lzkelley said:
Matter most certainly can be doppler shifted; but its much less exciting than you would think.
According to debroglie's equation, momentum and frequency are directly related (the equation says wavelength; but for our purposes i think frequency is easier to think about). If the electron is shot out, when the gun is moving towards us, it seems to be higher momentum than it is from the stationary gun frame.
This is exactly the same as throwing a baseball from a moving car - the speed of the car will add to the speed of the baseball from a stationary observer.
The doppler effect allows the debroglie relation to hold in front of relativity.
There are some issues for Doppler Shift on matter waves.
The wavelength is given by :-
\lambda = \frac{h}{p} where p is the momentum.
The phase velocity of the wave is:-
v_p = \frac{C^2}{v} where v is the velocity of the particle.
The wave is super luminal.
There are more fun to it.
What is exactly the source of this wave?
Assuming the particle to be the source, the wavelength measured should already be Doppler Corrected, as the source is moving with a velocity *v*.
There are loads of other fun, namely :-
\lambda_{compton} = \frac{h}{m_0C}
<br />
\lambda_{DeBroglie} = \frac{h}{mv} = \frac{h}{m_0 v \gamma } = \frac{h}{m_0 \beta C \gamma }<br />
where
<br />
\beta = \frac{v}{C}<br />
Hence,
<br />
\lambda_{DeBroglie} = \frac{ \lambda_{compton}}{\gamma \beta} <br />
Assuming that the Compton effect was done in a moving reference frame, [that is the beauty]
<br />
\lambda_{compton} = \frac{h}{m_0 \gamma C}<br />
And, hence
<br />
\lambda_{DeBroglie} = \frac{ \lambda_{compton}}{ \beta} <br />
Now, why there is a similarity here?
I am working on this for past 20 days, and I am overwhelmed by the amount of things taken to be granted about matter waves.
Some are linked here:-
http://www.springerlink.com/index/G862M2015L774784.pdf
http://linkinghub.elsevier.com/retrieve/pii/0375960184902913
I personally see, there is some issue with interpretation of matter waves.
In fact, that might unravel the idea why we need a complex probability amplitude for quantum systems.
Why I am saying this?
Consider this:-
<br />
E^2 = p^2C^2 + {m_0}^2C^4<br />
Assume this energy is a Complex number.
Check that , the last term is Compton effect. I earlier tend to go with the pC term for De Broglie wavelength, but I was wrong.
You can rewrite
E_{complex} = J pC + m_0C^2
J is the complex number `i' . Why I am assigning J to pC ? Because that is dependent on motion of a particle, which can be frozen out. [May be I am dead wrong].
Any thoughts?
