Q:

I can’t convince my friends that there aren’t a googol (1x10^100) atoms in the universe. I was just wondering if I was right before I carry on arguing

- nick (age 16)

sixth form college, uk

- nick (age 16)

sixth form college, uk

A:

Nick,

The short answer is "no, dont keep arguing". There is not a googol of anything in the universe. Even if you count neutrinos and photons as well as atom, you come up about a factor of 1x10^10 short. A great answer, provided by an astrophysicist here at the U of I, is included below.

Mats

-----------------------------------------

Hi Mats,

Thanks for the referral. I'm going to be lazy and first quote an answer I've written in the past regarding the related quantity of the number of atoms in the universe:

First of all, by "the universe" I'll assume we're talking

about the _observable_ universe, which is necessarily much smaller than the entire physical volume of space (which for all we know could be infinite). The size of the observable universe is basically spherical, with radius about equal to the "Hubble radius", d_H = c/H_0 ~ 4e9 pc = 1e28 cm ~ ct_0 (where 0 denotes present values). In fact, when you allow

for the expansion of the universe, the better value is more like 2d_H. I hesitate to bother you with such factors, but we are after all going to cube this. Anyway, the volume of

the observable universe is just that of a sphere with r=2d_H, or about 3e85 cm^3. The microwave background photons have T=2.728 K, and Planck tells you that this gives n_gamma = 411 photons/cm^3, so there are about 1e88 photons in the observable universe (not far from a googol in some sense, though 12 orders of magnitude is not negligible, even for astrophysics).

There are comparable numbers of each neutrino species.

Finally, big bang nuke (that's me) sez that the baryon to photon ratio in the universe is about n_b/n_gamma = 3e-10 (i.e., the photons and nu's outnumber [but not outweigh]

the measley baryons by a hefty factor) so the baryon number of the universe is about 3e78. So this is 21.5 or so orders of magnitdue off from a googol, but in some sense it's not so far off.

I'm sparing you worries over the dependence on various cosmological parameters; this just gives the basic ballpark figure.

As for the mass of the (observable) universe, this again needs some clarification, it turns out. If we are interested in just the (nonrelativistic) _matter_ (including dark matter), then it looks like Omega_matter = rho_matter/rho_crit ~ 0.3, so that M_matter = 0.3 M_crit = 3e21 M_sum = 6e51 kg. On the other hand, since the microwave background now gives strong evidence that Omega_tot = 1, then it looks like there is another component

to the universe, the "dark energy" (or cosmological

constant). If we include this too, then we don't need the factor of 0.3, and we get more like 2e52 kg.

More than you wanted to know, I'm sure, but I hope this

helps.

Brian

The short answer is "no, dont keep arguing". There is not a googol of anything in the universe. Even if you count neutrinos and photons as well as atom, you come up about a factor of 1x10^10 short. A great answer, provided by an astrophysicist here at the U of I, is included below.

Mats

-----------------------------------------

Hi Mats,

Thanks for the referral. I'm going to be lazy and first quote an answer I've written in the past regarding the related quantity of the number of atoms in the universe:

First of all, by "the universe" I'll assume we're talking

about the _observable_ universe, which is necessarily much smaller than the entire physical volume of space (which for all we know could be infinite). The size of the observable universe is basically spherical, with radius about equal to the "Hubble radius", d_H = c/H_0 ~ 4e9 pc = 1e28 cm ~ ct_0 (where 0 denotes present values). In fact, when you allow

for the expansion of the universe, the better value is more like 2d_H. I hesitate to bother you with such factors, but we are after all going to cube this. Anyway, the volume of

the observable universe is just that of a sphere with r=2d_H, or about 3e85 cm^3. The microwave background photons have T=2.728 K, and Planck tells you that this gives n_gamma = 411 photons/cm^3, so there are about 1e88 photons in the observable universe (not far from a googol in some sense, though 12 orders of magnitude is not negligible, even for astrophysics).

There are comparable numbers of each neutrino species.

Finally, big bang nuke (that's me) sez that the baryon to photon ratio in the universe is about n_b/n_gamma = 3e-10 (i.e., the photons and nu's outnumber [but not outweigh]

the measley baryons by a hefty factor) so the baryon number of the universe is about 3e78. So this is 21.5 or so orders of magnitdue off from a googol, but in some sense it's not so far off.

I'm sparing you worries over the dependence on various cosmological parameters; this just gives the basic ballpark figure.

As for the mass of the (observable) universe, this again needs some clarification, it turns out. If we are interested in just the (nonrelativistic) _matter_ (including dark matter), then it looks like Omega_matter = rho_matter/rho_crit ~ 0.3, so that M_matter = 0.3 M_crit = 3e21 M_sum = 6e51 kg. On the other hand, since the microwave background now gives strong evidence that Omega_tot = 1, then it looks like there is another component

to the universe, the "dark energy" (or cosmological

constant). If we include this too, then we don't need the factor of 0.3, and we get more like 2e52 kg.

More than you wanted to know, I'm sure, but I hope this

helps.

Brian

*(published on 10/22/2007)*