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The setup assumes a large mass(Earth?) an a photon launched from its surface initially. The wavelength of the photon on launch is known. Then the new energy of the photon is compared with energy it traveled a distance $d$.

Initially, $$E_1 = \frac{GMm}{R} + \frac{1}{2}mv^2$$ where m- mass of photon, M is mass of earth, R is radius. Finally $$E_2 = \frac{GMx}{R+d} + \frac{1}{2}xv^2$$ where $x$ is new mass of the photon. Putting $m = \frac{h}{Lc}$, $L$ is wavelength

Leet the new wavelength be $W$. we can equate the 2: $$\frac{GMh}{RLc} + \frac{1}{2}\left(\frac{h}{L}\right)^2 = \frac{GMh}{(R+d)Wc} + \frac{1}{2}\left(\frac{h}{W}\right)^2$$ Solving and putting $\frac{1}{W} = y$, we'll get,

$$\frac{h^2}{2}y^2 + \frac{GMhy}{(R+d)c} = \frac{GMh}{RcL} + \frac{1}{2}\left(\frac{h}{L}\right)^2$$ So this follows a quadratic equation $ax^2 + bx + c =0$

So that would mean that $y$ would keep decreasing as $d$ increases. Then it would mean that the mass of the photon increases as it rises.

Is this logic correct or am I making a really blockheaded mistake here?

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2 Answers 2

up vote 6 down vote accepted

A photon has zero mass, so you can't write it's kinetic energy as $\frac{1}{2}mc^2$.

You are thinking along the right lines. As the photon moves away from the earth it loses energy. However the loss of energy causes the photon to be red shifted i.e. it moves to a lower frequency. The photon energy is given by:

$$E = h\nu$$

and using your notation:

$$E_1 - E_2 = h\nu_1 - h\nu_2$$

You can use this to calculate the new frequency of the photon.

This frequency change has been measured using the Mossbauer effect. See for the details.

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I thought red shift occurred due to Doppler effect. Or are they both the same? – Likhit Feb 20 '12 at 10:11
The Doppler effect can certainly cause light to be red shifted. That's how we know distant galaxies are moving away from us. However light can also be redshifted by a gravitational field. – John Rennie Feb 20 '12 at 10:15
The effect was first successfully measured by Rebka and Pound, but it has been measured many times since then. The original paper is a nice piece for a upper division or early graduate student to puzzle over: very instructive. – dmckee Feb 20 '12 at 14:45

More correct post-newtonian equation is


where $R_2=R_1+d$ and $E=h\nu\;$, so

$ {E_2\Big(1-\frac{GM}{R_2c^2}\Big)=E_1\Big(1-\frac{GM}{R_1c^2}\Big)}$


$\large {\frac{\Delta\nu}{\nu_1}=\frac{\Delta E}{E_1}=\frac{E_1-E_2}{E_1}=1-\frac{E_2}{E_1}=1- \frac{\Big(1-\frac{GM}{R_1c^2}\Big)}{\Big(1-\frac{GM}{R_2c^2}\Big)}}=…=\frac{GM(R_2-R_1)}{ R_1(R_2c^2-GM)}$

Finally for $R_2\;>>\;R_S=\frac{GM}{c^2}\;$ and $\;d\;<<\;R_i:$

$\large {\frac{\Delta\nu}{\nu_1}\approx \frac{GM(R_2-R_1)}{ R_1R_2c^2} \approx \frac{\mathbb{g} d}{c^2} }$

where $\mathbf{g}= \frac{GM}{R_1^2}$

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