help me to derive E=MC^2

Discussion in 'Physics' started by dileepchacko, Nov 17, 2008.

  1. dileepchacko

    Thread Starter Active Member

    May 13, 2008
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    Can anyone help me to derive the most famous equation of Einstein, E=MC^2
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    I think you have to have Einsteins mind to do that :p

    You can search in some high level physics books or on the internet.

    Here is a link I found by a simple search in google

    http://www.adamauton.com/warp/emc2.html
     
  3. Dave

    Retired Moderator

    Nov 17, 2003
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  4. markm

    Member

    Nov 11, 2008
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    I had to learn that derivation to pass a Physics class once, but that was 35 years ago... IIRC, Einstein started with relativistic corrections others had already worked out to the Newtonian equation F=ma. That is, as the velocity nears the speed of light, the acceleration "a" (as viewed by an observer who isn't accelerating) due to a constant force "F" decreases. Attribute that decrease to an increase in "m" with velocity.

    Now, work out the energy input (F times distance) for a change in velocity, and compare that to the change in mass.
     
  5. steveb

    Senior Member

    Jul 3, 2008
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    Nobody can help you better than Einstein himself. He wrote a very nice book suitable for anyone who understands basic algebra.

    Relativity, the Special and General Theory (A clear explanation that anyone can understand)

    See the attached jpg picture of the front cover.

    He provides a very simple derivation of E=mc^2
     
  6. Dave

    Retired Moderator

    Nov 17, 2003
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    You do realise that the above paper I referenced was an English translation of Einstein's 4th Annus Mirabilis paper that embodied part of the work in that text. In fact the translation has maintained the notion Einstein used in 1905.

    Dave
     
  7. steveb

    Senior Member

    Jul 3, 2008
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    Nope, didn't realize that, but now I do.
     
  8. Dave

    Retired Moderator

    Nov 17, 2003
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    Apologies Steve, I think I came across a little brash in my previous post.

    Dave
     
  9. steveb

    Senior Member

    Jul 3, 2008
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    No problem. You pointed out some good information.
     
  10. Dave

    Retired Moderator

    Nov 17, 2003
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    I should actually have said here, "the translation has maintained the notation Einstein used in 1905."

    Dave
     
  11. blazedaces

    Active Member

    Jul 24, 2008
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    Try drawing a plane (or spacecraft) above the earth. In the plane, draw two parallel mirrors, with a photon moving in between the two mirrors. This is going to be your picture showing how the light travels to an observer on the plane (or spacecraft).

    Now, below it you're going to draw how the light travels to an observer on earth. To do this draw three different planes, starting from one side and going to the other. In the first picture, draw the light at the bottom (how it begins). In the middle picture draw the light at the top, and in the third picture draw it back on the bottom. Now draw a line from each point to the next, one at a time. It should look like a triangle.

    Now, equipped with the knowledge that light ALWAYS travels at the speed of light, c, start to try and equate the two observations. The plane is moving at a certain velocity v and the length of the triangle sides are dependent on this velocity, or the time it's traveling really. Remember that time and mass can change, but not the speed of light.

    Of course, all the sources already provided use this proof or a variant of it (I should probably say that this proof is probably just a variant of the original, just to be specific).

    Good luck,
    -blazed
     
  12. scubasteve_911

    Senior Member

    Dec 27, 2007
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    In a vacuum! Otherwise, this is not true.

    Steve
     
  13. steveb

    Senior Member

    Jul 3, 2008
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    Also, we should remember that there are some recent theories that suggest that even the value of c in vacuum is not truely constant!

    http://www.abc.net.au/lateline/stories/s347215.htm
     
  14. blazedaces

    Active Member

    Jul 24, 2008
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    Yes, I forgot to point out this whole proof should happen in a vacuum (theoretically)...

    -blazed
     
  15. Dave

    Retired Moderator

    Nov 17, 2003
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    I wouldn't be surprised. In a universe where every motive quantity is in reference to a particular inertial frame, why should we have one variable that is universally constant?

    In note the findings are that the speed of light is a variable of time and therefore is momentarily constant (at least tangibly in short-time frames such as the existence of humankind!).

    Dave
     
  16. studiot

    AAC Fanatic!

    Nov 9, 2007
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    What a pity the original poster has disappeared from the discussion.

    Actually there is no requirement for a vacuum, so far as we know the equation holds good, whatever the environment.

    Remember that c refers to a (supposed) universal constant which is referenced in vacuo and is the ratio of the permittivity to the permeability of free space.

    Of course the actual local speed of light itself varies from c according to its local environment, but if we wish to use the matter(mass) - energy equivalence relationship this always requires the use of c as a constant, not the local speed of light. Clearly if we have some matter and wish to calculate its energy equivalent, we don't have a vacuum.
     
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