I have been thinking a very long time on this subject and am unable to gain a clear answer. As we know, all of physics is based on conservation of fundemental properies, e.g Energy, Spin, Momentum etc, and when we find that a certain quantity is not conserved, we are able to imply the existence of new phenomema to balance this non-conservation (e.g. the neutrino) My question is this: According to special relativity, a bodies inertial mass will increase as v -> c, does its gravitational mass also increase in such a manner?, and if so is the gravitational force exerted by this body on another body increased. If so is gravitational potential energy invariant under relativistic translation? This would mean that the gravitational force exerted on a body would be dependant on the reference frame (inertial or otherwise). I will not produce my thoughts on what i believe to be the consequence of this, as I am probably wrong in my logic. Clarification on this would be well accepted.
If one uses the notion of "relativistic mass," then you just need to go into the reference frame of a cosmic ray, and look at the Sun. In the cosmic ray's reference frame, of course, it's at rest, and the Sun is moving at an insane fraction of the speed of light. Were collapse into a black hole simply a function of relativistic mass, then from the cosmic ray's frame, the Sun would be more than massive enough.
Then you are sayng that gravitational mass and inertial mass are not equivalent. My belief is the same as yours in that if we were to consider relativistic mass, then funny things would happen like you highlighted, but that does not mean that it is not true. I need a solid logical or mathematical arguement to push me over the line.
In general relativity, spacetime curvature is produced by <b>rest mass/energy</b> and momentum flows. If you change reference frame, the gravitational curvature does not change.
the source term for the gravitational field in GR is the energy-momentum tensor <B>T<sup><font face=sumbol>mn</font></sup></B>. In it are "jumbled" mass, momentum, energy densities, stress, pressure, etc. The form of the energy momentum tensor is going to look different to different observers: the energy density is defined as <B>T<sup>oo</sup></B>. But a general coordinate transform <B>T<sup><font face=sumbol>mn</font></sup></B> to <B>T' <sup><font face=sumbol>mn</font></sup></B> yields a<sup><font face=sumbol>m</font></sup><sub>o</sub>a<sup><font face=sumbol>n</font></sup><sub>o</sub><B>T<sup>oo</sup></B>, which is not <B>T<sup>oo</sup></B>. Thus what looks like energy density to one observer looks like a combination of momentum density and energy flux density to others. Typically the invariant trace of the energy momentum tensor <B>T<sup><font face=sumbol>mn</font></sup></B> is regarded as the rest mass of an object. However you should note that this isn't the complete source for gravitation (consider for example that <B>T<sup><font face=sumbol>mn</font></sup></B> = 0 for the electric field, which would imply that the field energy doesn't gravitate - which is not correct). When deciding which "energies" contribute as gravitating sources, a good test (I find) is to consider whether a general coordinate transform can "remove" the energy. This rules out kinetic energies as gravitating sources - they will disappear in rest frame transform. Hope that helps! The Chicken
There’s an intuitive way to view relativistic mass. From the cosmic ray’s perspective, due to time dilation the Sun’s rate of time is a fraction of its own rate, say 1/x. Instead of a gravitational attraction for the surface of the Sun of 270 m/s^2 as we on Earth observe, from the cosmic ray’s perspective the rate is 270/x m/s^2. But that clashes with the fact that all observers record the same path for the cosmic ray (all observers including the cosmic ray will agree as to what coordinates the ray “touched” along its path). If however you increase the mass of the sun by a factor of x, then the gravitational attraction of 270/x m/s^2 becomes 270 m/s^2 again and all is well. From the cosmic ray’s perspective the Sun is more massive than we on Earth observe, but the Sun’s rate of time is decreased to perfectly compensate for the extra mass such that the curvature of spacetime near the Sun does not change, hence the cosmic ray’s path is the same for all observers and the Sun is no closer to becoming a black hole.
Relativistic gravitational mass invariance of receeding bodies Relativistic gravitational mass invariance of receeding bodies I'm quite amazed that we were able to think of the same thing in such proximity of time, and that your post so quickly followed mine. Would you care to elaborate on when you thought of this?
If you’re referring to me, I didn’t think of it. I was paraphrasing from one of my books Understanding Einstein's Theories of Relativity. To which post of yours do you refer?
Shawn34m If you are refering to me, I have been thinking about this for a while. I have not seen your( Or believe I have not seen it). Also if you are refering to me, don't try to claim originality, this is physics not engineering.Please Register or Log in to view the hidden image!
