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The existence of Infinity.

  1. Oct 9, 2008 #1
    Is it true that 0.1% of the mass in an uncontrolled nuclear chain reaction gets converted into energy via E=Mc^2? And if this is so, then does the mass technically go to infinity in accordance with the Lorentz transformation before turning into energy?
     
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  3. Oct 9, 2008 #2

    JesseM

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    This isn't a case of a sublight particle being accelerated to lightspeed, but of one type of particle being annihilated and another being created, akin to what happens when antimatter and matter annihilates and produces high-energy photons. Particle creation/annihilation is something that is allowed by the rules of quantum field theory.
     
  4. Oct 9, 2008 #3
    Particle creation/annihilation takes place at the event horizon, where the laws of physics may be broken. Radiation is the inevitable result of such activities.
    Can it be argued that infinity is still playing a role somehow, since there are the same conditions set up that also exist at that place where physical laws may be broken (event horizon)?
    Can we apply the part of the Lorentz transformation that says mass tends to infinity when it turns into energy and get away with it, without introducing the particle/antiparticle pair?
     
  5. Oct 9, 2008 #4

    JesseM

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    No, it happens everywhere in quantum field theory. Particle creation/annihilation at the event horizon is just used as a way of explaining Hawking radiation.
    Physicists don't think GR breaks down at the event horizon, but only in the vicinity of the singularity. And it's not that all laws of physics fail there, it's that a theory of quantum gravity (which GR would just be an approximation to, the way Newtonian gravity is an approximation to GR) would be needed to deal with what's happening there.
     
  6. Oct 10, 2008 #5
    Now I went here:
    http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html
    And scrolled down to the third frame, under Relativistic Mass. I inputted the value of "1" for the speed of light, and M returned infinity. To this day I am having a hard time trying to get around this fact of Einstein's equation!
    HELP!!!
     
  7. Oct 10, 2008 #6

    Hurkyl

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    Actually, you inputted the value "1 c" for velocity.

    Actually, it returned "Infinity * m0". (In particular, it's an indeterminate form when m0=0)

    Why? What's the problem?
     
  8. Oct 10, 2008 #7

    atyy

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    I tried it and it said m=(infinity)mo.

    Would it make you happy if mo=0?
     
  9. Oct 10, 2008 #8

    JesseM

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    But no particle with nonzero rest mass can move at the speed of light. A photon (or any other particle moving at the speed of light) has zero rest mass, so if you try to use the relativistic mass formula you get 0*infinity which is undefined. But the formulas of relativity like the relativistic mass formula or the time dilation formula are only supposed to give the values for things in different inertial frames (which always move at less than c relative to one another), they're not meant to give useful answers when v=c. I don't know if physicists even talk about the "relativistic mass" of photons (most physicists prefer to avoid the concept of relativistic mass altogether and just talk about relativistic momentum and energy), but here's one way you could do it: if the relativistic mass of an object is M we have the formula for its energy E=Mc^2 (which is equivalent to the more common formula E^2 = m^2*c^4 + p^2*c^2, where m is the rest mass and p is the relativistic momentum p = m*v/sqrt[1 - v^2/c^2]), and from quantum mechanics we also have the formula for the energy of photons E=hf where h is Planck's constant and f is the frequency, so you could set these equal and get Mc^2 = hf, or M = hf/c^2.
     
  10. Oct 10, 2008 #9
    Thanks for the elaborate answer.
    I think I'm just having problems with this quote of yours:
    Now why not? Where is the limit set in the equations Einstein wrote that says v musn't equal c? There's not even an implied limit, as far as I can see, anyway.
    About what you stated earlier, which was this:
    I took this to mean that at near v=c, things get uncertain and statistical. Is this correct? Is it correct to say that as the mass becomes infinite as it equals c is impossible to observe, just as it's impossible to observe virtual particle pairs?
     
  11. Oct 10, 2008 #10

    Dale

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    In the part where you get division by zero when v equals c.
     
  12. Oct 10, 2008 #11

    JesseM

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    Because the equations are meant to translate between different inertial reference frames, and there are no inertial frames with a relative velocity of c. Also, as DaleSpam said, if you try to plug in v=c you get division by zero.
    No, where would you get that idea? In any case "near v=c" isn't really meaningful in an absolute sense, anything that's moving at close to c in one inertial frame is at rest in some other inertial frame, and the laws of physics are identical in every inertial frame.
    No particle with nonzero rest mass ever reaches c in any inertial frame, and a particle traveling close to c shouldn't be any harder to observe than any other particle.
     
  13. Oct 11, 2008 #12
    I got the idea from you. This is what you said:

    Then to that statement of yours, which introduced uncertainty and statistics (AKA quantum field theory), I replied:

    Do things get uncertain and statistical?
     
  14. Oct 11, 2008 #13
    Then you are denying that energy is being created from mass in nuclear chain reactions?
     
  15. Oct 11, 2008 #14

    Dale

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    No. How did you come to that mistaken conclusion?
     
  16. Oct 11, 2008 #15
    Because. The hyperphysics link says that when you input c for v, you get infinity for M as a result. This means that, according to the relativistic mass equation, the only way you're going to get E is by accelerating that v to c. And, as the link demonstrated, when you accelerate V to c, you get infinity for M.
    Therefore,
    This quote of yours proves you deny that energy is being created from mass in accordance with E=MC2, because you said you had a problem with getting division by zero when v equals c.

    1. E=MC2
    2. That means, M must equal C to get the E
    3. In order for M to get to C, it must become infinite (infinite mass), according to the Relativistic Mass equation on www.hyperphysics.com that I posted earlier.
     
  17. Oct 11, 2008 #16

    Dale

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    No, it just proves that you don't know what you are talking about.

    The equation e=mc² means that mass is a form of energy (e.g. you can measure the mass of a particle in kilograms or electron volts with a proportionality constant of c² for converting between the two). It does not in any way imply that matter needs to be accelerated to c in order to "get the E". It already has/is the E.
     
  18. Oct 11, 2008 #17
    My G-d, I'm dealing with another Fredrik (read here for my encounter with Fredrik: Relativistic Rod and Hole ( 1 2 3 4 5 ... Last Page)!
    Look. We know this:
    1. Before the chain reaction, there exists nothing but a certain amount of U235 mass
    2. During the chain reaction, something happens, and that something has to be related to E=MC2.
    3. After the chain reaction, there exists less mass than what we started with.
    4. It's a well-known fact of life that that mass went bye-bye in the form of energy.
    I'm sorry, DaleSpam, but I cannot get it any more simple than this. I believe I made it illiterate-proof, in fact (but that depends on subsequent posting).
    What is known is that an E=MC2 process took place. Do you agree that an E=MC2 process took place (Y/N)?
    Knowing that an E=MC2 process took place, we have to analyze the mechanism by which that M turned into E (M stands for mass and E stands for energy).
    Since the only equation we may go by is the relavistic mass equation I posted earlier, we have to presume the energy got to be energy by a direct mass-to-energy conversion somehow.

    But how?

    The equation tells us the energy got that way by the mass (labeled as "v" in the equation--"v" stands for the velocity of the mass). Since the only way the mass can get to E is through a v=c phenomena, we easily and dullardly conclude that there was an acceleration of that mass to the speed of light that took place in order for the E (energy) to be generated.
    The only way that can happen (for the v of the mass to turn into the e of the energy) is if the mass, labeled as v, went to infinity.
     
  19. Oct 11, 2008 #18

    JesseM

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    Because it's permitted by the rules of quantum field theory for some set of particles to annihilate and create new particles in their place, with the rule that the total energy beforehand must be equal to the total energy beforehand (there are additional rules governing these creation/annihilation events too of course, but that's the only one that's relevant here). This is not the same as a particle being accelerated to light speed, so the relativistic mass equation isn't relevant.
    It's just not true that the "only way the mass can get to E is through v=c", that's a wrong idea of yours that doesn't come from anything in relativity. Quantum field theory allows discontinuous creation/annihilation events, and although this is a quantum phenomenon and not specifically predicted by SR alone, it also doesn't violate SR in any way.

    By the way, it's not even clear what you mean by "the only way mass can get to E"--"energy" doesn't refer to a type of stuff distinct from ordinary matter, it's a property of all particles much like momentum, particles moving slower than the speed of light have energy just like photons have energy, if you think it makes sense to say photons "are" energy while sublight particles aren't, you're confused about what the term means. As I said, for an object moving slower than light its energy is given by the equation E^2 = m^2*c^4 + p^2*c^2, where m is the rest mass and p is the relativistic momentum p = mv/sqrt(1 - v^2/c^2). For an object moving at light speed, the rest mass m is always zero so p is undefined in this equation, so you must instead use the quantum equation E = hf, where f is the frequency and h is Planck's constant.
     
  20. Oct 11, 2008 #19
    Does everyone else agree? Is energy and mass indistinguishable in this regard?
     
  21. Oct 11, 2008 #20

    JesseM

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    Indistinguishable in what regard? I just said that energy was a property (as is mass), not a type of substance.
     
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