[Darawan (OP)]

Hi!

Imagine that we are in a rocket accelerating with some magnitude a₁ = dx²/d²y, also imagine that we have a stationary rocket ship in close proximity to ours, stationary relative to our reference frame, we will notice the stationary ship to measure time (If we have a clock onboard that ship which is visible to ours) at some time t₀.

Now the question leads to this, the time measured on our ship will be moving relative to the stationary ship, but the equivalence principle tells us that if we accelerate at some magnitude we can't tell it apart from gravity, but gravity bends spacetime in such a manner that time will be slowed relative to some reference frame, but if we compare the two clocks on these ships, ours and the stationary one, will our measurement account for the bending of spacetime due to the acceleration?

I live in Sweden so my English might not be perfect, excuses are made.

Now debate.

[PmbPhy]

Now the question leads to this, the time measured on our ship will be moving relative to the stationary ship, but the equivalence principle tells us that if we accelerate at some magnitude we can't tell it apart from gravity, but gravity bends spacetime...

A common misconception in general relativity is that Einstein demonstrated that gravity is a curvature in spacetime. That's totally wrong. Einstein never said or even implied it. Max Von Laue sent Einstein a copy of the book he wrote on relativity for him to look over. In that book von Laue wrote that gravity is a curvature in spacetime. When Einstein sent back his comments on the book back to him he wrote

... what characterizes the existence of a gravitational field from the empirical standpoint is the non-vanishing of the components of [the affine connection], not the vanishing of the [components of the Riemann tensor]. If one does not think in such intuitive (anschaulich) ways, one cannot grasp why something like curvature should have anything at all to do with gravitation. In any case, no rational person would have hit upon anything otherwise. The key to the understanding of the equality

The components of the affine connection describe how the particle is accelerating in spacetime. The components of the Riemann tensor describes how two particles accelerate relative to each other which when its non-zero the spacetime is curved. Important note: If the spacetime curvature is flat then you can't change to any set of coordinates in which the spacetime is curved.

You can read all about this in the article I wrote on the subject: Einstein's gravitational field by Peter M. Brown, http://xxx.lanl.gov/abs/physics/0204044

Does that answer your question?

[hamdani yusuf]

Equivalence principle according to wikipedia.

The equivalence principle was properly introduced by Albert Einstein in 1907, when he observed that the acceleration of bodies towards the center of the Earth at a rate of 1g (g = 9.81 m/s2 being a standard reference of gravitational acceleration at the Earth's surface) is equivalent to the acceleration of an inertially moving body that would be observed on a rocket in free space being accelerated at a rate of 1g. Einstein stated it thus:

we ... assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system.

— Einstein, 1907

So if we are inside a rocket and feel the pushing force of 1 g by the floor, we can't distinguish if we are stationary under gravitational acceleration, or accelerating in free space without looking outside. But if we wait long enough experiencing constant 1 g, can't we be sure that we are not linearly accelerating in free space because we would exceed the speed of light?

[Halc]

But if we wait long enough experiencing constant 1 g, can't we be sure that we are not linearly accelerating in free space because we would exceed the speed of light?

Proper acceleration (which is the 1g being felt by your guy) measures the rate of change in rapidity, not in velocity.  Rapidity is not limited by c, which is why I can get to Betelgeuse before I die if I go fast enough, despite it being over 600 light years away.

Acceleration (as opposed to proper acceleration) is frame dependent measures the rate of change in velocity. Velocity (relative to a given inertial frame) cannot exceed c, and thus acceleration is limited in frames where the velocity is already near c.

Assume a rocket with a very large but finite amount of fuel, accelerating at 1g. The occupant feels as though he is standing on a planet, but knows he is accelerating because his fuel gauge is decreasing.

Ah, but the same guy hovering with his rocket a meter above Earth is also watching his fuel gauge decreasing. Both measure the same thing if they're equivalently set up.

[Janus]

Equivalence principle according to wikipedia.

The equivalence principle was properly introduced by Albert Einstein in 1907, when he observed that the acceleration of bodies towards the center of the Earth at a rate of 1g (g = 9.81 m/s2 being a standard reference of gravitational acceleration at the Earth's surface) is equivalent to the acceleration of an inertially moving body that would be observed on a rocket in free space being accelerated at a rate of 1g. Einstein stated it thus:

we ... assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system.

— Einstein, 1907

So if we are inside a rocket and feel the pushing force of 1 g by the floor, we can't distinguish if we are stationary under gravitational acceleration, or accelerating in free space without looking outside. But if we wait long enough experiencing constant 1 g, can't we be sure that we are not linearly accelerating in free space because we would exceed the speed of light?

Exceeding the speed of light relative to what?.  There is no absolute reference to judge your speed against.  So, you must first choose one.   The easiest one to work with in this case is the inertial frame you were at rest with respect to before you began your acceleration.
So let's say that before firing your engines, you drop off a space buoy.  This gives you something to act as a marker to measure your progress.  After accelerating for some time, you find yourself moving at 0.1c relative to your first buoy.  You release a 2nd buoy which now maintains a speed of 0.1c relative to the first buoy.  You continue to accelerate, and after the same time interval, you are moving at 0.1c relative to the 2nd buoy. wat do you measure your speed as being relative to the first buoy?  Newtonian physics would say 0.1c + 0.1c = 0.2c.  But velocities really don't add up that way in our Relativistic universe, but instead like this: (0.1c+0.1c)/(1+0.1c(0.1c)/c^2) = ~0.198 c, just a tad under 0.2c.
You can do this again,dropping a 3rd buoy and accelerating to 0.1c relative to it, and you will now measure your speed relative to the first buoy as being ~0.292 c
You can keep doing this as many times as you want, and each time your speed relative to the first buoy will increase, but by a smaller amount each time, and no matter how long you do this, it will never equal c, let alone exceed it.

[hamdani yusuf]

Exceeding the speed of light relative to what?

Relative to the frame when the experiment began.

[Janus]

Exceeding the speed of light relative to what?

Relative to the frame when the experiment began.

If you read the rest of my post, that's exactly what I addressed.

[hamdani yusuf]

Exceeding the speed of light relative to what?

Relative to the frame when the experiment began.

If you read the rest of my post, that's exactly what I addressed.

I know. Halc's answer is clear enough.

[hamdani yusuf]

Proper acceleration (which is the 1g being felt by your guy) measures the rate of change in rapidity, not in velocity.

Does it mean that the kinetic energy becomes ½mw² instead of the usual  ½mv² ?

[hamdani yusuf]

No. Acceleration (coordinate acceleration in particular) is rate of change of velocity. The rate of change of momentum is force, at least for a constant mass.

Thanks for the correction. Is there relativistic term for acceleration, just like rapidity for velocity?

[Halc]

This thread has been split here: https://www.thenakedscientists.com/forum/index.php?topic=83621

Proper acceleration (which is the 1g being felt by your guy) measures the rate of change in rapidity, not in velocity.

Does it mean that the kinetic energy becomes ½mw² instead of the usual  ½mv² ?

Excellent question that I didn't see until now. Sorry.
I don't think there's any such thing as proper (linear) kinetic energy since the ship is always stationary in its own frame, but I think the answer to your question is still yes: Kinetic energy relative to the original inertial frame is equal to ½mw² where m is the proper mass of the ship, instead of ½mv² where m is the relativistic mass of the ship. They should equal the same thing.

No. Acceleration (coordinate acceleration in particular) is rate of change of velocity. The rate of change of momentum is force, at least for a constant mass.

Thanks for the correction. Is there relativistic term for acceleration, just like rapidity for velocity?

It's usually just called acceleration, which is different from proper acceleration. If you want to be non-ambiguous, you can call it coordinate acceleration.