# 'Mass' of an object at different positions in a weak gravitational field

P: 87
 Quote by Jonathan Scott As far as I can see, what you've calculated is the momentum in a coordinate system which is flat but locally coincides with the local coordinate system. By the time you get to a different potential, it can no longer match the local coordinate system, in that both time and space have changed in scale, so the coordinate speed of light has changed and the momentum defined by that process differs from local momentum. However, the change in the energy is only affected by the relative time rate, so that can be compared from any static point of view.
I don't think I've used a locally flat coordinate system, that would be one in free fall, the coordinate system I'm using is at rest relative to the surface of the earth, or at least that's what I've tried to do.
P: 87
 Quote by csmyth3025 I wont pretend to understand the formulas and equations that have been used in this thread. Conceptually, the OP seems to be asking in the mass of an object that is raised to a higher level in a gravitational field decreases. Since this section deals with Special and General Relativity, I'm guessing that the question boils down to whether the equivalent mass/energy of the object decreases as it is raised in a gravitational field. I would look at this problem from the perspective of a distant observer viewing a large colud of interstellar dust and ice unperturbed by any outside forces. This cloud certainly contains a set amount of mass in the form of dust grains and ice particles. It also contains a set amount of gravitational potential energy. In layman's terms, the constituent particles are far away from each other, but their natural tendancy is to "fall" together. If one were to observe this cloud over the course of several million years, there would be a contraction - very slowly at first, but gradually becoming faster and faster. The gravitational potential energy of the once aimlessly drifting particles is being converted into the kinetic energy of particles moving towards each other. Let's say, for the sake of simplicity, that this cloud eventually coalesces into a sub-stellar object of 10 Jupiter masses. The particles making up this object are the same particles that comprised the original cloud. To be sure, a lot of energy has been radiated away by the heat of all this stuff crashing in on itself. After 10 or 20 billion years most of this heat will radiate away and there will be a ball of stuff all mashed together as closely as the various inter-molecular forces will allow. The question posed by to OP is whether the mass of a piece of this ball will decrease if it is raised some distance from the surface. We can consider that there has been a lot of energy lost by the various particles making up this ball from where they originally started to their final resting place. To time-reverse this process (even for a small piece of this ball) will require the input of energy. Does this inputed energy reside in the particle's mass, or in the particle's position relative to the big ball of stuff? My take on this is that the answer is a qualified "yes" to both questions. Considering that every particle in the universe has some equivalent mass/energy relative to every other particle in the universe, the mass/energy of the piece of the ball that is raised above the surface will increase relative to the ball. I apologize for this unsophisticated reply - bereft of equations such as it is. I may be totally wrong about all this. If so, please feel free to correct me. Chris
I can't really correct you as I obviously don't know the answer to this question (that's why I made the thread). However, I think what you're getting at is the idea that the potential energy of a system should manifest itself in an increased mass of the particles? Although the gravity question is still unclear for me I am pretty confident that in electrostatics for example, the potential energy doesn't work this way. Instead the energy, and the mass that goes with it, is stored at every point in space with a density proportional to the square of the magnitude of the electric field at that point.
PF Gold
P: 1,159
 Quote by TobyC I don't think I've used a locally flat coordinate system, that would be one in free fall, the coordinate system I'm using is at rest relative to the surface of the earth, or at least that's what I've tried to do.
By "local" coordinates I mean static SR coordinates in the local vicinity of the static observer, who observes that light locally always travels at the standard c. Such a system can however only be approximate in a region where the gravitational potential varies, as local space-time is curved. However, we can instead use a coordinate system which is flat and which locally matches local coordinates, but begins to diverge a bit further away; there is more than one possible scheme for such coordinates.

As an analogy, consider a flat map of a large part of the earth. It can be exactly accurate in scale and shape at some chosen point, but the further away one gets the more the scale or shape has to change.

I think (although I'm not sure) that you've effectively started with an expression that defines momentum relative to the map, with the result that it has an extra scale factor because of the difference between the map and the local value.
P: 87
 Quote by Jonathan Scott By "local" coordinates I mean static SR coordinates in the local vicinity of the static observer, who observes that light locally always travels at the standard c. Such a system can however only be approximate in a region where the gravitational potential varies, as local space-time is curved. However, we can instead use a coordinate system which is flat and which locally matches local coordinates, but begins to diverge a bit further away; there is more than one possible scheme for such coordinates. As an analogy, consider a flat map of a large part of the earth. It can be exactly accurate in scale and shape at some chosen point, but the further away one gets the more the scale or shape has to change. I think (although I'm not sure) that you've effectively started with an expression that defines momentum relative to the map, with the result that it has an extra scale factor because of the difference between the map and the local value.
I don't think this is what I've done. The coordinate system I've used should accurately describe everything, not just local things, from the point of view of an observer stationary relative to the ground.
PF Gold
P: 1,159
 Quote by TobyC I don't think this is what I've done. The coordinate system I've used should accurately describe everything, not just local things, from the point of view of an observer stationary relative to the ground.
Yes, that's what a coordinate system is for, but it is NOT the same as the values a local observer would observe at each of the mapped points. As I've been saying, it is like a map. It can describe everything, but it cannot be the same scale and shape as the local view everywhere, because the local view is curved.

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