Math issue or data?

[budrap00]

I've been intrigued by this graph of the Ambient Cosmic Electromagnetic Radiation:

Fig. 9. Overall estimate for the continuous cosmic background radiation. The data in Fig. 8 were used to derive upper and lower limits at all frequencies, which is proportional to the spread in this splatter plot. Where the background is well measured the line is narrow, and it becomes thicker in regions where the uncertainties are larger.

--- Hill R, Masui KW, Scott D. The Spectrum of the Universe. Applied Spectroscopy. 2018;72(5):663-688. doi:10.1177/0003702818767133

I suppose there is some way of calculating the total power represented by a complex integration but that's above my pay grade. So I decided to get a baseline estimate of the power by simply multiplying the highest amount of power displayed that is common to all frequencies - $10^{-14} W m^{-2} sr^{-1}$ - times the total number of frequencies, $10^{26}$ which yields an unrealistic result of $10^{12}W m^{-2} sr^{-1} s^{-1}$.

The math seems pretty straightforward but the results don't make sense. Am I screwing up the analysis or...? I'm beginning to suspect that it is wrong to assume that electromagnetic radiation is emitted at all possible frequencies as implied in the plot. Any ideas?

[Francois-Zinserling]

The units are already in Watts per square meter per steradian so the power is in the curve.

I don't know if multiplying by 10^26 is sufficient because the power spreads across frequencies. From the basic E=hf, higher frequencies have linearly more energy. To get from frequency to power though you also need intensity. By using a log scale for frequency those values on the right are already partly 'integrated' over a wide spectrum.

What the graph doesn't show is intensity measured, but it is implied in the interpretation. If the intensities of all frequencies had hypothetically been equal (some fixed number) then the curve would have increased linearly up on a linear x-scale (E=hf), and logarithmic upward on a log x-scale like for this graph. Clearly it isn't because high-frequency intensities drop off significantly.

We could simply say it looks like a half-sine and integrate. Peak of the sine at value $$10^{-6}$$. That gives a crude total value of ~ $$2x10^{-4} W/m^2/sr$$
But it's not a sine curve though so the answer is off.

The fact that it briefly peaks and then drops away means the intensities of the higher frequencies are MUCH MUCH lower than the intensities of the lower frequencies. Intensity more likely follows an $$e^{-x}$$ downward curve or typical blackbody spectrum with ~ $$e^{-ax}$$ after the initial peak. That blackbody curve would be recognisable on a linear x-scale. If the frequency scale had been linear, those high frequency powers would be hitting near zero from after 10^15Hz.

The /sr part of the units is easy to get rid of - multply by 4π for spherical total. Then we have a graph of W/m^2 measured on the surface of a sphere of radius 1m. But we won't need that if we assume a black-body. We just need to know where is the peak.

For the rest - assuming a blackbody curve - we can use Wien's law: ν_max = 5.88 × 10¹⁰ × T. The graph visibly peaks at ~10^11 Hz, but contorted due to log-x plot. Known figures quote for CMB v_max as ~1.6x10^11. This gives the familiar T=2.75 from the above equation.

The power of the CMB is: $$P=σT^4$$ with σ=Boltzman and T=2.75 just calculated.

$$P=3.128×10^{−6} W/m^2$$

This is the total power arriving from all directions (the full 4π sky), not just from one direction.

[budrap00]

@Francois-Zinserling Thank you but your remarks don't address the point. All the frequencies of the spectrum have a minimum incoming power of $10^{-14} W m^{-2} sr^{-1}$ according to the plot. If that is accurate then the minimum total incoming power across the entire 10^{26} spectrum is $10^{12} W m^{-2} sr^{-1} s^{-1}$. This again seems unrealistically high - a trillion watts!

I kind of expected that the more mathematically inclined here would have jumped at the chance to point out some obvious error in my analysis or math but so far nothing. As with the calculated result I'm not sure what that means. In addition to the possibility that not all frequencies in the spectrum carry power, another thought has occurred to me.

Is it possible that the total power of a wide spectrum has less cumulative effect than the same total power concentrated in a narrow frequency band? Power is power so $10^{12} W$ should have the same physical effect whether it comes in a narrow or wide band shouldn't it, or should it? I guess it's time to revisit basic electromagnetic theory.

[RedshiftDrift]

@budrap00 note that the vertical scale of the graph is $\nu I_\nu \ {\rm W m}^{-2}{\rm sr}^{-1}$, not what you have in mind i.e. the specific intensity $I_\nu \ {\rm W Hz}^{-1} {\rm m}^{-2}{\rm sr}^{-1}$ which would be the right quantity to integrate over a frequency band.

Hill et al. explain their unusual scale in §2.1.

The math is straightforward but the graph is presented in a strange way (looks strange to me anyway...) You have to do the integration in narrow frequency bands.

E.g. for the CMB peak around $\nu_p = 1.5\times 10^{11}$ Hz, in a band between $\nu_a = 10^{11}$ Hz and $\nu_b = 3\times 10^{11}$ Hz, the graph gives $\nu_p I_\nu \approx 5\times 10^{-7} \ {\rm W m}^{-2}{\rm sr}^{-1}$.

Dividing by $\nu_p$ and multiplying by the bandwidth $\Delta \nu = \nu_b - \nu_a = 2\times 10^{11}$ Hz gives

$$ I_\nu \approx \frac{5\times 10^{-7}}{1.5\times 10^{11}} \ 2\times 10^{11} \approx 6\times 10^{-7} \ {\rm W m}^{-2}{\rm sr}^{-1}. $$

This is slightly overestimated (by a factor 2) because I used the peak value $5\times 10^{-7}$ rather than the real shape of the spectrum. But the number makes sense.

Another example for the CXB peak around $\nu_p = 10^{19}$ Hz, in a band between $\nu_a = 5\times 10^{18}$ Hz and $\nu_b = 2\times 10^{19}$ Hz, gives

$$ I_\nu \approx \frac{10^{-10}}{10^{19}} \ 1.5\times 10^{19} \approx 1.5\times 10^{-9} \ {\rm W m}^{-2}{\rm sr}^{-1}, $$

400 times less power than in the CMB peak.

Unfortunately the authors of the paper don't provide an excel spreadsheet of the curve. You have to do everything by hand for each band of interest.

However if you want the total power in that spectrum, the CMB is much more powerful than everything else combined! So the answer is given above by @Francois-Zinserling .

[Francois-Zinserling]

@budrap00 as @RedshiftDrift also says above: the confusion is in the scaling of the graph. It is an interesting way to find value in something that is almost zero though.

It appears to be a basic function of an x-y graph but the logarithmic x-scale makes it complicated because the power is not measured at discrete frequencies. It's measured across a band of frequencies. Narrow band for low frequencies but comparatively very wide band at extreme high frequencies. So if you want to spread the x-scale back to a linear scale, you will have to distribute the power of the graph amongst each new point that you add on the x-scale.

On the graph it looks like there is ~ the same amount of power at 10^9 as there is at 10^25 but if you break the 10^25 region into little 10^9 bands, each band will contain very little power.

[ericcosmos]

I got something wrong in my last message. Here's the better take: the y-axis label would be a bit clearer if it were written as Hz × W·m⁻²·sr⁻¹ / Hz, so you don’t need to calculate the traditional "area under the curve." So @budrap00, you don't need to multiply (10^-14 W·m⁻²·sr⁻¹) by (10^26 Hz - 10^7 Hz) in your example above. The 10^-14 W·m⁻²·sr⁻¹ already is the power per area you’d normally associate with that "bottom rectangle."

[budrap00]

No argument there.

[budrap00]

@HanDeBruijn
I know you are not a mathematicist and I don't believe I ever said you were. I am also aware that we actually have similar views in many ways. That said I would never presume to tell you what you think:

The bad news is that you think that you are the only one who is fighting against mathematics overruling modern physics. No Bud, you are NOT the only one.

If I were a sensitive soul like you I might take that as a personal attack. As it is I could not care less about such a stupid remark. Note, this is not a personal attack. I did not say you are stupid, just that your comment was - you quite obviously don't know what I think and probably don't care. If you actually wanted to know you could have done the polite and civilized thing - ask me.

[ericcosmos]

Anyone else think it’s kind of funny that Bud’s first question was, “Am I screwing up the analysis or…?”

[budrap00]

@ericcosmos Why do you think that's funny?

[HanDeBruijn]

[RedshiftDrift]

As for the carbon discussion, what's actually wrong with carbon as a possibility? Or iron, tungsten, and other heavier elements? Those are all ejected by dying stars. Couldn't carbon, iron, tungsten, and other metals exist outside the Milky Way, and carbon and those metals naturally be heated to 2.7 K? Just to be clear, though, I'm not arguing for DemystifySci's "CMB is the oceans" idea. I don't buy that explanation either. - @ericcosmos

There's nothing wrong with carbon as a possibility for generating the CMB. What's wrong is DemystifySci's position that only solid carbon can generate blackbody radiation.

A closer examination of the data shows that these metals, in the form of atoms or ions, is in the few 1000K.
Dust itself (a solid) is colder, perhaps reaching the low temperatures of 2.7 K in between galaxies. But to emit a blackbody spectrum the medium must also be opaque. Astrophysicists pretend to know the amount of dust present in the IGM, but I can't confirm their numbers. It is unlikely however that there is enough absorption by dust to produce a blackbody spectrum. We know that because we see objects at a large distances emitting at these wavelengths.

[budrap00]

@HanDeBruijn your red flag has been seen. But many other posts also deserve a red flag.

It seems I'm the only one left, out of three moderators, who is still reading this stuff. So: cut it out everyone.

@budrap00, here's some guidance in addition to the code of conduct:

As an American politician once remarked... - @budrap00

This is a scientific forum, the opinion of politicians is not welcome here.

If you find it offensive when I point out a misrepresentation of the facts... - @budrap00

That's not what you are doing. Instead, you are rejecting languages you can't read.

If someone writes $A = \oint_L f(s) {\rm d}s$, you reject it as mathematicism because it's above your pay grade. But being illiterate in the language of calculus doesn't mean that other people who can read it are wrong. This is not a matter of fact, it's a convention.

If someone shows a logarithmic and a linear scale

you reject it as being disingenuous because you can't read graphs. But being illiterate in the representation of graphs doesn't mean that other people who can read them are wrong. This is not a matter of fact, it's a convention.

If I say "traduire", but Han says "vertalen", it's not a sign of obfuscatory nonsense. Being illiterate in other languages doesn't mean that the people who use them are wrong. This is not a matter of fact, it's a convention.

What people answered here is that "you are screwing up the analysis" because you can't read the graph. It's not a math issue, it's not the data. In addition, numerical examples and various representations of the graph are given, with the hope that it would help. But despite these many ways to represent the same physics, you still are screwing up the analysis because you can't read those graphs.

Now that you've "brushed up your physics", it's time to brush up on a few more languages if you want to continue discussing the physics.

To everyone: when no progress is made because communication is impossible, it's time to stop sending replies.

The question has been answered, this discussion is closed.

[budrap00]

Anything to change the subject right Louis. I didn't label your graph lin-lin, you did, demonstrating that it is you who has trouble reading graphs and understanding the difference between linear and logarithmic scales. It's of a piece though with your difficulties with standard physics like claiming that you had an interferometer sensitive to $10^{-15}meters$ - because the laser frequency was $10^{15}Hz$.

The next time you go to label a graph you should first refer to this page. And try to think before you write - thinking is the hard part.