Handwaving to match orders of magnitude

swle

Hello

This is my first post here. Although I've often read posts in Physics Forums, I've never actually contributed anything - so here goes!

I always enjoy a good hand-waving argument on orders of magnitude; when I first started my astrophysics degree, one homework question was "estimate how many piano tuners there are in this town" - that was it, no other clues! Of course, it was designed to get us to think of ways of solving a problem by making realistic guestimations. Another argument I admired goes that there cannot be an infinitie number of stars in the universe otherwise there would be an infinite amount of light in the sky, so we would not expereience darkness at night.

Anyway, I was sitting in the library today (should have been working on a Solid State problem) and started wondering: if I took the energy of a photon as a "rest mass" and multiplied it by the number of photons in the universe would that be anywhere close to the estimated figure for dark matter? So I looked up some figures from not-very-reputable sources (this is just hand-waving, after all) and got this:


Number of photons in the universe: 10^91 (source: Science Blog)
"Rest Mass" of a photon: 4.2 * 10^-40 kg (source: Wikipedia)

So, total estimated photon "Rest Mass" = 4.2 * 10^52 kg.

Now compare to Dark Matter in the Universe: 3.5 * 10^52 kg (source: Wikipedia)

They match within 20% - well within an order of magnitude!


Well, I thought it was interesting even if doesn't mean anything!
Anyone got any other hand-waving examples that they can think of?

K^2

Photon has no rest mass, so I have no idea what you are doing. Or why.

Your estimates cannot be remotely right, however. They suggest that gravitational mass in radiation is significantly greater than gravitational mass of matter in the universe. That would, indeed, result in our sky being a touch on the bright side.

Actually, that's only true if distribution is uniform. If there is an infinite universe with star density is being a fractal, it's possible to achieve finite luminance of the sky. And since distribution of matter agrees with it on scales we do observe, the universe could still be infinite based on our observation of the sky.

jtbell

That is the upper limit on the mass of the photon. We do not actually have any unambiguous experimental evidence that the photon has mass. This upper limit is simply a measure of the finite precision of our best experimental results. The mass could be that big and still "look like" zero as far as our experiments are concerned.

Have you ever seen a graph with error bars on the data points? The upper limit is basically the top edge of the error bar, with the actual data point being at somewhere around zero.

H2Bro

I think estimating the strength of a nuclear blast using some bits of paper ranks pretty highly.

then again, I think both that example and the piano tuning are due to Enrico Fermi.

Heres a nice little story on it for those curious: http://sci.mercer.edu/handouts/fermi_calculations.htm

swle

Thank-you for taking time to reply!

Like I said, just a bit of fun!

I don't follow that argument - perhaps you cold explain?

I don't follow this argument either, perhaps you can explain this one too? The universe may well be fractally distributed, but as far as we know it is isotropic? Treatment of the improper integral is fairly straightforward?
Is your argument that a finite number of stars are visible (so the sky is dark at night) but an infinite amount of stars could exist at an infinite distance so that we would never see their light (i.e. a non-bigbang solution)?

swle

Brilliant - thanks!

swle

Thanks for reply!

It was just a bit of fun - I realise that most objections will centre around what is, or what is not, "rest mass". I took the view that if someone is prepared to quote a figure, then I am prepared to put it in to a calculation!

K^2

Suppose we have an infinite space filled with identical stars distributed with an average density of ρ. How much light are we going to receive from stars distance R away? N/R², where N is number of stars at that distance. Of course, if we take continuous limit, the total illumination becomes:

[tex]\int_0^\infty \frac{dN}{r^2} = \int_0^\infty 4\pi\frac{r^2 \rho dr}{r^2} = \int_0^\infty 4\pi\rho dr = \infty[/tex]

So we will get infinite luminosity. Now imagine, instead, that stars are grouped into clusters of density ρ. These clusters are spread out a bit more from each other, so if you take average density across groups of clusters, the new density is some ρ'<ρ. These groups are also clustered, and that gives you density on even larger scale to be ρ''<ρ'<ρ. Because we live near a star, which belongs to a cluster, which belongs to a group of clusters, etc, locally we have the highest density of stars, and it drops with distance. So ρ(r) is now a decreasing function. Say, ρ(r)=ρ₀exp(-λr) just for illustration. Now the total luminosity is different.

[tex]\int_0^\infty \frac{dN}{r^2} = \int_0^\infty 4\pi \rho_0 e^{-\lambda r} dr = \frac{4\pi \rho_0}{\lambda}[/tex]

We get a finite quantity.

Notice that this kind of structure is similar to what we really have. In our immediate vicinity, density of stars is very high. 1 star for a moderately small volume. As we look at the scale of the galaxy, the density is significantly lower. If we look at local group, the density of stars is much lower, on average, than in a galaxy. And so on. This sort of fractal structure allows for a finite amount of light to be reaching us from the sky despite effectively infinite number of stars out there.

Note that stellar dust would not make a difference. If stars were uniformly distributed, interstellar dust would reach thermal equilibrium with the stars and would glow just as hot.

mfb

The Cosmic Energy Inventory
Radiation is 4 orders of magnitude below dark matter.

The upper limit on a photon mass, according to the particle data group, is <10^-18 eV (or <10^-26 eV with a specific model). This corresponds to 2*10^-54kg as upper limit per photon.

swle

You got me! I chose numbers that, although widely available in the public domain, gave the me answer that I wanted.

btw, great links - thanks

swle

Thanks very much for the reply, but I have a problem with this line
Are you not arguing a flux here? (i.e. Gauss ) What happens to the photons once they have reached a destination?
While individual photons will be absorbed, won't the total number available will remain constant?
Sorry if that's a little confusing, but, again, thanks for the reply - it's very interesting.

cragar

the light from each ring that reaches earth will be the same. And we are adding up an infinite amount of rings. the intensity falls off as 1/r^2 but the further we go out
there will be r^2 more stars to add to the light so the light from each ring will be the same no matter the distance.