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E^2 = m^2

  1. Jul 15, 2009 #1
    hi, I have a very simple question. I am reading Bernard F. Schutz's A first course in general relativity. after completing the special relativity section I had a goal to sit down with paper and completely derive from square one the lorentz metric, four velocity and momentum, and finally come up with e=mc^2 without referring once to the book. Well I did all this quite quickly, and my result was E^2=m^2 at rest or E=m at rest.

    I wanted to corroborate my result with those who would know, is this a result of time being in different units involving c, or is it an error to simply come up with e=m, or both?

    If you need more details for my derivation let me know, and also if you can show how this turns into e=mc^2 with normal time please do so.
    sorry if there's a similar post
    thanks,
    matt
     
  2. jcsd
  3. Jul 16, 2009 #2
    Re: E=m

    Well in some situations people set constants like hbar or c to one in which case you would get E = m but in some retarded units. In general E = m is not an enlightening equation and so it's best to keep everything in. If you managed to get that far you've probably just missed a factor of c^2 along the way =P
     
  4. Jul 16, 2009 #3

    diazona

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    Re: E=m

    Natural units, they're called, and they're not "retarded," just inconvenient. It makes it easier to catch many mistakes when you keep the units and constants in the equations.

    Matt, I'd suggest going through your work line by line and looking for the place where the units start to be inconsistent. If you get stuck you can post your work and I'm sure someone here could identify the error.
     
  5. Jul 16, 2009 #4

    Fredrik

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    Re: E=m

    I strongly disagree with this. Natural units are neither retarded, nor inconvenient. They just make everything easier. Many of those mistakes that you can "catch" by keeping the c's are mistakes you make because you keep the c's. I always use natural units, and I really can't see any reason why anyone wouldn't.
     
  6. Jul 16, 2009 #5

    clem

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    Re: E=m

    People who don't like c=1 should calculate c in astronomical units of Light Years per year.
    Hint: There are 3.16X10^7 seconds in a year.
     
  7. Jul 16, 2009 #6
    Re: E=m

    I am working in natural units, and i think the biggest problem for me is getting out of them :).

    Here is my work starting from the four momentum vector B:
    B->(E,mvγ,0,0) where γ is the lorentz equation (1-v^2)^-.5
    I think the biggest problem is that I don't fully understand why the time component is E (and i don't quite expect to as a high school student), and that is one of the only exceptions I had in my derivation. My rationale was this though: the basis vector of the four vector was its derivative, therefore the basis vector of the four momentum is its derivative, ma. Then I went on to say that potential energy mgh is a form of ma since g is acceleration, and I decided to accept that the time component was E from there. is that totally incoherent?

    so I found the magnitude of the four momentum: -E^2+(B_2)^2 for rest particles is just -E^2

    Then looking at the four momentum magnitude as a whole: mass^2(UdotU)=m^2(-1)=-m^2 where U is the four velocity.
    then E^2=m^2 or E=m

    now I suppose if I understood completely where the E came from then I could convert everything back to normal time. Otherwise, if this is correct for natural time I would still feel somewhat accomplished.
     
  8. Jul 16, 2009 #7

    Fredrik

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    Re: E=m

    You can take the equation [itex]p=(E,\vec p)[/itex] as the definition of the symbol E, and then show that [itex]E^2=\vec p^2+m^2=\gamma^2m^2[/itex]. This gives us a reason to interpret E as energy, because the work required to accelerate a particle with mass m from speed 0 to speed v is equal to the change in the quantity [itex]\gamma m[/itex]. See #15 in this thread. (Ignore the first paragraph).
     
  9. Jul 16, 2009 #8
    Re: E=m

    ok thank you. I have completed the process, and found out that my relation to the potential energy was in fact decent reasoning :)
     
  10. Jul 17, 2009 #9
    Re: E=m

    Talking about the natural (or geometrical) units, may be the problem that somebody have with them (and says that this units are inconvenient) is they don't know the definition of them ([tex]c=1, G=1[/tex] for special relativity is only relevant [tex]c=1[/tex] and therefore we have to redefine time units to measure time in meters with the consecuence that velocity is adimensional, energy, linear momentum and mass are measured in kilograms) and they don't know how to reverse to SI units (to convert energy from natural to SI we have to multiply by [tex]c^{2}[/tex] (the SI value of [tex]c[/tex]), for linear momentum multiply by [tex]c[/tex], for mass you don't have to multiply by nothing).

    And in fact not only the natural units make things easier, much more... this units made more visible the fact of relativity as unifying theory: one that unifies space and time or energy and momentum.

    And who is saying that are retarded must know that almost all books about relativity in with modern approach or mathematical focused are in this units. Retarded is carrying [tex]c[/tex] in every calculation with the risk to make a mistake, as is retarded in some old texts to try making Minkowski spacetime a Euclidean one by introducing [tex]ict[/tex].
     
  11. Jul 17, 2009 #10
    Re: E=m

    This is definitely an important point. That is, understanding what natural time is. I did understand the meaning of natural time, unfortunately I did not figure out how to switch back until after a few replies to this thread.

    Here is something to think about: seconds is a manmade interval, and natural time is c (mother nature's most constant thing) over a given interval. Therefore when natural time is used and understood the results are indeed more enlightening, and far from retarded.
     
  12. Jul 17, 2009 #11

    Fredrik

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    Re: E=m

    "Natural time" is not a standard term. I don't see what you mean by it. :confused:
     
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