Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Does Mass disappear and become Energy?

  1. Mar 23, 2010 #1
    Every so often hear things similar to:

    The mass lost is due to the energy produced. Mass becomes energy from the equation E=mc2.​

    We seem to have two competing ideas:

    1) E=mc2
    2) E + mc2 = constant.

    A) In equation 1) energy and mass are equivalent (up to a units conversion factor).
    B) In equation 2) the sum is conserved--one is changed into the other.

    Any disagreement?
    Last edited: Mar 23, 2010
  2. jcsd
  3. Mar 23, 2010 #2
    What makes you think these ideas are 'competing'?

    They seem perfectly compatible to me.
  4. Mar 23, 2010 #3
    I dunno. What happens if you accept both equations and solve for E?
  5. Mar 23, 2010 #4


    User Avatar

    Staff: Mentor

    Can you give a specific example where this idea is used? I suspect that the E in this case is kinetic energy and the m is invariant mass ("rest mass").

    In general, you have to be careful about which energy and which mass you're talking about. Also whether you're talking about a single particle or a collection of particles.
  6. Mar 23, 2010 #5


    User Avatar
    Homework Helper

    The most general equation is E=constant, where E includes rest mass energy, potential energy, and every other form of energy you care to dream of.
  7. Mar 24, 2010 #6
    You get E = constant/2. What's the problem?

    This is of course assuming that the equations are correct, or even make sense, which itself shows a certain misunderstanding happening.
  8. Mar 24, 2010 #7
    I use the mass, m the same in each case, as inertial mass.

    I think I could ask around and get any number of specific examples where it it a commonly accepted belief that energy is converted to mass. Maybe I should take a poll...

    The Wikipedia authors of Mass-Energy Equivalence thought it something that required repeating in several different ways.

    "Rather, neither one appears without the other. Rather than mass being changed into energy, the view of relativity is that rest mass has been changed to a more mobile form of mass, but remains mass. In this process, neither the amount of mass nor the amount of energy changes."

    Have you noticed this opinion at all yourself?
    Last edited: Mar 24, 2010
  9. Mar 24, 2010 #8
    I assume that you're talking about special relativity ( otherwise, mass is conserved in newtonian mechanics).
    The newtonian concepts of mass & energy lose their individual meaning in Special Relativity. One has a compatible 4-vector (the energy-momentum vector) which looks like (0,0,0,mc^2) in the stationary frame of reference.Its (covariant)magnitude is defined as the energy.
  10. Mar 24, 2010 #9
    Here's one from USA Today


    My boldface.

    Physic Forum, 2008 https://www.physicsforums.com/showthread.php?t=254886"
    No one seems to have noticed.

    I focused on .edu websites:
    --Standford website

    --Harvard website

    --Cornell website

    --Stanford SLAC :eek:

    --Yale Astronomy website

    And my favorite Princeton
    I didn't make it up!
    Look, I am sane, I am sane like the rest of you! --Catch 22, the movie, via Joseph Heller, author
    Last edited by a moderator: Apr 24, 2017
  11. Mar 24, 2010 #10

    Doc Al

    User Avatar

    Staff: Mentor

    I'm not seeing the issue. When composite 'particles' break up or are formed, rest mass can be exchanged for kinetic energy (and vice versa) or other forms of energy (radiation). The total 'mass-energy' is conserved, of course.
  12. Mar 24, 2010 #11
    Scroll to the top of the screen. You are posting in a Special and General relativity folder.
    That is not the energy momentum 4-vector. In any case, not this mass: m=|-p,E|, but this mass: E=m is the topic that evades discussion.
  13. Mar 24, 2010 #12
    Do you see the difference between exchanged and conserved?

    We write this for exchanged.

    [tex]E \Leftrightarrow mc^2[/tex]

    which is the same as E + mc2 = const

    We write this for conserved


    E varies in the first. E changes form but is invariant in the second.
    Last edited: Mar 24, 2010
  14. Mar 24, 2010 #13

    Doc Al

    User Avatar

    Staff: Mentor

    Sure. Do you see the difference between rest/invariant mass and the old-fashioned 'relativistic' mass? (Perhaps that's what's messing things up.)

    Rest mass is not conserved. But total energy (including the rest energy, E = mc²) is conserved.
  15. Mar 24, 2010 #14


    User Avatar
    Homework Helper
    Gold Member

    According to the modern definition, the mass of a particle is an invariant quantity, and it doesnt depend on which coordinate system it is described from. Energy, kinetic energy and momentum are not invariant quantities, and they depend on the coordinate system.

    If a particle of mass m disintegrates into two particles that move away with velocity v, then the sum of the masses of the resulting particles is m√(1 - v2) (< m). Here mass is converted into kinetic energy.

    If two particles of mass m collide and stick together with a relative velocity 2v, then the mass of the resulting particle will be 2m/√(1 - v2) (> 2m), so here kinetic energy is converted into mass.
  16. Mar 24, 2010 #15
    I don't care to trade insults, if this is where this is going.
  17. Mar 24, 2010 #16


    User Avatar
    Homework Helper
    Gold Member

    Also, it may be useful to note that mass is in a sense analogous to proper time. In this analogy, the non-conservation of mass is equivalent to the "twin paradox".
  18. Mar 24, 2010 #17


    Staff: Mentor

    Hi Phrak,

    First, when you use the equation E=mc² it is important to know which E and which m you are talking about. Usually E in this context refers to the rest energy and m refers to the rest mass.

    By the conservation of four-momentum one particle of mass m may be exchanged for two or more photons of total energy E and each photon individually of rest mass 0. That is all that is meant by the equivalence of mass and energy.

    Note that although the sum of the masses of the photons is zero the system of photons has mass m, which is a subtelty that is usually overlooked.
  19. Mar 24, 2010 #18


    User Avatar

    Staff: Mentor

    When people talk about "converting mass into energy" or vice versa, they are really talking about converting rest energy into kinetic energy and vice versa. For example, consider nucleus A which fissions into nuclei B and C.

    If A is at rest initially, all its energy is rest energy which is associated with its rest mass:

    [tex]E_A = E_{A0} = m_{A0} c^2[/tex]

    whereas B and C each have both rest energy and kinetic energy:

    [tex]E_B = E_{B0} + K_B = m_{B0} c^2 + K_B[/tex]

    [tex]E_C = E_{C0} + K_C = m_{C0} c^2 + K_C[/tex]

    The total energy is conserved:

    [tex]E_A = E_B + E_C[/tex]

    [tex]E_{A0} = E_{B0} + K_B + E_{C0} + K_C[/tex]

    [tex]m_{A0} c^2 = m_{B0} c^2 + K_B + m_{C0} c^2 + K_C[/tex]

    I consider it incorrect (strictly speaking) to say that some of A's mass was "converted" into energy because that implies (to me) that the energy didn't exist before and that the total energy increases. But in fact, the energy did exist before, as (part of) A's rest energy; and the total energy remains constant, so no new energy was created.

    If you prefer to think in terms of "relativistic mass," then

    [tex]E_A = m_{A,rel} c^2[/tex]

    and similarly for B and C; and

    [tex]m_{A,rel} c^2 = m_{B,rel} c^2 + m_{C,rel} c^2[/tex]

    In this case the total "relativistic mass" is constant. Again it's incorrect to speak of some of it being "converted" into energy because that implies (to me) that the total "relativistic mass" should decrease.

    Nevertheless, the colloquialism "converting between mass and energy" is so common that it's certainly impossible to eradicate. I think the best we can do is to interpret it as "converting between rest energy and kinetic energy."
    Last edited: Mar 24, 2010
  20. Mar 30, 2010 #19
    Thanks Dale and jtbell for your thoughtful responses. I seem to simply have no more to add.

    I think you've switched over to intrinsic mass in this statement, right? In any case, I hadn't really gotten that far with it. I think this is something ZapperZ would call a one body solution to an n-body problem, or n one-body solutions to an n-body problem.
  21. Mar 30, 2010 #20


    Staff: Mentor

    You are very welcome. And sorry about the lack of clarity in that post. I was referring specifically to the invariant mass when I said "although the sum of the masses of the photons is zero the system of photons has mass m".

    Say one photon has a four-momentum (E/c,E/c,0,0) and another photon has a four-momentum (E/c,-E/c,0,0), then each one individually has a mass of 0. Now if we take the two four-momenta and add them together we get (2E/c,0,0,0) for the four-momentum of the system, which has a mass of 2E/c². Does that help explain it better that time?
  22. Apr 1, 2010 #21
    Yes, I'd thought that was what you were saying. I've been comfortable in assuming that the vector addition of (1/2)(E/c,E/c,0,0) and (1/2)(E/c,-E/c,0,0) should be plausible due to the indeterminate trajectory of a single particle, such as a photon in a laser cavity oriented along the x axis; and by extension groups of identical particles. (the mass would be the magnitude of the vector)

    Until you'd put it into words, the notion that sum of the masses was not necessarily equal to the whole hadn't managed to breach my preconcieved expectations. That helps a great deal! I'm curious. How did you manage it?
    Last edited: Apr 1, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook