Temperature of Solar Wind: Complexities of Measuring Heat in a Stream of Particles

  • #1
jefo
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I read that the solar wind is a stream of particles, primarily electrons and protons, flowing out from the sun at speeds as high as 900 km/s, and at a temperature of 1 million degrees. Using the equipartition theorem, I calculate that a proton moving at 900,000 meters per second exhibits a temperature of over 32 million degrees Kelvin, and an alpha particle would be 4 times as hot! What gives?
 
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  • #2
Why do you think something not in thermal equilibrium even has a temperature?
 
  • #3
jefo said:
electrons and protons, flowing out from the sun at speeds as high as 900 km/s
So this is the maximum observed particle speed ##v_{\text{max}}##.
jefo said:
and at a temperature of 1 million degrees.
Assuming thermal equilibrium, this temperature reflects average particle speed: ##v_{\text{avg}}\propto T^{1/2}##.
jefo said:
What gives?
At any fixed temperature, ##v_{\text{avg}}\ll v_{\text{max}}##.
 
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  • #4
Apart from the points above, a single proton does not have a temperature. Temperature is a statistical property of a system in thermal equilibrium.
 
  • #5
Okay, renormalize. Let me try again.

To determine the AVERAGE (proton) particle speed streaming out of the sun, I can rearrange the equipartition equation thus: v^2 = 3kT/m. Plugging in 1.38x10^-23 (Boltzman constant) for k, 1,000,273 degrees Kelvin for T, and 1.67×10^-27 (mass of a proton in Kg) for m, yields only 2.48×10^10 for v^2.

As I understand it, v is in meters per second. I get 1.57×10^5 for the square root of v squared, which is 157 kilometers per second, way less than the escape velocity. I guess I better read the articles posted by berkeman and jasonRF listed below.

Thanks to all who replied, facilitating my continuing education (I am 76 years old. I was 11 when Sputnik orbited, before the solar wind was noticed by anyone).
 
  • #6
jefo said:
I get 1.57×10^5 for the square root of v squared, which is 157 kilometers per second, way less than the escape velocity. I guess I better read the articles posted by berkeman and jasonRF listed below.
Yes, you might start by reading Wikipedia since even it knows more about the solar wind than I do! https://en.wikipedia.org/wiki/Solar_wind
In particular, that entry states:
"While early models of the solar wind relied primarily on thermal energy to accelerate the material, by the 1960s it was clear that thermal acceleration alone cannot account for the high speed of solar wind. An additional unknown acceleration mechanism is required and likely relates to magnetic fields in the solar atmosphere."
So sun escape velocity may not be so important since there is apparently an active force propelling the wind away from the sun.
 
  • #7
renormalize said:
Yes, you might start by reading Wikipedia since even it knows more about the solar wind than I do! https://en.wikipedia.org/wiki/Solar_wind
What a wonderful article. Although it helps answer my question, it also reveals at least two more. One new question is: Is there a sorting process going on, separating higher energy particles occurring in the distribution, or is there another acceleration force active here? Quoting from the article:

"The Sun's corona, or extended outer layer, is a region of plasma that is heated to over a megakelvin. As a result of thermal collisions, the particles within the inner corona have a range and distribution of speeds described by a Maxwellian distribution. The mean velocity of these particles is about 145 km/s, which is well below the solar escape velocity of 618 km/s. However, a few of the particles achieve energies sufficient to reach the terminal velocity of 400 km/s, which allows them to feed the solar wind. At the same temperature, electrons, due to their much smaller mass, reach escape velocity and build up an electric field that further accelerates ions away from the Sun."

Two criticizms: I really don't believe that fast moving electrons are able to "build up" more of an electric field than they already possess (at rest). And, I would prefer to see the plain English term, "heat up", in place of the article's somewhat cryptic "achieve energies". After all, the temperature of an item is a measure of the average kinetic energy of the item's constituent particles.

On the plus side, I can now answer my own question: What is the temperature of the solar wind (at least at it's source)? Using the equipotential equation (T=mv^2/3k), and plugging in 400,000 meters per second for v, I find a temperature of 6,500,000 degrees Kelvin for protons. The article also mentioned that 8% of the (non electron) particles have an atomic mass of 4 or more, which would increase the average kinetic energy, therefore the temperature of the wind as a whole, at least 25%. That puts the temperature of the solar wind at it's source (2 solar diameters above the photosphere), in excess of 8 million degrees Kelvin!

Now, the only remaining part of my question is: What is the temperature of the ballistic solar wind at 1 AU (in the vicinity of earth), after most of the kinetic energy has been converted to potential energy? Would that also be the temperature of the particles (ions) trapped in the earth's magnetic field?
 
  • #8
Vanadium 50 said:
Why do you think something not in thermal equilibrium even has a temperature?
That question reminds me of, "Why do you think a tree falling in the forest makes any noise if there is no one around to hear it?"

If you could stick a thermometer in (you name it), you will not only see indication of temperature, you will also see the rate of increase or decrease of that temperature, equilibrium or no equilibrium.
 
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  • #9
A thermometer reads a temperature when it is in thermal equilibrium with the thing it is measuring. If the thing it is measuring is not in thermal equilibrium, the reading on a thermometer is meaningless, and indeed two thermometers using different materials or principles of operation might well read two different values.
 
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  • #10
Orodruin said:
Apart from the points above, a single proton does not have a temperature. Temperature is a statistical property of a system in thermal equilibrium.
I have to disagree. Unless the proton is in a particle accelerator, it has a temperature, according to this definition: "In science, temperature is a measure of hotness or coldness, which in turn is a measure of the kinetic energy of particles."

Maybe I should have said 'typical' proton, or 'average' proton. However, even a single proton that is not moving has a temperature . . . . . Absolute zero. I submit that the temperature, in degrees Kelvin, of a proton moving a v meters per second is 4.03x10^-5v^2 (4 hundredthousanths of v squared).
 
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  • #11
the following is nitpicking, since I ASSUME (but you should SAY) that you are talking in all cases about motion relative to the sun.

jefo said:
even a single proton that is not moving
There is no such thing. It would imply a preferred frame of reference. If you mean not moving relative to the sun, or the Earth, or whatever, then say so. You can't just say "not moving".

Also, since, as you say,
jefo said:
In science, temperature is a measure of hotness or coldness, which in turn is a measure of the kinetic energy of particles.
the "temperature" of a single particle is whatever positive value you would like to call it since you can just choose a reference frame that has it at any speed up to just below c.
 
  • #12
Unfortunately, non-equilibrium thermodyanmics cannot be understood at the B-level or with high school physics definitions. However, to convince you it is a "thing", what is the temperature of a thermos bottle, if the inside contains hot coffee, but immediately outside the bottle is a raging blizzard?
 
  • #13
Re: Comment by phinds:

Good point. Motion is relative. In that case, so is energy. For that matter, temperature would also be relative to a frame of reference, although that is hard for me to visualize.
 
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  • #14
jefo said:
Unless the proton is in a particle accelerator, it has a temperature, according to this definition: "In science, temperature is a measure of hotness or coldness, which in turn is a measure of the kinetic energy of particles."
That is not a complete and accurate definition of "temperature", because it is leaving out several other conditions that must apply before we can get a temperature from the kinetic energy of particles. One of these is as @Vanadium 50 says being in equilibrium (and even that is an oversimplification).

Sites like sciencenotes.org are why PhysicsForums has its rule that acceptable sources must be textbooks or peer-reviewed publications. They are interesting, often helpful for getting started and giving a non-technical audience a sense for how things generally work... but in simplifying for that audience they leave out so much that they cannot be used uncritically.
 
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  • #15
jefo said:
I have to disagree.
Congratulations, you just disagreed with the actual definition of temperature. Good luck rewriting physics with your own definitions and having people actually understand what you mean.
 
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  • #16
jefo said:
I have to disagree. Unless the proton is in a particle accelerator, it has a temperature, according to this definition: "In science, temperature is a measure of hotness or coldness, which in turn is a measure of the kinetic energy of particles."
The problem with this is that it's assuming that there's a meaning to "the kinetic energy" of particles. There isn't - as you noted later on, energy is frame dependent.

If you have two or more particles, though, one can always pick a frame in which their total momentum is zero. In this frame the individual particles may be moving, but their center of mass isn't (this is like the case of still air in a room). Then there's a meaning to the temperature, which is to ask about the average energy of particles in the frame where the gas is stationary. But if you have only one particle, that's always zero - or, better said, it doesn't make sense to ask the temperature of one particle.

I'm afraid that's only the beginnings of the complexity around temperature in this kind of case.
 
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  • #17
Orodruin said:
Congratulations, you just disagreed with the actual definition of temperature
It is embarrassing to have revealed myself to be such a moron, by posting replies (and even this whole thread) prior to doing a little more research.

That said, my research has revealed a web page posting ongoing readings taken by a satellite hovering 1% sunward of the earth: swpc.noaa.gov/products/real-time-solar-wind. Among the readings displayed are continuous measurements of velocity, temperature, and density.

To sum it up: The velocity ranged between 300 and 400 km per second (around midnight UPC T was slightly higher). Temperature averaged around 100,000 K. Density was 2 to 3 per cc. Obviously, plugging those values (v, T) into the equipotential formula [m=4.14x10T/v2] yields nonsense. However, thanks to replies posted by ibix, phinds, orodruin, and others prodding me to specify the frame of reference of my temperature readings, I have grown a new brain cell that helps me deal with it.

Here is what I have learned since yesterday: The temperature posted on the noaa page, T ~ 100k K, measures the temperature reflecting the kinetic energies of particles within the frame of reference of the solar wind, which is moving toward the earth at a speed of 3x10^5 km/s.

Otherwise, it would be like me in a supersonic jet with a thermometer stuck out into the airstream, trying to measure the air temperature, instead of using a weather balloon like the pros do.

Now, does anyone know how to obtain temperature data from the Magnetosphere Multiscale Mission?
 
  • #18
I mean VELOCITY was slightly higher (than 400 km/second).
 
  • #19
Oops again. The equipotential equation for determining the average mass of particles is m=4.14×10^-23T/v^2. I missed the "^-23". Though even with the right equation, the result is still nonsense (the average mass of the particles is NOT 1/36th that of a proton).
 
  • #20
While looking for something else, I ran i to this paper ApJ 944:82 which shows data on the temperatures of the proton and electron components of the solar wind. You will notice they are not equal and not even close to equal.

It is also the case that the small component of alphas and heavier ions is not at the same temperature as either the protons or electrons.

"The temperature of the solar wind" makes as much sense as "the temperature of the United States." Miami is not Fairbanks. Just as the US is not in thermal equilibrium with itself, neither is the solar wind.

As far as thermodynamics and relativity, we do not have a good theory of that. Non-relativistically, there are several equivalent definitions of temperature. Relativistically, they are no longer equal. To have a sensible discussion, one needs to be very, very careful with terminology and definitions, rather than a fast and loose B-level discussion.
 
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  • #21
The key point is that collisions between particles of different mass share energy more slowly than particles of the same mass, so this is why electrons and protons can have a "temperature" without it being the same as each other. What is really meant is that in the frame where their average velocity is zero, they have something close to a Maxwellian velocity distribution with mean energy per particle of 3/2 kT, which defines what is meant by T for each constituent. But it's actually even more complicated than that, because the kinetic energy associated with motion along the magnetic field can be different than the kinetic energy associated with motion perpendicular to the magnetic field, so you will see references to "parallel T" and "perpendicular T", even though now you can see we have gotten extremely far away from any notion of thermodynamic equilibrium. Worse, there are nonMaxwellian components that involve a subset of the electrons that stream freely and are only held back by the electric field they themselves create when they try to get away from the gravitationally bound protons. So the forces and energy distributions in the solar wind are quite complex, yet the concept of individual particle temperatures both along and perpendicular to the magnetic fields still prove useful notions, since there are still mostly Maxwellian distributions in those different directions, due to the fact that Maxwellians are the most likely ways to distribute some fixed amount of available kinetic energy. You can think of this as a statistical approach to the concept of temperature, rather than a strict thermodynamic one.
 

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