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alwynj48

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About alwynj48

  • Birthday 03/11/1953

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  • Location
    Orlando Florida
  • College Major/Degree
    Electrical and Electronic Engineering HNC
  • Favorite Area of Science
    Biochemistry
  • Biography
    Born in Liverpool England now living in the USA

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  1. Perhaps this will help: Ignoring how you wiggle an electron then when you do wiggle the electron you create a EM field. The EM field is related to the probability of detecting a photon. You can not say the electron emits a photon at a particular place any more than you can say which slit the photon went through in a double slit experiment. The above is just part of what happens. In practice you going to use EM to wiggle the electron and the electron interacts with that EM and its own EM field but it does so via photons and probabilities. You have to get away from the idea that a photon (and the electron) is a particle with clearly defined history but thats just QM.
  2. I will try one more time then rest my case. I thing we can agree that KE is the work done in accelerating a mass to a given velocity ie the product of force times distance. Apparently you do not understand that is not the work done in throwing momentum out of the back of a spacecraft. That’s the equation I have given with the energy in units of mass. Specifically its the mass you must turn in to photons and direct out of the back of the spacecraft to generate the thrust to get to the given velocity. Previously I did think I could use KE by simple equating it to the energy of the fuel used. But that is incorrect for the reasons I have already given. I have already explained I did not find the equation I derived it. I claim no originality it has probably been derived a million time before and been named after someone. This is not rocket science oh wait a moment it is and what more its relativistic rocket science LOL Ok so it is rocket science but it only involves elementary physics, momentum and energy of a photon, laws of motion and momentum plus a bit e=mc^2. I have just thought of this, perhaps it will help you: If you want to use the relativistic KE equation you must do so from the perspective of the stationary frame. So let consider how the space craft will appear from that frame. Ok so in the frame of space craft its converting mass in to photons and projecting them out of the back of the craft. It accelerated and eventual approaches c let suppose that some how it acquires some extra mass and hence accelerates past c. From the perceptive of the stationary frame an observer might calculate the KE energy of the space craft as it approaches c and conclude that its KE is approaching infinite. He might wonder how the spacecraft is obtaining this almost infinite amount of energy. So he looks very closely at the space craft and notice that the craft is powered by a mass to energy convertor. He also notices that the space craft appears to him to be converting almost an infinite amounts mass to energy and this is where the almost infinite KE comes from. However this curious observer not only knows about the relativistic KE equation he also knows about Lorentz transforms. He’s smart dude is this observer. So he transforms the mass to the frame of the space craft. Everything is now ok. The infinite mass has disappeared and in this transformed perspective the space craft can accelerate up to c and beyond with out requiring an infinite amount of anything. Note I have previously explained the craft will run out of fuel before it gets to c. One more point suppose the statuary observer transformed his value of the KE of the space craft to the frame of the space craft??? Do you still thing KE is not dependent on your inertial frame? Rhetorical questions.
  3. You are probably not taking in to account we a referring to a craft's frame of reference in which the craft accelerates by throwing momentum out of the back to accelerate. I was assuming that the energy consumed (100% eff) was the usual kinetic energy E=1/2mv^2 to reach a given v. I believe your where making reference to the higher order terms of that equation . I suggested such terms can not be correct or physics would not be the same in all inertial frame. Well I was right about the higher order terms. In fact it must linear for the same reason. The equation is E=cmv where the terms have the usual meaning. The regular ke equation is simply not the correct one to use with or with out the high order terms. So we where both had the physics wrong LOL. The result given above is suprising. But again if you think about the context it is seems reasonable. For given impulse the craft must accelerate by the same amount no matter what velocity it has already achieved. An other surprise is only a finite amount of energy is required to get to c ie mc^2 the equivalent energy of the mass being accelerated. This result is a quick fiddle with the equations and although I have checked them there may still be a mistake. If anyone is interested I can clean up the derivation and send them a copy in mcad and possibly an other format that mcad can produce. So getting back to the original question. Can we accelerate to the velocity of light using on board fuel. I previously would have said no and even now I believe you can not jsut by throwing momentum out of the back because even with 100% efficiency you would must convert the total mass of the craft get to c. Speculating: having got very close to c a perhaps different thrust mechanism could propel a sub craft to c and beyond. Wow previously I did not think this was even theoretical possible. Though of cause there are huge practical difficulties.
  4. Don't forget I am refering to the accelerating frame. If the formula fails as c is approached I can determine my absolute velocity from the error. What ever velocity I happen to be moving wrt to some other frame will not change my phyics assuming relativity is right.
  5. Here is a link to one of the utube clips of super cooled water
  6. alwynj48

    charge

    I guess that must make black holes hell and white holes heaven LOL I always thought that the maths of QM and GR was hell does that make me a bad person LOL again
  7. My understanding is: the outside observer will not observe a particle crossing the event horizon but the converse is not true. In the time of the particle it can cross the event horizon in fininite time. As to the collaps of the wave function. We do not know how it works with out even horizons so its speculation with one. This is my specualtion: The particles are in cummuniction via an other dimension that appaently can bridge space and time. Given the forgoing it is not difficult for me to belive that this can be done across an event horizon. An event horizons is just an area of very curved space.
  8. 2X/Y=T which is just a short way of saying if its accelerating at Y m/s/s and the initial velocity is X m/s it will take X/Y to decelerate to 0 and the same time to accelerate back to X m/s. or say its starts at 100 m per s every second it decreases by 10m per s so it would take 10 s to stop and 10 s to fall back down. Don't forget what acceleration is ie the change in velocity per second If you want to calculate the distance covered you should know that the velocity profile is triangular so the average velocity is half the peak velocity. So multyply the time by half the velocity and give you the distance or in the case the distance travelled.
  9. Hi, Sounds like you want something that is no linear. Meaning at low light intensites its white or transparent and at high intensites it absorbs light so its black or at least colored. The obvous problem is if it's tranparent or white its not absorbing any light so it can't react to it. So it must have some absorbtion to start with meaning it can not be totally invisible. Here is wiki link on photochromics http://en.wikipedia.org/wiki/Photochromic_lens It also shows that major effort has already been expended on it
  10. It’s interesting to consider how much energy is required in the moving frame. This is given by 1/2mv^2 where m is the mass being accelerated and v is the velocity attained If the velocity required is c then the energy required is 1/2mc^2. The amount of mass that must be converted to generate that energy is given by E=mc^2. Ie the energy required is equivalent to half the rest mass of the object that’s accelerated. Note I may have missed out a factor due to the units but I think its ok if consistent units are used. Ok so far we have managed to accelerate to the velocity of light and only used a finite amount of fuel equal to half the mass of the object we are accelerating. Admittedly where using a 100% efficient mass to energy converter but this is a thought experiment so that’s ok. However the mass must be carried on board and accelerated which will require additional energy (mass) to accelerate it. I have not done the maths but I suspect it would show for a finite amount of fuel you can get arbitrarily close to c but never c unless you start with an infinite amount of fuel.. Note We don’t have any 100% mass to energy converters and the usual method of propulsion, by ejecting the products of an energetic chemical reaction out of a nozzle probably have efficiencies in nano percents. As a side note I am new here so hi all you science freaks.
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