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  1. Why geometrically four acceleration is a curvature vector of a world line?

    Geometrically, four-acceleration is a curvature vector of a world line. Therefore, the magnitude of the four-acceleration (which is an invariant scalar) is equal to the proper acceleration that a moving particle "feels" moving along a world line. The world lines having constant magnitude of four-acceleration are Minkowski-circles. (Wikipedia)

  2. And what is proper acceleration?

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3 Answers 3

this is why 4-acc is Geometrically curvature.

$$A^{\mu}=\frac {d^2x^{\mu}}{ds^2}+\Gamma_{\alpha\beta}^{\mu}\frac {dx^{\alpha}}{ds}\frac {dx^{\beta}}{ds}=0$$

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An object moving on an inertial path has a straight worldline in special relativity. The four-acceleration then measures how non-straight the worldline is.

When gravity is involved, an inertial (free-fall) path may be curved due to gravity. Four-acceleration still measures the acceleration relative to such a path, and proper acceleration is basically just that four-acceleration converted to a three-vector. For example, a free-fall path in the vicinity of Earth would be, well, one in which you freely fall towards Earth's center. Such a path has zero proper acceleration in the absence of other forces. If instead you're standing on Earth's surface, your proper acceleration is radially directed away from the center of the Earth, with magnitude $g$.

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Curvature of a plane curve is defined as the (magnitude) rate of change of the unit tangent vector over the length of the curve:

$$\kappa =\left | \frac{d\mathbf{T}}{ds} \right |$$

This is a natural definition, because for example the curvature of a circle is just $1/r$, so when the circle is small it has large curvature and when it's big it has small curvature. The unit tangent vector, as you know, is just the velocity vector divided by its magnitude:

$$\mathbf{T}=\frac{\mathbf{u}}{|\mathbf{u}|}$$

Do a bit of calculus and you find that $d\mathbf{T}/ds =\kappa (s) \mathbf{N}(s)$, where $\mathbf{N}$ is the unit normal vector. So you can define $d\mathbf{T}/ds$ as the "curvature vector" of the curve which points normal to the curve and is scaled by the curvature.

In SR, the four-velocity $U^\mu$ is defined such that its magnitude is always $c$. Also, we tend to work in units where $c=1$, so the four-velocity is actually a unit tangent vector to the worldline. The four-acceleration, which is defined as $A^\mu = dU^\mu /ds$, is therefore the curvature vector of the worldline, the magnitude of which is the curvature.

Proper acceleration is just a fancy way of saying "the acceleration that you would be able to measure with an accelerometer." The magnitude of four-acceleration is always proper acceleration, so geometrically the proper acceleration of an observer is the curvature of his worldline.

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Mostly a fair answer, IMHO; it could be more explicit on how to measure $U^\mu[ s ]$ and $s$ separately, in order to evaluate $d/ds[ U^\mu ]$. But the last paragraph seems backwards. Instead, "what you read off a good accelerometer" is at best a covert, unspecific way of saying "magnitude of four acceleration $A^\mu := d/ds[ U^\mu ]$". At worst, "what you read off some accelerometer" is "not even wrong", w/o considering the definition of the quantity to be measured, and thus without a way of determining whether or to which accuracy the given accelerometer had been "good" in any given trial. –  user12262 Jun 4 '13 at 20:24

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