What dates are accepted for the age of the Sun?

[Robittybob1]

Because your not listening. Instead your assuming I'm wrong simply because you don't understand it.

 

No I'm a resident expert, that grants me certain moderator leeway.

Well can you explain why the DDE adds more momentum to the dust particle when the photon comes from one side of the Sun?

Roflmao...

 

Our own motion relative to the CMB causes a dipole anisotropy to the CMB temperature measurements. Does that mean the universe is hotter in the direction of our motion?

 

Man spend a little time studying and less jumping to misguided conclusions.

 

Thats like saying I can burn you up by measuring the frequency your emitting and flying fast enough toward you while I measure your temperature.

Slow down and go to the topic for today DDE please.

wikipedia doesn't have much on it.

 

 

The Differential Doppler effect occurs when light is emitted from a rotating source.

In circumstellar environments it describes the difference in photons arriving at orbiting dust particles. Photons that originate from the limb that is rotating away from the particle are red-shifted, while photons emitted from the limb rotating toward the particle are blue-shifted.

This is not just for the benefit of a single dust particle but just about every photon emitted from the Sun is effected in some degree by this DDE.

http://adsabs.harvard.edu/full/1984CoSka..12...99K

"Nongravitational effects affecting small meteoroids in interplanetary space"

Mentions DDE

 

"The radiation force on small particles"

http://www.sciencedirect.com/science/article/pii/0032063381901446 you might have access to that paper.

"Radiation Forces on Small Particles in the Solar System" https://www.researchgate.net/profile/Steven_Soter/publication/23871263_Radiation_forces_on_small_particles_in_the_Solar_System_A_re-consideration/links/557c962708aeb61eae236401.pdf

 

 

 

Also briefly mentioned is a new

effect, the differential Doppler effect, discovered

by McDonough (1975), which may

be appreciable in other planetary systems.

IX. DIFFERENTIAL DOPPLER EFFECT page 34

Another, more subtle radiation drag,

identified by McDonough (1975), operates

because light emitted from the retreating

eastern half of the Sun will be red-shifted,

decreasing its momentum, while identical

photons from the approaching western half

will have increased momenta due to being

blue-shifted. This asymmetric delivery of

radiation momentum produces an additional

transverse force on an interplanetary particle

(Fig. 12).

We obtain a meaningful upper bound on

the effect by considering half of the solar

radiant energy to be emitted at the eastern

limb and the remaining half at the western

limb; that is, the particle (assumed on a circular

orbit) no longer sees a spherical Sun

but rather just two point sources, one

slightly red-shifted and the other equally

blue-shifted. The relative velocity between

the particle and the eastern "light bulb" is

Vrel = ~eRo - nr sin s r

= (to o-n)rsin~; (56)

the orbital angular velocity of the particle is

n, while the solar spin rate is too.

The transverse momentum resulting from

two photons of rest momentum po/2, emitted

on each side, striking the particle is then

P = (po/2)(1 + Vrel/C) sin

-- (p0/2)(1 - /)rel/c) sin

= po(vre~/c) sin ~. (57)

Following the radiation pressure derivation,

and substituting for sin ~ and Vr~, we find

FDDE ----- (QprSA/c)(R~/rc)(to O - n). (58)

This substantially overestimates the actual

FIG. 12. A schematic diagram of the differential

Doppler effect. (a) Photons arriving at the particle from

the retreating eastern hemisphere carry less momentum

on the average (because of red-shifts) than do

those from the approaching western hemisphere. The

particle is pushed forward. (b) The Sun is modeled by

point sources on each limb in order to use the relative

velocity between the limb and the particle.

force since most solar energy comes from

regions which have smaller Doppler shifts

than the limb values; nevertheless the correction

is less than an order of magnitude.

The force is small and positive for distant

particles but is a drag for particles within

the synchronous orbit position.

If this force were larger than the

Poynting-Robertson drag in certain regions,

then particles could be driven inward

when FDDE <Fp-R and outward when

FDDE > Fp-R and large dust concentrations

could accrue. However, from (17) and (58),

FDDE/Fp-R = (gJr)Z[(toJn) - l] (59)

and thus the differential Doppler force is

always less than the Poynting-Robertson

drag. It is important only if a particle is near

the solar surface (cf. Guess, 1962), where it

adds to the Poynting-Robertson drag; however,

it could be appreciable for particles

orbiting a contact binary system (S. J.

Weidenschilling, private communication,

1979). The effect is only significant because

it provides a more profound insight into the

phenomena that produce the PoyntingRobertson

effect itself.

 

I must admit there is a paucity of papers on the effect.

Edited February 20, 2016 by Robittybob1

[Mordred]

The problem isn't that a particle wouldn't be affected by DDE. The problem was thinking the Suns spin would be affected. The sun emits an average wavelength regardless of its spin. DDE effects won't change its spin rate as from the reference frame of the Sun as the emitter there is no DDE effect.

 

From the particle it's a different story. Frequency relates to energy, so a shorter wavelength will add more energy to the particle.

 

However as I mentioned the peak average wavelength is 500 nm on the 11 year cycle average. The DDE effect is plus or minus 0.00395 nm. The ratio difference between 500nm and 0.00395 nm is 1265822.78481.

 

Comparitively the 0.00395 nm would be a miniscule factor.

[Robittybob1]

Mordred did you access that paper https://www.researchgate.net/profile/Steven_Soter/publication/23871263_Radiation_forces_on_small_particles_in_the_Solar_System_A_re-consideration/links/557c962708aeb61eae236401.pdf

In there are some formulas for the calculation of the DDE. Would you be so kind as to help me with them (mainly getting the correct values for the variables? Each equation has a number so we could decide which ones are important and go from there.

It will be in a couple of days for I'm busy tomorrow.

The problem isn't that a particle wouldn't be affected by DDE. The problem was thinking the Suns spin would be affected. The sun emits an average wavelength regardless of its spin. DDE effects won't change its spin rate as from the reference frame of the Sun as the emitter there is no DDE effect.

From the particle it's a different story. Frequency relates to energy, so a shorter wavelength will add more energy to the particle.

However as I mentioned the peak average wavelength is 500 nm on the 11 year cycle average. The DDE effect is plus or minus 0.00395 nm. The ratio difference between 500nm and 0.00395 nm is 1265822.78481.

Comparitively the 0.00395 nm would be a miniscule factor.

I understood most of this. The only question I have is about the Sun. "From the reference frame of the Sun" is the Sun spinning on an axis or is it fixed? Is it a rotating frame of reference?

Edited February 20, 2016 by Robittybob1

[Mordred]

Good article its written in a good clear format, it essentially conforms to what Ive been explaining to you. Good study aid, I'll probably add it to my database.

 

I understood most of this. The only question I have is about the Sun. "From the reference frame of the Sun" is the Sun spinning on an axis or is it fixed? Is it a rotating frame of reference?

 

 

you can treat it as a rotating frame of reference

[Robittybob1]

Good article its written in a good clear format, it essentially conforms to what Ive been explaining to you. Good study aid, I'll probably add it to my database.

 

 

 

you can treat it as a rotating frame of reference

So what about that request for help in #128, will you help me tomorrow like in 24 plus hours time? We should go through that paper and give the variables some values and see what happens.

I find it confusing for they too say it could be a driving force or like the P-R effect a drag, and then one sentence later he says the P-R effect is always > DDE so he had contradicted himself in just two adjacent sentences IMO.

 

It could be that it cancels the P-R effect but never exceeds it, that could be the result (and what he is saying), but either way I want to see for myself using the predicted size and spin rate of the PMS Sun and see why it can never exceed the P-R effect.

 

Did you notice what seemed like an error in the paper?

....

you can treat it as a rotating frame of reference

I've never used math on rotating frames of references (RFoR) before. The last YT lecture (before dozing off) on them started talking about fictitious forces are we going to run into those sorts of problems?

Edited February 20, 2016 by Robittybob1

[Mordred]

yeah I can give you a hand later on, currently working 12 hour days in the field this week so my time scedule is tight.

[Robittybob1]

yeah I can give you a hand later on, currently working 12 hour days in the field this week so my time schedule is tight.

That will be fine just let me know when you are available.

 

Does this site have a way of selecting the proper symbols for variables? We'd need something like that. How do you get the Greek letters etc when discussing variables?

Edited February 20, 2016 by Robittybob1

[Mordred]

Here is a list of latex symbols for various characters.

 

When you do latex, type the word latex at the beginning and surround that word with [l.tex ] then type latex at the end surround it with [/l.tex ] use the forward slash in front of latex. (Replace the dot with the letter a)

 

If you quote this post look at this demo. ( I'll do a greek character and fraction combo)

 

[latex]w=\frac{\rho}{p}[/latex]

Edited February 21, 2016 by Mordred

[Robittybob1]

Here is a list of latex symbols for various characters.

 

When you do latex, type the word latex at the beginning and surround that word with [l.tex ] then type latex at the end surround it with [/l.tex ] use the forward slash in front of latex. (Replace the dot with the letter a)

 

If you quote this post look at this demo. ( I'll do a greek character and fraction combo)

 

[latex]w=\frac{\rho}{p}[/latex]

did you type all of that or use a LaTex editor?

Edited February 21, 2016 by Robittybob1

[Robittybob1]

In T Tauri stage if the only fusion reaction that happens is ⁷Li + ¹p -> 2 ⁴He then there should be no neutrinos produced. I think you're digging in a wrong place.

I have read that there is deuterium fusion during the T Tauri stage as well. I'll get references later, as I'm just re-reading the thread to see if there are gaps in the discussion.

 

Robbity, first of all, stars in T Tauri stage a proto-star already emits light, even though fusion hasn't started in the core. This is due to contraction heating. In fact, these stars have higher luminosity than what they will have in the future as main sequence stars, because they are simply larger. So you can't say that there are neutrinos coming, but there's no light. There is definitely light and there's solar wind and there might or might not be neutrinos.

 

Secondly, I've said it many times before and I hope you take a note of it this time. Even if there is fusion during T Tauri stage, it's only lithium burning, or pp II chain and there's not much lithium, so the total output of this fusion is probably tiny compared to the heat generated due to contraction.

 

And lastly, yes, based on what we've seen so far, Sun should have gone through T Tauri stage. It seems like it's a normal stage of development for all main sequence stars.

....

Lithium but nothing about deuterium. But this was a very important post. Thanks Pavelcherepan.

Not hotter. Just brighter. Effective photosphere temperatures of T Tauri stars are about the same as the sun ~5500 K but they are bigger and hence higher luminosity.

 

The more time paseses, the less of daughter isotopes you have and the less of mother isotopes. It gets increasingly harder to measure relationship between those and as a result the error increases exponentially (correct me if I'm wrong ) with the increase of the time period measured.

Appreciated 5500 degrees Kelvin bigger hence higher luminosity, and spinning faster so the DDE is "stronger???

from a quick google Poynting Robertson effect is due to asymmetries involving very small particles of dust in orbit about a star.

 

Very simplistically -

From the POV of the particle the sunlight hits it a tiny bit more on its leading side than its trailing side (it is running into the rain) thus there is a tiny loss of angular momentum as the inequality of pressure from radiation provides a net torque against its rotational motion around the star. From the POV of the star the light hits the particle at exactly 90 degrees BUT the re-emission of thermal radiation is stronger in the direction of travel than the reverse and thus the imbalance of photons emitted can be seen as lowering the angular momentum WRTo the star

The simplicity still makes a lot of sense. Thanks.

[Mordred]

did you type all of that or use a LaTex editor?

 

 

I type all my latex in, its easy with a little practice on the syntax

[Robittybob1]

 

 

I type all my latex in, its easy with a little practice on the syntax

OK and what do you do your calculations on? Like I tend to use Excel formulated worksheets for each problem. That way I can vary the magnitude of one variable at a time and drag the formulas down and graph the results and that way I can get a feel what the equation is doing. Then repeat that for the other variables.

With the DDE and P-R effect we should be able to see how each operates at the same time.

Do you do that?

[Mordred]

Normally I would use excel and scilab. (Similar to matlab). Except I have to replace my laptop. So lately I do the calcs by hand or on Wolfram.

[Robittybob1]

Normally I would use excel and scilab. (Similar to matlab). Except I have to replace my laptop. So lately I do the calcs by hand or on Wolfram.

So does your computer have excel, for then when we do the calculations if we both use it we will have similar looking equations.

[Mordred]

Before you worry about DDE specifically I would run through the Doppler shift calcs. Get a feel on those equations. Also run the calcs for Weins displacement law, convert wavelength to Blackbody temperature and back a few times.

 

For PR, a preliminary is luminosity, key note play with brightness and surface area of the emitter.

 

(This is based on the confusions you've had on this thread)

One bite of the Apple at a time as they say

Edited February 21, 2016 by Mordred

[Robittybob1]

Before you worry about DDE specifically I would run through the Doppler shift calcs. Get a feel on those equations. Also run the calcs for Weins displacement law, convert wavelength to Blackbody temperature and back a few times.

 

For PR, a preliminary is luminosity, key note play with brightness and surface area of the emitter.

 

(This is based on the confusions you've had on this thread)

One bite of the Apple at a time as they say

Take it softly micro bits at a time ... OK

I'll try looking at all of those at the same time

I just want to make sure we are using the same values as inputs into the equations, the initial values of the variables and the exact meaning of the symbols used. Can you help me on those please?

How is the Wien's Displacement law (WDL) going to help us? Wikipedia on WDL

 

Wien's displacement law states that the black body radiation curve for different temperatures peaks at a wavelength inversely proportional to the temperature. The shift of that peak is a direct consequence of the Planck radiation law which describes the spectral brightness of black body radiation as a function of wavelength at any given temperature.

We will just be using the one temperature of the Sun to calculate the P-R effect and the DDE. Were there temperature dependent terms in any of those effects?) We shouldn't compare them by varying the temperatures but using the same temperature at any one time. Temperature of the PMS Sun is a variable but we are not going to vary it while making the comparisons.

Sure the Doppler shift calculations will be important but wouldn't the DDE formulas already have them included?

Edited February 21, 2016 by Robittybob1

[Mordred]

what temperature you choose will be up to you, I'll show you how the equations work but you'll be doing the main calcs.

 

for now we won't worry about the differential rotation of the Sun. It spins faster at the equators than any other latitude. So we will just use the equator rotation rate at the very edge.

 

Take it softly micro bits at a time ... OK

I'll try looking at all of those at the same time

I just want to make sure we are using the same values as inputs into the equations, the initial values of the variables and the exact meaning of the symbols used. Can you help me on those please?

How is the Wien's Displacement law (WDL) going to help us? Wikipedia on WDL

We will just be using the one temperature of the Sun to calculate the P-R effect and the DDE. Were there temperature dependent terms in any of those effects?) We shouldn't compare them by varying the temperatures but using the same temperature at any one time. Temperature of the PMS Sun is a variable but we are not going to vary it while making the comparisons.

Sure the Doppler shift calculations will be important but wouldn't the DDE formulas already have them included?

 

 

Weins law is going to help, because not all datasets will have the info your looking for. You may just get the luminosity or blackbody temp. Then its up to you to calculate the rest.

 

 

so for starters lets assume you have a blackbody temperature from peak wavelength from the Sun at 5000 degrees. You can use Weins law to now calculate the average wavelength.

 

https://en.wikipedia.org/wiki/Wien%27s_displacement_law

 

 

now use the formula on this page to calculate the emitter wavelength. The Weins displacement constant is on that page as well. Should be straightforward.

[Robittybob1]

what temperature you choose will be up to you, I'll show you how the equations work but you'll be doing the main calcs.

 

for now we won't worry about the differential rotation of the Sun. It spins faster at the equators than any other latitude. So we will just use the equator rotation rate at the very edge.

 

 

 

Weins law is going to help, because not all datasets will have the info your looking for. You may just get the luminosity or blackbody temp. Then its up to you to calculate the rest.

 

 

so for starters lets assume you have a blackbody temperature from peak wavelength from the Sun at 5000 degrees. You can use Weins law to now calculate the average wavelength.

 

https://en.wikipedia.org/wiki/Wien%27s_displacement_law

 

 

now use the formula on this page to calculate the emitter wavelength. The Weins displacement constant is on that page as well. Should be straightforward.

By average do you really mean peak? From the link there is no mention of averages just peaks.

There is a full range of wavelengths so when you say emitter wavelength are you still talking about the peak?

 

Wien's displacement law states that the black body radiation curve for different temperatures peaks at a wavelength inversely proportional to the temperature.

Edited February 21, 2016 by Robittybob1

[Mordred]

Yes in the wavelength values in our Sun the peak is in the yellow green spectrum. However there is considerable power in the ultraviolet.

 

https://en.m.wikipedia.org/wiki/Black-body_radiation

[Robittybob1]

wavelength max = b/T

b≈2900 μm·K

T = 5000 K

wavelength max = 0.58μm

 

Did I get that right?

 

This "visible light" corresponds to a wavelength range of 400 - 700 nanometers (nm) and a color range of violet through red. The human eye is not capable of "seeing" radiation with wavelengths outside the visible spectrum.

So that is a peak right in the middle of the visible light range 580 nanometers.

Edited February 21, 2016 by Robittybob1

[Mordred]

yes in the case of the spectrum of our sun. and yes correct.

 

now take the suns diameter calculate the radius and roatation speed to calculate the blueshift on one side then redshift on the other. Redo the above calc at 5800 k.

 

when you do that you just calculated DDE influence from our sun today.

 

post what values your going to use for rotation velocity, diameter and temp. first do a static observer.

Don't worry this will get more complex as we go. Edited February 21, 2016 by Mordred

[Robittybob1]

yes in the case of the spectrum of our sun. and yes correct.

 

now take the suns diameter calculate the radius and roatation speed to calculate the blueshift on one side then redshift on the other. Redo the above calc at 5800 k.

 

when you do that you just calculated DDE influence from our sun today.

 

post what values your going to use for rotation velocity, diameter and temp. first do a static observer.

Are you using these values Rotational periods

25.379995 days at the equator or 2192831.568 seconds

 

To convert that to radians per second 2 Pi radians in 360 degrees so that is 2 pi()/2192831.568

In radians per sec = 2.86533E-06

at 5800 wavelength is 500 nm.

 

Radius of Sun 696.3 million m or 6.963E+8 meters

 

What is the formula for the Doppler effect? Are you going to use the ones for sound or is the one here OK

https://en.wikipedia.org/wiki/Doppler_effect#General

[Mordred]

This one will work to start with. We will need to adapt this one later.

 

Those values will work were just training on how to model build atm

Are you using these values Rotational periods

25.379995 days at the equator or 2192831.568 seconds

 

To convert that to radians per second 2 Pi radians in 360 degrees so that is 2 pi()/2192831.568

In radians per sec = 2.86533E-06

at 5800 wavelength is 500 nm.

 

Radius of Sun 696.3 million m or 6.963E+8 meters

 

What is the formula for the Doppler effect? Are you going to use the ones for sound or is the one here OK

https://en.wikipedia.org/wiki/Doppler_effect#General

[Robittybob1]

I'll try this from the reference to Doppler effect above:

If the speeds v_\text{s} \, and v_\text{r} \, are small compared to the speed of the wave, the relationship between observed frequency f and emitted frequency f_\text{0} is approximately[5]

Observed frequency Change in frequency
f=\left(1+\frac{\Delta v}{c}\right)f_0
\Delta f=\frac{\Delta v}{c}f_0

 

 

The formulas didn't paste very well but they were the ones where the speeds are much lower than the speed of the wave in the medium e.g the Sun's rotational speed is low compared to the speed of light.

They are also working in frequency not wavelengths so we need to change the wavelength back to frequency.

Do you agree?

 

Also the velocities are not in radians but something like instantaneous tangential velocities. So we don't seem to have calculated the right values to begin with.

We need the difference in velocity of both limbs of the Sun (just at the equatorial regions where one side is going away and the other coming toward an observer at a distance).

We need the wavelengths converted to frequencies.

Edited February 22, 2016 by Robittybob1

[Mordred]

You can for the purpose of learning choose rounded off values. At the moment were more interested in the steps themselves.

I'll try this from the reference to Doppler effect above:

 

The formulas didn't paste very well but they were the ones where the speeds are much lower than the speed of the wave in the medium e.g the Sun's rotational speed is low compared to the speed of light.

They are also working in frequency not wavelengths so we need to change the wavelength back to frequency.

Do you agree?

 

Also the velocities are not in radians but something like instantaneous tangential velocities. So we don't seem to have calculated the right values to begin with.

We need the difference in velocity of both limbs of the Sun (just at the equatorial regions).

We need the wavelengths converted to frequencies.

Very good, remember to learn latex you can quote a post with a formula so do that on my post (it will help learn the rules and syntax)

 

Here is some useful relations.

 

[latex]\frac{\Delta_f}{f} = \frac{\lambda}{\lambda_o} = \frac{v}{c}=\frac{E_o}{E}=\frac{hc}{\lambda_o} \frac{\lambda}{hc}[/latex]

 

 

Remember baby steps we will modify this formula for dust later.

 

[latex]f=\frac{c+v_r}{c+v_s}f_o[/latex]

c=velocity of waves in a medium

Vr is the velocity measured by the source using the sources own proper-time clock(positive if moving toward the source

vs is the velocity measured by the receiver using the sources own proper-time clock(positive if moving away from the receiver)