关于光子的 Doppler Shift

作者: optic_css | 来源:发表于2019-01-07 23:52 被阅读0次

关于光子的 Doppler Shift
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A breif introduction to the RDS

__with some subtle method

Let's assume that there are two frames (G1, G2, they respect the observer and the light source respectively) related with the velocity u, ie. Frame G1 is moving with the velocity V to frame G2. All below is at G2's view.

The relativistic particle $m_{0}$ with the velocity as $\vec{V} [ v(x), v(y), v(z) ]$ respect to frame G1 thus have energy of relativisty as

$E_{r}= \frac{E_{r0}}{\sqrt{1-(\frac{v}{c} )^2} }$ ,

from the Lorenz transform of the velocity, the particle have

$v_{r} (i)= \frac{v(i) }{\sqrt{1-\frac{uv} {c^2}}}$ ,

hence $V_{r}=\sqrt{\sum_{i=x,y,z}v_{r }^2(i)}$ , substitude the Lorenz transform into $V_{r}$ .

$1-(\frac{v_{r}}{c})^2=\frac{[1-(\frac{u}{c})^2][1-(\frac{v}{c})^2]}{1-\frac{uv}{c^2} }$ ,

The energy (of course, the gross one),

$E_{r}= \frac{1-\frac{uv}{c^2}}{\sqrt{1-(\frac{u}{c})^2}} = \gamma E(1-\frac{uv}{c^2})$ .

The reason to do this with out any restrict to $v$ and $v_r$ , is to let it fit all the particles, but not only for photon.

from the quantum mechanics, the energy of photon is $E=h\nu$ , besides the velocity of light is $v=v_{r}=c$ (of course for all the frames).

Hence,

$\nu _r=\gamma\nu (1-\frac{uv}{c^2} )$ ,

When the dimentions go up to 3, have that

$\nu _r=\gamma\nu (1-\frac{u cos\theta }{c} )$ .

The equation above is what is called the "Relativistic Doppler Shift".

You can form your own figure like below from this equation, and it is widely used in cosmo_......

doppler shift.

Referance: "Baidu.baike" , "Wikipedia_Relativity Theory".

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