Justification for a new doppler radar formula
in [Relativity]
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From: Surfer on 20 Feb 2010 22:08 The formula for radar Doppler shift can be derived without invoking Einsteins' theory of special relativity, because all observations are made in the same frame of reference. The result with c as the speed of light, V as the target velocity and Ft as the transmitted frequency, gives the shifted frequency Fr as: Fr = Ft (c+V)/(c-V) (1) However I will show below that this formula would be inaccurate if the one way speed of light were to be isotropic only with respect to a preferred frame and justify an alternative formula. Suppose a frame of reference is identified in which the one way speed of light is 'truely' isotropic, referred to below as the 'isotropic frame'. Suppose that relative to such a frame a radar system is moving with speed vi in the direction of a target, and that the target is moving towards the radar system with speed V relative to the radar system. (So vi, V and radar signals are collinear.) ------------ ------------ | Radar | ----------------c-vi------------>|Target | | System |<---------------c+vi------------ | | ------------- signal ----------- ---vi--> speeds <--V-- Radar system relative Target speed speed relative to radar relative to to isotropic system radar system frame To ensure consistent representation of distance and time, let time in the radar system frame be synchronized with time in the isotropic frame and let distance measurements be derived from the spatial coordinates of the isotropic frame. These conditions require the frames to be mapped to each other via Galilean transforms. Then in the frame of the radar system the transmitted signal will have a speed of c-vi and the reflected signal will have a speed of c+vi. Notes: 1) It would be usual to synchronize clocks in the frame of the radar system such that signal speeds in this frame APPEAR to be c in both directions. This would be consistent with mapping the frames to each other via Lorentz transforms. But doing so here would cause representation of signal propagation timing in the radar system frame to differ from that in the isotropic frame. That would prevent derivation of correct results. 2) The above speeds will give a calculated two way speed in the frame of the radar system of c (1 - vi^2/c^2) using the measures of length and time of the isotropic frame. This would convert to a value of c if converted to the measures of length and time that would apply in the radar system frame, if the mapping was via Lorentz transforms. So the way the Galilean transform is being used here does not conflict with observed constancy of the two way speed of light in inertial frames. In the radar system frame of reference, let the transmitted signal have frequency Ft, then the corresponding outgoing wavelength is, Lt = (c - vi)/Ft This signal will impinge on the target with period T = Lt/(c - vi + V) or frequency F = (c - vi +V )/Lt. The reflected signal has the same frequency, and so has wavelength Lr = (c + vi - V)/F, and is received by the radar system with frequency Fr = (c + vi)/Lr. Then overall we obtain, (c + vi) (c - vi + V) Fr = --------------- ---------------- Ft . (2) (c + vi - V) ( c - vi) This formula was derived using the measures of length and time of the isotropic frame, but since changing the measures at this point would change the denominators and numerators in equal proportion, such a change would not affect the ratio of Fr to Ft. So the formula can also be validly used with the measures that would normally apply in the radar system frame, that is, if the mapping between it and the isotropic frame had been performed with Lorentz transforms. When vi is zero the formula reduces to, Fr = (c+ V)/(c - V) Ft which is equivalent to the standard formula (1) above. But if the one way speed of light is truely isotropic only wrt to a preferred frame, then in general vi will not be zero and the standard formula will give inaccurate results. -- Surfer
From: artful on 20 Feb 2010 22:16 On Feb 21, 2:08 pm, Surfer <n...(a)spam.net> wrote: > The formula for radar Doppler shift can be derived without invoking > Einsteins' theory of special relativity, because all observations are > made in the same frame of reference. No .. because for there to be a shift it has to be adifference between what one frame measures to what another measures. You can't have a shift when there is just one value > The result with c as the speed of light, V as the target velocity and > Ft as the transmitted frequency, gives the shifted frequency Fr as: > > Fr = Ft (c+V)/(c-V) (1) Which is not what we find experimentally > However I will show below that this formula would be inaccurate if the > one way speed of light were to be isotropic only with respect to a > preferred frame and justify an alternative formula. There is no preferred frame > Suppose a frame of reference is identified in which the one way speed > of light is 'truely' isotropic, referred to below as the 'isotropic > frame'. That is EVERY and ANY inertial frame > Suppose that relative to such a frame a radar system is moving with > speed vi in the direction of a target, and that the target is moving > towards the radar system with speed V relative to the radar system. > (So vi, V and radar signals are collinear.) > > ------------ ------------ > | Radar | ----------------c-vi------------>|Target | > | System |<---------------c+vi------------ | | > ------------- signal ----------- > ---vi--> speeds <--V-- > Radar system relative Target speed > speed relative to radar relative to > to isotropic system radar system > frame > > To ensure consistent representation of distance and time, let time in > the radar system frame be synchronized with time in the isotropic > frame In other words: "cheat" by not synchronising in the frame where you are taking measurements > and let distance measurements be derived from the spatial > coordinates of the isotropic frame. In other words: "cheat" by not synchronising in the frame where you are taking measurements > These conditions require the > frames to be mapped to each other via Galilean transforms. Which we know experimentally are only an approximation at v << c > Then in the frame of the radar system the transmitted signal will have > a speed of > c-vi Except we know experimentally that light speed is NOT dependent on the source speed > and the reflected signal will have a speed of > c+vi. Except we know experimentally that light speed is NOT dependent on the source speed > Notes: > 1) It would be usual to synchronize clocks in the frame of the > radar system such that signal speeds in this frame APPEAR > to be c in both directions. Because it IS c > This would be consistent > with mapping the frames to each other via Lorentz transforms. > But doing so here would cause representation of signal > propagation timing in the radar system frame to differ from > that in the isotropic frame. That would prevent derivation of > correct results. No .. what you are doing is cheating [snip rest unread]
From: Androcles on 20 Feb 2010 22:33 "Surfer" <no(a)spam.net> wrote in message news:7s81o5d6d521r6ps5fda2kgvt4gfh7aobk(a)4ax.com... > The formula for radar Doppler shift can be derived without invoking > Einsteins' theory of special relativity, because all observations are > made in the same frame of reference. > > The result with c as the speed of light, V as the target velocity and > Ft as the transmitted frequency, gives the shifted frequency Fr as: > > Fr = Ft (c+V)/(c-V) (1) > Incorrect, even for sound, which would be Fr = Ft (c+Vt)/(c-Vr) where c is the velocity of sound relative to the medium (air), Vt is the velocity of the source relative to the medium and Vr is the velocity of the receiver relative to the medium. All velocities are relative. Since there is no medium for light, Doppler's equation reduces to Fr = Ft (c+V)/c > However I will show below that this formula would be inaccurate if the > one way speed of light were to be isotropic only with respect to a > preferred frame and justify an alternative formula. > > Suppose a frame of reference is identified in which the one way speed > of light is 'truely' isotropic, referred to below as the 'isotropic > frame'. > > Suppose that relative to such a frame a radar system is moving with > speed vi in the direction of a target, and that the target is moving > towards the radar system with speed V relative to the radar system. > (So vi, V and radar signals are collinear.) > > ------------ ------------ > | Radar | ----------------c-vi------------>|Target | > | System |<---------------c+vi------------ | | > ------------- signal ----------- ------------ ------------ | Radar | --------------- (-c)------------>|Target | | System |<---------------c+vi------------ | | ------------- signal ----------- There is no vi from the radar system. > ---vi--> speeds <--V-- > Radar system relative Target speed > speed relative to radar relative to > to isotropic system radar system > frame > > > To ensure consistent representation of distance and time, let time in > the radar system frame be synchronized with time in the isotropic > frame and let distance measurements be derived from the spatial > coordinates of the isotropic frame. These conditions require the > frames to be mapped to each other via Galilean transforms. > > Then in the frame of the radar system the transmitted signal will have > a speed of > c-vi No, it has a speed of -c. If another car is passing the first with velocity Vj that cannot affect the speed of the signal from the radar system. > and the reflected signal will have a speed of > c+vi. Yes, that is correct. The two return signals from two cars travel at c+vi and c+vj.
From: Surfer on 20 Feb 2010 23:39 On Sat, 20 Feb 2010 19:16:42 -0800 (PST), artful <artful_me(a)hotmail.com> wrote: >On Feb 21, 2:08�pm, Surfer <n...(a)spam.net> wrote: >> The formula for radar Doppler shift can be derived without invoking >> Einsteins' theory of special relativity, because all observations are >> made in the same frame of reference. > >No .. because for there to be a shift it has to be adifference between >what one frame measures to what another measures. You can't have a >shift when there is just one value > >> The result with c as the speed of light, V as the target velocity and >> Ft as the transmitted frequency, gives the shifted frequency Fr as: >> >> � � � � Fr = Ft �(c+V)/(c-V) � � � � � � � � � � � � � � � � � � � � (1) > >Which is not what we find experimentally > Sorry, you are wrong. The above formula works well enough for normal situations. It's described here, http://en.wikipedia.org/wiki/Doppler_radar but is presented in the form Fr = Ft �(1+V/c)/(1-V/c)
From: Dono. on 21 Feb 2010 00:16
On Feb 20, 7:08 pm, Surfer <n...(a)spam.net> wrote: > The formula for radar Doppler shift can be derived without invoking > Einsteins' theory of special relativity, because all observations are > made in the same frame of reference. I don't know what gave you this idiotic idea. This is a simple exercise in SR: 1. Since the object is approaching at V, the frequency hitting the incoming object is F_obj=F_source*sqrt((1+V/c)/(1-V/c)) 2. The object acts as asecondary source approaching the radar detecot at V, so, the frequence at the radar detecort is: F_radar=F_obj*sqrt((1+V/c)/(1-V/c)) So, according to SR: F_radar=F_source*(1+V/c)/(1-V/c) Einstein rules. You suck. PS: a simple exercise dor you, what happens if the object is moving away from the taerget. Can you find the formula? (hint: you need to use SR). |