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Pulsar
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the emitter and the receiver move with constant velocities relative to an inertial frame and $v$$V$ is the constant velocity of the receiver relative to the emitter and away from it.

No, both the emitter and the receiver are accelerating, and the receiver has gained an extra velocity $v$$\Delta v$ between the time the photon was emitted and the time it was received.

In other words, consider an emitter and a receiver, a distance $h$ apart, and both accelerating with a constant acceleration $g$, and suppose the emitter is a distance $h$ behind the receiver. Also, suppose that $g$ is low enough so that relativistic effects can be ignored.

If the (trailing) emitter sends out a photon with wavelength $\lambda_0$$\lambda_e$, it reaches the (leading) receiver after a time $t=h/c$$\Delta t\approx h/c$ (ignoring the little extra distance that the photon has to travel because the receiver has accelerated). 

During this time, the receiver has gained an extra velocity $v=gt=gh/c$$\Delta v=g\Delta t \approx gh/c$. SoIf $\Delta v$ is small, the standard Newtonian Doppler effect applies, and the wavelength of the received photon has changed as $$ \frac{\Delta\lambda}{\lambda_0} = \frac{v}{c} = \frac{gh}{c^2}. $$$$ \frac{\Delta\lambda}{\lambda_e} = \frac{\Delta v}{c} \approx \frac{gh}{c^2}, $$ or equivalently, the frequency has changed as $$ \frac{\Delta\nu}{\nu_e} = -\frac{\Delta v}{c} \approx -\frac{gh}{c^2}. $$ According to the EEP, the acceleration of the emitter and the receiver is equivalent to a gravitational field.

the emitter and the receiver move with constant velocities relative to an inertial frame and $v$ is the constant velocity of the receiver relative to the emitter and away from it.

No, both the emitter and the receiver are accelerating, and the receiver has gained an extra velocity $v$ between the time the photon was emitted and the time it was received.

In other words, consider an emitter and a receiver, a distance $h$ apart, and both accelerating with a constant acceleration $g$. If the (trailing) emitter sends out a photon with wavelength $\lambda_0$, it reaches the (leading) receiver after a time $t=h/c$. During this time, the receiver has gained an extra velocity $v=gt=gh/c$. So the standard Doppler effect applies, and the wavelength of the received photon has changed as $$ \frac{\Delta\lambda}{\lambda_0} = \frac{v}{c} = \frac{gh}{c^2}. $$ According to the EEP, the acceleration of the emitter and the receiver is equivalent to a gravitational field.

the emitter and the receiver move with constant velocities relative to an inertial frame and $V$ is the constant velocity of the receiver relative to the emitter and away from it.

No, both the emitter and the receiver are accelerating, and the receiver has gained an extra velocity $\Delta v$ between the time the photon was emitted and the time it was received.

In other words, consider an emitter and a receiver, both accelerating with a constant acceleration $g$, and suppose the emitter is a distance $h$ behind the receiver. Also, suppose that $g$ is low enough so that relativistic effects can be ignored.

If the (trailing) emitter sends out a photon with wavelength $\lambda_e$, it reaches the (leading) receiver after a time $\Delta t\approx h/c$ (ignoring the little extra distance that the photon has to travel because the receiver has accelerated). 

During this time, the receiver has gained an extra velocity $\Delta v=g\Delta t \approx gh/c$. If $\Delta v$ is small, the standard Newtonian Doppler effect applies, and the wavelength of the received photon has changed as $$ \frac{\Delta\lambda}{\lambda_e} = \frac{\Delta v}{c} \approx \frac{gh}{c^2}, $$ or equivalently, the frequency has changed as $$ \frac{\Delta\nu}{\nu_e} = -\frac{\Delta v}{c} \approx -\frac{gh}{c^2}. $$ According to the EEP, the acceleration of the emitter and the receiver is equivalent to a gravitational field.

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Pulsar
  • 14.9k
  • 3
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  • 85

the emitter and the receiver move with constant velocities relative to an inertial frame and $v$ is the constant velocity of the receiver relative to the emitter and away from it.

No, both the emitter and the receiver are accelerating, and the receiver has gained an extra velocity $v$ between the time the photon was emitted and the time it was received.

In other words, consider an emitter and a receiver, a distance $h$ apart, and both accelerating with a constant acceleration $g$. If the (trailing) emitter sends out a photon with wavelength $\lambda_0$, it reaches the (leading) receiver after a time $t=h/c$. During this time, the receiver has gained an extra velocity $v=gt=gh/c$. So the standard Doppler effect applies, and the wavelength of the received photon has changed as $$ \frac{\Delta\lambda}{\lambda_0} = \frac{v}{c} = \frac{gh}{c^2}. $$ According to the EEP, the acceleration of the emitter and the receiver is equivalent to a gravitational field.