Does the force needed to produce unit acceleration in say a 1kg in free space object vary depending on the location?
Basically is inertia stronger or weaker in other parts of our universe or is it uniform throughout?

Does the force needed to produce unit acceleration in say a 1kg in free space object vary depending on the location?
Basically is inertia stronger or weaker in other parts of our universe or is it uniform throughout?
If inertia is dependant upon mass and mass increases due to relative speed?
For someone who claimed to have "spent a few years studing Einstien exclusivly" you're remarkably uninformed.
It must have taken a great effort to remain so ignorant.
Sir Daffystein duck, actually it takes appsolutley very little effort. Sorry for my above stupid question.
I am corrected. Thank you Deacon.
Inertia  being the resistance to change in the state of motion  is a property of all forms of energy; it thus applies to relativistic mass, and not just to rest mass. This simply means that as an object approaches the speed of light, it becomes harder and harder to accelerate it more, or, in other words, that a constant acceleration has less and less effect as you get closer to c.
It depends on the way you view it. For example, Newton's second law can be relativistically expressed in terms of fourdimensional vectors:
( is the absolute differential operator, the covariant form of the ordinary differential operator)
Here, is the invariant (rest) mass.
Alternatively, a force density can be defined in terms of the covariant divergence of the energymomentum density tensor :
Here, has to be only a part of the total energymomentum density because for the total energymomentum density:
which may be considered as a form of Newton's third law. Nevertheless, the individual components of the force density is the result of the individual components of the energymomentum density tensor (i.e. relativistic mass).
Actually, it's the notation used in my first textbook on Tensor Calculus. It also uses the semicolon notation for the covariant derivative (i.e. ), but I dislike that notation, so I use the nabla operator (i.e. ) which I got from another textbook. On forums where symbol fonts are unavailable, I would use for the covariant differential operator and for both the partial and ordinary differential operators.
An alternative way of expressing the covariant form of the ordinary differential operator is:
which is an application of the chain rule for the total derivative ()
I tend to prefer this way as it doesn't introduce any new notation to a set of expressions that already uses the covariant differential operator ().
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