This is a reply to a question by atyy in post #40 of another thread about photons
https://www.physicsforums.com/showthread.php?t=251161&page=3 but I have put the reply here as I will not not be focusing just photons and do not want to be accused of hijacking that thread.
atyy said:
This statement requires:
1) An object has inertial mass
2) An object has gravitational mass
3) Inertial mass equals gravitational mass
In the statement, "the mass of a photon is zero", is the inertial or gravitational mass being referred to?
First, any statement like "the mass of a photon is zero" in any modern text where the type of mass is unspecified, then it should be assumed thay are talking about rest mass. So the statement could be stated as "the rest mass of a photon is zero". However, modern usage is that there is no other kind of mass, other than rest mass, so there is no need to use any term for mass other than simply mass.
Next I want to focus on the 3rd statement:
3) Inertial mass equals gravitational mass
Here is a simple thought experiment that demonstrates that statement is not true.
An observer is on an asteroid of mass (M) that has a tower and releases a test particle of mass (m) from the top of the tower and times how long it takes to fall a short distance (d). The distance d is sufficiently small relative to the combined height of the tower and radius of the asteriod (R) that the acceleration due to gravity can be considered to be constant over the distance d. The mass of the asteriod is also considered to be suffiently small that gravitational time dilation is insignificant relative to the kinetic time dilation of the extreme relativistic velocities that will be considered. Now say our observer on the asteroid measure a time interval of one second for the test object to fall 1 meter. The acceleration of the particle is calculated as
a = 2d/t^2 = 2 m/s/s
which also equates to
a = GM/R^2
That is a Newtonian equation but the velocity attained by a particle in falling one meter in the gravitational field of a small asteriod does not require a relativistic equation.
A second observer moving horizontally relative to the tower at 0.8c records the time for the particle to fall as
t' = \frac{t}{\sqrt{1-v^2/c^2}} = 1.666 seconds.
By his calculations the acceleration of the test particle due to gravity is
a' = 2d/t^2 = 2/(1.6666)^2 = 0.72 m/s/s
and so
a' = GM/R^2 * (1-v^2/c^2)
If we assume the gravitational constant G is constant and the R is taken to be (near enogh) constant since there is no transverse length contraction then the gravitational mass of the asteroid can be taken to be as M(1-v^2/c^2).
In other words the active gravitational mass of a moving object is less than when it is at rest by a factor of gamma squared.
The inertial (relativistic) mass of the asteroid is calculated as M/sqrt(1-v^2/c^2) so the gravitational mass is not the same as the inertial mass.
Since the rest mass of the asteroid is simply M then the gravitational mass is not the rest mass either.
So we have established that
3) Inertial mass equals gravitational mass
is not always true and
4) Rest mass equals gravitational mass
is not always true either.
Earlier, I suggested that the quantity M(1-v^2/c^2) that appears in the equation a' = GM/R^2 * (1-v^2/c^2) could be described as the active gravitational mass of the asteroid which I will denote as M_g so the equation becomes:
a' = G \frac{M_g}{R^2}
Now iif we define the inertial mass of the test object as m_i then the force of gravity acting on the stationary suspended test mass is:
F' = m'a' = m_i a' = G \frac{M_g m_i}{R^2} = G \frac{M m}{R^2} * \sqrt{1-v^2/c^2} = F * \sqrt{1-v^2/c^2}
which agrees with the already known result that transverse force in Special Relativity transforms as F' = F * \sqrt{1-v^2/c^2}
In summary, if you want to think in terms of different kinds of masses, then they could be defined as:
Inertial mass = \frac{ m_o}{\sqrt{1-v^2/c^2}}
and Active gravitational mass = M_o * (1-v^2/c^2)
From the above it should be apparent that the inertial mass is acting as the passive gravitational mass.
The equation for the active gravitational mass is specific to the thought experiment described here with all the necessary aproximations and does not claim to cover the general case. It is used here, just to illustrate that there is clear example that inertial mass and rest mass are not equal to active gravitational mass.
