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Questions about e=mc2 and probably other equations

  1. Nov 29, 2012 #1

    So, I was wondering something about the units used with other units in a equation like e=mc2

    For example;

    m = 1.67x10^24 grams
    c = 292.792.458 m/s

    that makes e about 1.437x10^-8
    but what is the unit used to represent e in this context?
    and what if I used kg in m and km/s in c, what would the unit for e be then?

    An explanation would be really appreciated.

    Side note:
    I just started to get an interest in physics like 2 weeks ago,
    and never had any prior education in school or whatever about these sort of things.
    nor had any education in maths used with most physics equations (algebra and stuff)
    I really find it all very interesting, seen various coursed on youtube by Walter Lewin from MIT, and other channels like Sixty Symbols, veritasium1 and numberphile (and others)

    Kind regards and thanks for reading,
  2. jcsd
  3. Nov 29, 2012 #2


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    For one thing the equation is correctly written ##e=m c^2## or e=mc^2. And the value your for e given m and c is WAY OFF. Finally, kg*m^2/s^2 is a unit of energy called a Joule. Other combinations like g*m^2/s^2 or kg*km^2/s^2 don't have names. If you are given other units, I'd convert them to kg, m and s first.
    Last edited: Nov 29, 2012
  4. Nov 29, 2012 #3

    Ray Vickson

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    Not only do you multiply the numerical values, you also multiply the units just as though they were algebraic objects. So, c^2 has units m^2/s^2, hence E = m c^2 has units
    kg m^2/sec^2; this is also called the Joule and is denoted as J. (Unfortunately, the symbol 'm' for mass is the same as 'm' for meter, and that can cause confusion; that is why sometimes we write out the word 'meter' or 'metre' in full.)

    Note that I used 'E' instead of 'e', because that is more usual in Physics (E = energy, e = magnitude of electron charge, or base of natural logarithms).
    Last edited: Nov 29, 2012
  5. Nov 29, 2012 #4


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    Stick to (or convert to) SI units:
    kg for mass
    m for distance
    s for time

    Then you find that all the equations give you the SI unit for whatever, and you can look that up on Wikipedia.
  6. Nov 29, 2012 #5
    That confused me even more just now.

    What I get is; use Meters in combination with kilograms to get energy as Joules
    What I don't get is; km^2/s^2, as km/s is a short representation of kilometers per second, right? 1 seconds squared would still be 1 right?

    So, to calculate E with the numbers given above should be like
    m = 1.67x10^-27 kg (i think?)
    c = 292.792.458 m/s (is still meters per second)
    then E = (1.67x10^-27)*(292,792,458^2) = 1.431x10^-10 J

    Am I going the right way with this now?

    Thanks for the replies :)
  7. Nov 29, 2012 #6


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    The calculation looks right now. Yes, 1.67x10^(-24)g=1.67x10^(-27)kg. You didn't have the minus sign on the exponent in the original post. That's why I thought it was so far off.
  8. Nov 29, 2012 #7
    Ah lol indeed, i forgot to put the minus sign in the first post, my bad :p
    Thanks for clearing this up :)
    Now I can sleep without this swarming through my brain lol :p
  9. Nov 29, 2012 #8


    Staff: Mentor

    Numerically, it's 1, but the units would be sec2.
  10. Nov 29, 2012 #9
    Lewin's videos are for people who have already taken calculus so those may not be the best ones out there. You might just want to look for some conceptual videos.

    Also I wouldn't start with trying to understand Einstein if you're completely new to physics. Go back to Newtonian physics and progress from there if you want a deeper understanding of physics.
  11. Nov 29, 2012 #10


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    The CGS unit of energy, 1 g*m^2/s^2 , is called an erg. For historical reasons, most astronomers and astrophysicists use ergs instead of Joules. 1 Joule = 10^7 ergs.
  12. Nov 29, 2012 #11
    I actually find Lewings lectures quite easy to understand, and his lectures got me started in physics :)
    But the reason for this thread was not to understand einsteins e=mc^2, but more about the actual maths behind it, and it actually got a lot more clear to me now :)
  13. Nov 30, 2012 #12
    You may find it hard t comprehend units like "per second squared" but you can think of it like "per second per second."

    If Chevrolet says their new Corvette goes 0 m/s to 20 m/s in 4 seconds, that is an acceleration of 5 meters per second per second or 5 meters per second squared. This is kind of irrelevant but I hope you get the meaning of some of these units.
  14. Nov 30, 2012 #13
    An erg is 1 g*cm^2/s^2.
  15. Nov 30, 2012 #14


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    My mistake. Thanks for the correction.
  16. Nov 30, 2012 #15


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    You are going to confuse yourself as well as others if you "mix" numerical notations. The United States, England and about half of the world use "." to separate the integer part of the number from the fraction part and "," to separate groups of thousands. That is the notation you use in "m= 1.67 x10^24 grams". Most of Europe and the other half of the world reverse those. That is the notatation you use in "c= 292.792.458 m/s".

    Use either
    m= 1.67 x 10^24 grams
    c= 292,792,458 m/s

    m= 1,67 x 10^24 g
    c= 292.792.458 m/s

    But not both!

    No, it doesn't. Without checking the actual arithmetic, m and c both has positive powers of 10 so their product cannot have a negative power of ten.

    In this contex "gram-meters squared per seconds squared" which does not have a standard name because it mixes "MKS" and "gms" systems. (You keep doing that!)[/quote]

    That would make more sense. You are now working completely in the "MKS" (meter-kilogram-second) system. Then e would be in "kilogram-meters squared per second squared" which is called "Joules".

    If you measured mass in grams and speed in centimeters per second, you would be working in the "cgs" (centimeter-gram-second) system. Then e would be in "gram-centimeters squared per second squared" which is called "ergs".

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