Confused about E=mc²

Discussion in 'Science and Technology' started by Lt. Uhura-Brown, Jul 18, 2013.

  1. Lt. Uhura-Brown

    Lt. Uhura-Brown Lieutenant Commander Red Shirt

    Joined:
    May 14, 2013
    Location:
    New Zealand
    I was thinking about the famous equation E=mc² the other day, when I realised I didn't actually understand it, or rather, the units used.

    It seems to me that Joules = (volume multiplied by density) multiplied by (300 million meters divided by 1 second) multiplied by (300 million meters divided by 1 second)?

    I know there's something I must be missing, but I can't wrap my head around how it all fits together.
     
  2. Christopher

    Christopher Writer Admiral

    Joined:
    Mar 15, 2001
    Mass is a basic quantity that doesn't have to be broken down; kilograms are generally the preferred unit. A joule is a newton-meter, and a newton is a kilogram-meter per second squared; so that means joules are (kg x m/s^2) x m = kg x m^2/s^2 = kg x (m/s)^2, which is mass (kg) times velocity (m/s) squared, with the velocity being the speed of light.

    So a mass of 1 kg, if converted to energy, would come out to 1 kg x (3 x 10^8 m/s)^2 = 9 x 10^16 J.
     
  3. Asbo Zaprudder

    Asbo Zaprudder Admiral Admiral

    Joined:
    Jan 14, 2004
    Location:
    Rishi's Sad Madhouse
    ^What Christopher says. Also, in particle physics and astrophysics, the units commonly used are electron volts (eV) for energy and eV/c² for mass -- the latter usually being abbreviated to eV if natural units are used which set c=1.

    One volt (V) is defined as being the potential difference when one coulomb (C) of charge crosses that potential resulting in a change of one joule (J) of energy. One electron volt is defined as the magnitude of the energy change of an electron moving through an potential difference of one volt. So 1 eV = 1.602x10^−19 J given that the magnitude of the charge on the electron is 1.602×10^−19 C (the charge is actually a negative quantity for electrons and positive for positrons, of course).
     
    Last edited: Jul 19, 2013