Solar Science

A blog of solar physics

Is the Earth's magnetic field a cosmic ray funnel?

with 25 comments

Portrait of Henrik Svensmark

Portrait of Henrik Svensmark

In this world of rampant climate alarmism, its to be expected that theories and hypotheses which do not support the AGW theory will get the full treatment of bad analysis and character assassination. After all,  where’s the funding going to go if there’s an alternative theory that bombs the bridge in front of the gravy train?

One such is Dr Henrik Svensmark‘s hypothesis on the modulating effect of the solar magnetic field on the Earth’s climate. In the recent article in the New York Times on the solar cycle that I recently mentioned, we have this:

The idea that solar cycles are related to climate is hard to fit with the actual change in energy output from the sun. From solar maximum to solar minimum, the Sun’s energy output drops a minuscule 0.1 percent.

But the overlap of the Maunder Minimum with the Little Ice Age, when Europe experienced unusually cold weather, suggests that the solar cycle could have more subtle influences on climate.

One possibility proposed a decade ago by Henrik Svensmark and other scientists at the Danish National Space Center in Copenhagen looks to high-energy interstellar particles known as cosmic rays. When cosmic rays slam into the atmosphere, they break apart air molecules into ions and electrons, which causes water and sulfuric acid in the air to stick together in tiny droplets. These droplets are seeds that can grow into clouds, and clouds reflect sunlight, potentially lowering temperatures.

The Sun, the Danish scientists say, influences how many cosmic rays impinge on the atmosphere and thus the number of clouds. When the Sun is frenetic, the solar wind of charged particles it spews out increases. That expands the cocoon of magnetic fields around the solar system, deflecting some of the cosmic rays.

But, according to the hypothesis, when the sunspots and solar winds die down, the magnetic cocoon contracts, more cosmic rays reach Earth, more clouds form, less sunlight reaches the ground, and temperatures cool.

“I think it’s an important effect,” Dr. Svensmark said, although he agrees that carbon dioxide is a greenhouse gas that has certainly contributed to recent warming.

Dr. Svensmark and his colleagues found a correlation between the rate of incoming cosmic rays and the coverage of low-level clouds between 1984 and 2002. They have also found that cosmic ray levels, reflected in concentrations of various isotopes, correlate well with climate extending back thousands of years.

All well and good. But then there’s always someone who produces a sophisticated argument why you shouldn’t believe your own lying eyes.

But other scientists found no such pattern with higher clouds, and some other observations seem inconsistent with the hypothesis.

Terry Sloan, a cosmic ray expert at the University of Lancaster in England, said if the idea were true, one would expect the cloud-generation effect to be greatest in the polar regions where the Earth’s magnetic field tends to funnel cosmic rays.

“You’d expect clouds to be modulated in the same way,” Dr. Sloan said. “We can’t find any such behavior.”

Still, “I would think there could well be some effect,” he said, but he thought the effect was probably small. Dr. Sloan’s findings indicate that the cosmic rays could at most account for 20 percent of the warming of recent years.

Would the Earth’s magnetic field “funnel cosmic rays”? This was taken up by Stephen Ashworth in an email to Benny Peiser’s CCNet mailing list:

Dear Benny,

Kenneth Chang in the New York Times reports that some observations seem inconsistent with the solar magnetic field–cosmic ray–cloud formation hypothesis.  He wrote (CCNet 113/2009 — 21 July 2009, item 3):

Terry Sloan, a cosmic ray expert at the University of Lancaster in England, said if the idea were true, one would expect the cloud-generation effect to be greatest in the polar regions where the Earth’s magnetic field tends to funnel cosmic rays.

“You’d expect clouds to be modulated in the same way,” Dr. Sloan said. “We can’t find any such behavior.” Still, “I would think there could well be some effect,” he said, but he thought the effect was probably small. Dr. Sloan’s findings indicate that the cosmic rays could at most account for 20 percent of the warming of recent years. [sic — he clearly means the *reduction* in cosmic ray influx to the Earth in recent decades of the more active Sun — SA]

I am skeptical about Dr Sloan’s claim.  The reason is as follows.

A few years ago I read a suggestion that an interstellar space probe might be able to do a flyby of the star Sirius, and use its gravity to redirect itself to a subsequent flyby of Procyon, in the same way that Pioneer, Voyager and other probes have used the gravity of Jupiter to redirect themselves to Saturn and beyond.  I have a formula for the change in direction caused by a flyby of a massive body, so I was able to check this idea numerically.

It turned out that if the interstellar probe was travelling at a speed that was a significant fraction of the speed of light, say 0.1c — which it would have to if it was to reach Sirius in only a few decades flight time — then the deflection of its trajectory even on a flyby which grazed the star’s atmosphere was only in the region of one degree, totally insufficient to redirect it to Procyon.

The lesson was that the gravitational fields of planets and even stars (Sirius is more massive than our Sun) are almost imperceptible to a vehicle if it is travelling at such a high speed.

Cosmic ray particles come into the Solar System at a significant fraction of the speed of light.  I would therefore expect them to be largely immune to our local gravitational and magnetic fields.  I would not expect Earth’s magnetic field to funnel them towards the poles, as it does with the lower-energy solar particle flux.  (Presumably someone has already checked this numerically?)

It would seem that Svensmark’s cosmic ray–cloud formation hypothesis depends on the difference in strength between the Sun’s and the Earth’s magnetic fields: the Sun being strong enough to modulate the cosmic ray flux in the inner Solar System over its longer-term cycles of activity, while the Earth is too weak to redistribute incoming particles geographically during their last second or so of flight before hitting the atmosphere.

Best wishes,

Stephen Ashworth
23 July 2009

Stephen Ashworth, Oxford, U.K.
http://www.astronist.demon.co.uk/

Quite so. Cosmic rays travel at significant percentages of the speed of light and wouldn’t be deflected significantly by the Earth’s weak magnetic field.

I wonder if Terry Sloan would care to answer? Someone should ask him.

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Written by John A

July 25, 2009 at 4:08 pm

Posted in Uncategorized

25 Responses

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  1. It is the magnetic fields generated by the solar wind which affects cosmic rays (particularly the very low energy ones) – gravity has an insignificant effect. The changes in the solar wind during the 11 year solar cycle cause changes in the cosmic ray rate. (Look at http://cosmicrays.oulu.fi. You can make yourself a graph of the rate from say 1970 to the present and you will see the cosmic ray rate varying with the 11 year solar activity).
    The effect of the solar wind on cosmic rays is then bigger near the poles than at the magnetic equator. This is not a funnelling effect but an effect of the Earth’s magnetic field which deflects away from the Earth lower energy cosmic rays at the equator than at the poles. The effect of this is that the 11 year solar cycle causes bigger changes in cosmic rays and hence in the ionization in the atmosphere at the poles than at the equator.
    Hence if clouds are caused by ionization one would see a bigger 11 year cycle in cloud cover at the poles than at the equator. We did not see such an effect so we failed to find evidence to corroborate Svensmark’s hypothesis that changing ionization from cosmic rays causes the changes in low level cloud cover (he observed a dip in some data on low level cloud cover in the solar cycle peaking in 1990).
    Since Svensmark published his hypothesis we have completed another solar cycle and the change in clouds which he observed in the solar cycle peaking in 1990 is not present in this recently completed solar cycle (peaking in 2000), further evdience which fails to corroborate his hypothesis. (see http://isccp.giss.nasa.gov/climanal7.htm the last graph on the IR low level cloud. You can see a dip in 1990 but very little sign of one in 2000).

    Terry Sloan

    July 27, 2009 at 3:00 am

  2. I like how the naysaying scientists begrudgingly “admit” that cosmic rays are only responsible for ~20% of observed warming.

    Oh really. Just a totally insignificant 20% huh? And after the science was settled on CO2 . Imagine that. I can really respect their honesty in admitting this very insignificant point. I mean we all know that 20% of something is an utterly insignificant fraction of the whole, but they showed their commitment to accuracy and scientific integrity by admitting it after thoroughly invalidating it’s relevance through supposition.

    It warms my heart to see such a rational scientific debate taking place. More than cosmic rays could ever warm it…

    Dublds

    July 27, 2009 at 3:23 am

  3. I am a little confused about what you are asking.

    Are you saying the a magnetic field will have less of an effect on higher velocity charged particles?

    The force exerted on a charged particle by a magnetic field is proprotional to the part of the particle’s velocity that is perpendicular to the magnetic field. Mathmatically this is expressed as:

    F= q(V x B)

    where F, V, and B are vectors
    F is the force on the charged particle
    q is the charge of the particle
    V is the velocity of the particle
    and B is the magnetic field

    (V x B) is the vector cross product of the particle velocity and the magnetic field

    The force will always be at right angles to the particle’s velocity, and conseqently will not change the absolute value of the velocity (the speed), but will change the particle direction.

    The greater the particle’s velocity, the greater the deflection.

    If this is already obvious to you, then please disregard my comment. I am simply not sure what you are asking here.

    Best Regards
    Tom Moriarty
    ClimateSanity

    Tom Moriarty

    July 27, 2009 at 9:20 am

  4. Tom

    I think what is germaine is the deflection of a high energy cosmic ray in a very weak magnetic field like the Earth’s. If I read correctly, Terry Sloan was arguing that the Earth’s magnetic field should measureably alter the direction of the cosmic rays before they impact the atmosphere and what Stephen Ashworth is saying is that the deflection for such high energy particles would be negligible.

    Clearly there is a solar magnetic effect, but is there a measureable geomagnetic one?

    John A

    July 27, 2009 at 2:21 pm

  5. or perhaps a much simpler answer, not based on solar science but rather on physical geography/climatology.

    The theory states “When cosmic rays slam into the atmosphere, they break apart air molecules into ions and electrons, which causes water and sulfuric acid in the air to stick together.

    The distribution of water vapor in the atmosphere is not even and the Polar regions are extremely dry – water vapor is almost zero. The trace amount of water vapor would likely result in very few droplets forming and thus only trace cloud formation.

    Fred from Canuckistan . . .

    July 28, 2009 at 8:07 am

  6. Does the strength of the magnetic field affect the amount of microwaves entering earth’s atmosphere? Microwaves excite water molecules, hence the Microwave Oven. Doesn’t the practical application of microwaves for cooking lend itself to a straight forward conclusion that as the sun increases it’s Microwave output (F10.7 flux), that increase is going to directly affect the water vapor in the earth’s upper atmosphere by heating it? Occam’s Razor would suggest that any increase in microwaves coming from the sun must in some measure heat (add energy) water that it strikes. There are many forms of energy the sun generates, each form or wavelength heats a molecule in a specific manner unique to that compound. Microwaves don’t heat O2, N2 or CO2 to a significant degree, but it does heat H20 very well, that’s why microwave ovens heat food and minimally the air surrounding it.

    dscott

    August 2, 2009 at 12:33 pm

  7. The model of a larger mass causing a slingshot of a smaller mass does not apply to the cosmic rays affected by Earth’s magnetic field. As Tom Moriarity explains above, the force on the charged particle will be determined by the level of its charge times the magnitude of the magnetic field and the velocity of the particle. There are no such multipliers in the gravitation equation. Ignoring relativistic effects, the gravitational force on a mass will be independent of the relative speed of the two masses. Cosmic rays are all ionized so the faster they move, the greater the path bending force. The more they are ionized, the greater the path bending force. This force will be much much larger than the force of gravity. A general physics class attempting to calculate the path of a charged praticle in a magnetic field would normally ignore the force of gravity in the solution because it will be orders of magnitude smaller than the electromotive force acting on the cosmic ray. Protons would be bent the most but some cosmic rays are high energy iron atoms that are fully ionized. The path of an iron atom would bend less than a proton because its charge to mass ratio is lower. But, it would have higher impact energy if it hit the atmosphere because of its proprotionately higher mass.

    My explanation does not provide an answer to the problem. Because of the cross product of the velocity vector with the magnetic field vector, a cosmic ray approaching from one direction will have its path bent towards the earth by the magnetic field. Whether it hits the earth will be determined by how its velocity vector relates to the earth’s position and the Earth’s magnetic field, the mass of the particle, and its ionization state. On the other hand, the same particle in the same condition hitting the same magnetic field line at exactly at the same point with the same velocity but coming from the opposite direction will be bent away from the earth. This means that the “funnel” effect for particles coming in perpendicular to the Earth’s axis will only occur on one hemisphere but this spatial situation will switch to the opposite hemisphere in the opposite season. Cosmic ray particles coming in parallel (from the north or south) to the Earth’s axis will generally be bent very little because their paths will be mostly parallel with the magnetic field lines. Therefore, I would not be surprised to find that the “funneled” cosmic rays affect the Earth mostly between +/-60 degrees from the equator and not at the poles.

    NASA no doubt has documented the shape and relative strength of the Earth’s magnetic field as well as the general composition, flux, energies, and direction vectors of cosmic rays. It seems that someone somewhere has probably run a calculus solution or a Monte Carlo solution to this problem and published the results. Such a study would be a good starting point for this discussion.

    Joe

    August 10, 2009 at 3:54 pm

  8. One more comment. The Earth’s magnetic field is stronger than the Sun’s in our location in space. Otherwise, all compasses would point to the sun!

    Joe

    August 10, 2009 at 3:57 pm

  9. Me again. I was watering my yard late on a cool August night (Imagine that! A cool August night!) when I realized that I had misjudged the geometry of the cosmic ray path bending by the Earth’s magnetic field. The hemispherical aspect I mentioned in my earlier post does not exist. Instead, it is a left/right rule! When a charged particle crosses a magnetic field line, it feels a force at a right angle to the both its direction (X) and the direction of the magnetic field (Y). With these designations, the particle will gain momentum in the Z direction. But which Z? Wikipedia gives a good rule: if the thumb of your right hand (not your left hand) points in the direction the particle is moving and the fingers point in the direction of the magnetic field (towards the north pole), then the palm of your hand faces in the direction of the force applied to the particle.

    With this directionality in mind, we can now visualize what happens to cosmic particles passing through the Earth’s magnetic field. Imagine that there is no magnetic field around the Earth. In that case, the collision cross-section for cosmic rays would be the diameter of the Earth’s atmosphere. Next, imagine a perfectly spherical magnetic field around the earth and that we are directly over New Orleans at the same distance as the moon, looking down on New Orleans. It is dawn in New Orleans so it is noon in London and midnight at the International Date Line. Ignoring the offset between the Earth’s rotational axis and its magnetic field axis, lets view three cosmic rays heading in parallel formation line-abreast. The center particle is initially heading directly through New Orleans to the center of the Earth while the other two particles are flying parallel with the center particle but further away than the radius of the Earth. If no magetic field were present, they would miss the Earth. Because of the geometry with which I have set up this problem, all three particles perpendicularly encounter a Earth magnetic field line of the same magnitude pointing in the same direction, up to the north pole. Applying the right hand rule, all three particles would bend in the same direction: towards the east in this case. The center particle would bend east towards Biloxi. The right hand particle passing above Europe would bend away from the Earth. The left hand particle would bend towards the Earth. With no asymmetry in the geometry, the collision cross-section would still seem to be the same diameter as the earth with the exception that the circle of the capture cross-section would be offset to the left from our viewpoint relative Earth’s actual position to accommodate the rightward bending. The only change would be a longer path distance through the atmosphere due to the perpendicular rays now coming in diagonally.

    We can now break the Earth into five directions and look for asymmetries. The sixth direction faces the Sun (London at noon) and it is assumed that no cosmic rays can hit the Earth from that direction. From our original position at the 90 degree lattitude, there is a second position at 180 and another at 270. Add to that the cross-sections coming from the north and from the south. The three capture cross-sections along the equator would seem, on first order, to have equal collision cross-sections as the Earth without the magnetic field but offset “left” from the spherical center of the Earth. The north and south directions would not see an effect from the magnetic field because of bending symmetry near the poles and parallel travel with the field lines away from the poles.

    The Earth’s magnetic field is not spherical but looks like a tear-drop with a roughly spherical hemisphere towards the sun and a tear-drop tail away from the sun. This tear-drop changes its size with the activity of the Sun. As well, it is highly asymmetrical. Particles coming in north and south see a slightly different geometry but I suspect the change will be minimal. Particles coming from the three equatorial directions described above would see completely different geometry with respect to the directionality of their respective bending moments. Particles coming in from 90 degrees lattitude (at noon in London) would bend away from the tail. Particle coming in at 270 degrees lattitude would bend towards the tail. Particles coming in from 180 degrees lattitude would be running directly up the tail.

    Would there be some focusing by all this asymmetric geometry? It is hard to say without more thought and some pencil and paper. However, there is another complication. Cosmic rays come in from outside the Sun’s system and the Sun has its own magnetosphere. It no doubt has its own tear drop shape due to the galactic wind. This asymmetric magnetosphere would certainly affect the density of cosmic rays coming our way. As well, the relative orientation of Earth’s magnetosphere and the Sun’s magnetosphere are probably not symmetricly oriented along a single axis and that relative orientation probably changes on a yearly basis as well as on longer time scales due to precession, eccentricity, and obliquity. Add to that periodic variations in the strength of the Sun’s magnetic field and this becomes a very complex geometry problem.

    The description above ignored the Sun’s magnetic field around the Earth outside of Earth’s magnetosphere. It is the Sun’s magnetic field carried by plasma (the solar wind) that forms the Earth’s magnetic field into the tear drop. The Sun’s magnetic field reverses direction every 11-year sunspot cycle. Certainly, when it is “aligned” in the same direction as the Earth’s field (which is remarkably constant) the Earth’s tear drop magentosphere will probably be a different size than when the Sun’s magnetic field is aligned in the opposite direction. If the magnetosphere affects cosmic rays and the cosmic rays affect albedo, there should be a 22-year variation in the cloud record.

    Joe

    August 10, 2009 at 6:04 pm

  10. I have done a little research on the subject of cosmic rays and found that I was mostly right and little wrong in my modeling.

    1. There is a parameter called the “vertical rigidity cutoff, an old term, which essentially describes the amount of energy a cosmic ray requires to reach to within 20 miles of the Earth’s surface. It is as high as 15 gigaelectron volts at the equator and drops very low above 60 degrees lattitude. ( A lucky guess on my part.)

    2. Cosmic rays emitted by the sun are low energy. Intergalactic cosmic rays are usually very high energy, >15GeV.

    3. There is a directionality to the rigidity cutoff value. Standing on any point between +/-60 degrees lattitude, cosmic rays coming from your east must have higher energy to make it to the 20 mile point than does a ray coming straight down on you while cosmic rays coming from your west require less energy.

    4. I interpret #3 above to mean that I had the bending direction reversed. The rays must bend to the left. My error probably arose from determining if the Earth’s magnetic is positive pointing from south to north or positive pointing from north to south. I got that wrong but it does not affect the shape of the asymmetry, just which way the asymmetry shifts.

    5. Cosmic ray astronomers long ago mapped the cosmic ray flux across the Earth’s surface to determine the best place to put observatories. They found that the large variations in cosmic ray flux by location because of the geometry of the Earth’s magnetic field plus the direction and energy from which the rays approach. There are also seasonal changes in the flux plus sudden reductions that last for only a few hours. Both of these types of changes I assume are related to the sun.

    6. Basil Copeland in a guest blog today on What’s Up With That (http://wattsupwiththat.com) describes a model of the historical surface temperature using “random walk” mathematics. Using the statistical tools from this model, he is able to pull out the solar influence on the apparent random walk of the surface temperature sincer 1860. He finds that this influence is stronger every other solar cycle. Might that be caused by the parallel/anti-parallel relationship of the Sun’s magnetic field to the Earth’s magnetic field over every solar cycle pair affecting cosmic ray behavior?

    Joe

    August 12, 2009 at 3:49 pm

  11. Basil Copeland in a guest blog today on What’s Up With That (http://wattsupwiththat.com) describes a model of the historical surface temperature using “random walk” mathematics. Using the statistical tools from this model, he is able to pull out the solar influence on the apparent random walk of the surface temperature sincer 1860. He finds that this influence is stronger every other solar cycle. Might that be caused by the parallel/anti-parallel relationship of the Sun’s magnetic field to the Earth’s magnetic field over every solar cycle pair affecting cosmic ray behavior?

    The weak/strong behaviour is a well known feature of the approx. 22 year Hale cycle which has two solar maxima within it.

    Although I am still thinking about how to model the behaviour, it seems to me likely that the [tex]F = v \times B[/tex] behaviour of cosmic rays is probably overwhelmed by the low value of B in the case of the Earth. Thus although the velocities of cosmic rays are very high, the Force experienced by them due to the Earth’s field is simply too small to cause significant deflection.

    I’d like to think about how to model this behaviour in any case.

    John A

    August 12, 2009 at 6:43 pm

  12. John A,

    I am just starting to learn this subject area but the existence of the “vertical rigidity cutoff” value (going back at least to the 50’s) indicates to me that the Earth’s magnetic field seems to be quite effective at turning away cosmic rays. 15GeV is a lot of energy. Have you considered the effect of the long distances over which it acts on an incoming ray? Would that make up for its weak magnitude?

    Joe

    August 13, 2009 at 5:28 pm

  13. Joe,

    he existence of the “vertical rigidity cutoff” value (going back at least to the 50’s) indicates to me that the Earth’s magnetic field seems to be quite effective at turning away cosmic rays. 15GeV is a lot of energy. Have you considered the effect of the long distances over which it acts on an incoming ray?

    I doubt it. Theoretically a magnetic field line can have infinite length.

    I’m going to actually admit to a small but stupid error in my previous comment. The actual equation for the Lorentz force is

    [tex]F = q(v \times B)[/tex]

    Note the q refers to the charge on the cosmic ray particle (which are usually protons or helium nuclei (alpha particles) but can be electrons (beta particles))

    Because of the vector form, the Force operates on the particle in the direction perpendicular to both the velocity vector and the magnetic field.

    So the maximum force on a particle is F = qvB but only if the particle is happens to be moving perpendicular to the magnetic field of the Earth (which is actually like a dipole field). If the particle is moving in the direction of the B field there is no force at all.

    If a particle were to be moving exactly perpendicular to the Earth’s magnetic field then

    [tex]F = qvB = \frac{mv^2}{R}[/tex] where R is the radius of curvature.

    This leads to [tex]R = \frac{mv}{qB}[/tex]

    We can substitute some figures for the mass (which we’ll assume to be a proton), the velocity (which we can work out by looking at the energy of the cosmic ray) and the strength of B for the Earth.

    It’s late at night for me and I’m tired so I’ll pick this up with some figures tomorrow. If you want, you can find some sample figures and see if the radius of curvature for a cosmic ray isn’t a lot bigger than the radius of the Earth.

    John A

    August 13, 2009 at 11:05 pm

  14. John,
    I had noticed the missing “q” but I figured it was late. Besides, as I get older, I try to type faster and my messages start to look like they have been encrypted.

    I will await your calculation at your convenience!

    I do think, though, that the issue is more complex than just the radius of curvature. The real question is whether the diameter of the capture cross-section for the cosmic rays becomes larger than the Earth’s radius and, if so, under what conditions. The problem becomes quite complex because a passing cosmic ray does not have to have an initial path with any relationship to a radial from the Earth’s center for it to be turned, even if only slightly, into a collision course with the Earth. I am not sure how to construct a linear set of integrals to cover all possible initial orientations and velocities a cosmic ray might have.

    I do agree with your assessment about paths parallel vs perpendicular to the magnetic field lines. This makes the polar regions qualitatively different than the equatorial and temperate regions.

    Joe

    August 14, 2009 at 3:10 pm

  15. OK. Cosmic rays are 90% protons, nearly 10% helium nuclei, the rest electrons and ionized elements heavier than hydrogen or helium

    Energies for cosmic rays can be anything up to [tex]10^{20}[/tex]eV but normally [tex]10^{15}[/tex]eV are usually quoted

    The charge of a proton is [tex]1.602 \times 10^{-19}[/tex]C and the rest mass is [tex]1.672 \times 10^{-27}[/tex] kg

    The maximum strength of the Earth’s magnetic field is a maximum of 0.6 gauss over the magnetic poles.

    If I use the energy applied in eV as the [tex]\Delta V[/tex] potential difference, and use the relativistic form of the equation then the radius of curvature is

    [tex]R = \frac{\gamma mv}{qB}[/tex]

    which becomes

    [tex]R = \sqrt{\frac{(\gamma +1)m \Delta V}{qB^2}}[/tex]

    where [tex]\gamma[/tex] is the Lorentz factor [tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    More in a while (why do I always do this at the end of the day?), when I calculate the Lorentz factor and try to work out what the minimum possible radius of curvature of a cosmic proton is when it encounters the Earth’s magnetic field.

    John A

    August 14, 2009 at 11:58 pm

  16. Just as an aide memoir, the Kinetic Energy is [tex]K.E. = q \Delta V = ( \gamma -1)mc^2[/tex]

    so therefore [tex]\gamma = \frac{q \Delta V }{mc^2} +1[/tex]

    John A

    August 15, 2009 at 12:06 am

  17. OK.

    The Lorentz factor is therefore

    [tex]\frac{1.602 \times 10^{-19} \times 10^{15}}{1.672 \times 10^{-27} \times (3 \times 10^8)^2} +1[/tex]

    which I calculate to be

    [tex]\gamma = 1064593.3 +1 \backsimeq 1064594[/tex]

    John A

    August 15, 2009 at 10:44 am

  18. Therefore the minimum radius of curvature is

    [tex]R = \sqrt{\frac{1064594 \times 1.627 \times 10^{-27} \times 10^{15}}{1.602 \times 10^{-19} \times (6 \times 10^{-5})^2}[/tex]

    which I calculate to be

    [tex]R=5.48028 \times 10^{10} metres[/tex]

    which is 0.366 astronomical units (or about a third of the way to the sun)

    or 8620 times the radius of the Earth.

    That’s the absolute minimum radius of curvature of a cosmic ray proton under the most favourable of circumstances.

    I cannot see how cosmic ray photons would be measureably deflected by the Earth’s magnetic field.

    John A

    August 15, 2009 at 11:04 am

  19. John,
    I will peruse your numbers!

    Joe

    August 16, 2009 at 4:39 pm

  20. […] Sloan has written his thoughts on my calculations as to whether Svensmark’s process should cause sufficient deflection towards the […]

  21. John,
    I verified your Lorentz factor and your result, coming up with the number 5.56E10 meters for the radius of curvature. Close enough!

    How much is that? The magnetosphere from the earth extends to 10 earth radii towards the sun and about 14 away from the sun. I’ll use 12 as a reference. The earth radius is 6.4E6 meters, making the distance a cosmic ray travels through the Earth’s magnetic shield equal to approximately 7.7E7 meters, 1/714 of the 5.5E10 radius for a 90 degree bend. So, the Earth’s field should result in much less than 1 degree of bending. You appear to be right. (Note that I am learning this as I go. I am a coulomb engineer, not a magnetics engineer.)

    This little exercise just emphasizes the point that the sun’s magnetic field is probably the primary contributor to the variation in the cosmic ray flux shown in the Terry Sloan’s posting today. Are there other contributions?

    Joe

    August 17, 2009 at 5:04 pm

  22. […] Sloan has written his thoughts on my calculations as to whether Svensmark’s process should cause sufficient deflection towards the […]

  23. This little exercise just emphasizes the point that the sun’s magnetic field is probably the primary contributor to the variation in the cosmic ray flux shown in the Terry Sloan’s posting today. Are there other contributions?

    Just have a look at the two new posts.

    BTW I was astonished at the Lorentz factor for a 10^15 eV proton when I calculated it. I never realised it was that high.

    John A

    August 17, 2009 at 8:30 pm

  24. John,
    Thanks for the links. I too was amazed by the Lorentz factor. My first impression was that you had calculated it incorrectly, but, no, you did not unless we both made the same mistake. One interpretation is that the mass of the energetic proton appears to us to be over 1 million times heavier than its rest mass. What must we look like to the proton since it sees us as 1 million times heavier. (Should we recalculate based on that fact? Does apparent G increase by the Lorentz factor? I have never seen such a statement for constant velocities.) I attempted to derive the velocity associated with such a Lorentz factor. The velocity was so close to “c” that I simply used the speed of light in the radius of curvatuve calculation. How does a physicist separate out the properties of a 1E20 cosmic ray vs a 1E15 cosmic ray?

    Joe

    August 18, 2009 at 4:35 pm

  25. Hello Solar Science,

    I am sending a news release regarding Earth’s magnetic field that you can use on your blog. Please consider using it. You can find images on my website.

    Dennis

    CRAM SCHOOL

    FOR IMMEDIATE RELEASE

    Contact: Dennis Brooks
    Phone: 1-808-566-0654
    Email: dennisbroo@gmail.com

    Earth’s Magnetic Field Is Produced By An External Dynamo System, Not An Internal Dynamo.

    Researcher finds that Earth’s magnetic field is not produced by an internal dynamo. Nor is it produced by ocean current. The dynamo is outside the Planet! New findings by independent researcher, Dennis Brooks, show that Earth’s magnetic field and the planet itself are components of a complex dynamo system, which surrounds the planet. The planet and its magnetic field are part of the dynamo.

    According to this new theory, no internal dynamo or ocean current helps in producing or maintaining the magnetic field because other planets with magnetic fields do not have ocean currents or iron cores.

    Image by NASA

    Each planet does not have a unique way of producing its magnetic field. The magnetic field of each planet is produced by a planetary dynamo system and its ring current.

    For many years researchers thought that a similar dynamo system was within the planet and that this internal dynamo generated the magnetic field. However, we know now that it is too hot inside the planet to produce and maintain a magnetic field there.

    The planetary dynamo system is composed of a magnetosphere, the planet, the magnetic field, radiation belts, ring current, and charged particles from the solar wind. The planet is the central component of the system and its rotation plays an important part in operating the dynamo and generating ring current. The magnetic field is generated by the system’s ring current, which is made up of charged particles. The magnetic field captures even more charged particles and brings them into the dynamo system as fuel. Everything works together.

    Earth’s inner and outer core simply cannot provide the fuel a dynamo system needs. If earth’s dynamo had to depend on energy from the planet for fuel, the entire planet would have been completely consumed many years ago.

    To learn more about Earth’s magnetic field, Visit
    http://sites.google.com/site/earthsmagneticfield/

    Dennis Brooks

    October 26, 2009 at 4:39 am


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