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Concerning E=mc2

  1. Jul 6, 2012 #1
    E=mc2. I am an avid reader of physics and science in general and a question popped into my head concerning the squaring of the speed of light in the equation E=mc2. It is my understanding that the speed of light is constant it does not accelerate. What rule or mechanism or whatever allows for the squaring of the speed of light in this equation.
  2. jcsd
  3. Jul 6, 2012 #2


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    The speed of light is used in this equation as a conversion between units of mass and energy. As energy is measured in units of Joules (kg*m^2/s^2) and mass is measured in kg, one needs to multiply a mass by units of velocity^2 in order to get units of energy. It turns out that the required conversion factor is exactly c^2 (requires a bit of special relativity to derive).

    One should view the factor of c^2 as simply a conversion factor. I has absolutely nothing to do with light actually moving around or accelerating or some such.
  4. Jul 7, 2012 #3


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    The mechanism is "arithmetic". It has nothing to do with physics. In this formula, c is just a number- there are plenty of numbers larger than "c" and [itex]c^2[/itex] happens to be one of them.
  5. Jul 7, 2012 #4
    So, the "energy" that any "matter" has is equal to that "matter" multiplied by the speed of light sqaured?

    Id love to see that equation worked out using some real number to come up with exact amounts and then recreate them perfectly in reality...I suppose I am a doubter too...E=mc2 has always seemed a bit dreamy to me.
  6. Jul 7, 2012 #5
    What does 1kg of matter equal in energy? And how do I muliply matter and c2 in reality? a match?
  7. Jul 7, 2012 #6


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    1 × 299,792,458 × 299,792,458 joules.
  8. Jul 7, 2012 #7


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    Or, more pedantically,

    1 kg × 299,792,458 m/s × 299,792,458 m/s

    = (299,792,458)2 kg·m2/s2
    = (approximately) 9 x 1016 joules
  9. Jul 7, 2012 #8


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    Which is the energy released in the explosion of a 20 megaton thermonuclear weapon...
  10. Jul 8, 2012 #9
    Do not confuse "matter" with "mass". They are not the same thing nor interchangeable. (This is actually a pretty common mistake that many people make.)

    The proper way of expressing the above would be "the energy that any matter has is equal to its mass multiplied by c squared". Big difference.

    As already mentioned, the formula would not work as "E = mc", and one clear-cut way of determining why that formula is incorrect is to see that the units on both sides of the equation do not match. This is always a good test to check if a physics equation is invalid: If the units on both sides of the equation do not match, then you know that the equation is bogus. (Of course note that the same does not hold in the inverse case. In other words, just because the units match doesn't necessarily mean the equation is correct. The units matching is a requirement, but of course it's not all in itself sufficient to determine the validity of the equation.)

    Could someone post a short summary on how the "E=mc^2" equation is derived?
  11. Jul 8, 2012 #10
    Good info...thanks!

    I'm more interested to see the equation proven in the real world...

    Someone mentioned that 1kg of mass creates the energy of a 20 megaton bomb...why wouldnt it create the energy of a 1 kilogram bomb?
  12. Jul 8, 2012 #11


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    A nuclear fission or fusion bomb converts only a small fraction of its original mass into energy. (but a much larger fraction than with TNT)
  13. Jul 8, 2012 #12


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    All the "tonnes" in bomb making are given with respect to TNT bombs. E.g. 20 million tons of TNT.

    The largest thermonuclear bomb ever detonated was the Tsar bomba which had a yield of 50 million tons of TNT. Obviously the bomb itself did not weigh anywhere near 50 million tons.
  14. Jul 8, 2012 #13
    Thanks everybody for the replies. E=mc2 looks a lot like F=ma to me. Since electromagnetic waves to not accelerate I'm curious as to why squaring the speed of light works in the equation. But correct me if I'm wrong please about the EM waves. It's my understanding that photons move at 300,000 km/s from the moment of thier creation.
  15. Jul 8, 2012 #14
    They are completely different equations, with completely different meaning. The only thing they have in common is the mass.

    Square of speed has nothing to do with acceleration. They have exactly zero in common.

    In vacuum, approximately, yes. The exact value for vacuum is 299,792,458 m/s and is denoted c.
  16. Jul 8, 2012 #15
    You are correct that electromagnetic waves (or photons) always travel at the speed of light c. However E=mc^2 is a (static) relation concerning the energy contained in a body with mass m, i.e. the energy due to it having mass. It is not a dynamical equation telling us how things move (like F=ma). Therefore it is totaly alright to have c^2 in the equation.
  17. Jul 10, 2012 #16
    Disabuse yourself of the notion that squaring c has anything to do with acceleration. c is a constant, and c squared is another constant. It has absolutely nothing to do with acceleration.

    (Besides, m2/s2, which is the unit of c2, is not the unit of acceleration. It's still the unit of speed.)
  18. Jul 10, 2012 #17


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    To avoid confusion, no, [itex]m^2 / s^2[/itex] is not a unit of velocity. It simply is what it is. One might say it's a unit of Energy per mass.
  19. Jul 10, 2012 #18
    Are you sure? Is there any difference between saying "that car is traveling at 4 [itex]m/s[/itex]" and "that car is traveling at 16 [itex]m^2/s^2[/itex]" (other than the latter being quite unconventional and needlessly complicated but, ultimately, the same thing as the former)?

    Where is that coming from?
  20. Jul 10, 2012 #19
    It's not the same. Why would it be? The latter is a square of the former.
  21. Jul 10, 2012 #20


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    Yes there is quite a difference. The first way is correct, the second way is wrong.

    Would you say "the length of this table is 9 m^2"? No you would not, m^2 is a unit of area not length. You can only say "the length of this table is 3m".
  22. Jul 10, 2012 #21
    Assuming the energy of a body is linear correlated to its mass (as used by Newton):

    [itex]E = k \cdot m[/itex]

    Than the change of its energy is

    [itex]dE = k \cdot dm = F \cdot ds = \left( {m \cdot \frac{{dv}}{{dt}} + v \cdot \frac{{dm}}{{dt}}} \right) \cdot ds = m \cdot v \cdot dv + v^2 \cdot dm[/itex]

    Integration of the resulting differential equation

    [itex]\frac{{dm}}{m} = \frac{{v \cdot dv}}{{k - v^2 }}[/itex]

    leads to

    [itex]m = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{k}} }}[/itex]

    The constant of integration m0 is the mass of the body at rest and as this equation has real solutions for v²<k only the factor k must be the square of a maximum velocity that can not be reached or exceeded by the body. Experiments show that this speed limit is the speed of light in vacuum. Thus there can be no proportionality between energy and classical inertial mass except:

    [itex]E = m \cdot c^2 = \frac{{m_0 \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}[/itex]
  23. Jul 11, 2012 #22
    Center O Bass , thank you for the insightful information.
  24. Jul 11, 2012 #23
    That's a great idea. Here is the essense of Einstein's E = mc2 equation.

    Consider a body, at rest in the inertial frame S, which emits two photons, each of frequency f, one in the +x direction, the other in the –x direction. The total energy of the two photons is E = 2hf. For energy to be conserved there must be a decrease in the energy of the body. This implies that the body previously contained energy. There must have been an internal change that resulted in a physical state having lower value of energy. Since the total momentum of the photons is zero the emitting body must remain at rest otherwise total the total momentum of the system would not be conserved.

    Now consider the process from the inertial fram S' which is in standard configuration with respect to S and is moving in the +x direction with speed v. In S’ the body is moving in the –x’ direction with velocity v’ = -v ex. An observer in S’ observed the body emit two photons. One photon is emitted in the +x’ direction and the other in the –x’ direction. The photons have frequencies f+ and f- respectively. The velocity of the body remains unchanged. Due to Doppler shift the photon moving in the +x direction is red shifted from f and the photon moving in the –x direction is blue shifted f. The shifted frequencies, i.e. f+ and f- are related to f by

    f+ = sqrt{[1 - (v/c)]/[1 + (v/c)]}f

    f- = sqrt{[1 + (v/c)]/[1 - (v/c)]}f

    The total momentum in S’ before the photons are emitted is the initial momentum of the body given by

    p'i = m'ive_x

    where m’i is the initial mass of the body as measured in S’. If m’f is the final mass of the body as measured in S’ then, since the velocity of the body remains unchanged, the velocity remains unchanged so that the momentum of the body after emission is

    p'f = m'fve_x

    The momenta of the photons in S’ is given by

    p'+ = (hf+/c)ex

    p'- = -(hf-/c)ex

    The total energy of the photons as measured in S’ is

    E = hf+ + hf+

    Conservation of momentum requires

    p'i = p'f + p'+ + p'-

    Substituting the values above gives

    -m'iv ex = -m'vf ex + (hf+/c)ex - (hf-/c)ex

    Upon equating the components on each side we get, upon rearranging terms and substituting the values in Eq. (19)

    Δm'v = 2γβhf/c = Δm’v = 2γhfv/c^2

    It can be shown that

    E’ = 2γhf

    Substituting the expression for E’ into the expression for the change in momentum

    Δm’v = 2γβhf/c = E’v/c2

    E’ = Δm’c2

    From this Einstein deduced (he used L where I use E')
    Therefore E = mc2 as was to be shown.
    Last edited: Jul 11, 2012
  25. Jul 11, 2012 #24
    Now that's a weird conclusion! :wink:
  26. Jul 11, 2012 #25
    Oops! My bad! I forgot the c after the m. Must be the painkillers. :biggrin:
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