Does an accelerated rocket curve spacetime?

Question

I am not a physicist, I am an observer, a student of the theory of relativity. I have a doubt about the principle of equivalence. I know that there are two different acceleration phenomena, one is free fall and the other is Non-Uniform acceleration(between bodies, not the non-uniform only body changing in time acceleration), like a rocket with a person inside for example, the question is: if space around an accelerated rocket curves because of the equivalence principle, like it happens with gravity around large bodies, would space-time curves because of rocket's increasing speed?, I believe that space not really curves around an accelerated rocket but the space-time changes its form someway, because a curvature points to the center of mass and this is not happening with acceleration, I think there is some gravitational effect around bodies not uniformly accelerated with respect to each other. A clock in the accelerated rocket also goes slower like in some planet at a given heigh(I am not telling about the speed effect on time dilation which happens to for example two different moving rockets at uniform speed,But the gravitatory effect on time dilation which affects time in different proportions between systems of reference with different acelerations and masses)

Answer 1

This is an excellent example of how acceleration and gravity are different things. It is a common misconception that the Equivalence Principle states they are one and the same. Actually, it states that they are the same locally, i.e., at a single point. By doing experiments at a single point in spacetime, you can't distinguish between gravity and acceleration. However, if you move around sufficiently (where "sufficiently" depends on the precision of your experiment and the conditions you're performing the experiment in), you will be able to distinguish gravity and acceleration.

Notice that all observers must agree on whether spacetime is curved. That is because curvature means, for example, that two straight lines will get closer to each other as time passes (in GR, free bodies move on "straight lines", the technical term being "geodesics"). This is similar to how all maps of the Earth agree that the Earth is round, regardless of which projection they use to represent it (azimuthal, cylindrical, conic, etc).

In short, an accelerated observer won't see curved spacetime in Special Relativity.

Remark: in practice, to produce acceleration you'll need a source of energy and energy does bend spacetime, so in this sense there will be curvature. But that is due to practical implementations, not due to the equivalence principle.

Answer 2

Fictitious forces are not really due to curved spacetime, but to curved/moving coordinate systems.

Suppose we have a particle position $x$ in an inertial coordinate system where $F=m\ddot{x}$. We apply a coordinate transform $X=Tx$ where the transformation $T$ can change over time. When we differentiate this, we get extra terms because of the coordinates. The first derivative is $\dot{X}=T\dot{x}+\dot{T}x$. The second derivative is $\ddot{X}=T\ddot{x}+2\dot{T}\dot{x}+\ddot{T}x$. Multiply both sides by $m$ to get $m\ddot{X}=Tm\ddot{x}+2\dot{T}m\dot{x}+\ddot{T}mx$. The first term on the right is $TF$, the coordinate transformation of the force vector, and what we might call the 'true' forces. The other two terms are corrections due to using moving coordinates - a velocity-dependent Coriolis term and a position-dependent centrifugal term (if T is rotating) - which are appearing on the same side of the equation as the coordinate-transformed true force. These are called 'fictitious' forces. They have the notable property that they are proportional to the mass.

The thing about spacetime being curved is that it forces you to use curved coordinate systems. You can't draw flat xy coordinates on a sphere. And being curved in time as well means the coordinate system has to be moving. You can always find a flat one locally close to any given point, like you can always approximate a small patch of a sphere with a flat plane. But as you try to extend it, you find your coordinate system bending, creating the appearance of extra forces. This is where gravity comes from.

On the surface of the Earth, we can make gravity 'disappear' by choosing freefall coordinates. If you are in a falling lift, everything you can see falls at the same rate, and you appear to be 'weightless' in an inertial coordinate system. But you can't extend that flat freefall coordinate system to the rest of the world. You have to bend it around the planet.

The relationship of the accelerating coordinate system on a rocket to the inertial coordinate system of an observer watching it zoom by is exactly the same as the relationship between the flat local coordinate of a physics lab sat on Earth experiencing a uniform gravitational field, and the flat local freefall coordinate system of the experimental apparatus the lab assistant has just dropped.

We only experience this so-called 'gravity force' in the lab because we use lab-fixed coordinates and not freefall coordinates.

If you start with a uniform gravitational field (constant force in a constant direction, everywhere in the universe), and consider it's freefall coordinates, this turns out to be identical to flat inertial space - like an empty universe with no gravity. You only need curved spacetime to deal with non-uniform gravity fields.

But to have a gravityless freefall coordinate system that is everywhere falling inwards towards a central point, we have to curve spacetime. We can't make gravity 'fictitious' without doing so. It is in that sense that 'gravity is due to curved spacetime'.