Electric Sun: Energy Budget from KNOWN Sources

Conservation of energy is Nature's version of the 'balanced budget'.  It is also much more rigorously enforced that the human financial version. 

In Nature, you can occasionally engage in some deficit spending, but only for a very short time (via the Uncertainty principle, see UVa: The Energy-Time Uncertainty Principle: Decaying States and Resonances).  Because we know these conservation laws are rigorously enforced, the Patent Office automatically rejects any device proposed for patent that tries to violate it, such as perpetual motion machines (Wikipedia: Perpetual Motion:Patents).

One method popular among scientists and engineers when examining systems is to explore their power requirements, also known as their energy budget.  If power requriements to run a system dramatically exceed the available power sources, the system is unworkable.  Many a scam has been perpetrated on scientifically naive consumers (see AGU Blog: The Latest Perpetual Motion Fraud) and investors (see NY Times: There's One Born Every Minute) with the notion of energy from nothing.

Conservation of Energy raises one of the major questions for any hypothesis of stars being powered by external sources:

Where does the energy come from?

Now there is one  prime candidate as a possible exterior energy source for the Sun and other stars - cosmic rays.  Cosmic rays consist of energetic charged particles - electrons as well as protons and heavier ions, that seem to approach the Earth roughly uniformly from all directions. 

This is the largest source of energy moving towards the Sun, so the obvious question is:

How Much Power Can Come from Cosmic Rays?

And the related question:

Is there enough energy in incoming cosmic rays to power a star? 

To determine this energy, we need to know the particle flux, number of particles striking the surface per time, and the amount of kinetic energy carried by each particle.  We'll assume the energy comes from the kinetic energy of the particle rather than the full mass-energy which would require matter-antimatter annihilation.  A good starting point for information about cosmic rays would be the Cosmic Ray entry on Wikipedia.

We can do a simple estimate of the power available using the measured energy flux of cosmic rays at the orbit of Earth (assuming they were all directed towards the Sun, which they are not) using the basic equation for power and some dimensional analysis:

Power = Energy per time 
      = (Energy per particle) * (number of particles per area per time) * area

It is a little more complex than this because cosmic rays have a range of energies and there are a different number of particles at each energy.  This is represented by a spectrum as seen on the wikipedia page. 

However, because the flux of cosmic rays is roughly constant at low energies and then decreases at higher energies, we can probably get a reasonable estimate.  There are occaionally very high energy cosmic rays, but these are very rare and contribute little to the TOTAL power in cosmic rays.  In space, we'll use the measured particle flux of about

20,000 particles/m^2/s = 2e4 /m^2/s for energies of 1 GeV 

(1 GeV = 1 billion electron volts).  This value will cover most of the cosmic-ray population.  As noted above, at higher energies, the flux (number of particles per area per second) decreases rapidly, so they will probably make only a small contribution to the total power available.

Since we are measuring these particles from a detector at Earth orbit, to get the total number of particles heading towards the Sun, we need to multiply by the area.  The area needed in the calculation is the area of the spherical surface defined by the earths orbit.  We'll make the simplifying assumption that all the particles crossing the orbit of Earth inbound towards the Sun converge on the Sun.

Using the radius of the Earth's orbit as 1 Astronomical Unit (149e9 meters) and the area of the sphere as 4*pi*r^2, we find:

Area of surface = 4*pi*(149e9 m)^2 = 2.79e23 m^2

Now we complete the computations

Power = 1 GeV         * (2e4/m^2/s) * 4*pi*(149e9 m)^2 
      = 1e9*1.6e-19 J * (2e4/m^2/s) * 4*pi*(149e9 m)^2 
      = 1.6e-10J      * (2e4/m^2/s) * 2.79e23 m^2
      = 8.93e17 Watts

With the solar luminosity of 3.84e26 Watts, we see that the incoming cosmic rays have only

(8.93e17 Watts) / (3.84e26 Watts) ~ 2e-9.

Cosmic rays can provide only 2 billionths of the energy needed to power the Sun!

This is the inbound particle population we can detect, and it is woefully insufficient for the task of powering the Sun. 

So where are all these incoming particles EU 'theorists' claim can power the Sun?

How do they hide? 

One popular EU excuse is it's a 'dark current'.  But 'dark current', in the plasma community, is an archaic term left over from the days when experimenter had to rely on their eyes as the detector.  'Dark current' meant there were particles moving through the region, they just didn't emit visible light.  Such 'dark currents' can be detected by their emission in wavelengths beyond visible light, or by particle detectors.  This excuse is not applicable.

Exercise for Readers:
From the cosmic ray review at the Particle Data Group, the cosmic ray spectrum above 1 GeV can be approximated by

I(E) = 1.8e4 (E/1GeV)^-alpha  nucleons/m^2/s/sr/GeV

where the power-law index, alpha, is equal to 2.7.

Compute a more accurate power from cosmic rays by integrating over the spectrum over the energy range.  How does it compare to the estimate above?  Hint: it's still not enough to provide a substantial part of the solar luminosity.

Historical Note:  In the late 1800s, before we knew about nuclear energy, it was actually hypothesized that infalling meteorites would provide some energy for the Sun.