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I was thinking of it, If I say: "I'm moving at a velocity $v_1$ relative to a reference frame $M$ then the acceleration will be the derivative of $v_1$ relative to the reference frame $M$." In other words, from the perspective of my brother at home, I'm travelling at a velocity $v_1$ and I have an acceleration $a_1$. But from my perspective, he is travelling at $v_1$ (and I'm standing still) and thus his acceleration is $a_1$. But General Relativity tell us that acceleration is not relative, so why?

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@Qmechanic ♦ What does it have to do with the equivalence-principle? –  user29727 Sep 27 '13 at 23:14
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Note that what GR says is detectable is acceleration relative to a free-falling frame of reference. –  Ben Crowell Sep 28 '13 at 0:55

4 Answers 4

There are 2 different concepts.

The first concept is "relative" acceleration. This could make sense, for instance, if we consider an inertial frame $F$, an accelerating particle $P$ relatively to this inertial frame. We could choose, at any instant $t$, a inertial comoving-frame $G(t)$ (different for each $t$) which has, at any time, the same speed that the particle (and with same origin). We could consider, at this instant $t$, the acceleration of the particle $P$ relatively to $G(t)$, and we may consider that the acceleration of $G(t)$ relatively to $P$ is the opposite.

[EDITED]

(added Precisions about the locality of the experiment)

However, feeling an acceleration in a frame is an other thing. It has to do with some local experiment made in this frame. If I am in an inertial frame, and I drop an apple with initial zero speed , the apple will always have a zero speed. This is not the case in a non-inertial frame (for instance, at the surface of the Earth, our frame is a non-inertial frame because, even if the apple has initial speed zero, the apple fall, and so get a non-zero velocity). Each observer or frame has in fact a local feeling of the word, which is dictated by the metrics "felt" locally by this observer. For a inertial frame (for instance a free-falling observer), this is a Minkowski metrics. For a non-inertial frame, this is not a Minkowski metrics. If I want to know the equation of the movement of the apple, it is a geodesics :

$$ \frac{du^i}{ds}+\Gamma^i_{jk}u^ju^k=0$$

In a Minkowski metrics, the Christoffel symbols are zero, so the equation of the movement of the apple is simply $ \frac{du^i}{ds}=0$, so with an initial speed zero, the apple will keep this zero speed. Of course, it is no more true in a general non-inertial frame.

Of course, all that is only true "locally".If the apple is far from the free-falling observer, it is no more true. For instance, in a spherical metrics, the trajectories of 2 radial free-falling observers are converging.

So, finally, inertial frames are very specific frames, which "felt" locally the Minkowski metrics, and this is an "absolute" specificity.

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How do you know there is no "real" force acting on the apple? Do you just apply Ockham's razor? –  jinawee Sep 28 '13 at 9:16
    
Apples are following geodesics. In the world seen by a free falling observer, and which corresponds to an inertial frame, these geodesics correspond to $\frac{du^i}{ds}=0$ –  Trimok Sep 28 '13 at 9:20

From inside a closed lab one cannot determine its speed , no matter what experiment is attempted. By contrast, from inside a closed lab one can determine that the lab is accelerated or not with a simple accelerometer. This is why, speed is relative while acceleration is absolute.

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Can the measured acceleration change relative to something other? –  user29727 Sep 28 '13 at 8:56
    
If we have acceleration we can measure velocity? Also the acceleration caused by the gravitational force of the earth is a relative quantity. –  user29727 Sep 28 '13 at 9:23

Absolute velocity cannot be discerned by any experiment - you and your brother can each say that it is the other that is moving. Absolute acceleration, however, is readily discerned by pseudo-forces. If you were accelerating away from your brother, only you would feel a pseudo-force. Your brother would not feel a pseudo-force. Acceleration is, therefore, absolute, unlike velocity, which is relative.

When I say that acceleration is absolute, I mean that it can be determined that you are accelerating, without reference to another person's perspective. In contrast, velocity, being relative, is always relative to another person's perspective.

You can test for pseudo-forces by dropping a test mass, e.g. a tennis ball. If, relative to you, the test mass moves, you are experiencing a pseudo-force.

Were you to jump out of a window (which I wouldn't recommend), you would feel no pseudo-forces, for your tennis ball would fall with you. If you were to stand on the earth (much safer), you would feel a pseudo-force of the earth on your feet, for the tennis ball would fall to the earth relative to you. In General Relativity, confusing though it is at first, you are not accelerating if you jump out of a window, but you are accelerating if you stand on the earth.

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But it is involved in velocity, and velocity is relative? –  user29727 Sep 27 '13 at 23:04
    
Yes, velocity is relative. Absolute velocity cannot be discerned by any experiment. Absolute acceleration, however, is readily observed by pseudo-forces. –  innisfree Sep 27 '13 at 23:10
    
What are pseudo-forces for velocity? –  user29727 Sep 27 '13 at 23:16
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If you jump out of the window, are you accelerating or feeling a force or pseudo-force? Secondly, e.g. in your car, how do you differentiate between pseudo-force and force? And what does is mean for a quantity (velocity/acceleration) to be absolute? –  NikolajK Sep 27 '13 at 23:17
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@innisfree: You can still edit the answer and elaborate on these points. –  NikolajK Sep 27 '13 at 23:32

If I say: "I'm moving at a velocity $v_1$ relative to a reference frame $M$ then the acceleration will be the derivative of $v_1$ relative to the reference frame M. In other words, from the perspective of my brother at home, I'm travelling at a velocity $v_1$ and I have an acceleration $a_1$."

This supposes that "reference frame M" is an inertial frame, i.e. a set of suitable participants (including the "brother at home") who determined that were at rest to each other throughout the experimental trial under consideration.

But from my perspective, he is travelling at $v_1$

This supposes that "you" belong to an inertial frame, i.e. "you" are member of a set of suitable participants who determined that were at rest to each other throughout the experimental trial under consideration.

However, this condition is mutually exclusive to the condition of the members of inertial frame $M$ determining "your" acceleration $a_1$ as non-zero.

In consequence, speed is determined mutually equal between inertial frames (at least in regions with refractive index $n = 1$); while non-zero acceleration cannot be determined mutually at all.

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