stevendaryl said:
I don't agree. The "induced gravitational field" that Einstein was talking about is EXACTLY SR in noninertial coordinates. There is no difference.
The SR equations of motion for a test mass are simplest when you use inertial cartesian coordinates and parametrize using proper time. Then it's just:
m \frac{d U^\mu}{d\tau} = F^\mu
where F is the 4-force. So it looks just like Newton's F=ma.
If you switch to using curvilinear, noninertial coordinates and use some other parameter s besides proper time, you have, instead:
m (\frac{d \tilde{U}^\mu}{ds} + \Gamma^\mu_{\nu \lambda} \tilde{U}^\nu \tilde{U}^\lambda + \frac{d log(f)}{ds} \tilde{U}^\mu) = \tilde{F}^\mu
where \tilde{U} and \tilde{F} is rescaled versions of U and F, and where f = \frac{ds}{d\tau}
The "induced gravitational field" due to acceleration just amounts to moving terms from the left-hand side to the right-hand side, and writing:
m \frac{d \tilde{U}^\mu}{d\tau} = F_{eff}^\mu = \tilde{F}^\mu + F_{grav}^\mu
where
F_{grav}^\mu = - m(\Gamma^\mu_{\nu \lambda} \tilde{U}^\nu \tilde{U}^\lambda + \frac{d log(f)}{ds} \tilde{U}^\mu)
Whether you put the terms F_{grav} on the left side, and call them connection coefficients, or put them on the right side, and call them gravitational forces, is just a matter of taste, but it doesn't change the physics.
Is F_{grav} a real force, or not? Well, it's not real, in that it's not due to any source. People talk about it being "induced by acceleration", but that's not true, really. They are induced by the choice of the noninertial coordinate system. That choice isn't forced on you by the fact that you're in an accelerating rocket. A person inside a rocket can use inertial coordinates just as well as someone floating inertially. Anybody can use any coordinates they like; you don't have to use coordinates in which you, personally, are at rest.
On the other hand, F_{grav} is real, in the sense that it is measurable, to the same extent that coordinate acceleration is.
