Robittybob1 Posted February 19, 2016 Author Share Posted February 19, 2016 Anyway, how is this relevant to the topic of discussion? What does this have to do with the age of Sun? it was to work out how quickly the dust from the debris disk can be cleared. If the dust stays longer the inner planets would be shielded from the radiation coming from the pre main sequence sun. So we are trying to work out if there are 3 mechanisms to clear the dust and in which direction the dust goes. Now we are bogged down on the mechanism of Poynting - Robertson effect. It's not too hard to calculate if you know the direction the particle is moving. The angular velocity of the body at the same orbit as the Earth will be: [latex]\omega=\frac{2*\pi}{365*24*60} = 1.19*10^{-5} radians/min[/latex] We observe the particle an then aim the laser to 16 minutes ahead of it's observed position, because it was there 8 minutes ago and it will take another 8 minutes for the laser beam to reach it. Then we should aim at 1.19*10-5*16 = 1.9*10-4 radians ahead of it's observed location. And if the aim is good enough, we should hit it.... That sounds like you are firing from a stationary Sun. The dust particle in the thought experiment is always directly overhead. Link to comment Share on other sites More sharing options...
pavelcherepan Posted February 19, 2016 Share Posted February 19, 2016 Shielded? How much dust do you think was there? Take a guess in particles/m3. I don't understand what you're doing. Are you trying to perform your own research of the topic and come up with your own number? In that case you really-really need to read up on planetary and star formation more and be ready to provide calculations. Otherwise, why would you think that existing research on the topic is inadequate? Do you think that a whole lot of scientists got it all wrong? And if that's the case, what is your reasoning behind it? Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 19, 2016 Author Share Posted February 19, 2016 (edited) Shielded? How much dust do you think was there? Take a guess in particles/m3. I don't understand what you're doing. Are you trying to perform your own research of the topic and come up with your own number? In that case you really-really need to read up on planetary and star formation more and be ready to provide calculations. Otherwise, why would you think that existing research on the topic is inadequate? Do you think that a whole lot of scientists got it all wrong? And if that's the case, what is your reasoning behind it? How much dust do we need between the PMS and planet Mercury to allow Mercury to be within the habitable zone? That is the amount of dust and that dust must not clear too quickly so that abiogenesis can occur on the inner planets. I have no idea how to express that in particles per m^3 Edited February 19, 2016 by Robittybob1 Link to comment Share on other sites More sharing options...
pavelcherepan Posted February 19, 2016 Share Posted February 19, 2016 That sounds like you are firing from a stationary Sun. The dust particle in the thought experiment is always directly overhead. Well, then you don't understand orbital mechanics. After all discussions we've had it's not surprising. How much dust do we need between the PMS and planet Mercury to allow Mercury to be within the habitable zone? That is the amount of dust and that dust must not clear too quickly so that abiogenesis can occur on the inner planets. You've already had a 300-page discussion of physforums on that topic. If you want to talk about habitability of Mercury create another topic. Stop hijacking your own discussions. Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 19, 2016 Author Share Posted February 19, 2016 (edited) .... You've already had a 300-page discussion of physforums on that topic. If you want to talk about habitability of Mercury create another topic. Stop hijacking your own discussions. No it was your point that the PMS being too large and too luminous that made me have to rethink the whole idea. Can we time the development of the planets to the stages of Sun development that is the purpose of this thread. It is not hijacked by my answer to your question but that is the ultimate intention of my study (in a general sense) and that is what you asked for. Well, then you don't understand orbital mechanics. After all discussions we've had it's not surprising. Why do you say that? I'm thinking the same about you. It is not the way to progress. Point out where I'm going wrong if you think I'm wrong, that's more likely to yield results. Edited February 19, 2016 by Robittybob1 Link to comment Share on other sites More sharing options...
pavelcherepan Posted February 19, 2016 Share Posted February 19, 2016 Why do you say that? I'm thinking the same about you. It is not the way to progress. Point out where I'm going wrong if you think I'm wrong, that's more likely to yield results. OK, let's try. You have a particle and from your POV on the Sun it's hanging in the same spot in the sky all the time. From the POV of the particle you're also in the same spot all the time, but for an external observer both you and the particle will be moving. Back to your reference frame. You see particle in the zenith. Is it there now? No. It was there 8 minutes ago when light reflected off of it. So now you try and shoot your laser at the particle. Should you aim at it's apparent position in the sky? No. Like I said before, it was there 8 minutes ago. Then should we aim at where the particle actually is at this moment in time, i.e. 8 minutes ahead of it's apparent position? Again no. By the time your signal reaches it, it will be ahead and you'll miss. This all boils down to the fact that you'd have to aim ~16 minutes of travel ahead of it's current apparent position and then you'll have a chance of hitting it. Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 19, 2016 Author Share Posted February 19, 2016 That was pretty good, and the terminology shows me you do know your stuff, zenith and signal very good. So you were on the Sun firing your laser up to a point that is always at the zenith (copied) Why not just fire it straight up knowing that the momentum of the Sun will go with the signal and carry it to the right place. You know this will happen for at any stage of the signals journey one could look along it one way looking down you will see the origin and looking up you will see the target. It is a series of triangles. Too tired to think - I'll try again tomorrow. -1 Link to comment Share on other sites More sharing options...
pavelcherepan Posted February 19, 2016 Share Posted February 19, 2016 (edited) Why not just fire it straight up knowing that the momentum of the Sun will go with the signal and carry it to the right place. Why?! Why would Sun impart any momentum on a photon? What is the mechanism of it? P.S. That was pretty good, and the terminology shows me you do know your stuff, zenith and signal very good. Stop patronizing me. Edited February 19, 2016 by pavelcherepan Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 19, 2016 Author Share Posted February 19, 2016 (edited) P.S. Stop patronizing me. Doesn't even look like me. Stop picking on me! Didn't you pick on me before I patronised you? Now that we've got past that let's get back to the science please. Why?! Why would Sun impart any momentum on a photon? What is the mechanism of it? That is the most important question for the day. Parallel to doing this thread I am also studying relativity and the conservation of energy and I'm curious as to what is going on for some terms are not conserved when you change inertial frames. I think we might have uncovered a situation, the odd one out, that can't be explained by relativity. (Note I often get ahead of myself.) Let us do this; if it looks like truly an unusual situation I will start a thread in a more appropriate section of the forum but in the meantime let's see if we can keep it related to the way the Sun cleared the inner solar system during the PMS stage. Agree? That was pretty good, and the terminology shows me you do know your stuff, zenith and signal very good. So you were on the Sun firing your laser up to a point that is always at the zenith (copied) Why not just fire it straight up knowing that the momentum of the Sun will go with the signal and carry it to the right place. You know this will happen for at any stage of the signals journey one could look along it one way looking down you will see the origin and looking up you will see the target. It is a series of triangles. Too tired to think - I'll try again tomorrow. I hope I only got negative rep points for patronising and not for lack of effort. If someone disagreed with the physics let they may speak to the forum. So we have this thought experiment going on with the star and this dust particle, and there is no relative motion between them as far as you can tell. You look at the dust and it is not moving, the dust looks at you and it sees you not moving. The thought experiment could be happening in deep space. Now you wouldn't have a problem of two space rockets flying in empty space, with zero relative motion between them, firing laser beams at each other would you? But how do you know those two rockets weren't in actual orbit around each other? OK the gravity will be extremely weak but it must seem weak between the Sun and the particle and because it is orbiting the forces are balanced. Is there a connection here? For in one situation you take aim and fire directly at the target but in the other you figure you need to fire 16 minutes ahead of the target! http://www.scienceforums.net/topic/93429-what-dates-are-accepted-for-the-age-of-the-sun/page-5#entry906944 Which is it? Edited February 19, 2016 by Robittybob1 Link to comment Share on other sites More sharing options...
Ophiolite Posted February 19, 2016 Share Posted February 19, 2016 Too tired to think - I'll try again tomorrow. Eventually it might take. Link to comment Share on other sites More sharing options...
swansont Posted February 19, 2016 Share Posted February 19, 2016 But how do you know those two rockets weren't in actual orbit around each other? You can tell if you are accelerating. That's not true of inertial motion. Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 19, 2016 Author Share Posted February 19, 2016 (edited) You can tell if you are accelerating. That's not true of inertial motion. How can you tell? I know it is not true inertial motion but how can one tell whether they need to take the acceleration into account? (That is how can you tell from within the thought experiment) Did you try and figure out what direction one would fire the laser from the Sun to hit the dust particle to get the P-R effect to work? Later we will see what happens when the dust is not heliocentric, does that make any difference if the Sun is spinning faster, does it get somewhat harder to hit the target? Edited February 19, 2016 by Robittybob1 Link to comment Share on other sites More sharing options...
swansont Posted February 19, 2016 Share Posted February 19, 2016 How can you tell? I know it is not true inertial motion but how can one tell whether they need to take the acceleration into account? (That is how can you tell from within the thought experiment) How you do it depends on the situation, but acceleration is not relative. You can tell if you are accelerating. Did you try and figure out what direction one would fire the laser from the Sun to hit the dust particle to get the P-R effect to work? Later we will see what happens when the dust is not heliocentric, does that make any difference if the Sun is spinning faster, does it get somewhat harder to hit the target? It's already been explained to you. You have to fire ahead of the target by 2x the time it take light to get there. The path light will take will look curved to you (if you can see the path), just like throwing a ball on a carousel/merry-go-round, because you are in a rotating frame. 1 Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 19, 2016 Author Share Posted February 19, 2016 (edited) How you do it depends on the situation, but acceleration is not relative. You can tell if you are accelerating. It's already been explained to you. You have to fire ahead of the target by 2x the time it take light to get there. The path light will take will look curved to you (if you can see the path), just like throwing a ball on a carousel/merry-go-round, because you are in a rotating frame. Thanks. I remember those demonstrations now. @Swansont - So does the position (distance from the center) and the rotational speed of the thrower make no difference to the path it takes? (Especially in a situation where the rotational speed of the thrower (not at the true center) can be varied independently from the carousel.) If I said "Increasing the thrower's rotational velocity is the same as changing the angle of firing the laser" i.e. firing it further ahead. Is that true? Note: I had not rejected Pavelcherepan's suggestion and I definitely wasn't sure of mine but now I recall the demonstrations of what you mention and the balls would definitely take a straight path but they looked curved to the thrower who was sitting on the carousel. Edited February 19, 2016 by Robittybob1 Link to comment Share on other sites More sharing options...
Mordred Posted February 19, 2016 Share Posted February 19, 2016 (edited) No the rotation speed makes no difference in the case of light which always travels at c. No matter what speed the Sun rotates the speed of light will remain at c. The radiation pressure it will exert will depend on its wavelength when it arrives at the dust particle. However due to the size of the particles not all the radiation pressure due to the full wavelength is involved. This is why you have the the Mie scattering terms in the equations Edited February 19, 2016 by Mordred Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 19, 2016 Author Share Posted February 19, 2016 (edited) No the rotation speed makes no difference in the case of light which always travels at c. No matter what speed the Sun rotates the speed of light will remain at c. The radiation pressure it will exert will depend on its wavelength when it arrives at the dust particle. However due to the size of the particles not all the radiation pressure due to the full wavelength is involved. This is why you have the the Mie scattering terms in the equations Is this a true statement? If you are firing forward and rotating in the same direction, since the speed of light is limited to c, those photons (that leave the gun and hit a particle directly overhead) are going to be blueshifted compared to those fired straight up or worse still fired backward. There was a name for this effect, i'll see if I can find it again, but it relates to the hemispheres of the sun. The side coming toward you produces blueshifted light compared to the light coming from the side going away from you. So would this be a true statement? "That the degree of blueshift is dependent of the orbital speed (and particularly the tangential speed on the surface) of the Sun". Edited February 19, 2016 by Robittybob1 Link to comment Share on other sites More sharing options...
Mordred Posted February 19, 2016 Share Posted February 19, 2016 (edited) Blueshift and redshift affect wavelength, now consider a spinning star emitting light will blueshift on one side and redshift on the other. The qauntity of the shift will be the same on both sides of the star. So the resulting shift total will be the same as though it were emitted from the center of the star. In other words the various shift at each point in the star averages out to be the same value. Closest point on the star, particularly over long distances. Edited February 19, 2016 by Mordred Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 20, 2016 Author Share Posted February 20, 2016 Blueshift and redshift affect wavelength, now consider a spinning star emitting light will blueshift on one side and redshift on the other. The qauntity of the shift will be the same on both sides of the star. So the resulting shift total will be the same as though it were emitted from the center of the star. In other words the various shift at each point in the star averages out to be the same value. Closest point on the star, particularly over long distances. So did you agree with both statements in #91? Wavelength and frequency are inversely related -- so yes Averaged the red and blue shift with be the same regardless of spin rate - I think you're right. Unfortunately we are not considering long distances but the region between the Sun and the first few planets. True or false? "The higher energy photons will becoming from the point behind the dust for the dust and the star are rotating in the same direction. so the radiant pressure will be higher driving the dust to higher orbits." Is this a true statement? If you are firing forward and rotating in the same direction, since the speed of light is limited to c, those photons (that leave the gun and hit a particle directly overhead) are going to be blueshifted compared to those fired straight up or worse still fired backward. There was a name for this effect, i'll see if I can find it again, but it relates to the hemispheres of the sun. The side coming toward you produces blueshifted light compared to the light coming from the side going away from you. So would this be a true statement? "That the degree of blueshift is dependent of the orbital speed (and particularly the tangential speed on the surface) of the Sun". The name of the effect is the differential Doppler effect. thanks wikipedia: https://en.wikipedia.org/wiki/Differential_Doppler_effect 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. Link to comment Share on other sites More sharing options...
Mordred Posted February 20, 2016 Share Posted February 20, 2016 (edited) The differential Doppler between the edges of our Sun at the equator works out to 0.0079 nm. Or plus or minus 0.0079/2. From the centre. It's extremely difficult to detect from Earth. The only time it will have any effect on dust is if the dust is close enough to the Sun that the Suns diameter blocks the opposite side or close enough to get a measurable difference in distance between the two edges. That would be well within the radius of Mercury. ( close enough that the average luminosity wavelength relation is more applicable). It's a handy principal to measure the Suns rate of rotation but for overall radiation pressure equations not so much. For the overall Poynting Robertson metric which applies to a far larger region it's too circumstantial to apply to the overall metric. You'd have to be extremely close to the Sun to have any measurable effect, at least as far as it's overall effect on radiation pressure. Key note the formula uses luminosity. Which is the stars radiated power over ALL wavelengths and is the amount of energy emitted by a star each second. If you wish to determine the average wavelength you can apply Weins law. Which will give the peak and wavelength via the Suns blackbody temperature. https://en.m.wikipedia.org/wiki/Wien%27s_displacement_law Our Suns peak wavelength via Weins law is 500 nm. ( on average). Our Sun has an 11 year cycle. Compare that to the differential Doppler value of 0.0079 nm Edited February 20, 2016 by Mordred Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 20, 2016 Author Share Posted February 20, 2016 (edited) I've been looking for papers that compare The differential Doppler (DDE) effect to the P-R effect in pre main sequence stars. DDE has many uses but there was nothing so far that I saw on it function in PMS dust disk. I am thinking if the dust disk is thick the star are harder to see and harder to study for there was a study on the debris disk of MS stars where the dust remaining is called the debris disk. But the two effects P-R and DDE both move the particles in opposite directions so that could contribute in the overall slowing the clearing of the dust in the inner parts of the dust disk once separation from the PMS star has occurred. As it was pointed out earlier the PMS sun would be much larger and spinning a lot faster so from the way that the DDE works the distances it operates at would be proportional greater and the strength of the blueshift greater. i.e. I'm not sure how much larger the PMS sun was but if it was 12 times as wide as one estimate (but that was too early IMO) you could estimate that the distance it operates at could be 12 times as far (I'm not sure now whether the size was volume or diameter, sorry, but the concept remains true the wider it is the more distant will its effect be). There is such a vast size reduction of the protosun in that period (starts off 1000 AU as a nebula and ends up PMS <1 AU just as rough concept). Then what is the size reduction rate in the PMS stages? Do you know the rate of contraction of the PMS sun during the PMS stage? Does anyone know of these estimates please? Edited February 20, 2016 by Robittybob1 Link to comment Share on other sites More sharing options...
pavelcherepan Posted February 20, 2016 Share Posted February 20, 2016 PMS sun PMS Sun was moody and irritable. Do you know the rate of contraction of the PMS sun during the PMS stage? Does anyone know of these estimates please? You can check this paper. http://adsabs.harvard.edu/full/1993ApJ...418..414P 1 Link to comment Share on other sites More sharing options...
Mordred Posted February 20, 2016 Share Posted February 20, 2016 Your forgetting one aspect of PMS stars. Let's assume our star in PMS stage was 12 times its current size. It's rate of rotation would be considerably slower. A stars rotation involves conservation of angular momentum. They spin faster as they shrink. Taking that into consideration you will have less differential Doppler. As they haven't reached the fusion stage, it's blackbody temperature is far lower thus the Poynting Robertson effect is also reduced. PMS Sun was moody and irritable. Roflmao Link to comment Share on other sites More sharing options...
Robittybob1 Posted February 20, 2016 Author Share Posted February 20, 2016 (edited) Your forgetting one aspect of PMS stars. Let's assume our star in PMS stage was 12 times its current size. It's rate of rotation would be considerably slower. A stars rotation involves conservation of angular momentum. They spin faster as they shrink. Taking that into consideration you will have less differential Doppler. As they haven't reached the fusion stage, it's blackbody temperature is far lower thus the Poynting Robertson effect is also reduced. Roflmao No I believe it to be different. They spin faster as they shrink but somehow that momentum has been lost now. (We will need to look into this again and clear it up, for I want to be accurate and sure.). I covered that earlier (in this thread) the Sun's rate of rotation appears to be slowing, for it was faster early on in the PMS. The temperatures are right up there Pavelcherepan has given some estimates of the surface temperature of the PMS sun. I can't remember exactly but they were in the order of 5,000 - 10,000 degrees C. It was being heated by gravitational compression, energy has to be lost to allow it to collapse. (we will have to revise this as well). Edited February 20, 2016 by Robittybob1 Link to comment Share on other sites More sharing options...
Mordred Posted February 20, 2016 Share Posted February 20, 2016 (edited) No it has nothing to do with appearance but basic physics. Take a figure skater, spinning with arms spread out. When she tucks her arms in she spins faster. Same physics. Conservation of angular momentum. http://www.einstein-online.info/spotlights/angular_momentum Edited February 20, 2016 by Mordred Link to comment Share on other sites More sharing options...
pavelcherepan Posted February 20, 2016 Share Posted February 20, 2016 (edited) No I believe it to be different. What you believe is irrelevant. Can you prove it? I covered that earlier (in this thread) the Sun's rate of rotation appears to be slowing, for it was faster early on in the PMS. Yeah, that would be because Sun is tidally interacting with planets, notably Jupiter, and loses angular momentum as a result. Edited February 20, 2016 by pavelcherepan Link to comment Share on other sites More sharing options...
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