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Relativistic Mass and Gravity

  1. Mar 2, 2013 #1
    Imagine you're standing on a long, straight and smooth road. You place a block on the road.

    The normal reaction force acting on the block is equal to its weight. Then you give the block a little push. Since the block is moving from your frame, its relativistic mass increases.

    Will its weight and hence the normal reaction force acting on it also increase?
     
  2. jcsd
  3. Mar 2, 2013 #2

    Bill_K

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    This sounds like an easy question, and the snap answer is yes, Equivalence Principle (mumble, mumble) But there's more to it than that!

    Because what you're asking about is a pretty small effect. The rest energy of a particle is mc2, while its kinetic energy is ~ mv2. In other words, the effect is O((v/c)2). And so to answer the question correctly, you have to include everything that size, and that brings in GR as well as SR.

    Instead of a supported particle, consider a freely falling one. In linearized GR, the force on a particle has three parts: the "electric" force at rest, the "magnetic" force linear in v, and a third part which is quadratic in v. The equation of motion of a freely falling particle is the geodesic equation, dvμ/dτ + Γμαβvαvβ = 0, or in terms of the momentum,
    dpi/dt = - γm(c2 Γi00 + 2c Γi0j vj + Γijk vj vk)
    For a stationary linearized gravitational field,
    Γi00 = - ½ h00,i
    Γi0j = - ½ h0j,i + ½ h0i,j
    Γijk = - ½ hjk,i + ½ hij,k +½ hik,j
    and in particular for the linearized Schwarzschild solution,
    Γi00 = M/r2
    Γijk = - M/r2 δjk
    The net result is dp/dt = (-M/r2c2)(mc2 + ½ mv2) (1 - (v/c)2) er
    where er is the unit vector in the radial direction. The first factor is what you'd get if all that were involved was the equivalence principle. The second additional factor arises from GR and the nonuniformity of the Schwarzschild field.
     
    Last edited: Mar 2, 2013
  4. Mar 2, 2013 #3

    Vanadium 50

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    Relativistic mass is an obsolete idea, partially because it makes it easy for people to make the same mistake you did: thinking that you get relativistic gravity by plugging relativistic mass into Newtonian gravity.
     
  5. Mar 2, 2013 #4
    Whatever adds energy to a system changes how much it "weighs" whether one considers that to be due to added energy or whether utilizing the deprecated concept relativistic mass.

    A hotter brick weighs more. A coiled spring weighs more. Presumably a moving brick weighs more and thus requires a greater normal force...
     
  6. Mar 2, 2013 #5

    Bill_K

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    This is what I meant by "Equivalence Principle (mumble, mumble)" :wink:
     
  7. Mar 2, 2013 #6
    So the normal reaction force is only proportional to the object's "rest weight" so to speak?
     
  8. Mar 2, 2013 #7
    Let's see if I'm getting this:

    Granted these are very small differences: when the sun heats up a planet, the rest mass of the planet is not increased, but the energy is, and this increased energy causes spacetime to warp more/differently, increasing the weight that would be measured for each object on the planet's surface, thus increasing the normal force keeping each in place. Any errors there? Thanks.
     
  9. Mar 2, 2013 #8

    Bill_K

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    1977ub, It's clear that heat causes the energy to increase, and this leads to an increase in weight exactly as given by the Equivalence Principle. But what's not so clear is the additional increase due to GR, because to such a high precision as we are discussing here, the action of GR on moving objects is different from that of Newtonian gravity.

    For a single moving object, as we said above, there's a gravitational effect that's quadratic in the velocity. This would apply equally well to the kinetic energy of the molecules in a warm object. Generally speaking, an object with internal degrees of freedom contains stress as well as energy, and GR also couples to stress.
     
  10. Mar 2, 2013 #9
    That is correct. Plugging the relativistic mass, as normally formulated in SR, into the Newtonian expression for gravitation will only generate one third of the anomalous perihelion shift. However, it is entirely feasible to invent a more complex implicit time-dependence of the mass, that can be plugged into the Newtonian expression for gravity, thus getting a behaviour that is very close to what is expected from the Schwarzschild solution.
     
  11. Mar 2, 2013 #10

    Dale

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    The difference is that those other things do not add momentum. Momentum often has the opposite effect of energy. So I don't think that line of reasoning is necessarily correct.
     
  12. Mar 2, 2013 #11

    pervect

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    There was a fairly detailed analysis of this in another thread with the gravity replaced by the artifical gravity of an accelerating rocket. By the equivalence principle, this should be more or less the same (of course, for actual gravity from a planet, the "straight" road will curve with the planet).

    There's also some literature on the publicized version of the paradox, known as "Supplee's paradox", which is elaborated a bit more.

    https://www.physicsforums.com/showthread.php?t=646874&highlight=supplee

    The answer to how it appears in both frames isn't as simple as I'd like. Howeer, one can say that one will need more force to make the moving brick follow the straight road than the non-moving brick.

    One non-intuitive result is that the the road (or rather the floor of the accelerating spaceship) will no longer appear straight in a frame comoving with the brick.

    This effect, plus time dilation, should explain everything from the bricks point of view, but I'm not sure how to do so clearly and simply.

    This is a different effect from the fact that the road on a planet would be curved, it's due to the relativity of simultaneity.
     
  13. Mar 3, 2013 #12
    more of a digression I guess. thinking through simpler issues. I had e=mc^2 and GR/gravity in different boxes in my mind.
     
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