- #1
epsilon
- 29
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I have been studying the "fallen" photon experiment, in which the frequency of a photon changes as it falls through a height H.
[itex]f'=f_0(1+\frac{gH}{c^2})[/itex]
It is often stated that this is a huge factor in the application of GPS. However, I do not understand why.
I understand that the photon will arrive at Earth with a different frequency, however I do not see why the time taken should vary - it will simply arrive in a different part of the electromagnetic spectrum.
The distance is not changing and the photon is traveling at the speed of light too. Surely this also rules out the effects of special relativity, as the time dilation, length contraction and mass increase equations all incorporate the Lorentz factor:
[itex]\gamma=\sqrt[]{1+\frac{v^2}{c^2}}[/itex]
However as [itex]v=c, \gamma=1[/itex] for all 3 of the relativistic effects: [itex]t'=t_0, l'=l_0, m'=m_0[/itex].
So why is the gravitational frequency shift important for GPS?
[itex]f'=f_0(1+\frac{gH}{c^2})[/itex]
It is often stated that this is a huge factor in the application of GPS. However, I do not understand why.
I understand that the photon will arrive at Earth with a different frequency, however I do not see why the time taken should vary - it will simply arrive in a different part of the electromagnetic spectrum.
The distance is not changing and the photon is traveling at the speed of light too. Surely this also rules out the effects of special relativity, as the time dilation, length contraction and mass increase equations all incorporate the Lorentz factor:
[itex]\gamma=\sqrt[]{1+\frac{v^2}{c^2}}[/itex]
However as [itex]v=c, \gamma=1[/itex] for all 3 of the relativistic effects: [itex]t'=t_0, l'=l_0, m'=m_0[/itex].
So why is the gravitational frequency shift important for GPS?