E=MC2

Discussion in 'Science' started by ronmatt, Oct 6, 2014.

  1. ronmatt

    ronmatt New Member

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    This is for those science pros.
    If energy is equal to mass traveling at the speed of light squared, why doesn't the speed of light have to observe the universal speed limit of 186,000mph in a perfect vacuum. How does mass attain the speed of light squared (186,000 x 186,000)? What is mass before it attains that speed? Say, 163,000 mph. squared.
    I know that there's a good answer for this, there must be. So can I get it in terms that I can possibly understand and not some formula?
     
  2. HonestJoe

    HonestJoe Well-Known Member Past Donor

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    This equation doesn't represent anything moving. It's just establishing that an object's energy is equivalent to its mass. The speed of light (c) in the equation is just a constant, a raw number that happens to be the factor of that mass/energy equivalence. There will be fundamental reasons why that constant is the speed of light but I'm not going to pretend I could explain what they are.
     
  3. tecoyah

    tecoyah Well-Known Member Past Donor

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    In a nutshell....the speed of light squared in this equation is a multiplier used primarily to demonstrate and incorporate the supposition that matter and energy are the same thing. It is an important aspect of the equation, but not to be taken out of the context in which it was used. Thing of it as the comma, in this sentence.

    The speed of light is a constant, and can only be postulated in mathematics and theory.
     
  4. AlpinLuke

    AlpinLuke Well-Known Member

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    I cannot say there is a simple way to explain this. Let's start from some points, then you can ask for further details.

    That square is a math formalism, it indicates, for example, also the energy factor of velocity in the kinetic energy of a moving mass.

    When you calculate the kinetic energy of a moving mass at not the speed of light you measure the work necessary to stop it [and it is E =1/2MV2, when V=C it's 1/2 of the energy of Einstein's formula].
    Einstein's formula represents the total potential energy of the mass in its ground state [since the speed of light is a constant and it's the maximum limit, the kinetic energy of a mass moving at that speed is just 1/2 of the total potential energy, so it's undetermined like the mass].

    From the perspective of light ... light is firm. There is no apparent motion, in the sense of the motion itself, for an observer traveling at the speed of light, since distance becomes equal to null and time substantially infinite.
    There are two equations describing this, which usually accompany the equation you have mentioned: the Lorentz transformations.

    For lengths it is: L1=L0*[(1-V^2/C^2)^0.5]

    And this is rational: it's evidently impossible to accelerate further an infinite mass and so at speed of light a mass cannot move [in poor words the space time becomes a 2D ring surrounding the observer].
     
  5. ronmatt

    ronmatt New Member

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    You'll correct me if I'm wrong? What you're saying is; In the physical universe, mass does not (can not) travel at the speed of light squared thus converting to energy. That this occurs only in a mathematical equation?
     
  6. ronmatt

    ronmatt New Member

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    Whew...I'm so pleased with my chosen vocation.
     
  7. tecoyah

    tecoyah Well-Known Member Past Donor

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    Not quite, the implication would be that for all intent and purpose, mass is energy and vice versa in this physical universe and as what we see as light is pure energy it obviously travels at the speed of itself. I was pointing out the realities of mathematics and thus equation. This one is a thought experiment meant to allow for observation and explanation of scientific understanding of what was at the time invisible.
    In this way we can create theory and build upon it or dismiss it. If this energy gains mass, it is forced to follow the laws of physics like everything else and will thus be limited in the velocity it achieves.
     
  8. ronmatt

    ronmatt New Member

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    I will have to allow myself time to digest that. Thanks
     
  9. Prunepicker

    Prunepicker Well-Known Member

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    There isn't an easy answer. You'll do good to search for yourself, as
    those who have responded have done, on Google.

    Here's a article that offers some important basics that can help you
    understand the equation. Once we understand the fundamentals we
    can move on to the next level.

    http://profmattstrassler.com/articl...tter-etc/matter-and-energy-a-false-dichotomy/
     
  10. ronmatt

    ronmatt New Member

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    In my estimation, that's one hell of an article. And you're correct that there isn't an easy answer (either to my initial question or any future questions I may conjure up). I will print it out and read it slowly at my leisure (I hate to read things on a computer screen). And I will follow the provided links...then, possibly get back in a year or two with more questions. OR simply come to grips with being too damn stupid to understand any of it and divert my future inquiries to things like spaghetti sauce recipes.
    One last question though; How do theoretical physicists ever communicate with one another with so many 'ifs, ands or buts'?
     
  11. Prunepicker

    Prunepicker Well-Known Member

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    Don't ever feel stupid about this. There are physicists that don't agree with
    themselves. They're still learning because things change, well, except
    matter and energy.
    Oooh, I see how you're bewildered. They use a series of profound explanans
    that are comprised of ifs, ands or buts to communicate with each other. They
    always have. Some will communicate by sending a code of antilogarithms by
    banging their heads on a hard surface. This can be quite quaint.

    Up is sometimes down but, and it's a big but, what is always "huh?"
     
  12. cupid dave

    cupid dave Well-Known Member

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    They use Mathematics,...
     
  13. cupid dave

    cupid dave Well-Known Member

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    You are reading the equation improperly here.

    The equation says that a particular mass which is changed into Energy will equal its value multiplied by C^2.

    It is not saying that the mass is actually moving,
    Nor does it mean the mass is traveling at C^2, either.

    When atoms split into two different kinds of other atoms, their masses when added together are less than the starting mass of the atom which is split.
    That difference, between the starting mass and the ending mass, become Energy.
     
  14. Poor Debater

    Poor Debater New Member

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    In classical (Newtonian) dynamics, there is a way to compute the kinetic energy of a body in motion. That equation is:

    E = 1/2 mv²

    Note that energy, the left hand side of the equation, can be expressed in terms of mass times velocity squared, the right-hand side of the equation.

    When you add energy to a body at rest, its mass does not change but its velocity does. As that body accelerates to relativistic speeds, things change: the mass of the object begins to increase as a consequence of its motion. As the body continues to accelerate, its mass continues to increase, which makes it harder and harder to accelerate the object. So as you add energy to the moving object, more and more of that energy goes towards increasing its mass, and less and less of that energy goes to increasing its velocity. At the limit, which is the speed of light, all of the energy you add to a body goes to its mass, and none of it goes to velocity. Thus no object can ever accelerate beyond lightspeed.

    Newton answers an important question: when we add energy to a body at rest, it gains velocity. So what is the "rate of exchange" between energy and velocity? In other words, how much energy do we need to add, to obtain a given velocity? And the Newtonian equation above tells us that.

    Einstein extends that question by asking another, related question: when we add energy to a body at lightspeed, it gains mass. So what is the "rate of exchange" between energy and mass? How much energy do we need, to gain a certain amount of mass? And the answer is:

    E = mc²

    ... where c is the speed of light. Note that in Einstein's equation, just like Newton's, energy (on the left hand side) can be expressed as the product of mass and velocity squared (on the right hand side).
     
  15. Phoebe Bump

    Phoebe Bump New Member

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    Unless I'm missing the concept, whenever there is a small fission explosion, mass equal to the approximate weight of a dollar bill gets converted to energy. It's not really pure mathematics anymore.
     
  16. 10A

    10A Chief Deplorable Past Donor

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    There's kind of two different masses at work. There's the rest mass and the relativistic mass (which is really just energy). As velocity is increased, the relativistic mass does increase according to M=E/C^2 (even at slow speeds, but the energy and relativistic mass delta is miniscule). The faster you go, the more energy is required to make you go even faster. Essentially inertial mass is increased. The rest mass doesn't increase, e.g. the gravitational field of the object remains the same (otherwise fast moving objects would turn into black holes, or warp space-time).

    Another way to look at Einstein's equation is that to accelerate an object to the speed of light, you have to convert all its rest mass to energy. But then you don't really have an object anymore. Anything traveling at light speed has zero mass.
     
  17. fifthofnovember

    fifthofnovember Well-Known Member

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    To kind of simplify this explanation, E=MC^2 IS the kinetic energy formula (E=1/2 MV^2). They are the same. The V in the kinetic energy formula is just given a definite value of C because an atom is actually made up of a bunch of particles moving at the speed of light, which creates the energy fields such as electron shells. The mass thus contains that amount of moving energy while "sitting still", because there really is that amount of movement going on inside of it. The reason that there is no "divided by two" in E=MC^2, like there is in the non-relativistic kinetic energy formula, is because normally velocity can only be measured as the difference in speeds between two bodies. If you are sitting in a chair, you are not moving compared to the chair, but you are moving very fast relative to the sun. Your velocity is meaningless except as compared to exactly one other object. But the speed of light is different. It is the same speed compared to both the chair and the sun. Therefore, instead of dividing by two, you divide by one.
     
  18. Gelecski7238

    Gelecski7238 Well-Known Member Past Donor

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    Space is thought to be granular, down to the Planck size, which is mighty small. Some sources call the granules "cells" or "reality cells." Transmission occurs from cell to cell as each cell changes state. The speed of light is the fixed rate at which the cells can change state. One "explanation" offered for the value of the fixed rate is that it is assigned by the design of this particular universe.
     
  19. fifthofnovember

    fifthofnovember Well-Known Member

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    Actually, I think I can clarify a bit more. There is ONE formula which appears to be two formulas, because of a "hidden variable". That variable is "frames of reference". The formula would be stated "(Energy) equals (Mass) times (Velocity squared) divided by (Frames of Reference)". But there are only two possibilities to plug into the "frames" variable: one, in the case of lightspeed, and two, in the case of everything else.
     
  20. 10A

    10A Chief Deplorable Past Donor

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    They really are 2 formulas, they are not equivalent (other than units). One has a constant (for speed) and one has a variable (for velocity). Rest mass converts to energy according to E=mc^2 no matter what speed an object is traveling. c is not a vector, v is. c is a constant, v is not.
     
  21. fifthofnovember

    fifthofnovember Well-Known Member

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    No, velocity is not a vector. A vector contains both speed AND direction information, velocity does not. And yes, C is a constant, whereas V is a variable, but anytime you go to actually solve the formula, you plug in the actual number (constant) for V. C is the number you plug in to the variable in this case. E=MC^2 is just a specific case wherein the velocity is known.
     
  22. Herby

    Herby Active Member Past Donor

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    There is some confusion around here. Let me try to clear things up a little.

    E = m c^2

    Is the energy E of a particle with mass m at rest. This is a special case. There is a more general equation for the energy contained in a particle which is moving (with momentum p, velocity v).

    E^2 = (m c^2)^2 + (pc)^2
    or
    E = m c^2 / sqrt(1 - v^2/c^2)

    Note that the special case of a massless particle m = 0 (like a photon of light) is also covered by the above equation.

    E = pc

    The classical kinetic energy formula is an approximation which holds for v << c. The mathematical tool used to get this insight is called a Taylor expansion.

    E = m c^2 + 1/2 m v^2 + 3/8 m v^4/c^2 + ...

    The first term mc^2 is the energy at rest (v = 0). The second term 1/2 m v^2 is the classical kinetic energy. You might ask yourself now how Newton and everybody else using "old school physics" could possibly miss the mc^2 part which is usually much larger than the 1/2mv^2 part in everyday life. Well, energy is a rather abstract concept. Only the transfer and conversion of energy is measurable. The mass of the bodies involved does not change (much) in our usual interactions with matter though. Even the most energetic chemical reactions result in very minute changes in mass (near impossible to measure) because mc^2 is very large compared to the energy stored in chemical bonds. Changes in mass due to nuclear reactions, however, have been measured.

    The link below shows a plot of the energy contained in a moving particle according to the theory of relativity and its classical approximation.
    http://www.wolframalpha.com/input/?i=plot+1/sqrt(1-v^2)+and+1+v^2/2+from+v+=+0+to+0.9

    Here is the Taylor series.
    http://www.wolframalpha.com/input/?i=taylor+series+in+v+of+m*c^2/sqrt(1-v^2/c^2)
     
  23. Aleksander Ulyanov

    Aleksander Ulyanov Well-Known Member

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    I have one gram of matter. How much energy will it convert to? (In ergs and/or joules) Show your work and that should clear the matter up, yes?
     
  24. AlpinLuke

    AlpinLuke Well-Known Member

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    If you are able to convert a gram of matter into energy you will obtain about 90.000 billions of joules or, if you prefer a more common measure, 25,000,000 Kwh.

    This is not so easy to get: an atom bomb converts in energy only a poor 0.5% of the mass ...
     
  25. One Mind

    One Mind Well-Known Member Past Donor

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    Speaking of energy, I have heard physicists say, that in the vacuum between the nucleus and the electron of a hydrogen atom, a single atom, that it contains more energy than all of the objects in space for something like a billion light years.

    Now that is completely mind blowing. A human mind has trouble grasping that, or mine does. LOL
     

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