John Morrell said:
Gamma is just a ratio between the effect of a force on the particle while at rest and the effect of a force on a particle with a given velocity. I don't see why this is in any way not a description of a very real and fundamental "relativistic mass". Could you explain?
You can call ##\gamma m## the "relativistic mass" if you'd like. It's simply total energy in different units. But remember: total energy is the sum of rest energy ##E_0## and kinetic energy ##E_k##, so if you call ##\gamma m## "relativistic mass," you must understand that you're talking about the sum of "rest mass" (the ##m## that everyone agrees on) and the "kinetic mass" (which is just kinetic energy in different units). Today, few physicists use the word "mass" to mean anything other than "rest mass."
You're mistaken about ##\gamma##. There are a few ways to think about what ##\gamma## is, but fundamentally it's exactly what I said it is before: ##(1 - v^2 / c^2)^{-1/2}##. You can think of it as the time-dilation factor (##t = \gamma t_0##, where ##t_0## is proper time). You can also think of it as the "kinetic energy factor" (in the sense that ##E = \gamma E_0 = E_0 + E_k##).
John Morrell said:
So it makes sense that gravity is a curvature of space, not a frame-variant quantity. But one thing I still don't understand is the argument against mass being dependent on a frame of reference. We define mass as the ratio between a force and an acceleration, right? I mean, mass has no other meaning as far as I know. And to an observer, it will take a much greater force to accelerate a body moving near relativistic speeds than it would to accelerate a body at rest with relation to the observer. A body moving at relativistic speeds will take less force to accelerate a given amount in its own reference frame than in another. Eventually, as the relative velocity approaches the speed of light, an observer would say that it takes nearly infinite force to accelerate the body any appreciable amount, while the body itself would not experience this effect.
I know my language is not precise, but this seems like sound reasoning to me. The main reason I can think of for this not being true is if my understanding of special relativity doesn't apply to cases with force and acceleration, as some of you have maybe suggested.
The definition of mass you've cited ("ratio between a force and an acceleration") only holds in the classical limit.
In special relativity, momentum and velocity aren't directly proportional. Instead of being related by a constant multiplicative factor like they are in Newtonian physics (mass), they're related by a
function of speed (total energy):
##\vec p c = E \vec \beta = \gamma E_0 \vec \beta##,
where ##\vec \beta = \vec v / c##.
Force is the time-derivative of momentum. The time-derivative of that equation requires the chain rule, and you end up with three terms on the right side! When the rest energy (mass) is taken to be constant, that reduces to two terms (I'm using overdots to indicate ##ct##-derivatives, not ##t##-derivatives):
##\vec f = \gamma^2 (\beta \cdot \dot{\vec \beta}) E \beta + E \dot{\vec \beta}##,
or if you prefer using ##m##, ##\vec v##, and ##\vec a##:
##\vec f = \gamma^3 \left( \dfrac{ \vec v \cdot \vec a }{c^2} \right) m \vec v + \gamma m \vec a##.
We're pretty damn far from ##\vec f = m \vec a##.
In short, there simply is no such quantity as the "ratio between a force and an acceleration," because the velocity vector figures in there, too.
In the special case that the force is parallel or anti-parallel to the velocity, we obtain:
##\vec f = \gamma^2 E \dot{\vec \beta} = \gamma ^3 m \vec a##,
and we can call ##\gamma^3 m## the ratio between force and acceleration. But note that if the force is perpendicular to the velocity, we instead get:
##\vec f = E \dot{\vec \beta} = \gamma m \vec a##.
I hope this helps correct some of your misconceptions.
