conservation of momentum II

Discussion in 'Pseudoscience Archive' started by DRZion, Jun 21, 2010.

  1. AlphaNumeric Fully ionized Registered Senior Member

    And that's your excuse for completely ignoring books? There's a difference between learning while taking everything as not gospel accurate truth and simply refusing to learn. Besides, you've been applying non-relativistic equations as if they are sarced truths, precisely when they aren't! Yes, its sometimes said that the education system is a series of smaller and smaller lies but that doesn't negate the worth of each step. You can't get to the complicated stuff till you've learnt the slightly incorrect simplification of the complicated stuff.

    The 'easy' stuff has already been done, as I've been telling you. You have no idea where the 'boundary' of current knowledge is, its much much much further out than you think. There's nothing in this area which would be new and 'easy', regardless of someone's knowledge and ability, its all already done. This is why particle physics involves things which move at 99.999% the speed of light or are at temperatures like 0.001K, everything 'easy' has been examined to exhaustion.

    Coulomb forces is the \(\frac{1}{r^{2}}\) force associated to electromagnetism. Bremsstrahlung is a specific kind of interaction. When doing QED you compute scattering cross sections for bremsstrahlung and you can extract the associated effective potential using Fourier transformations. Its all the same model, not separate ones!

    I really don't get how you think you have any idea what has or hasn't been done or what is or isn't relevant to accelerator physics when you don't know either the theory or the experimental results. Energy losses due to acceleration applied to charged particles is the major reason the LHC collides protons rather than electrons/positrons. It's not 'insignificant'.
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  3. DRZion Theoretical Experimentalist Valued Senior Member


    Actually now that I think about it the momentum of the object will in fact increase. If the object loses mass, but keeps the same kinetic energy after the radiation event, it would mean that there is more kinetic energy/mass and so higher momentum/velocity.

    Still, it doesn't seem to account for all of the momentum since loss of mass will only be m=e/c^2 , while momentum of the photon will be m=e/c ... but I haven't checked the math. Its a possibility.

    You got me there! I'm working with absolute truths .. am I so wrong?
    -Any charge which accelerates emits radiation
    -due to conservation of energy the source of photon energy is kinetic energy
    -photons do not carry much momentum/energy , considerably less than all massive particles going at non-relativistic speeds

    These are broadly speaking TRUTHS

    I disagree, I think that most smart people have exactly the attitude you do and they are unwilling to work on easy gain projects in fear of looking foolish. I don't like the idea that we have to do new science to get new results. There are a lot of old methods (maybe not in physics) which, if applied correctly, will get you great results. But sometimes what is new is sexy and people don't want to get behind the times, or they want new fields to become experts in. Also, thats the nice thing about working on grandiose projects - you know no one has done it before so its not really necessary to read the literature.

    So youre saying its possible to calculate potential from the cross section of bremsstrahlung?

    Energy loss sure, but not momentum loss! Its easy enough (for you at least

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    ) to figure out the change in trajectory of the electron due to bremsstrahlung - the loss of momentum should be imparted on the stationary magnet..
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  5. Pete It's not rocket surgery Registered Senior Member

    The object's kinetic energy will change as well. The radiation energy doesn't come only from the object's rest mass - it's a combination.

    The numerical value of c depends on your choice of units, so c^2 is not necessarily numerically greater than c. And you can't directly compare e/c with e/c^2, they're like distance and speed, apples and oranges.
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  7. DRZion Theoretical Experimentalist Valued Senior Member

    With a bit of whiskey in me I feel completely capable of convincing everyone that the braking radiation process known as bremsstrahlung could be expanded on and explored if one had the desire.

    It will change, but according to conservation of energy it will have to go somewhere. But in order to make conservation of energy and conservation of momentum agree you will have to do some extra digging - it just doesn't come from the basic equations..

    Right.. I am not quite sure how one would translate c into any amount of momentum , but this is fairly basic unit conversion from high school chem. it boils down to some paper and pencil-work. Momentum is kg * m/s while c is m/s , and mass is in grams. You then have to write out all your values precisely and just see what cancels with what.

    To spin this thing one more time-

    You have a charged particle travelling along the x axis with m velocity, y mass and q charge. Its easy to compute its momentum and kinetic energy. This particle then loses some of its kinetic energy as a photon travelling on the x axis in the -x direction. You are saying that in this system momentum and energy are conserved. This is not true using the basic equations. p=e/c and p=mv ... (its silly to consider these two in one scenario exclusively, right? ... anyhow, it does seem to be to be a GOOD ENOUGH approximation when the given particle is travelling at 1,000 m / sec) At non-relativistic speeds it newton's equations are sufficient to accurately describe the scenario. Using newton's equations we can experimentally find the momentum of a photon. Combining the two the above scenario will be impossible. Of course, there is no reason for a free-travelling charged particle to emit a photon, BUT this is essentially what our argument boils down to. You are saying that in the above scenario momentum will be conserved.
    Last edited: Jun 30, 2010
  8. Pete It's not rocket surgery Registered Senior Member

    Of course it goes somewhere - it goes into the photon energy, along with the lost mass.
    Your problem is that you are using relativistic equations for photon momentum and energy, but newtonian equations for object momentum and energy.
    Use the basic relativistic equations, and it all balances.

    There's two problems here. Firstly, p=mv is an approximation.
    Try \(p=\frac{mv}{\sqrt(1-v^2/c^2)}\) instead.

    Also, I'm not sure whether the object gains or loses kinetic energy when the photon is emitted in the opposite direction to the object's velocity. I suspect it could be either, depending on the particular situation.

    Only in the same way that photon momentum and energy = zero is a GOOD ENOUGH approximation.

    Yes I am. Use consistent equations, and you'll say it as well.
    Last edited: Jun 30, 2010
  9. Pete It's not rocket surgery Registered Senior Member

    Try this.

    You have an object mass m travelling along the x axis with momentum v. It's easy to compute its momentum and total energy:
    \(E^2 = m^2c^4 + p^2c^2\)

    This object emits a photon of momentum p, energy pc travelling on the x axis. The object now has mass n and velocity u:
    \(n = \sqrt{m^2 - 2mp/c}\)
    \(u = \frac {v - pc^2/\sqrt{n^2c^4 + p^2c^2}}{1 - vp/\sqrt{n^2c^4 + p^2c^2}\)

    I'll continue when I have time. Or, you can try it yourself. It's just ugly algebra.
  10. DRZion Theoretical Experimentalist Valued Senior Member

  11. AlphaNumeric Fully ionized Registered Senior Member

    When its known they aren't true, yes.

    I'd say your second one is dubious and even then they don't imply your conclusion. Yes, they are closely associated to the claim you're making but they don't imply your claim.

    Foolish in whose eyes? Each other? There's enough people in physics (or any academic area) who never do anything other than pick the simplest and easiest things they can find to work on, due to the fact too many people are stuck in the 'publish or die' mentality to an extreme level.

    I really don't think you're in a position to judge the mentality of researchers, given you're not one of them, never have been and haven't even got experience in even learning any decent amount of physics.

    Besides, bremsstrahlung processes are worked on and are tested. You're not talking about some unexplored corner of physics yet, you're talking about (quantum) electrodynamics 101.

    If someone said to you tomorrow that the Sun goes around the Earth what would you say? I imagine you'd disagree and then make some comment about astronomy and space craft demonstrating otherwise. People have worked out what gravitational models say, done observations and compared with the model. The fact we can send probes through the solar system accurately kills the notion of a geocentric universe. You're claiming the QED version, that a model (QED) cannot account for something we know it can. Yes, we don't know if QED is 100% accurate but its at least 99.9999% accurate in terms of observed bremsstrahlung processes. If you are right it'd be impossible for QED to be that accurate, as your saying another particle (which QED doesn't include) is needed to carry the momentum. Compton scattering is likewise, an explicit demonstration a photon can pack a huge punch.

    Yes, its important not to say "Forget testing QED every again, it works", its important to retest each time our instruments get better. In fact we know QED involving only electrons and photons is an incomplete model both from the theoretical approach (it contains a Landau pole) and the experimental one (can't account for muon contributions which are measurable etc). But the fact remains that QED says bremsstrahlung processes where electrons and photons interact conserve momentum and energy and can model the world to a high level of accuracy.

    If, according to models, a photon couldn't pack much of a punch and another particle was needed to account for momentum transfer then it would mean tree level calculations would be way off. They aren't, they usually account for the vast majority of an interaction.

    Where did I say otherwise? I'm talking about the 'coal face' of research. Every physicist has a 'tool box' of methods used in previous models (you know, all that stuff people learn in university!) so the methods for old models are not just thrown out.

    Bullshit. Utter bullshit. Firstly the 'literature' is basically the sum knowledge of all previous researchers so to avoid looking at their methods and ideas and even mistakes is just stupid. Secondly the literature is essential if you're going to jump into an area someone else has developed, you can't do some new funky quantum field theory model if you don't know quantum field theory. But to know quantum field theory you need to know quantum mechanics and relativity, which need linear algebra, complex analysis, geometry, vector calculus, electromagnetism, group theory and functional analysis. The only way you can possibly avoid the literature is to reinvent everything. Even if you want to build up a physics model from the ground up you're going to have to make quantitative predictions, which means producing numbers, so you need mathematics. Now you can either avoid all mathematics anyone else has ever done and basically reinvent the wheel by starting with something like "I know 1+1=2 and that's it" or you can learn some calculus from the literature and save yourself a long time. It took the greatest mathematicians who ever lived the entirety of their lives to develop the material which is expected knowledge for any physics graduate so I don't imagine you'll get far if you don't look in the literature.

    Preventing yourself just doing what everyone else is doing by not sticking too close to the mainstream is one thing. Binning the entirety of scientific knowledge from 5000 years of recorded history, amounting to millions of man years of work, is just idiotic.

    I'm saying the same model which allows you to calculate the \(\frac{1}{r^{2}}\) behaviour of the Coulomb force is the same model which predicts accurately bremsstrahlung scattering cross sections. QED gives accurate cross sections and it has a \(\frac{1}{r^{2}}\) Coulomb effective theory.

    Thus, given it conserves momentum and energy too, QED is an explicit example of a model which does what you claim can't be done.

    Energy and momentum are lost from the beam due to photon emission, which then hit magnets/detectors which then increase in the relevant amount of energy and momentum and the quantities are conserved.

    For the momentum to go from the beam to the magnets photons must be exchanged, thus the photons carry all of the momentum being exchanged.
  12. DRZion Theoretical Experimentalist Valued Senior Member

    The example of a charged particle travelling around and spontaneously emitting radiation is not like bremsstrahlung - in that case it would be a 'two particle decay', which has distinct values for kinetic energies and momenta of the two product particles, as is said in the hyperphysics link.

    I'ts hardly dubious. The chart clearly shows that the maximum energy of an emitted photon is equal to the full kinetic energy of the incoming electron. This would account for 99.99% of the energy - the rest might come from loss of mass energy; I'll consider it insignificant for now.

    It seems to me that this is better than not publishing anything.. in some way by publishing an author is showcasing his skills (at least) and interests so that are on the radar. Unless you mean something else.

    Okay, but I CAN make blanket statements which will be true in at least some cases..

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    And yes, I am an undergrad researcher in biology as well as an investigator for any number of kooky projects.

    Isn't the point to find new things about old models in an attempt to flesh those out? Its like when maxwell's equations were reworked to make them more accessible - nothing new was really created but it was certainly progress.

    Does QED really have no need for new particles??

    Compton scattering does conserve momentum by ejecting an electron which compensates for any 'new' momentum in the system so that it is all conserved. In my case the only thing which can be ejected according to theory is photons, and I have already said that these photons will carry away more energy than momentum.

    Thats not the point - you're talking about the momentum carriers being intermediates.. although i suppose virtual photons don't transfer much energy do they? Is it even experimentally verified that 'virtual photons' exist??

    By publishing easy articles physicists and other researchers are expanding and refurbishing their current 'tool box' and growing closer to being experts in whatever they are doing.

    All my claims are based on the works of two giants - newton and tesla. Skip all the other drivel !

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    I don't consider it impossible that I am wrong..

    All of the momentum is exchanged by 'virtual photons' . But these will not ever lock up or store momentum to my knowledge - they will act as short lived intermediates.
  13. AlphaNumeric Fully ionized Registered Senior Member

    Ah, gotta love internet 'debates' when one side doesn't actually know anything and falls back on Google. The fact you're trying to tell me, literally, my business seems all the more laughable. Yes, why should I believe the books I've read, the papers I've read, the people I worked along side for years when someone online tells me that some website told him that the people I know personally aren't doing what they say they are and that work I have done personally doesn't exist!

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    You should have quit while you were only mildly behind.

    A charged particle undergoing acceleration, for whatever reason, will emit photons. Any kind of interaction of this kind you wish to name, be it acceleration due to positive nuclei, individual protons, individual positrons, accelerators, anything, is accurately modelled by QED and its extension electroweak theory. You're claiming photons can't account for such massive momentum changes. You are wrong. Demonstrably wrong. Physicists didn't spend $10 billion on building CERN because they sorta understood electrodynamics, they did it because they've examined all the kinds of particle interactions you can get cheaper and their models passed.

    Actually the energy can go higher, the electron ends up going back the way it came. You only get your result if the electron goes to a stand still, which of course is a frame dependent quantity. Individually momentum and energy are frame dependent quantities, only the combination \(E^{2}-p^{2}\) is not.

    Go back to hyperphysics. Or open a book.

    Then you accept that your excuse no one will publish certain things for them being 'too easy' was baseless. Thanks.

    Yes, a statement is always true, except when it isn't. Got any other pointless tautologies you want to come out with?

    'Researcher'? Sure you are. Given your demonstrated skills for 'researching' physics before declaring the past 70 years didn't happen are zero I doubt 'researcher' is a label you can have applied to you by anyone other than perhaps creationists. Perhaps I wouldn't be so vitriolic if I weren't aware of what research really involves and you have no clue.

    You're not 'fleshing things out', you're not 'finding something new', you're denying the existence of work which exists and has existed for so long its in all the textbooks. You know textbooks, they are those big things with lots of words in them that other people at your university have.

    Where did I say anything about 'need'?

    Energy and momentum is conserved in Compton scattering, precisely as QED says. Precisely as QED is constructed to say. And you do realise there's factors of c all over the place to convert between momentum and energy quantities right? You do realise that often physicists work in units of c=1 so sometimes equations aren't in SI units, right?

    Oh god, this is actually painful. I honestly just did a 'face palm' having read that. Please please stop digging. Look up what a 'virtual particle' is. They don't obey \(E^{2}-p^{2} = m^{2}\), they can transfer as little or as much energy for a given amount of momentum (or vice versa) as they wish. But I wasn't referring to them, they are loop level contributions. Intermediate tree level photons aren't virtual, they obey \(E^{2}-p^{2} = m^{2} = 0\) and everything still works.

    Seriously, I suggest you learn how to do the bloody calculations before trying to tell people who can that the calculations don't say what every single person who can do them has seen them say. You're not talking about high level stuff here, its homework for 4th years. People have written computer programs which can evaluate thousands of these things, as some processes involve many thousands of contributions, all in an attempt to test the theory as much as possible.

    You know you don't know about virtual particles or quantum field theory calculations or special relativity's role in quantum field theory or any of the mathematics needed so why why why are you daft enough to tell people who have done such things that things which have been calculated and experimentally tested can't be done? There's stupidity and then just flat out denial of reality.

    Neither of which are sufficient to do quantum mechanics, special relativity or the combination of the two, quantum field theory and guess what, QED is a quantum field theory. Besides, my statements are based on the sum total of all the work of all the quantum physicists for the last 100 years, all of whom knew about work by Newton and Tesla. And Einstein and Dirac and Heisenberg and Poincare and Bohr and Born and Feynman and Schrodinger and ....

    Get the picture? You're dropping two names out of hundreds. Hundreds known to all people who study quantum field theory. I bet you don't even know much, if any, Newtonian mechanics & gravity or electromagnetism, you're just dropping names. The work of the people I just mentioned I am familiar with, as is everyone else in the particle physics community. Are daft are you when you're a biology student and you're trying to play the "I've heard of more physicists than you" game with a physicist?

    To do so would mean further denial of reality on your part.

    I've said 'tree level' several times in this post and more before. If you knew any quantum field theory you'd know what that meant. And you'd know it completely negates what you just said.

    And the problem is that 'your knowledge' is very very lacking.

    Seriously, put a sock in it.
  14. DRZion Theoretical Experimentalist Valued Senior Member

    Sometimes knowing less is better. Soldiers storming omaha beach didn't know their abysmal chances of survival and that was the only way the beachhead was established.

    There are already asymmetries in physics - noether's theorem says only if there is symmetry there will be conservation. Maybe there isn't symmetry in every radiation process!

    Tell me AN, does QED have a proof for the second law of thermodynamics? I would love to have at least some certainty that QED is wrong on some points. Also, have you heard of Godel's work? He has mathematically proven (using math most god fearing christians avoid) that any complex mathematical system will contradict itself. I am sure QED is no exception.

    The maximum energy released through bremsstrahlung is the kinetic energy of the incoming electron. If the particle is flung backwards it loses little kinetic energy and so the bremsstrahlung has no energy source.

    If bremsstrahlung really arises from every acceleration, I would like to know if it arises from gravitational acceleration as well. If it does it may be interesting to look for black holes which are directionally sucking up large amounts of charged particles and in the process gaining some mysterious momentum.

    BTW, moderator, its completely absurd this is in the pseudoscience forum. we are not making up quirky facts here but debating a scientific principle.
    Last edited: Jul 8, 2010
  15. Pete It's not rocket surgery Registered Senior Member

    Rarely, and even when it is, it's not as good as knowing even more.

    Prepare to be educated! (Cue furious typing from AN. Or maybe he'll cut'n'paste from an old response to Jack. Or maybe he'll just give up on you.)
    That's a quirky made up fact. It looks close to Godel's theorem, but is actually completely off the mark.

    This is in the right place, DZ. The problem is that you are not equipped to debate the scientific principle in question, and you are making up quirky facts about the underlying principles.

    It would be in the right place if you were willing to acknowledge that you do know less, and to discuss science at that level, rather than being assertive ouside your domain.

    I tried to engage you at the level of relativistic mechanics, because I suspect that's the right level (it extends Newtonian mechanics and doesn't rely on quantum mechanics), and also because it's the extent of my own domain. But I wouldn't pretend to be able to discuss bremsstrahlung, because I don't understand the models that describe it. Similarly, you'll never be able to intelligently discuss bremsstrahlung based solely on work of Newton and Tesla, because according to their models bremsstrahlung can't happen!
    Last edited: Jul 9, 2010
  16. Pete It's not rocket surgery Registered Senior Member

    DZ, you seem to want to extend newtonian physics to describe bremsstrahlung radiation by proposing a failure of conservation of momentum.

    That could be a valid (if questionable) scientific endeavour, but would be much, much harder than you seem to think. To be worthwhile, the extension you develop would have to be able to quantitatively explain a lot of experimental results that have been found to support other extensions.

    It's not enough to just say "Hey, maybe momentum isn't conserved!"
  17. AlphaNumeric Fully ionized Registered Senior Member

    And soldiers don't improve their fighting abilities by knowing about mathematical analysis of their situations. Physicists and mathematicians do. Even if you think that doing a degree would 'indoctrinate' you you can still obtain the information from books without the whole oversight of a lecturer and thus get the information without 'indoctrination'. There is no excuse for being deliberately wilfully ignorant of science you're attempting to discuss.

    Stop going to Wikipedia, you don't obtain understanding by reading the definition. I'm quite familiar with Noether's theorem and its applications to physics thanks, enough to see you are not. A specific calculation in QED will not be invariant under boosts or rotations because you're considering a specific set of particles, with defined momenta. However, this doesn't render Noether's theorem irrelevant, else it'd be impossible to have any notion of energy or mass conservation whenever you go from theory to application. The calculations are done by obtaining equations of motion from a Lagrangian, which itself possesses symmetries which Noether's theorem links to conserved quantities. Putting particles 'in' can be thought of as having boundary conditions. The equations themselves have the required symmetries and thus momentum and energy conservation is obtained and yet a model of some particle interaction is also obtained.

    I've told you several times now, you're arguing that things I have done with my own two hands don't exist, that the wealth of books, the tested models, the billion dollar accelerator experiments haven't happened.

    Where did I claim it did. How does it in any way have anything to do with your claim that there's an issue with a particular kind of radiative process?

    Again, stop getting your information from Wikipedia. Godel's work said that for any 'sufficiently complex' system (ie that which contained the Natural numbers) it is possible to formulate true statements which cannot be proven true within the axiomatic system which constructed them. He also proved that if a sufficiently complex system could prove its own consistency then it is actually inconsistent. He did not prove that any sufficiently complex system is automatically contradictory.

    Not only are you stupid enough to try to argue with a mathematician about mathematics you've never done but you're stupid enough to not even bother to find out what the mathematics you're name dropping actually says.

    I suggest you learn the equations so you can understand them. You clearly have no understanding of particle mechanics.

    If you looked at the equations you'd see terms like \(\ddot{x}\), it doesn't matter how that \(\ddot{x}\) term is obtained, only that it is.

    There'd be no mysterious gain of momentum for an object falling into the black hole, its a standard interaction, momentum before is momentum after. And yes, that follows in precisely the same way as any other area of physics, Noether's theorem.

    Its where it belongs because you have absolutely no knowledge of any relevant physics, you don't want to listen to people who do and you clearly have no interest in being intellectually honest.

    I've commented on this before but this thread really demonstrates it. You once PM'd me to ask if I wanted to work with you on 'Einstein manifolds'. Your PM was worded in a way which made me think you didn't know any relativity or differential geometry and since then you've demonstrated your ignorance is much deeper than that, you've shown ignorance is special as well as general relativity, electromagnetism and now we're going into full blown hack territory where you just name drop as many things as you can in an attempt to dig yourself out of the hole you've made. Give it a rest. I know things like Noether's theorem seem mind blowingly complicated to you, seeming as if its so complicated it must be beyond all but the best professors but you're wrong. It's bread and butter stuff for particle physicists, no one who has done quantum field theory has not done Noether's theorem. You tried to aim over my head and ended up shooting yourself in the foot.
  18. Guest254 Valued Senior Member

    Having read the last few post, and this latest one, I can see why AlphaNumeric is becoming annoyed.
    This comment demonstrates a complete lack of understanding. It is clear the author has no clue what Noether's theorem is and how it related to QED.

    And to top it off, you do what lots of idiots do, and mention Godel's work in an attempt to add credence to your position. Like the aforementioned group of people, you demonstrate that you have absolutely no idea what Godel's mathematics means.

    I did enjoy the comparison between soldiers on Omaha beach and ignorance of basic science. Good form.
  19. DRZion Theoretical Experimentalist Valued Senior Member

    Not true, you can certainly never know all the facts so sometimes its wiser to pick and choose.

    I can certainly debate by using scientific facts only. They may be outdated facts but I don't think that they are obsolete.

    But you do understand bremsstrahlung - its very simple. Charged particles emit radiation when accelerated. You can't argue with that!

    I did look at the relativistic mechanics and it still seems that the momentum goes nowhere ... i might have to go to the physics help room and work through these with somebody on paper. A problem is that the relativistic equation for energy is very inaccurate at low velocities while the newtonian equations are inaccurate for high velocities.

    Yeah they do - its called morale. If a soldier knows he is cannon fodder he won't even show up on the field.

    The indoctrination is hard to avoid - I've tried speaking to academics without using the jargon - its harder than it sounds. Its because of respect and the academic hierarchy that indoctrination is necessary. I try not to buy into the hype at the lab and for this reason I don't always get along so well with my co workers. I am not a sheep.

    Stop going to the university! Just because your books say so doesn't mean its not cooked up anyways ... although it does mean you can get a job ..

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    I admit I can't argue with you about this stuff. But you don't respond directly to my very simple queries using any logical statements. You are saying that no matter what QED is right and momentum is conserved.

    So now you are an expert on black holes?

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    I know that black holes accumulate accretion disks, which means that matter falls into the black hole symmetrically and so any momentum gained would be cancelled out.

    Also, how would a black hole emit bremsstrahlung? Isn't the 'charged part' of a black hole already inside the event horizon? BUT, you are saying that even if the black hole doesn't emit bremsstrahlung the incoming electron would conserve momentum (considering it's gravitational interaction with the black hole and the consequent momentum gained by both electron and black hole) by emitting photons which clearly have a different energy/momentum ratio.

    And I mean CLEARLY, (courtesy of wiki)

    "The relativistic mass is defined as the ratio of the momentum of an object to its velocity, and it depends on the motion of the object"

    I WILL NOT listen to people who are stating and restating status quo when clearly I am aiming for something revolutionary!!

    Okay, I will try to shift away from technical arguments because clearly those are futile. I never wanted to get into this kind of argument. I ask anyone to please refer to the original scenario which only involves the MOST BASIC rules of physics with which even AN can't argue.

    -momentum should be conserved
    -energy should be conserved
    -radiation is emitted when a charge particle accelerates

    And I am looking for contradictions with just those three statements by creating a very simple scenario which is easy to grasp.

    If you have two protons trapped very close to one another, and you let them go, they will accelerate and emit radiation. Clearly. They both emit equal amounts of radiation and so they both lose equal amounts of kinetic energy (going back to the three rules we know that the radiation emitted has to have some source of energy).

    Now, take a clump of matter which is a proton and a neutron stuck together and another clump which is a proton and a proton stuck together. Their masses are 99.99% identical and they will undergo similar acceleration. However the proton-proton clump has twice the charge of the proton-neutron clump. Repeat the scenario above and you will CLEARLY have one clump emitting twice the radiation of the other.

    Now, if both clumps lose kinetic energy as a result of radiation, then the proton-proton clump will have less kinetic energy, and hence less momentum than the other.

    Of course, the emitted photons can't be ignored and they also carry momentum. But, the momentum carried by the photons has a lower momentum/energy ratio than the momentum carried by the clumps. Inarguably. It is common knowledge that optical photons carry very little momentum (per photon) - this stems from the fact that photons do not have any rest mass and so all of their mass stems from their energy using e=mc^2.

    In conclusion, one of the clumps ends up with more momentum and the other while photons alone cannot possibly be enough for conservation of momentum. So, there is either violation of conservation or another particle has to be emitted whenever charged particles accelerate, or there is creation of energy and the photons carry away enough momentum but get energy from nowhere.

    And to appease you AN, I now know that photons are very important momentum carriers and that they in fact carry most of the momentum of the universe it seems - BUT, they carry very very little momentum per photon.

    It takes great spirit to overcome great odds!!

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    Last edited: Jul 9, 2010
  20. DRZion Theoretical Experimentalist Valued Senior Member

    Here is an example to illustrate the point that photons carry very little momentum.

    Say you stuff a fellow astronaut into a garbage can while aboard a space station. You kick the can off the station giving it a certain momentum

    The astronaut in the garbage can only has a flashlight to build momentum (photons after all carry momentum), one that is powered by a hand-crank. Who do you think will expend more energy to create that same momentum, the guy that kicked the can or the guy with the flashlight?

    Its the same with electron momentum and photon momentum. The electron has rest mass just like the garbage can while the photons have no rest mass and carry very little momentum for every joule of energy they carry.
  21. AlphaNumeric Fully ionized Registered Senior Member

    Ah crap, I just had an entire reply eaten by Firefox and my crappy broadband.

    I'll reply tomorrow when I can bring myself to go through your utter profound dipshit ignorance again. Suffice to say I suggest you go reread all those physics sources you claim to have looked as because relativity is valid at high and low velocities. The fact you're unaware of this demonstrates you don't know anything relevant and your whining about anything beyond Newtonian mechanics is based on nothing but your ignorance and desire to convince yourself its not your fault you're crap at physics.

    And you completely skipped over my bit about how you got Godel's work utterly wrong. Couldn't bring yourself to admit you simply name dropped things you don't understand? Man up and accept you're simply too damn ignorant and its your fault.
  22. funkstar ratsknuf Valued Senior Member

    I'd like a reply to this bit as well. If nothing else, then just to establish character.
  23. Pete It's not rocket surgery Registered Senior Member

    You haven't in this thread. For example, your statement about Godel.

    I don't understand the models that predict it, and neither do you.

    Rubbish. Another quirky made up 'fact'. Yor pretence at understanding is why this thread is in pseudoscience. Be honest with yourself about what you know and what you don't know.

    The relativistic equation for kinetic energy (which must be distinguished from total energy) reduces to the classical equation when v/c is very small. There's a clear proof on Wikipedia:

    \(KE = E - m c^2 = \frac{1}{2} m v^2 + \frac{3}{8} \frac{m v^4}{c^2} + \frac{5}{16} \frac{m v^6}{c^4} + \ldots ;\)

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