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Question on Mass-energy equivalence

  1. Apr 17, 2012 #1

    jpo

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    Hello All,

    is the following in principle, correct:

    Scenario A:------------

    1A) A box with mass M contains mass m, their weight is g(m+M)
    2A) the mass m is (somehow) converted to energy E=mc[itex]^{2}[/itex]
    3A) at this moment, the box still has weight g(m+M)

    Scenario B: ----------

    1B) A box with mass M has weight gM
    2B) external force imparts kinetic energy E to the box, such, that E=mc[itex]^{2}[/itex]
    3B) at this moment, the box has weight g(m+M)


    Many thanks!
     
  2. jcsd
  3. Apr 17, 2012 #2
    Consider the total energy in each scenario.

    A) E = (M+m)c2, both before and after the 'conversion'

    B) The total energy of the moving box is E = Mc2 + mc2 i.e. (M+m)c2

    Now in scenario (A), the box and contents isn’t moving and hasn’t gone anywhere, so its total mass – its rest mass - both before and after is, as you say, (M+m).

    In scenario (B), the box is moving, and many, if not most, relativists prefer to avoid talking about moving objects having greater mass. They tend to use the term ‘mass’ to mean exclusively ‘rest mass’. But the term ‘relativistic mass’ is presumably what you’re exploring and it can be used. So, if you want, you can say this about scenario (B): the rest mass is unchanged and still M, and the relativistic mass is (M+m).
     
  4. Apr 17, 2012 #3

    jpo

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    in Scenario B -

    if one weighs the moving system, will its weight be g(M+m)?
     
  5. Apr 17, 2012 #4
    Yes, if by weight you mean the amount of gravitational force the Earth exerts on the object. However, unless you impart a rather large amount of energy, the equivalent [itex]m[/itex] is going to be rather small.
     
  6. Apr 17, 2012 #5

    jpo

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    Thank you Goodison Lad and Steely Dan for your replies.

    When is the weight of the moving frame equal to g(M+m)?
    Is it when the weight is measured in the COM reference frame?
     
  7. Apr 18, 2012 #6
    In scenario (B), the mass would only be determined as (M+m) from the 'stationary' frame. The relativistic mass is frame-dependent.

    If you changed frames, so that you were moving along with the box and were in its COM reference frame, you would measure its mass to be just M - its rest mass.

    Interestingly, if you'd left the box alone in scenario (B), and it was you who decided to move yourself so that you were travelling away from it at a constant velocity, you would still determine its relativistic mass to be larger than its rest mass. You'd still determine that the box had kinetic energy, because it is moving relative to you.
     
  8. Apr 18, 2012 #7

    jpo

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    Say a disk with mass M rotates so that its periphery has velocity close to c. A stationary observer is in the COM frame. The peripheral sections of the disk will look more massive, but what about its total mass - it should still be M, right?
     
  9. Apr 18, 2012 #8
    If the rest mass of the unspinning disk were M, the spinning disk, because of the rotational energy, would have a greater mass.

    So if the disk were supported at its centre on set of scales, the weight of the spinning disk would be greater than the weight of the unspinning disk.
     
  10. Apr 18, 2012 #9

    jpo

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    This might need clarification...

    "If the disk were supported at its centre on set of scales"

    This means its weight is measured in the COM frame - thus, its mass should be M, right?
     
  11. Apr 18, 2012 #10
    The COM we can use is the frame in which the point of support is stationary. But the disk won't just have its original, unspinning rest mass, M. You've actually added energy, and hence mass, to it by making it spin. So the rest mass of the spinning disk is greater than the rest mass of the unspinning disk, due to input of energy to get it rotating.
     
  12. Apr 18, 2012 #11

    jpo

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    Here is a quote from a Wikipedia article
    http://en.wikipedia.org/wiki/Center_of_momentum

    "In relativity, COM frame exists for a massive system. In the COM frame the total energy of the system is the "rest energy", and this quantity (when divided by the factor c[itex]^{2}[/itex], where c is the speed of light) therefore gives the rest mass of the system:

    m = E/c[itex]^{2}[/itex]

    The invariant mass of the system is actually given by the relativistic invariant relation:

    [itex]m^{2}=( \frac{E}{c^2} ) ^2 - ( \frac{p}{c} ) ^2[/itex]

    but for zero momentum the momentum term (p/c)[itex]^{2}[/itex] vanishes, hence the total energy coincides with the rest energy."

    Do you suggest this quote is true for translational motion only?
     
  13. Apr 18, 2012 #12

    PeterDonis

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    Let me try to specify these scenarios more carefully so that the underlying issues in analyzing them will be clearer. I think what is really meant here is:

    1A) A box with invariant mass M contains some substance with invariant mass m, which is at rest with respect to the box. (Note that we specify "invariant mass" to make it clear what we are talking about.) When sitting at rest on the surface of the Earth, where the local "acceleration due to gravity" is g, the box with its contents registers a weight g(m + M) on a scale. (However, note carefully that this "weight" is due to the box being accelerated, i.e., not in free fall. If we dropped the box from a height in vacuum, along with the scale, the scale would read zero.)

    2A) The substance inside the box is converted to energy, say in the form of radiation. The net motion of the radiation is still zero with respect to the box, so it still has invariant mass m considered as a single system (even though the individual particles of radiation, the photons, each have zero invariant mass--the fact that they are bouncing around inside the box instead of escaping to travel freely is what gives the total system a nonzero invariant mass).

    3A) As long as the radiation remains confined inside the box, and the box remains sitting at rest on the surface of the Earth, then yes, the reading on the scale will be unchanged.

    This one is more complicated, but *not*, as you will see, because of any effect of kinetic energy:

    1B) Again, we have a box with invariant mass M sitting at rest in a gravitational field g, so it registers a weight gM on a scale.

    2B) We apply a force to the box that imparts a kinetic energy E to it (where we write m = E/c^2 to give a "mass equivalent" to this energy), as measured in its original rest frame (i.e., as measured in a frame at rest with respect to the surface of the Earth). You didn't specify in what direction the box moves, and different cases will give different answers for the weight:

    2Bi) The box moves horizontally. In this case, the reading on the scale is unaffected by the box's motion, since the motion is perpendicular to the direction of gravity. 3Bi) So the scale still reads gM, even though the box's total energy, as measured by an observer at rest with respect to the Earth, is Mc^2 + E--meaning its "mass equivalent" is M + m.

    2Bii) The box is lifted upward by an elevator at a constant speed, relative to the Earth. (We'll assume the upward distance is small enough that the acceleration due to gravity, g, is unchanged.) In this case, the reading on the scale *is* affected by the box's motion, but not in the way you appear to think. 3Bii) When the box first *accelerates* to its constant upward speed, the reading on the scale *increases*. Assuming the upward acceleration is a, the scale reads (g + a)M during the period of acceleration. Once the box reaches its constant upward speed, however, the reading on the scale goes back to gM. (And if we bring the box to a stop again, with acceleration - a, the scale will read (g - a)M during the period of deceleration.)

    Note, by the way, that in the second case, the total energy of the box is *more* than just M + m, because it is being lifted to a higher gravitational potential energy. Its total energy *increases* during the lifting, even if it moves upward at a constant speed.

    But of course the key point is that in *neither* case above does the kinetic energy of the box affect what the scale reads. Now let me describe a scenario that I think captures what you were really trying to ask about whether "kinetic energy has weight":

    1C) A box has a bunch of stuff at rest inside it; the total invariant mass of box plus stuff is M. The box is sitting at rest on the surface of the Earth, in a gravitational field g. The box with all the stuff inside registers weight gM on a scale.

    2C) The stuff inside the box is somehow put in motion (perhaps it is dust that is induced to fly around inside the box; perhaps it is small marbles that are jiggled around). The total kinetic energy of the stuff, as measured in the rest frame of the box, is m. The box as a whole remains at rest on the surface of the Earth, with the (now moving) stuff contained inside. The scale now reads g(M + m).
     
  14. Apr 18, 2012 #13

    PAllen

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    The 'system rest mass' or invariant mass of the rotating disk in its COM frame is the total energy / c^2, as suggested by the article. I think your misunderstanding is thinking that this is the sum of the rest mass of each 'piece' of the disk. Wrong. For figuring the invariant mass of the rotating disk, you add the total energy of each piece measured in the COM frame. Thus pieces on the rim contribute a large KE as well as their individual rest mass. What the COM frame allows you to do is ignore momentum in figuring invariant mass - add up all energy (not just rest mass) and divide by c^2. The COM frame does not allow you to ignore energy - you must account for all forms of energy.

    Result: the rotating disk has higher system rest mass than the stationary disk. It would weigh more if you supported it by its axis on a scale.
     
    Last edited: Apr 18, 2012
  15. Apr 18, 2012 #14

    jpo

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    Many thanks to all and to PeterDonis for the kind reply and the detailed analysis of the box problem. PAllen, I highly appreciate your comments.

    This is one amazing forum.
     
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