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Neil9327

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About Neil9327

  • Rank
    Baryon
  • Birthday 08/05/1970

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  • Location
    London, UK
  • Interests
    Wacky science and engineering ideas
  • College Major/Degree
    Cambridge University, UK
  • Occupation
    IT
  1. That's true - but slightly misses the point - probably because I haven't explained it properly. Climate change. Could you grow plankton or other biomass on the surface of the oceans to cover them completely, with the aim being to absorb carbon dioxide from the atmosphere at a much greater rate than algae can under the water? For two reasons: 1. At the surface there is greater availability of CO2 from the air compared to the water, because fresh air carrying new CO2 arrives constantly by the wind, whereas dissolved CO2 has to make its way to the plankton via diffusion, which is a slower process. 2. There is more power from the sun at the surface, power to drive photosynthesis, compared to under the water (due to as stated heating of the water). If the nutrient availability problem could be solved, could mankind cover the oceans with floating biomass? And leave it there to "hoover up" some or even all of the excess CO2 that is being produced by mankind's burning of fossil fuels? Basically to leverage the huge power from the sun that is currently being wasted on heating up water in the ocean.
  2. That's interesting. Quoting from that Wikipedia page, it says: "Phytoplankton account for about half of all photosynthetic activity on Earth". Now, I wonder how efficient that process is? It occurs to me that for ocean phytoplankton to photosynthesize light, that light has to first travel through a certain distance of water. But (from: https://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water), we know that a fair amount of the light will instead be absorbed by the water and dissipated as heat, before it has a chance to reach a phytoplankton cell. I wonder whether the efficiency of this process could in principle be improved, perhaps quite substantially, by artificially placing phytoplankton in high concentrations, in a thin layer on the surface of the oceans so that all of the light reaches it?
  3. Maybe. How would you hold the sheets in position though, and stop them either rising to the surface (where they might dry out) or moving deeper into the ocean?
  4. Ah yes, nutrients. That would explain it. Thanks! Now that example you have given, about lava added to water giving rise to chlorophyll, is interesting. "Once the lava stopped flowing into the ocean, the patch dissipated within a week" I wonder why this was? One reason might be that the plant growth used up all of the nutrients, leaving none left to support additional growth. But I think the more likely explanation, perhaps, is that the nutrients diffused away in every direction until they became too dilute for the plants to use - and growth stopped. I wonder whether there is a way to provide nutrients onto the ocean surface, in such a way that it is kept concentrated? One way might be to manufacture tiny spheres made of a material that allows water to diffuse through, but nutrients not. But this might be a big ask given the high degree of salinity. Another approach might be to grow the types of surface-free-floating seaweed that survive in the oceans, but to grow them initially in a lab in water that is artificially very high in nutrients. This might make their internal composition very high in nutrients too. Or you could even inject the nutrients in somehow. Then you could release them in the ocean and they might perhaps grow, say, to 10 times their current weight before the concentration of nutrients available to use becomes too low for growth to continue. I appreciate this is just a thought experiment and probably wrong in most respects!
  5. This is a question that has puzzled me for some time. It is that on land in every part of the world where there is abundant water and sunlight, there is vegetation everywhere (rainforests typically). But on the oceans in the same parts of the world (tropical and equatorial regions) there is typically very little, if any, floating seaweed. Given that floating seaweed is able to live in salt water, has access to abundant amounts of water (obviously), and has the same access to the air to support photosynthesis, why hasn't floating seaweed grown to the same extent as the plants on land? Now, there is one type of seaweed that does seem to accumulate - Sargassum. But not to anything like the extent that plants do on land. And then only in one part of the Atlantic Ocean - the Sargasso sea (named after it). But even then, judging from this photo of it, there doesn't seem to be very much of it as a percentage of the ocean surface (less than 5% I think): https://en.wikipedia.org/wiki/Sargasso_Sea
  6. Is there a link here? So on the one hand, the energy stored in the electromagnetic field created by a (constant DC) current passing through a coil (from its inductance L), and the kinetic energy of the electrons themselves passing around the link. And a follow-on question (assuming no such link) is whether there is any effective contribution to the inductance of a wire from the kinetic energy of the electrons in the (constant DC) current flowing through it. What are the equations?
  7. I think I've got it now, after reading this .doc file. It is all to do with the fact that the accelerating twin starts to receive faster return time pips immediately after the change in velocity (the turnaround) whereas the stationary twin has to wait for <distance>/c until they see the turnaround and start getting the faster blips. It is weird yet so simple to conceptualise once you've "got it". I wrote my own version of how this happens using my Andy&Brian scenario without using worldlines, but I now see that Janus's post above explains it more succinctly than I have. Still I've posted it here (in italics) just so I can feel I haven't wasted 20 minutes of typing (OK I'll admit it - I wasted 20 minutes of typing) Thanks very much for all your replies - much appreciated! Andy and Brian start at time=0 at the start point. Andy is stationary the whole time. Each has a beacon that sends a signal (a pip) every hour. At time=0, Brian moves out at half the speed of light towards a distant location. While Brian is in steady state motion travelling out, due to dopplar shift alone, each receives one pip from the other every 1.5 hours. Looking at the journey from Brian's perspective: After 90 hours (according to Brian), Brian has received 60 pips from Andy. Brian has travelled a distance of 45 light-hours. Brian then stops and starts back towards Andy at the same speed. Brian takes 90 hours (according to Brian) to get back to Andy. While Brian is travelling back, he can see pips more frequently from Andy - every 40 minutes to be precise, due to dopplar shift alone. So from Brian's perspective Brian receives a further 135 pips (90*1.5) from Andy. So total number of pips received by Brian from Andy = 60 + 135 = 195 pips. This is for Brian's own 180 pips. Looking at the same journey from Andy's perspective: On the outbound journey, Andy sees pips at the lower rate of one pip every 1.5 hours. However since it takes 90hours * 0.5*speed of light for the light showing Brian's arrival at the turning point to reach Andy, then from Andy's perspective, it has taken 135 hours of his time (Andy's time) to reach that destination. Over that 135 hours Andy receives 90 pips from Brian. And of course 135 pips from himself. Once Andy sees Brian reach the destination and start back, then Andy starts receiving pips from Brian at the quicker interval (every 40 minutes). We know that for the return journey that Andy must receive a further 90 pips from Brian because whatever relativity effects there are, the number of pips from Brian's machine in total for the trip must be the same whether measured by Andy or Brian. The time taken (from Andy's perspective) to receive the 90 pips from Brian is 90/1.5 = 60 hours. Add this to the 135 hours already spent watching the outbound journey gives a total number of hours for Andy from Andy's perspective of 195 hours (195 pips) Total number of pips received by Andy from Brian = 90 + 90 = 180 pips. This is for Andy's own 195 pips. So in both of the above cases Andy has aged 15 days more than Brian. Cool! And the reason why the situation is not symmetrical is, as indicated by the document you referenced, that the one that accelerates (Brian in this case) starts to see the change in frequency of receipt of pips as they complete the acceleration. But due to the distance between the two at that point, the one that has not accelerated (Andy in this case) has to wait for a few hours more perceiving the remote object to be still moving at its previous speed, before the light reaches him and he starts receiving pips at the faster rate. Since he thus receives them for a shorter period of time, he sees fewer pips, hence perceives the remote object to be ageing slower than him. (Andy sees Brian travel a distance back of 45 light-hours in only 60 hours. This would make Andy appear to be travelling at 45/60*C = three quarters of the speed of light on return.)
  8. So are you saying that during these one-second periods of acceleration of the travelling twin, that time speeds up? (the time of the stationary twin as observed by the travelling twin) Below are some notes in italics I wrote, to follow up my question - am I on the right lines here? OK so we are agreed that in my scenario that at the end of the journey, Brian and Andy (now standing next to each other) will look at their respective clocks and see that Brian's only shows 100 days (100 pips), and that Andy's shows 200 days (200 pips). And that this is because Brian's movement and heavy accelerations have caused Brian to age only half the time of Andy. This is where it gets interesting. So does this mean that to make this happen, Brian must have had to take the decision to turn back after only 50 days (50 pips) (as he perceives it) such that his clock has only got to 100 by the time he returns? And when he returns, he sees that Andy's clock has got to 200 days (this being consistent with Andy's own experience of Brian taking 200 days for the whole trip). Am I right about this so far? Assuming I am, then the following perhaps make sense: That for the portions of the journey when Brian is not accelerating, then for almost all of (from Brian's perspective) the 2 * 50 day journey, that Brian would look back at Andy, and see Andy's clock going at half speed, so resulting in only 25 days (25 pips) elapsing for the outbound journey, and 25 days elapsing for the return journey. 50 days total - as seen by Brian. Now since we know that on his return to Andy, Brian will see that Andy's clock has passed 200 days, there is a missing 150 days here. 150 days that needs to be added on to Brian's perception of Andy's clock at some point in the journey. Would I be right in thinking that these 150 days are accounted for during Brian's three 1-second periods of heavy acceleration? i.e. during each of these short bursts of acceleration, Brian will see Andy's clock go forward by (say) 50 days? - that Brian will receive 50 pips from Andy's clock over the course of this one second (one every 20ms) - i.e. much speeded up? If this is correct, then it is clear that while relative velocity of a target object (Andy in this case) will cause time at the target to slow down, acceleration of the observer can cause the observed time to speed up. Following on from this, then perhaps one can see a scenario whereby you would see a clock running at normal speed where these two effects exactly cancel each other out; where you are (momentarily) travelling at a different velocity to the clock under observation, at some distance, and accelerating. Is there a formula that describes this? If I am wrong, where above does my explanation depart from the reality?Thanks
  9. Special relativity says that a clock on a spaceship moving relative to a stationary observer appears to be running more slowly than if it was stationary. So if this is true, what would you get in the following scenario? Stationary observer Andy and moving observer Brian are both initially stationary and in the same location, and each has an atomic clocks measuring time in days, and displaying what they can see on a large monitor visible to both. In addition both clocks emit an electronic "pip" whenever one day passes to the next, and each records the number of pips made since reset. Day = 0, and the pip counter is set to 0. Brian rapidly accelerates towards a distant star, and it takes him one second to reach around half the speed of light. After 100 days of travelling Brian stops in 1 second and accelerates to the same speed back towards Andy (in one second), and arrives back at Andy in a further 100 days. Before they got out their clocks to compare what they showed, they each gave their opinion as to what they were expecting them to show. Brian reports that Brian's clock shows that 200 days has elapsed, and he reported having received 200 pips from it. Brian reports that Andy's clock appeared to be running at half the speed on both the outbound and return journeys, and he had correspondingly received 100 pips from it. Andy reports that Andy's clock shows that 200 days has elapsed, and he reported having received 200 pips from it. Andy reports that Brian's clock appeared to be running at half the speed on both the outbound and return journeys, and he had correspondingly received 100 pips from it. When they both look at their clocks, and read their pip counters, what do they see? Thanks
  10. I asked this question a week ago or so on Yahoo Answers (OK don't scoff). Got 10 replies http://uk.answers.yahoo.com/question/index;_ylt=AgMMhwzI6WpSsLJz.m3smFAhBgx.;_ylv=3?qid=20100212145607AAtahRr
  11. http://en.wikipedia.org/wiki/Trinity_(nuclear_test) On the right of the link is a photo taken a few milliseconds into the Trinity atom bomb test. You can see the smooth surface of the fireball shock wave expanding out, and you can see the rough dusty base where it (presumably) is interacting with the ground. But what is the paler band between the two? Also there are some blotches on the surface of the fireball. Are these as a result of the tower or bomb casing perhaps?
  12. Interesting. This raises a question: Discounting the moon's gravity, is there any centrifugal force experienced by someone on the moon's surface by virtue of the rotation of the moon?
  13. If the entire mass of the universe was compressed into a single point, it would form a black hole. What would be the radius of its event horizon? I calculated this from a mass of the universe of 10^52 kg, and it came out as around 10 billion light years. Is this right? If it is right then perhaps the whole universe is inside its own black hole. In which case where abouts is the singularity, and why aren't we sucked into it?
  14. You say that rocket B only has to give rocket A 1000 joules to accelerate B from 10 m/s to 20 m/s relative to earth (0 to 10 relative to it). But this isn't true, due to the conservation of momentum: What would happen is that rocket B would be slowed to 5 m/s and rocket B to 15 m/s because the effort of pushing forward A would equally push back B. So the combined kinetic energy of the two rockets relative to earth would be: 0.5 * 20 * (0.5)^2 + 0.5 * 20 * (1.5)^2 = 0.5 * 20 * 0.25 + 0.5 * 20 * 2.25 = 2500 Joules. Hmmm seem to have got something wrong here. But there is more to it than you have said. Got to go - friend here.
  15. wow I didn't realise that the rate of flow was proportional to the square root of the pressure differential. That definitely makes it more complicated. This all started with an arguement I was having with my mother about whether you can water plants faster holding the watering can still, or whether to swing it to and fro. I'm still not sure. Could someone please do an integration of the square root function - that must surely reveal the answer to the question - it was too long ago when I studied integration. Thanks:D
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