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Raw equal to E=mc^2?

  1. Sep 13, 2013 #1
    A long time ago I visualized mass accelerated to the speed of light(or theoretical maximum to this speed) as an upside-down parabola touching the y=0 line.

    Now, today I thought of the following:

    if you would represent a certain parabola(equation according to ^2 of the mass/energy ratio required) in 3d (not a cone) (maximum at y=0), and place the surface of a circle ∏r^2 within that '3d parabola' (horizontally) and move this circle upward towards the y=0, until ∏r^2 'within' the 3d parabola reaches y=0 ( so, ∏r^2) 'within' the 3d parabola = 0 = maximum acceleration to the point of black hole formation = 'point of black hole maximum mass' = speed of light limit ).

    And, ∏r^2 'within' 3d parabola moves downward so that ∏r^2 'within' 3d parabola = infinite = no mass = no acceleration (according to E=m(c)^2)

    Is it so that these '2 infinite limits'(infinite energy required and infinite ∏r^2) define the E? so that the above explanation is equal to M(c)^2? If so, what would the formula be for this 'system? (non simplified, that is all possible positions of 'horizontal ∏r^2 within 3d parabola' equation)


    I was intrigued when I read about Einsteins discovery that gravity is equal to acceleration (with equal consequences), so after some thinking I came up with the story above. Is it flawed?? as you can see, my math skills are VERY basic, but I think it should be simple to understand with some visualization.
     
  2. jcsd
  3. Sep 13, 2013 #2

    UltrafastPED

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    I couldn't follow what you were saying ... so I suggest that _you_ provide the visualization.

    I'm not even sure what your point is!
     
  4. Sep 13, 2013 #3
    My point is that this visualization is a kind of intuitive approach to E=mc^2.

    Just like a sine wave can be represented in 3d. (like a spring stretched out, from the side it will look like a sine wave). By using trigonometry within this structure, it is easier/more intuitive doing sine calculations.

    Similarly, I believe, E=mc^2 can be explained more intuitively with the explanation above, But I only have basic college math skills, so I suck at explaining it.
     
  5. Sep 13, 2013 #4

    UltrafastPED

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    E=mc^2 isn't even the complete equation ... it is the special case of a mass at rest. If it is moving you need

    E^2 = (pc)^2 + (mc^2)^2

    You might this "Minute Physics" video:
     
    Last edited by a moderator: Sep 25, 2014
  6. Sep 13, 2013 #5
    How come is it a 'special case'? Isn't E=mc^2 simply the acceleration and kinetic energy part taken out of E^2 = (pc)^2 + (mc^2)^2? It doesn't really change my question
     
    Last edited: Sep 13, 2013
  7. Sep 13, 2013 #6

    UltrafastPED

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    Nope. For clarification, watch the video.
     
  8. Sep 13, 2013 #7
    The video only confuses things; The core issue is relativistic E=mc^2. Great video for unique hits (there's more to e=mc^2 guys!!) but they don't seem to get it themselves.

    E^2 = (pc)^2 + (mc^2)^2 is about mass @ certain speed, which adds extra mass into the equation that is arbitrary, it's a side track not important to the core E=mc^2. This must be about mass-issues of things like photons (mass or no mass) In that case, to clarify things, E^2 = (pc)^2 + (mc^2)^2 could be needed.

    My question was about the interrelated-ness of energy/mass, not how it should be applied to objects.
     
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