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Basic questions on the speed of light

  1. Jan 29, 2012 #1
    Why do gravitons travel slightly slower than photons (99%) and why do neutrinos travel slightly faster than photons (101%)?

    What is an intuitive interpretation of why a joule is a gram multiplied by the speed of light squared? For a layman like me, this seems very unintuitive - first of all what role the speed of light plays in this relationship and why it's squared (m^2/s^2). I also see the speed of light to some power N>2 in other physics formulas. How can I intuitively interpret this speed of light to some power?
     
    Last edited: Jan 29, 2012
  2. jcsd
  3. Jan 29, 2012 #2
    In the formula E=m(c)^2, the c^2 doesn't really physically represent anything. Its a constant of proportionality, the formula tells us that mass is just another facet of energy, the c^2 is just there to tell us how much mass to energy there is. As for why things travel different speeds, well I don't really know.
     
  4. Jan 29, 2012 #3
    why does it happen to be c^2? or why is this at least the limiting case? why is it not 10^23*2.5.234624563456....it if was just another proportionality constant it seems to be a big coincidence to be the speed of light exactly squared....
    there is a releashionship, i jsut dont see it
     
  5. Jan 29, 2012 #4
    Both of these statements are not correct. The latter seems to have some experimental support of late, but keep in mind that 1. the experiment has not be independently repeated and verified and 2. we have been observing supernova neutrino for many years now and they are found to be slower-than-light.

    Graviton is predicted to travel at the speed of light. It is consequences of gravity being a long range force just like electromagnetism.
     
  6. Jan 29, 2012 #5
    The numerical value of c is irrelevant. It is just a factor for unit conversion. The universe does not distinguish energy and mass like we do with our language. So historically we invented two units for them, but in fact there is a natural conversion factor between them that we did not discover until early 20th century.
     
  7. Jan 29, 2012 #6
    nothing in the joule unit contains the speed of light and nothing in the gram unit contains the speed of light...
     
  8. Jan 29, 2012 #7

    DrGreg

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    Actually the metre is defined to be the distance travelled by light in (1/299,792,458) seconds, and the joule is defined in terms of the metre, so the speed of light does come into it.
     
  9. Jan 29, 2012 #8

    Janus

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    It evolves from the that fact that c happens to be a special speed in our universe, in that it is invariant. Invariant means that everyone measures it as having the same value relative to themselves regardless of their own motion. For example, if something was traveling away from you at c, and you tried to chase after it, no matter how fast you traveled, whenever you measured the speed of the object it would still be traveling away from you at c, while a person who stayed put and didn't tried to chase it would still measure it as traveling at c away from them.

    What this leads to is the conclusion that observers with relative motion with respect to each other measure time and space differently. There is a set of equations that can be used to relate these measurements between the two. ( the reason that no one noticed this before Einstein is that at the type of speeds with which we normally deal with, this difference is so small as to be unmeasurable.)

    If you apply this to determining the energy of a moving object, you come up with a formula that is different than the Newtonian equation. In the Newtonian equation, the energy goes to zero if the velocity of the object is zero, however the new formula it does not. If you set the velocity to zero you are still left with

    E=mc^2

    which indicates that even an object at rest has energy associated with it and that mass and energy have a relationship to each other.
     
  10. Jan 29, 2012 #9
    Yes, but it only comes into play once. Why do you divide twice by the speed of light?

    I was already familiar with the basic premise of special relativity but it's always good to get a refresher on it. Thanks! My real motivation, however, is to determine the significance of the speed of light to some power n (=2, 3, 4). I see it all the time and i dont have an intuitive explanation for it.
     
  11. Jan 29, 2012 #10

    atyy

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    If the left side has energy, and the right side has mass μ, then we must have the units on both sides be the same. One way to get that is to multiply the mass by the speed of light squared. This is just the same why the Newtonian kinetic energy is mv2/2. The only difference there is that that velocity is the body's velocity. In relativity, even rest mass has energy, and the only velocity that is the same for all bodies at rest is the speed of light, because the speed of light is a property of spacetime, and all bodies are in spacetime.

    If there is another mass M in the theory that is a fundamental constant, then we could make dimensionless quantities μ/M and multiply with it N times and still have the right units: E=μc2(μ/M)N. If we are willling to allow quantum mechanics (with the universal constant h) and gravity (with the universal constant G) into special relativity, then √(hc/G) has units of mass and we'd have to answer why not E=μc2(μ√(G/hc))N. However, since special relativity holds classically and in the absence of gravity, we could argue that we don't allow formulas with h and G in them.
     
    Last edited: Jan 29, 2012
  12. Jan 30, 2012 #11
    that hit the spot :approve: thx
     
  13. Jan 31, 2012 #12

    atyy

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    Just a couple more details. So we got to the point where E=μc2 for an object at rest, where μ is the rest mass. But we believe from Newtonian physics that the energy should also depend on its velocity. Now v/c is dimensionless, so we can imagine that velocity produces corrections to the rest energy that keep left and right hand units the same: E=μc2(a0 + a1(v/c) + a2(v/c)2 + a3(v/c)3 + a4(v/c)4 + ...)

    Since the energy is always positive (ie. moving left with velocity -v doesn't make the energy negative compared to moving right with velocity +v), we set a1, a3, a5 ... to zero. The a0 term has the form of rest mass energy, and the a2 term has the form of Newtonian kinetic energy. The interesting thing in special relativity is that the higher order terms a4, a6, a8 ... are not zero. All the aj are determined by the constancy of the speed of light in all reference frames and conservation of energy. The entire series sums up to E=μc2/√(1-(v/c)2), which reduces to E=μc2 for stationary objects.
     
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