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gordonblur

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I have read a number of threads on acceleration and special relativity, but can't find what I'm looking for. I would like to know, in the context of special relativity, how to compute the acceleration "felt" by an observer, and how to transform this acceleration into different inertial frames.

Given a particle's world line as a 4-vector [tex] x(\tau) [/tex] parameterised by proper time and relative to some inertial frame, then it's 4-velocity and 4-acceleration are

[tex] \dot{x}=\frac{dx}{d\tau}=\gamma(e_0+V) [/tex]

[tex] \ddot{x}=\dot{\gamma}(e_0+V)+\gamma^2 A [/tex]

where [tex] V\cdot e_0=A\cdot e_0=0, \gamma^{-2}=1+V\cdot V [/tex] and [tex] \dot{\gamma}=-\gamma^4 A\cdot V [/tex].

I appologise for the slightly informal notation that follows. [tex] V=(0,V_1,V_2,V_3) [/tex] and [tex] A [/tex] are the instantaneous 3-velocity and 3-acceleration.

Let the Lorentz tranform that maps into the particle's instantaneous co-moving frame frame be [tex] \Lambda_V [/tex], s.t. [tex] \Lambda_V \dot{x}=e_0 [/tex] and [tex] \Lambda_V e_0=\gamma(e_0-V) [/tex]. My initial thought was that the "local" acceleration experienced by the particle was simply [tex] \Lambda_V \ddot{x} [/tex].

However, consider the same world line expressed in a different inertial frame [tex] y=\Lambda_U x [/tex], [tex] \dot{y}=\Lambda_{U} \dot{x}=\delta(e_0+W) [/tex] and [tex] \ddot{y}=\Lambda_{U} \ddot{x} [/tex]. Then the local acceleration is either [tex] \Lambda_W \ddot{y} [/tex] or [tex] \Lambda_V \Lambda_{-U} \ddot{y} [/tex], but these are not equal since in general [tex] \Lambda_W \neq \Lambda_V \Lambda_{-U} [/tex] (although I think [tex] \Lambda_W e_0 = \Lambda_V \Lambda_{-U}e_0 [/tex]).

I guess the solution lies in splitting the velocity into two components, one parallel and one perpendicular to [tex] A [/tex], but can't find a mathematical justification. Any help would be greatly appreciated.

Thanks, Gordon.

Given a particle's world line as a 4-vector [tex] x(\tau) [/tex] parameterised by proper time and relative to some inertial frame, then it's 4-velocity and 4-acceleration are

[tex] \dot{x}=\frac{dx}{d\tau}=\gamma(e_0+V) [/tex]

[tex] \ddot{x}=\dot{\gamma}(e_0+V)+\gamma^2 A [/tex]

where [tex] V\cdot e_0=A\cdot e_0=0, \gamma^{-2}=1+V\cdot V [/tex] and [tex] \dot{\gamma}=-\gamma^4 A\cdot V [/tex].

I appologise for the slightly informal notation that follows. [tex] V=(0,V_1,V_2,V_3) [/tex] and [tex] A [/tex] are the instantaneous 3-velocity and 3-acceleration.

Let the Lorentz tranform that maps into the particle's instantaneous co-moving frame frame be [tex] \Lambda_V [/tex], s.t. [tex] \Lambda_V \dot{x}=e_0 [/tex] and [tex] \Lambda_V e_0=\gamma(e_0-V) [/tex]. My initial thought was that the "local" acceleration experienced by the particle was simply [tex] \Lambda_V \ddot{x} [/tex].

However, consider the same world line expressed in a different inertial frame [tex] y=\Lambda_U x [/tex], [tex] \dot{y}=\Lambda_{U} \dot{x}=\delta(e_0+W) [/tex] and [tex] \ddot{y}=\Lambda_{U} \ddot{x} [/tex]. Then the local acceleration is either [tex] \Lambda_W \ddot{y} [/tex] or [tex] \Lambda_V \Lambda_{-U} \ddot{y} [/tex], but these are not equal since in general [tex] \Lambda_W \neq \Lambda_V \Lambda_{-U} [/tex] (although I think [tex] \Lambda_W e_0 = \Lambda_V \Lambda_{-U}e_0 [/tex]).

I guess the solution lies in splitting the velocity into two components, one parallel and one perpendicular to [tex] A [/tex], but can't find a mathematical justification. Any help would be greatly appreciated.

Thanks, Gordon.

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