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  4. gravitational mass vs. rest mass
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gravitational mass vs. rest mass

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Offline flr (OP)

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gravitational mass vs. rest mass
« on: 06/04/2013 17:51:02 »

1.
 The energy E of an object having the momentum p and the rest mass m0 is:
 
 E^2 = (m0*c^2)^2+(p*c)^2

 To define both E and pc, I need a frame and these quantities depends on the frame.
 m0*c^2 is the same in all frames.

 2.
 The gravity law says that
 F = G*M*m / r^2; where M is the mass of (say) Earth and m is the mass of then object moving, and having its rest mass m0.

If the object does not move relative to Earth, gravity law becames:
   F = G*M*m0 / r^2;

---

 My question is: If objects move relative to Earth, which mass should I use in:
  F = G*M*m? / r^2;
 still the rest mass m0 or the relativistic mass  m_relativitic  = E/c^2 = sqrt((m0*c)^2+(p)^2) = gamma*m0*c^2
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Offline Pmb

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Re: gravitational mass vs. rest mass
« Reply #1 on: 06/04/2013 19:17:30 »
See http://home.comcast.net/~peter.m.brown/gr/grav_force.htm
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Re: gravitational mass vs. rest mass
« Reply #2 on: 06/04/2013 19:40:31 »
Thanks for the link. Interesting, but a bit too mathematical. Maybe in time you will explain more in words. Nevertheless, very interesting indeed.
 
Back to my question: From your Equation 21, it follows that the relativistic mass: m=sqrt((m0*c)^2+(p)^2) = gamma*m0*c^2  and not the rest mass m0, should be used in |F| = G*M*m? / r^2;.

Do I understand this correctly? Could you confirm it?

---------
If so, it means that the relative speed between objects determine how much gravitational attraction there is. Is that right?
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Re: gravitational mass vs. rest mass
« Reply #3 on: 06/04/2013 20:40:59 »
Quote from: flr on 06/04/2013 19:40:31
Thanks for the link. Interesting, but a bit too mathematical. Maybe in time you will explain more in words. Nevertheless, very interesting indeed.
 
Back to my question: From your Equation 21, it follows that the relativistic mass: m=sqrt((m0*c)^2+(p)^2) = gamma*m0*c^2  and not the rest mass m0, should be used in |F| = G*M*m? / r^2;.

Do I understand this correctly? Could you confirm it?

---------
If so, it means that the relative speed between objects determine how much gravitational attraction there is. Is that right?
For particle moving in a gravitational field the mass is function of both speed and position (i.e. gravitational potential). See Eq. (16) in http://arxiv.org/abs/0709.0687

I've never worked it out for a spherical body like the earth, only for a uniform gravitational field. See Eq.(12) at http://home.comcast.net/~peter.m.brown/gr/uniform_force.htm

The gravitational force has the form F = -mg where m = relativistic mass = "time component of 4-momentum"/c = P0/c.

Does that help?
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