- #76

samalkhaiat

Science Advisor

- 1,710

- 971

1) There is no such thing as “Still one question remains. If there is no unique general way of defining mass in GR (I mean gravitational mass, because as Sam demonstrated, inertial mass can be defined and turns out to be equal to gravitational mass), what is that thing people use when they construct the stress-energy tensor of a matter distribution(example here, where a mass-energy density [itex] \rho [/itex] appears in the definition of stress-energy tensor of a fluid)? If we need to know the spacetime geometry in order to define what we mean by mass, how can we have mass before actually finding out what the geometry is and then use it to find geometry?

**mass of gravity**”. Gravitational field is

**massless**, spin-2 field.

2) The mass of an object at rest is determined entirely by the number of its constituent

**Nucleons**. And the mass of each Nucleon is due (almost entirely) to

**QCD**and

**QED**effects. Since we do not

**yet**have a consistent theory of

**QG**, we cannot determined (

**if any**) the gravitational contribution to the Nucleon mass. In principle, however, if the mass of the Nucleon is defined by, [tex]m = P^{ 0 } = \int d^{ 3 } x \ \Theta^{ 0 }{}_{ 0 } ( x ) ,[/tex] then it is reasonable that

**ALL**of the

**4**fundamental interactions contribute to the

**TOTAL SYMMETRIC**energy-momentum

**TENSOR**[itex]\Theta_{ \mu \nu }[/itex].

3) The “gravitational mass” (also called heavy mass) which I talked about, was an historical concept. You see, Newton had no reasons to equate the inertial mass of an object (the one which enters in [itex]F = m_{ I } a[/itex]) with its gravitational mass (the one which appears in [itex]F = G \frac{ m_{ G } \ M_{ G } }{ r^{ 2 } }[/itex]). We now know that the equality [itex]m_{ I } = m_{ G }[/itex] is consistent

**only**with LOCAL (CAUSAL) field theory based on

**the principle of equivalence**(recall that Newton’s theory is an

**action-at-a-distance**theory).

Sam