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"Proper Reference Frame"-Accelerated observer
Hi guys. This is regarding section 13.6 (p.327) in MTW. Here the authors consider an arbitrary accelerated observer in any space-time and construct a set of local coordinates carried along the entire worldline of the oberver with the origin of the coordinates comoving with the observer; they term this the "proper reference frame" of the accelerated observer. Note that they do not assume the observer's spatial basis vectors are Fermi-Walker transported along his/her worldline.
In ex.(13.14) they consider an accelerated observer and a freely falling particle that at some event is coincident with the origin of the observer's "proper reference frame". They say to show that the 3-acceleration of the freely falling particle relative to the "proper reference frame" at that event is given by ##\frac{\mathrm{d} ^{2}x^{j}}{\mathrm{d} x^{0^2}}e_j = -a - 2\omega \times v + 2(a\cdot v)v## where ##j = 1,2,3##, ##(e_j)## are the spatial basis vectors carried by the observer, ##\omega## is the angular velocity of rotation of the spatial basis vectors, ##v## is the 3-velocity of the freely falling particle, and ##a## is the acceleration of the observer him/herself.
Because this event is on the worldline of the observer (i.e. the origin ##x^j = 0## of the observer's "proper reference frame"), the non-zero christoffel symbols are all given in (13.69a) and (13.69b). Now the freely falling particle satisfies the equations of motion ##\nabla_u u = 0## as usual. Taking again ##j = 1,2,3## in the above coordinates, we simply calculate ##\frac{\mathrm{d} ^2 x^{j}}{\mathrm{d} \tau^2} = -\Gamma ^{j}_{\mu\nu}\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}\frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}\\ = -\Gamma ^{j}_{00}(\frac{\mathrm{d} x^0}{\mathrm{d} \tau})^2 - 2\Gamma ^{j}_{k0}\frac{\mathrm{d} x^{k}}{\mathrm{d} \tau}\frac{\mathrm{d} x^0}{\mathrm{d} \tau} \\= -a^j(\frac{\mathrm{d} x^0}{\mathrm{d} \tau})^2 + 2\omega^{i}\epsilon_{0ijk}\frac{\mathrm{d} x^{k}}{\mathrm{d} x^0}(\frac{\mathrm{d} x^0}{\mathrm{d} \tau})^2\\ = (-a^j - 2(\omega \times v)^j)(\frac{\mathrm{d} x^0}{\mathrm{d} \tau})^2##
because all other Christoffel symbols vanish as per (13.69a) and (13.69b). I don't get where that third term ##2(a\cdot v)v## is coming from. Could someone point it out? Thanks in advance!
Hi guys. This is regarding section 13.6 (p.327) in MTW. Here the authors consider an arbitrary accelerated observer in any space-time and construct a set of local coordinates carried along the entire worldline of the oberver with the origin of the coordinates comoving with the observer; they term this the "proper reference frame" of the accelerated observer. Note that they do not assume the observer's spatial basis vectors are Fermi-Walker transported along his/her worldline.
In ex.(13.14) they consider an accelerated observer and a freely falling particle that at some event is coincident with the origin of the observer's "proper reference frame". They say to show that the 3-acceleration of the freely falling particle relative to the "proper reference frame" at that event is given by ##\frac{\mathrm{d} ^{2}x^{j}}{\mathrm{d} x^{0^2}}e_j = -a - 2\omega \times v + 2(a\cdot v)v## where ##j = 1,2,3##, ##(e_j)## are the spatial basis vectors carried by the observer, ##\omega## is the angular velocity of rotation of the spatial basis vectors, ##v## is the 3-velocity of the freely falling particle, and ##a## is the acceleration of the observer him/herself.
Because this event is on the worldline of the observer (i.e. the origin ##x^j = 0## of the observer's "proper reference frame"), the non-zero christoffel symbols are all given in (13.69a) and (13.69b). Now the freely falling particle satisfies the equations of motion ##\nabla_u u = 0## as usual. Taking again ##j = 1,2,3## in the above coordinates, we simply calculate ##\frac{\mathrm{d} ^2 x^{j}}{\mathrm{d} \tau^2} = -\Gamma ^{j}_{\mu\nu}\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}\frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}\\ = -\Gamma ^{j}_{00}(\frac{\mathrm{d} x^0}{\mathrm{d} \tau})^2 - 2\Gamma ^{j}_{k0}\frac{\mathrm{d} x^{k}}{\mathrm{d} \tau}\frac{\mathrm{d} x^0}{\mathrm{d} \tau} \\= -a^j(\frac{\mathrm{d} x^0}{\mathrm{d} \tau})^2 + 2\omega^{i}\epsilon_{0ijk}\frac{\mathrm{d} x^{k}}{\mathrm{d} x^0}(\frac{\mathrm{d} x^0}{\mathrm{d} \tau})^2\\ = (-a^j - 2(\omega \times v)^j)(\frac{\mathrm{d} x^0}{\mathrm{d} \tau})^2##
because all other Christoffel symbols vanish as per (13.69a) and (13.69b). I don't get where that third term ##2(a\cdot v)v## is coming from. Could someone point it out? Thanks in advance!
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