the observable universe, not just hydrogen. Plus, the sun is unique, so using the mass of the sun and saying it's just hydrogen seems just wrong to me to estimate the number of atoms. We would need the average actual number of atoms in stars. Or the average composition and mass of stars. His estimation seem very wrong to me.
We can call it an estimation, but one that is very far from the actual number.
@@jojolafrite90 Well, I thinks it's a good estimation and splitting it up to consider every single type of atoms is very insignificant because of their masses which are practically the same comparing to the result value. If you're so stubborn, you may also add quantum fluctuations, fission energy, etc, etc... But why do it if there's almost no difference?
Enjoyed this video a lot ❤️
Very good work
Hats off Sir..
Keep going..
Thank you. Nicely done.
Thanks!
Good job. Well presented.
Thanks for sharing :-)
Amazing work, well done!
Thanks a lot!
Really excellent calculations. Superb
this is why i love physics/maths. when bored at work i do silly calculatoins like how many times to the sun and back would a teaspoon of water be if you stacked up each molecule.
1 teaspoon=5ml=5g
1mol = 18g
5ml=1/4 mol
1/4 mol = 1.5x10^23
size of a wate molecule = 3 angstroms
1 angstrom = 10^-10m
(1.5x10^23)x(3x10^-10)=4.5x10^13m
distance to the sun = 1.5 x10^8Km=1.5x10^11m
(4.5x10^13)/(1.5x10^11)=300times
This is amazing. Thanks
I calculated the number of tiles 233280000=360(number of degrees in a circle)*60(number of minutes in a degree)*180(number of degrees needed to rotate a circle to make a sphere)*60(number of minutes in a degree)
Nice. Impressed old UK duffer here :)
@@tim40gabby25 now I'm looking back to my solution, I see this one has more errors, but simpler to understand I could say
Amazing video, with minimum number of assumptions,
And it gives you a good perspective on our daily life problems :)
Also it would be fun to estimate the number of atoms we physically encounter during our life span (well, the upper bound is Earth surface layer :).
That would be very wrong. Breathing makes you use way more atoms than there are on the surface of an Earth-sized sphere. In fact, I suspect that the water you drink in your lifetime could coat the Earth as well.
I found 10^80 atoms but in a bit of a weird way. And a way I don't really understand because I am not qualified in physics. Basically the number of planck areas on the event horizon of a black hole is given by the radius of the black hole squared divided by the planck length squared. So for an earth mass black hole with radius 0.00891m the number of planck areas is 3.04x10^65. For the observable universe it is about 10^122 planck areas. What I did was to divide 10^122 by 1.5x10^43 which gives you about 10^80, i.e. the number of atoms in the observale universe.
The 1.5x10^43 came from dividing the planck energy,1956150349 joules, by 1.305x10^-34, which I got from the energy density multiplied by the volume of any space we could be considering - I derivide this from the black body radiation formula for the energy density for the peak frequency output related to a particular temperature (1.305x10^-34 = (32hpi^2)/(3(e^2pi - 1)) ). It's basically the energy density (given in joules per volume) multiplied by the volume (i.e. the volume related to the radius gotten from the wavelength for a particular frequency of peak output). So it's not the total energy you'd get from integrating the black body formula over all wavelengths. So I don't know? May you're not supposed to do that? But what you get is a constant value of 1.305x10^-34 joules for any region of space.
So I basically just divided whatever energy I was considering by 1.304x10^-34. I then took the number of planck areas corresponding to a black hole for whatever mass I was consider and divided by the number I got. For an average temperature of the earth of 5000 Kelvin this resulted in about 10^50 atoms (which is correct), and for the sun with an average temperature of 7.5 million kelvin I got 10^57 atoms (again which is correct). I don't know what the average temperature of a galaxy is so I couldn't apply this method to that. And I don't know why it works for the planck energy to give 10^80 atoms in the obersevable universe. But using the CMB of 2.75 Kelvin I got about 10^111 atoms for the observable universe instead, which is a lot larger than 10^80 atoms.
So there you go I guess.
This is fantastic. I would have been lazy and just looked it up on Wikipedia.
Great job brother
This does not take into account of vast hydrogen gas clouds which haven't formed stars yet. So probably underestimates atom count assuming all other assumptions are fine.
Black holes as well?
@@decam5329 How you can calculate number of atoms inside black hole? When they claim that black holes have infinite density and infinite gravity.
@@oldi184 no idea.
It was a genuine question, not a challenge.
The mass in a black hole is made from what was atoms. Is it not possible to calculate the mass of, say, 1 cm³ of black hole to 1cm³ of star.
As I say, I'm not trolling here, I genuinely don't know.
@@decam5329 Its ok. Anyway I don't believe in black holes neither did Einstein .
The idea that you can have an object with
a. finite size
b. finite mass
c. infinite density
d. infinite gravity
is silly. Density is directly related to mass.
The more something is dense the more it will have mass.
Black holes do not exist. I can recommend you this lecture by S. Crothers. He dismantle the black hole existence using math.
Here
ua-cam.com/video/jINHHXaPrWA/v-deo.html
@@oldi184 you have some good points but it's an overreach to suggest that if an entity with a set of defining attributes has some incoherent attributes then no entity must exist. It's possible to refine "infinite density" and "infinite gravity" while preserving an entity that is heavy, dark and practically inescapable.
Nice video. I like your derivation of the mass of the Sun. I think you should've shown one more calculation - estimate of error in the number of atoms calculation. You used mass of the Sun representing mass of every star in the observable universe. What is the certainty that on average a star's mass is the mass of the Sun? Also, but I'm not sure, whether all of the atoms that went into the formation of black holes in the galaxies should have been counted. But I'm not sure whether adding the number of atoms in all of the black holes is going to significantly add to the humongous number 10^80. My guess: doesn't significantly add to 10^80.
Just to be difficult...these calculations don't include the atoms making up all the dust and gas that hasn't yet coalesced into stars, nor the mass of black holes, nor the mass of all planets in the observable universe....so the total number of atoms in the observable universe has to be significantly higher than the final number arrived at here. None of this detracts, of course, from the very interesting and entertaining estimations and math calculations shown in this video.
A fantastic effort overall and kudos to the wonderful mathematician and narrator!
A million second exposure? Wow! nearly 2 weeks. That's impressive.
You have really nice handwriting
Great vid! Approximation that is palpable to most viewers, with principles that are taught in high school, great execution too! The logic behind the numbers is way more interesting than the numbers itself
Thanks
I enjoyed the video :-)
and the magic of math evolves, and blossoms. I have seen flowers in the universe!!!
Great video and explanation! What about dust within and between galaxies? If we knew the mass of dark matter particles we could use the ratio between the mass of dark matter to visible matter to estimate the total particles of everything... can you do photons in another video? Thanks!
Yes, you are right about the dust. This is only a very rough approximation.
I've added dust found under teenagers beds, and it only increases to 10^81..
Odd thing about UA-cam or viewers. These are such well-made videos. But why are there so few comments in the past year, merely because new videos haven't been made? Aren't others discovering these for the first time?
This math is so beautiful.
isn't there a lot more matter in form of dust than in stars ? something like 40-50 times ?
The model is using the galactic rotation speed to arrive at a mass estimate and under Newtonian rules there's nothing terrible about that. He then divides the galactic mass estimate into an equivalent number of a known quantity, in this case, the sun. There's nothing wrong with that because he could have divided it into any known quantity, number of earths or number of battleships. He arrived at twice the number of equivalent suns as there are total stars in the Milky Way. So between smaller stars, larger stars, dust and everything else, he's probably about order of magnitude correct, that is, roughly within a multiple of times ten or a hundred (10²). When the final number is 10^80, the galactic error is probably not the biggest problem.
The biggest problem is that we don't know how many galaxies there are, we only know what a Hubble sample gave us.
Intergalactic matter is left out but even if it's half of atoms out there, the final answer would only go from 10^80 to 2×10^80. That's a lot but it's not relatively a lot for an approximation based on estimates.
Here's a thought.. in case you're feeling lockdown lonely, imagine one materialises randomly somewhere in the Universe. How far to the nearest object that could be held in the hand or stood upon? Answers on this modern postcard, please :)
[My estimate: 2m lightyears]
Depends if you materialized in a galaxy or not in which case it would be 4 light years going by proximal distance. Your guess is decent though because andromeda is roughly two million light years distant. Interesting thought
@@dt5072 Indeed. Thanks for the reply. If truly random, materialising within a galaxy is remarkably unlikely. It puts things into perspective, after stubbing my toe.. it's wonderful to be able to communicate via the Net, whilst I walk riverside, warmed by our nearby star, in a small Northern English town :)
@@tim40gabby25 haha im in a southern canadian town but our weather is probably similar :) i feel connected to humanity as we are able to communicate and discuss incredible things :) blessings to you random friend
@@dt5072 Blessings accepted. Have a lovely day, DT :)
I'm jealous, check out that neat handwriting
Honestly, at one of the steps, I would have added an overestimate of a couple orders.
Given that I'm sure these numbers are all underestimates by an order or two.
Newtonian orbital mechanics is known to not work on galactic scales. The speed of outer stars is much higher than what can be explained by gravitation of the visible matter in the galaxy. Dark matter is what is used to explain that discrepancy.
So at 12:20, the Mgalaxy you got includes dark matter, which is probably not what you want if you want to count atoms (normal matter).
Of course, this will only skew the result by roughly one order of magnitude - maybe the result should be 10^79, instead of 10^80, so you know, no difference really ;=)
Can you estimate the size of the computer needed to simulate our universe and all of us in it?
For the CPU, that's easy. You need 1 and 1/2 bit to calculate anything and a huge amount of memory. But then you got a problem: You need at least that CPU and memory to simulate that first CPU and so on. If that is too much hardware and you want to watch it in realtime, just look up in the sky.
Would it be wrong to say it takes 365.25 days to complete a full orbit? Because we make up for this extra. .25 days with a leap year
That would change the time estimate by 0.25/365 or about 1/15 of one percent. Take the 4th root of that since a factor of T^2 was used twice, and that would affect the estimation by around 16 percent. Given that the final estimate is a power of 10, it doesn't change the final answer here.
I'm confused by the part in step 3 where you relate the Hubble Deep Field to the SURFACE of a sphere. Surely that 10,000 figure represents the data we capture for galaxies at all DEPTHS, so it's not really a 'tile'? That count is for a 3 dimensional pie slice, and even that's not allowing for galaxies hidden behind others. Therefore, the question is, how many more captures can we make with our telescope? We can scan 360 degrees in one plane, then we have 180 of those to cover the entire sky. So there are 360x180x(60x60) slice = 233x10^6 slice in a full scan of the sky. That's about 60% more than the 147x10^6 figure you use.
You could probably use solid angle concept. Number won't differ much I suppose
60%
+COM Func Your slices overlap, because you have 360 * 180 = 64800 captures, but the area of the sky is just 41253 square degrees. The "tiles" are correct; the method is to project all of the galaxies in your slices onto the plane of the sky, so the depth doesn't matter. Thus each tile contains all of the galaxies in a slice.
@@pseudorandomly Thanks, I see that pie slices overlap now, and that the calculation has to be done as a fraction of surface area. Then the depths don't matter either, because its a data capture for all depths in that tile. This video shows nicely how much the scanned tiles overlap at the 'poles'. ua-cam.com/video/ANNbt_8ewuM/v-deo.html
Imagine all the uneaten pips in a watermelon left on the sliced rind? Boom :)
About the same number as the US debt
8:12 Assuming the mass of the galaxy is concentrated as a small sphere in the center of the galaxy is a major flaw as we are immersed within the mass of the galaxy. There is much mass outside and to the rear of our position in the galaxy. Measuring or calculating the rotational inertia around our position in the galaxy would be very much different (and larger ), than if the mass were actually concentrated within a very small radius at the center of the galaxy, or if we measured or calculated the rotational inertia of the galaxy from the position of the center of the galaxy.
(This is just as if the force of gravity of the earth is measured at some distance below the surface of the earth, even to the point of there being ZERO gravity at the center of the earth, due to the great amount of mass "OVER" your head as opposed to below your feet. The rotational inertia as measured or calculated from the center of mass, would be at a minimum, and would increase substantially as the point of calculation moved from the center of mass out to some radius from the center of mass)
Anyway, I think your estimate of the number of stars in the milky way might be short by 25 percent, a very minuscule error compared to the orders of magnitude of ten to the eightieth. IS BEEN 50 years since I last thought of calculation of rotational moment of inertia... Thanks for tickling my memories ... Yes, it WAS GOOD FOR ME!!! Very good.
I came for the handwriting. And daym boi, I was not disappointed.
Yes
I already cited this result somewhere but can't help wonder about the size of its standard deviation (+/-?) which could be easily some millions of galaxies 😱😁
What value for G did he use? at 7:00. I've tried 6.6742e-11 but i can't get 1.77e30
Never mind I'm such an idiot, I just solved for G and got the answer. G ≈ 9.49486e-5. Does anyone know why he used this value for G? When I use the value for G of 6.67e-11 I get a number 4 orders of magnitude from his
The sun contains a fair amount of helium which has a mass 4 times higher than hydrogen. This may reduce the answer by a small value... much much smaller than the uncertainty of every other estimates in that formula.
We know the rotation speed of stars around the center of galaxy is NOT given by Kepler law. I wonder if that would alter the final answer as the mass of the galaxy assuming Kepler law is probably wrong. But if order of magnitude is what you are after, I think the answer will still be the same.
Counting the points on the paper by numbers?
A good estimate for 5% of the Universe.
We still don't know what dark matter or energy are made of, or if they exist at all.
I'm not the only one that sees that. Good. The sun is unique, and it doesn't count nebulae, big clouds, planets, every other variation of matter density, the fact that atoms vary and there isn't just hydrogen in the universe. It's a very rough estimation, VERY, VERY crude.
@@zwz.zdenek Oh, we know it "exists". We lack understanding of what this means. To me, dark matter could be a manifestation of some "border" effect of living in a multiverse (particularly, the "virtual" mass of our anti-universe made of anti-matter).
Exponentials are deceiving, you know, if you multiplied the number by 100, you would get 10^82, so it's safe to assume that the actual exponent is somewhere in that range
Step 2 assumes galactic rotation rates are governed by the mass of atoms. The work of Vera Rubin and others shows this math doesn’t add up, hence a theory of dark matter. This would make your estimate too high. But your estimate is probably too low for other reasons - interstellar gas, black holes. So, bingo!
I think this shows the number of atoms which the universe must at least contain and not like the averege number of the atoms in the universe. Planets and dust are made up also from atoms. Very intresting video btw.
Step 1.Have neat pleasant handwriting that can be easily referred back to without causing aggravation.
Step 2. Hold on I'm still working on step 1.
It would be even neater if he held the pen correctly
Elegant...
This estimation is a bit off what you'd otherwise expect because you assumed that the milky way is representative of the mass of the average galaxy, when in fact it's about 1,000 times larger. And so the number of atoms in the observable universe should have an upper bound of around 10^77 rather than 10^80.
so it's a number of atoms inside a star in the observable universe right? how bout number of atoms in the planets? or living things? or any other object in the universe? does it counted too?
Any given "night sky" is only half of the observable universe.
Also, I have heard it said that you cannot square a circle... meaning there are no circles whose area can be exactly replicated in a square (and vice versa?). So how can we take a square section of the sky and apply it evenly across a dome or sphere when calculating numbers of galaxies in the observable universe?
Stick some square stamps on a child's globe, then get back to me :)
@@tim40gabby25 lol and I'll show you some islands that got missed
Does this account for nebulae gas, quasars, and the like? Better triple your estimates.
I am not sure how well you can estimate the matter in the universe by assuming it is mostly in stars. Every star came from the dust and gasses that were around it when it formed, that dust and gas exists in every galaxy but also exists between galaxies. It might be more concentrated in a galaxy but the distance between galaxies is so massive that the dust and gas between galaxies could very well outnumber the amount inside of galaxies.
Also the Milky Way Galaxy is on the large size for a galaxy and not a "typical galaxy".
The visible universe is only 5% of all the mass. What about dark matter/energy?
So add a couple of zeros, or 3 for luck :)
This calculation remind me of a small child who says, “When I grow up, I’m gonna be president of the US.” Realizing that you do not know is much less hazardous than thinking you do know when you really do not. The math here is fine, but the assumptions and estimates are quite unscientific, as pointed out by commenters, who makes valid points.
@proteusx "I would recommend to him to learn to hold his pen properly"
Lol that person has a better writing than you can dream of
It's certainly unscientific to not recognise one's assumptions - though I'm assuming you know this - so that approximations are declared to be just that. Old UK duffer here :)
It is amazing, indeed. For another massive number, calculate the number of beings that are related to you from the past. Even assuming just 5 generations in 100 years, the number of your ancestors that had to exist in the last 1000 years is 1,125,899,906,842,624. And, for 2000 years? 1,267,650,600,228,229,401,496,703,205,376 or ~1.2 x 10^30th human beings. And last but not least, the last 100,000 years. 1.4124670321394260368352096670161 x 10^1505. Note that life is estimated to have been around for over a billion years at the very least, with the generations per 100 years sky-rocketing the further you go back.
I like the approach and the method of estimation. The graphs and handwriting were well written and the explanation was very clear. Well done and a great job in presenting it. However, for all of your work on the estimation, I still think there is a 99.99999...% margin of error. I tend to think about what you have discovered is only 0.00000000....1% of it! Why? Because space is not actually empty! This doesn't account for all of the gas clouds, neutron stars or pulsars, black holes, and other entities that we don't even know about yet.
Consider this: what's the area between two vectors with some arbitrary angle theta? The amount of area is infinite because it has only 2 bounds by those vectors and extrudes out forever for there is no 3rd vector that is intersecting them! Also if you are talking about the night sky just from a circle's perspective, there are 360 degrees of rotation with an infinite amount of radii. Now if we consider a sphere we would have 360 x 360 x 360. Why do I multiply this 3 times? One for each 3D axis for the coordinate system which since we perceive vision in 3D, and this would give (360)^3 where 360 is already in units squared. Yet this only accounts for 3 dimensions of space...
So the question then becomes, how many perpendicular surface normals are there on the surface of a sphere? There are an infinite amount of them, yet when they protrude out away from the sphere over some distance x, they are all a delta degree of some arbitrary angle. The angle never changes but the farther away you protrude out let's say 10 light-years away which is nothing in terms of galactic distance nevermind inter-galactic distances only give a small percentage of the extruding area, volume.
Those surface normals that were once only nanometers apart that are now 10 light-years out will have an ever-increasing distance of separation. So if we observe at every one of these infinite points outwards and total up all of that mass even that is still less than 1% because of all of the space in-between those vectors that we didn't account for. And this phenomenon is only in 3D space. We didn't even consider higher levels of dimension. For that, we can take vector notation, geometrical properties, algebraic polynomial forms, and physics (transformations) to show the first few basic spatial dimensions. Then compare that to an arbitrary spatial dimension of size N where N is the number of orthogonal unit vectors within that dimension.
0D = arbitrary point, static starting point, point of reflection, point of rotation, point of symmetry, also the zero vector and the null or empty set! v, {0} ...
1D = addition, scalar, distance, linear transformations specific to translation and scaling within the confines of a line, there are 2 directions as they are opposing in vectors in nature +/-, N/S, E/W, U/D, I/O being positive and negative, north and south, east and west, up and down, in and out, etc. They never have any relative relation to other sets or lines. It is only the set of points within that line defined by y = mx +b. and the vector v. Physics as in translation.
2D = product, angle, rotation, area, and planes. I'll use a square for simplicity reasons: vector . Algebraic or polynomial being the quadratic is y = ax^2 + bx^2 + c and in geometrical terms, the relationship to an area via the square we will only consider y = x^2 where x > 0. As for transformations, they can be all of 1D as well as scaling in multiple dimensions and rotation. Physics as in velocity.
3D = power, multiple angles, surface area, volumes, and multiple planes. I'll use the cube for simplicity: vector . Algebraic polynomial y = ax^3 + bx^2 + cx + d. Geometrical volume we'll consider y = x^3 where x > 0. Physics as in acceleration.
Let's put these into a simple table to see if there is a pattern.
// Dimension - Geometric name - A Polynomial - Vector notation - Derivative - Integral
0D: arbitrary points f(x) = x^0 v f`(x) = 0 F(x^0)dx = x or C, some constant
1D: linear equations f(x) = x^1 v f`(x) = 1 F(x^1)dx = (x^2)/2 + C
2D: quadratics f(x) = x^2 v f`(x) = 2x F(x^2)dx = (x^3)/3 + C
3D: cubic functions f(x) = x^3 v f`(x) = 3x^2 F(x^3)dx = (x^4)/4 + C
As you can see with 3D and higher every component vector within the vector notation of that spatial dimension is 90 degrees from each other axes as they are all orthogonal - perpendicular to each other. Consider this polynomial:
10,000D hyperfunctions f(x) = x^10,000 v f`(x^10,000) = 10,000x^9,9999 F(x^10,000)dx = (x^10,001)/10,001 + C. As you can see there is nothing from stopping us in continuing with this process, and in that spatial dimension that we can not fathom to imagine except for conceptually based on mathematical notations, every one of those 10,000 independent axes would all be 90 degrees from each other! Don't try to visualize it because even with 4D it would give you a headache! So we can write this in a generic form:
ND: inf funcs {f(x^n):n->inf} {v:n- inf} {f`(x^n):n->inf = nx^(n-1):n->inf} {F(x^n)dx:n->inf = (x^(n+1) /(n+1)):n->inf}
As we have seen above just in 3D alone we can protrude an infinite amount of surface normals that are arbitrary close when individual points of perpendicular tangency are considered and how much they diverge even at a distance of 10 lightyears which is considered a short distance across the cosmos in how far they diverge. We also have seen that this only accounts for less than 1% of the night sky and this was only in 3D! What about the other Infinite Dimensions?
If you want my answer which is the absolute correct answer then it is this! The cosmos is Infinite In Size, Shape, Dimensions, Levels of Zooming In, and Out for it is an Infinite Singularity that is a Perfect Fractal with Perfect Symmetry. So how many atoms are there in all of that? Infinite! So your attempts of estimation are really moot at best because you have only approximated for less than 0.000000000....1% out of 0.000000000....1% of what is truly out there!
For granted I'll give you some lead way because you did State Observable. And infinity goes out beyond what we deem as observable, yet if we were to teleport instantly to the end of that spectrum we would see all of that again yet with all unique planets, stars, gas formations, galaxies, black holes, and other entities we haven't yet discovered. So, I am considering it observable because I was able to observe it just by the confines and laws of mathetics alone! I don't limit my mind to have boundaries that were trying to be forced or imposed on to me by the status quo, those who wrote the textbooks: I'm a free thinker and I'll think for myself thank you! Don't get me wrong, I truly enjoyed watching this video, the content was great and your presentation was awesome, but I believe that your assumptions are far off that's all!
That's a long comment, but the estimation of 10^80 is what I also heard before, so it seems pretty legit
@@Kokurorokuko In truth "we" meaning all of humanity truly doesn't know. 10^80 is astronomical in size yet compare that to 1,000,000,000^1,000,000,000. Then it becomes microscopic.
We don't know how old the universe truly is. People have estimated based on what they call the "observable universe" and the "theory of relativity" based on the supposed "speed of light" and other various phenomena.
I know... many people will claim, we've tested it again and again. Yet, I still think that we truly don't know, we just speculate and make many assumptions. I've heard the arguments before, "the numbers match and they all line-up".
I'm not trying to claim that they don't as many times they do, however, we can manipulate numbers and functions to make them match.
If the universe and the number of atoms, stars, and galaxies are finite or fixed in size, time, and age, then it would it really be possible for there to be natural constructs to exist that are fractals?
We must consider the fact that the estimated size of the visible universe is 93 billion miles, yet the actual size is unknown because we can not see any further or farther past that due to the limitations of our current technologies.
Let's suppose that we could teleport a device to that edge almost instantly say in a manner of a few seconds to a few minutes using the same devices that allowed us to see there in the first place. Also, that device has the capability to teleport over and over again as well as being able to record the data. On top of that, after so many iterations, it would have the mechanism to backtrack to here so that we can analyze that data.
Then we would need to ask. Would the equipment see that much more... another 93 billion lightyears away? Now, ask yourself this, if we were able to do that safely and were able to record that data and witnessed another plethora of galaxies and superclusters within a single line of direction, would it also be true for us to see that in every other possible direction that every infinite line a sphere project outwards?
If that's the case, then we could have chosen any arbitrary point direction and done the same. With that being said, then ask this: if we were able to do that once, how many more times could we repeat that? Ten times? A hundred times? A million times?
Suppose we could have this device teleport out 90-100 billion lightyears in let's say 5-10 minutes. We equipped it with functionality to make sure that it doesn't stop in the middle of a star, black hole, planet, etc. as it is designed to avoid such things.
On the first iteration, it ends up either close to the edge of what we call the observable universe or it is already past what we can now see. It then begins recording data long enough to see that far again. Once it is done recording and confirms the data's accuracy, it then teleports again.
We programmed this to continue on a straight path out without making any turns as it never changes its direction or heading. It avoids collisions with objects by a variable distance within a 10 lightyear span.
Let's say that it takes 1 hour to record the data and to begin the next jump including the time for the initial teleportation. We also programmed it to continue making jumps for 3 years. Then it backtracks and makes one jump back if possible or as few as possible close enough that we could easily retrieve it or its data without disrupting or doing any damage to our solar system.
Let's say it successfully made the trip back from a 3-year journey out and it took at most 1 year to get back. Then we analyzed the data and built models based on it. Let's say that the models that were built were each unique yet nearly an equivalent replica of what we can already see now. In 3 years' time, it made an approximate 95 lightyear jump ever hour on the hour every day.
If we were to do that math alone, that would be approximately 26,300 jumps. Let's say that every iteration showed more of the same, galaxies upon galaxies. Now, if there are approximately 200 billion - 2 trillion estimated galaxies now, let's take the upper bound and find out how many more galaxies there are. This would be 52,560,000,000,000,000 galaxies across 26,300,000 billion lightyears. Then how many atoms would there be?
You see, we can always divide by 2 and never reach 0. We can always multiply by 2 and it will always double, provided that the initial value is not 0. I say there is an infinite amount of atoms, stars, galaxies, etc. because I believe the Cosmos is a Fractal!
Nice point. Wrong, in my estimation.
We should remember, that most probably in the center of every galaxy there is a HUGE black hole, that can "store" the majority of galaxy's mass. This is a disputable question: do we consider interiors of black holes as the "observable universe"? How can we be sure, that inside of black holes there are still atoms an not something like a quark-gluon soup or any other form of matter/energy? Or maybe there is just pure energy - only electromagnetic radiation, like at the beginning of our universe?
That's not true. For example Sagittarius *A is approximately 4 million solar masses. So it's a very tiny fraction of the Milky Way mass.
Nevertheless. He did just a very rough approximation. Nobody knows what is the mass of the universe with accuracy.
Black holes make a sizeable contribution. 10 billion solar masses or more and most galaxies have a supermassive one and probably countless smaller ones.
all mass is included, we look at the period of rotation.
Well that's right but i don't know, if you can use Newtons law such easy, for the sun isn't really at the outer edge of our galaxy. There is still mass outward of the rotation circle of the sun around that black hole in the middle of the galaxy.
@@bracinggreen3785 If the mass of the galaxy exterior to the Sun's orbit was distributed evenly, it would all cancel out gravitationally, so the calculation done in the video properly only accounts for the mass inside the radius of the Sun's orbit. But that's probably going to get you within a factor of 2 or so -- certainly good enough for this calculation.
What's not being taken into account here is that some of the mass included in the calculation is probably dark matter, which is not atomic. But still and all, the video arrives at a pretty good value for the number of stars in the Milky Way, so I think it should get a pass on that.
12:20 But what about dark matter? We know that it is present, and that its gravitational effects are not trivial, from studies of galaxy rotation rates. This means that 85% of the mass of the galaxy we derived from Kepler's third law is not made of atoms, so the number of atoms in the galaxy should be 85% smaller.
What about the enormous swathes of gas known to exist between stars?
thank you I was trying to find out what would happen if every atom in the observable universe turned into a banana
Does that include gases?
So.. how many atoms are there per cubicmeter in the universe?
A simple method!
The sun is not a typical star though you get that right? Copernican principle perhaps? Arthur Eddington once said the sun was a middle class citizen of the steller world. Now he was wrong, but in his defence it was the 1930's.
Now immagine that the size of the observable universe compared to the whole universe is close to the size of an atom compared to the size of the observable universe.
So.. that's 10^160, then?
But the sun isn't really that standard for a star, it's size, density, composition is relatively unique. So, do we assume all stars are copies of our sun? That has to be very biased, quite a bad estimation of the actual number of atoms, isn't it? Since most stars are different vastly different in many ways to our sun.
?
Due to Being able to calculate the Mass of a Galaxy directly, inaccuracy from estimating the avarage star plays no role in the final equation. (You could Substitute no of atoms per star * no of star per Galaxy by deviding Mass of the Galaxy by Mass of an Atom)
Hope that helps
If R is known with a good degree of accuracy then you could equally started with “luckily the mass of the sun is known with a good degree of accuracy so we don’t need to estimate it”
When I listen to this, I picture the lecture being given by Augustus Caesar from the second season of HBO's Rome.
Theres about 10^27 atoms in a human. Theres about 10^57 atoms in a star. At first glance the difference between 27 and 57 at first seems kind of small.
The power of exponents.
@@underpowerjet True just consider binary 2^1 = 2 states okay 2^2 = 4 states okay not much of a difference yet 2^32 = billions of states and this is just in log base 2 math. Yet we are working in log base 10 math. So compare 2^32 versus 10^32. Now consider this number system with this level of dimensionality n^n where n->infinity which gives the expression: infinity^infinity... hmm seems like a fractal to me. How many atoms are within a fractal? An Infinite amount! That's My Answer and I'm sticking with it!
@@skilz8098 I'm not entirely sure what you mean. But, yea I guess increasing the base does make the curve more steeper.
@@underpowerjet 2^1 = 2, 2^2 = 4, 2^3 = 8. 10^1 = 10, 10^2 = 100, 10^3 = 1,000 ... It's an exponential curve in the increase of the value as the base of the logarithm increases. The larger the base of the number system, the steeper its curve becomes. As the base approaches infinity, the curve almost becomes a vertical line.
Agreed, only in that 'tiny' and 'huge' both have 4 letters :)
There are also free floating atoms in the vacuum of space. Count those.
I thought we couldnt use the conventional newtonian/kepler formula for the mass of the Galaxy due to dark matter issues. Why not simply use astronomical and diverse imaging methods to count the stars/bh/mass in the Milky Way ?
The mass was an estimation path to get to the number of atoms, modeled as hydrogen atoms. So far as we know, black holes and dark matter aren't composed of atoms. Imaging methods aren't going to do much better, we've only seen a small fraction of the number of galaxies in the observable universe. The Eddington number of protons in the universe, based on the fine structure constant, is about 10^80 by present estimates. The effort in the video isn't perfect but it's good enough to show how to model a problem that might seem otherwise out of reach, and get an answer that's not ridiculous so far as we know.
Notice that he arrived at 200 billion Sol-like stars in the Milky Way and observations put the estimate at half that. If we add in missing dust, etc, it still appears to be order of magnitude correct overall.
Going, for example, by the galaxies in the Local Group, the Milky Way is rather large to be taken as a typical galaxy. But for a back-of-the-envelope calculation like this, it's OK.
Would you not agree that, as an estimate, we are very likely to be residing within a typical galaxy? If you agree with that assumption, then the sums are fine :)
Damn!, I was 18 atoms off my guess🤨
Very good except that most of those galaxies and hence the atoms may not be there.
but the radial speed of the galaxy is constant, so the speed of the stars near the center and the peripheral stars are the same, this does not invalidate your calculations ???
I mean that the stars increase in speed almost linearly in relation to the radius of the galaxy, so the gravitational laws fail for the Galaxy, so I think your calculations are wrong, what do you think !?
Correct, this does not invalidate his estimate.
the numer of atoms in the universe is.... over 9000!!!!!!!!! ;))
Arthur Eddington arrived at a figure of 10^79.
Interesting. Have you ever calculated the total number of ancestors you have in just 2000 years (with an average of 5 generations per century)? That is a mind-boggling number as well.
2^100, minus overlaps. We're all related, bro :)
What do we mean by atom? Because stars don't primarily consist of atoms but of plasma. Plasma consist of ion's which are atoms that are missing a few electrons. But stars primarily consist out of hydrogen plasma which is just literally a soup of protons and electrons. Are those atoms?? What about the helium part of the plasma??
sorry I don't understand. Why you count only atom from the star? How the other objects which all made of atoms in the universe excluded.
This method only provides an approximate answer. Consider our solar system, for example. Approximately 99.9% of the mass of the entire solar system is accounted for by the sun. The mass of all the planets combined accounts for only 0.1% of the mass of the solar system. Assuming that this is true more generally, you can assume that most of the mass resides in stars, and that we can therefore neglect the mass of the planets in our calculation.
@@andrewlow54 Ok I see. Thk for detail explaination!!
The number of atoms in all of the non-star celestial bodies is inconsequential? What percentage of the mass of a galaxy is contained in the supermassive black hole in its center?
We have no evidence that neutron stars, black holes (including quasars), and dark matter are comprised of atoms and that's the target here. If dark matter happens to include dark atoms and dark chemistry then that's another issue entirely, but it would be a fantasy to include it now. He estimated 200 billion stars for the Milky Way and included nothing else like dust or any attempt to account for heavier and lighter stars. However so far as we know today, the Milky Way has about 100 billion stars so he's order of magnitude correct and the estimate errs in the right direction. We have no idea how many galaxies there are in the visible universe nor the right proportion to add for intergalactic material. The estimate was a swag at mass to demonstrate a method to estimate the number of atoms. The Eddington number is the estimate of protons in the universe and its based on the fine structure constant so that's an entirely different approach. Presently the Eddington number estimates about 10^80 protons so this estimate is not far from that.
Is either number accurate? No reason to believe it is. Neither does the video say anything other than estimate and approximation repeatedly so it's not as haphazard as you may have thought.