Pearson's chi-squared test

In the theory of relativity, four-acceleration is a four-vector (vector in four-dimensional spacetime) that is analogous to classical acceleration (a three-dimensional vector). Four-acceleration has applications in areas such as the annihilation of antiprotons, resonance of strange particles and radiation of an accelerated charge.

Four-acceleration in inertial coordinates

In inertial coordinates in special relativity, four-acceleration is defined as the change in four-velocity over the particle's proper time:

$\mathbf{A}=\frac{d\mathbf{U}}{d\tau }=({\gamma }_u{\dot{\gamma }}_{u}c,{\gamma }_u^2\mathbf{a}+{\gamma }_u{\dot{\gamma }}_u\mathbf{u})=({\gamma }_u^4\frac{\mathbf{a}\cdot \mathbf{u}}{c},{\gamma }_u^2\mathbf{a}+{\gamma }_u^4\frac{(\mathbf{a}\cdot \mathbf{u})}{c^2}\mathbf{u})$,

where

$\mathbf{a}=\frac{d\mathbf{u}}{dt}$

and

${\dot{\gamma }}_u=\frac{\mathbf{a}\mathbf{\cdot }\mathbf{u}}{c^2}{\gamma }_u^3=\frac{\mathbf{a}\mathbf{\cdot }\mathbf{u}}{c^2}\frac{1}{{(1-\frac{u^2}{c^2})}^{3/2}}$

and ${\gamma }_u$ is the Lorentz factor for the speed $u$. A dot above a variable indicates a derivative with respect to the coordinate time in a given reference frame, not the proper time $\tau$.

In an instantaneously co-moving inertial reference frame $\mathbf{u}=0$, ${\gamma }_u=1$ and ${\dot{\gamma }}_u=0$, i.e. in such a reference frame

$\mathbf{A}=(0,\mathbf{a})$

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 i.e. hyperbolas (see hyperbolic motion)

The scalar product of a four-velocity and the corresponding four-acceleration is always 0.

Even at relativistic speeds four-acceleration is related to the four-force such that

$F^{\mu }=m{A}^{\mu }$

where m is the invariant mass of a particle.

In special relativity the coordinates are those of a rectilinear inertial frame, so the Christoffel symbols term vanishes, but sometimes when authors uses curved coordinates in order to describe an accelerated frame, the frame of reference isn't inertial, they will still describe the physics as special relativistic because the metric is just a frame transformation of the Minkowski space metric. In that case this is the expression that must be used because the Christoffel symbols are no longer all zero.

When the four-force is zero one has gravitation acting alone, and the four-vector version of Newton's second law above reduces to the geodesic equation. A particle executing geodesic motion has zero value for each component of the acceleration four vector.This conforms to the fact that Gravity is not a force. The radial component four-acceleration of a body under free-fall,incidentally, is zero.This identifies the fact that the spatial part of four-acceleration is different from what we understand by acceleration in common day to day experience for example in the case of an apple falling from a tree.

Four-acceleration in non-inertial coordinates

In non-inertial coordinates, which include accelerated coordinates in special relatrivity and all coordinates in general relativity, the acceleration four-vector is related to the four-velocity through an absolute derivative with respect to proper time.

$A^{\lambda }:=\frac{D{U}^{\lambda }}{d\tau }=\frac{d{U}^{\lambda }}{d\tau }+{\mathrm{\Gamma }}^{\lambda }{}_{\mu \nu }{U}^{\mu }{U}^{\nu }$

In inertial coordinates the Christoffel symbols ${\mathrm{\Gamma }}^{\lambda }{}_{\mu \nu }$ are all zero, so this formula is compatible with the formula given earlier.

See also

• four-vector
• four-velocity
• four-momentum
• four-force

References

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• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534

External links

• Curvature vector on Britannica