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I E=mc^2 problem

  1. Oct 29, 2016 #1
    I've just watched a video by PBS spacetime on how an object with more potential energy has more mass, or less, in some cases (mass defect). But there is a problem.
    Quantum mechanics tells us that a particle has a chance of appearing at difderent places ( wave function), and that it also decreases with distance. So a particle has a chance (REALLY small, I know, but not 0) of appearing 1m from the event horizon of a black hole. There is also a chance that all the particles in a ball teleport thar way an arrange themselves in a ball. It isn't 0.
    So the ball has the gravitational potential energy of...yeah.
    Since it can appear almost everywhere, its potential energy should be huge, perhaps infinite, and therefore, it's mass.
     
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  3. Oct 29, 2016 #2

    phinds

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    I think you've got cause and effect reversed. An object with more mass has more potential energy for the same height in a gravitational well. This does not imply that the same object gains mass by virtue of being raised higher in a gravity well, it just gains potential energy.
     
  4. Oct 29, 2016 #3

    bhobba

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    Potential energy in QM is equally as nebulous and the equations governing it and a black hole are rather complicated and cant quite be viewed like that.

    What can be said however is we actually have a theory to describe that one:
    https://arxiv.org/abs/1209.3511

    Thanks
    Bill
     
  5. Oct 29, 2016 #4

    Nugatory

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    OK, I think I see the problem..... :)
    And seriously, kidding aside, you cannot learn physics from video documentaries - they leave out all the important stuff that you need to know to understand what's really going on. For example.....
    Before we even get into quantum mechanical stuff, we have to understand how ##E=mc^2## works in problems of this sort. When we lift the ball in the earth's gravitational field, we're doing work, adding energy and increasing the potential energy - not of the ball but of the entire system consisting of the earth and the ball. The mass of the ball doesn't change, and the mass of the earth doesn't change, but if we could put the whole thing in an enormous box we would say that we're adding energy to the box when we move the ball farther from the earth. Thus, by ##E=mc^2##, we find that the box weighs more when the ball is farther away from the earth. The total mass of the box is the mass of the earth (which doesn't change), plus the mass of the ball (also doesn't change), plus an additional contribution from the potential energy.
    However, we can never get the energy to be infinite, no matter how far away we move the ball. This is because the force of gravity becomes weaker with distance so as the ball gets farther from the earth it takes less energy to move it yet farther away. (This is an I level thread, so you should able to verify this for yourself - evaluate ##\int{r^{-2}}dr## from the surface of the earth to infinity).

    OK, on to the quantum mechanical treatment of the energy of this system.....
    For that, you will want to start with this thread: https://www.physicsforums.com/threads/conservation-of-energy-in-quantum-measurement.871313/

    Just remember that the system whose energy we're considering is not the ball in isolation, but rather the entire ball+earth system. It makes no sense, even classically, to talk about the potential energy of the ball in isolation - the potential energy comes from the earth-ball interaction.
     
  6. Oct 29, 2016 #5

    jtbell

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    This is true if whatever is moving the ball is outside the box. If the ball is moved by something inside the box (presumably including us :oldwink:), then that "something" loses energy equal to the work done, so the total energy and the mass of the system inside the box both remain constant.
     
  7. Oct 29, 2016 #6

    Nugatory

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    Yes, that is an important correction - thanks - we have to add energy from outside to increase the total amount of energy in the box.
    In particular, if the ball is moving freely under the influence of the earth's gravity, the total energy in the box won't change as we're just trading potential energy for kinetic energy.
     
  8. Oct 29, 2016 #7

    vanhees71

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    Also note that, strictly speaking, to treat gravity correctly within relativity you need to use the general theory of relativity.

    One should also stress that mass is nowadays understood as "invariant mass" or the "rest mass" (if we don't deal with massless objects). Einstein has been much more precise in his original statement of "##E=m c^2##", as the formula got public in a somewhat unfortunate way, talking about the "dependence of inertia on a body's energy content".

    Indeed, although he has not known the modern mathematical treatment of special relativity (Minkowski 1907) when he wrote this paper (shortly after the famous one on Special Relativity as a follow-up paper), this is how to understand the physics of its content: The invariant mass of a body, defined in its center-momentum-frame, depdends on the "energy content" of this body. E.g., if you heat up a body of mass ##m##, adding heat energy ##\Delta q##, the body gets not only hotter (i.e., the temperature increases) but also it's mass changes to ##m'=m+\Delta q/c^2##.

    Another example is a capacitor in its rest frame. When uncharged it's invariant mass may be ##m##. Then if you put a voltage ##U## at it an electric field builds up which adds the electromagnetic field energy ##E_{\text{em}}=C U^2/2##. Thus now the mass of the capacitor is ##m'=m+C U^2/(2c^2)##.

    On the other hand, if your body of invariant mass ##m## just moves at a velocity ##\vec{v}##, its invariant mass doesn't change (that's why this definition of mass is called invariant, i.e., it's a Minkowski scalar) at all. It's the energy that changes, and it is convenient to include the "rest energy" ##E_0=mc^2## in the total energy. Then the energy of the body is ##E=m c^2/\sqrt{1-\vec{v}^2/c^2}##. The reason why this is convenient is that then energy and momentum of a body make up the components of a Minkowki four-vector,
    $$(p^{\mu})=(E/c,\vec{p}),$$
    with simple transformation properties under Lorentz transformations.
     
  9. Oct 29, 2016 #8

    Strilanc

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    Small caveat:

    PBS-related YouTube channels like spacetime/veritaseum/physicsgirl/minutephysics do still leave things out. But they are lightyears better in quality compared to tv documentaries. "They leave stuff out" still applies... but not as much.

    This is the video the poster is talking about. It's probably not as bad as you're imagining. It might even surprise you how many times he mentions that something is a common misconception.
     
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