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

Mass equivalence solved?

  1. Feb 18, 2015 #1
    In formulating GR, Einstein was motivated by the otherwise strange coincidence of the equivalence of gravitational and inertial mass. He removed this by discovering that trajectories determined by gravitation are actually inertial. However, gravity's source is still the tensor T, which is closely related to mass. Isn't this the same coincidence?
     
  2. jcsd
  3. Feb 18, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    It is not the same.
    In general, you can distinguish between "active gravitational mass" (objects that attract other objects), "passive gravitational mass" (objects that get attracted) and inertial mass.

    In Newtonian gravity, all three could be different if you do not care about other conservation laws: ##m_i a = \frac{GM_a m_p}{r^2}## where i=inertial, a=active, p=passive.
    It is a very reasonable assumption that active and passive gravitational mass are the same - otherwise physics gets really weird (violation of energy conservation and so on). There is no special reason why this should be identical to inertial mass, however, especially if you compare it to electromagnetism where this equivalence does not exist (electric charges correspond to gravitational mass there).
    General relativity requires inertial and passive gravitational mass to be the same.
     
  4. Feb 18, 2015 #3
    Yes, I understand all that. But the question of why M(a) is the same as M(i) still seems to demand a more ontological answer than "otherwise we would have violations of CoE". In fact, GR seems to make this worse: Newtonian gravity is a force interaction between the bodies, so it is intuitive that it depends on both masses symmetricallly. In GR, active and passive gravitational mass seem unrelated!
     
  5. Feb 18, 2015 #4

    atyy

    User Avatar
    Science Advisor

  6. Feb 18, 2015 #5

    martinbn

    User Avatar
    Science Advisor


    What is the formulation of the equivalence principle in terms of minimal coupling, or do I have to read these papers? By the way, when you say usually in GR, what do you mean, in most standard texts or something else?
     
  7. Feb 18, 2015 #6

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    What GR via the POE does is make M(i) and M(p) the same (really the same - they cannot be separated). Then, the question is between M(p) and M(a). This equivalence is less fundamental, especially in GR. To motivated it, I don't see anything better than conservation laws being true locally.
     
  8. Feb 18, 2015 #7

    atyy

    User Avatar
    Science Advisor

    I cannot remember but it's something like one should write the Hilbert action then add the matter action, and the matter action contains the fields and their first derivative, and the metric, but no derivatives of the metric. This is found in Hawking and Ellis, section 3.3.
     
  9. Feb 19, 2015 #8

    martinbn

    User Avatar
    Science Advisor

    I still don't understand the connection with the equivalence principle. There is nothing about this in Hawking & Ellis, section 3.3. That section explains how to get the stress energy tensor of a field from the Lagrangian of the field. The following section has a remark that Einstein's equations can be obtained from a variation of a curtain action, but nothing more.
     
  10. Feb 19, 2015 #9
    Come to think of it, I've read that pressure, along with mass, is a source of gravity. So if two bodies of equal mass but different pressure are interacting, won't that result in a a violation of CoE?I'll bet the answer to this is relevant to my first Q.
     
  11. Feb 19, 2015 #10

    atyy

    User Avatar
    Science Advisor

    Would you accept comma goes to semicolon as an expression of an equivalence principle?
     
  12. Feb 19, 2015 #11

    martinbn

    User Avatar
    Science Advisor

    It is not about me being stubborn and not accepting things. I am just trying to understand the connection between EP and minimal coupling that you mentioned. Now you mention something else. Would you elaborate on it?
     
  13. Feb 19, 2015 #12

    atyy

    User Avatar
    Science Advisor

    The minimal coupling is comma goes to semicolon, and the form in Hawking and Ellis is the Lagrangian form of it.
     
  14. Feb 19, 2015 #13

    martinbn

    User Avatar
    Science Advisor

    But what is the connection with EP? I reread that part of Hawking and Ellis and still don't see it.
     
  15. Feb 19, 2015 #14

    atyy

    User Avatar
    Science Advisor

    The EP means that locally we can use Minkowski coordinates, and the "laws of physics that don't couple to curvature" will be the same as in flat spacetime.

    So for the EP to hold, there have to be some laws that don't couple to curvature. If we take the Lagrangian as the fundamental expression of the laws, then that means the fundamental laws don't couple to curvature. The derived laws can couple to curvature.
     
  16. Feb 19, 2015 #15

    martinbn

    User Avatar
    Science Advisor

    I think I see what you mean. I suppose I misunderstood you, I thought you were talking about a formulation of the EP. I would say the minimal coupling (comma to semicolon) is a condition on the theory for the theory to satisfy the equivalence principle. The reference to H&E is not really relevant.
     
  17. Feb 19, 2015 #16

    atyy

    User Avatar
    Science Advisor

    Yes, it's just the closest "standard reference" I knew about for the Lagrangian form of comma goes to semicolon, which is how the EP is satisfied within GR. So for example, if one had curvature coupling in the Lagrangian, then it is possible for the EP to fail. The tricky part is that sometimes even though minimal coupling fails, there is a change of variables so that the EP holds, which is what the papers I linked to discuss.
     
  18. Feb 19, 2015 #17

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    The correct statement is that pressure, along with energy density, is a source of gravity. "Mass" in this context would include a contribution from pressure, as well as energy density.

    No, because, as above, the "mass" includes whatever contribution there is from pressure. If the two bodies have equal mass but different pressure, they must also have different energy density, so that the contributions from energy density and pressure result in the same mass in both cases.
     
  19. Feb 19, 2015 #18
    Sorry for my ignorance. I just got through elementary SR and we learned that mass is defined as
    the magnitude of the energy four-vector for the system. How is that different from the energy density in a CMRF?
     
  20. Feb 19, 2015 #19

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Yes, and in that context, pressure is not a source of gravity because there is no gravity. In SR, spacetime is flat, and the stress-energy tensor (which is the source of gravity) is zero everywhere. If you're talking about pressure as a source of gravity, you're talking about GR, not SR, and that's what I was talking about in my previous post. In GR, "mass" is defined differently (in fact, there is no single definition that always applies; you can only define "mass" in certain kinds of spacetimes), and includes a contribution from pressure as well as energy density.

    The length of the energy-momentum 4-vector has units of energy, not energy density. You can convert it to an energy density by dividing by the object's volume, but that won't be invariant under Lorentz transformations (unlike the length of the 4-vector, which is).
     
  21. Feb 19, 2015 #20
    Got it. Thanks for clearing that up.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Mass equivalence solved?
  1. Mass-energy equivalence (Replies: 17)

Loading...