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*Spacetime Physics*by Taylor and Wheeler at @PeterDonis's suggestion. In chapter 3, they say:

Ok, so force is not an invariant. Fine.Apply force to a moving object: Its velocity changes; it accelerates. Acceleration is the signal that force is being applied. Two events are enough to reveal velocity; three reveal change in velocity, therefore acceleration, therefore force. The laboratory observer reckons velocity between the first and second events, then he reckons velocity between the second and third events. Subtracting, he obtains the change in velocity. From this change he figures the force applied to the object.

The rocket observer also measures the motion; velocity between the first and second events, velocity between second and third events; from these the change in velocity; from this the force acting on the object. But the rocket-observed velocities are not equal to the corresponding laboratory-observed velocities. The change in velocity also differs in the two frames; therefore the computed force on the object is different for rocket observer and laboratory observer. The Principle of Relativity does not deny that the force acting on an object is different as reckoned in two frames in relative motion.

Then I go to http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html, where the term ##a## appears in various equations and is defined as acceleration. Is it an invariant?

The authors didn't state how they computed the force. I would assume they used the standard ##F = ma##. If ##F## is not invariant, then either ##m## or ##a## is invariant, or both, correct?

My mental model was of someone on a rocket accelerating at 1g. When they step on a scale calibrated for the Earth's surface, a 1 kg object would generate a 1 kg reading, which I thought was a measure of force. The accelerated rocket has an instantaneous moving frame, one that changes each instant, but the 1 kg object would continue to generate a 1 kg reading.

As I'm not in school and can't ask the teacher a question, I thought I'd ask the group here to straighten me out. Thanks.