What is the relationship between classical and relativistic Doppler formulas?

[ghwellsjr]

Classical Doppler coeficient $$1-\beta$$ multiplied by $$\gamma$$ becomes relativistic Doppler $$\sqrt{\frac{1-\beta}{1+\beta}}$$. Is this independent of the second postulate?

Where did you get that formulation for the classical Doppler coefficient? It has only one speed in it. It needs to have two, one for the source relative to the medium and one for the receiver relative to the medium. Here's the formula from wikipedia:

 

[sweet springs]

Any receiver can regard that the medium of light rests with him. Put $$v_r=0$$ for IFR of the receiver. It is beyond a pure Newtonian view, I admit. Further, relativistic time dilation factor $$\gamma$$ should be multiplied. I emphasized this factor.

 

[ghwellsjr]

Any receiver can regard that the medium of light rests with him.

If that is true, then isn't it also true that any receiver can regard that the medium of light rests with the source? Or that the medium of light is traveling at any speed with regard to the receiver?

Put $$v_r=0$$ for IFR of the receiver.

If I do that, I don't get your Classical Doppler coefficient that you posted in #30.

It is beyond a pure Newtonian view, I admit.

Then why are you pursuing this?

Further, relativistic time dilation factor $$\gamma$$ should be multiplied.

Why? Where is it declared that you should multiply by gamma instead of dividing or doing something else?

I emphasized this factor.

Did you come up with this argument on your own or can you provide a reference for it?

If you want to see how Special Relativity calculates the Relativistic Doppler factor, you can look in Einstein's 1905 paper in section 7:

http://www.fourmilab.ch/etexts/einstein/specrel/www/

I think you are misunderstanding what I am saying. Please go back and read my post #12 and see if there is anything in it that you disagree with or don't understand.

 

[sweet springs]

Thanks. I should be more careful.

Following the classical way obeserver at rest in media observe light from the source moving with speed beta (positive for leaving) as

$$\nu\ at\ rest=\frac{1}{1+\beta}\nu_0\ in\ motion$$

Observer moving in media observe light from the source at rest is

$$\nu\ in \ motion=(1-\beta)\nu_0\ at\ rest$$

In order these two coincide

$$\nu_0\ in\ motion=\sqrt{1-\beta^2}\nu_0\ at\ rest$$
$$\nu\ in\ motion=\sqrt{1-\beta^2}\nu\ at\ rest$$

Now we know there is no media. Choose one, at rest or in mortion, and delete the other.

$$\nu=\sqrt{\frac{1-\beta}{1+\beta}}\nu_0$$

Improved, I hope.

 

[sweet springs]

PS FIrst order approximation for both the formula is $$\nu=(1-\beta)\nu_0$$. I confused it be classical Doppler effect of light.

 

[sweet springs]

PS2 Formula $$\nu \ in \ motion = \sqrt{1-\beta^2}\ \nu\ at\ rest$$ was derived as for motion against light media, however SR allow us to interprete it apply for any relative motion.

 

[ghwellsjr]

I'm sorry, I can't follow your logic. If you look in the wikipedia article on classical Doppler or in Einstein's paper, they both define two frequencies, one emitted at the source and one measured at the receiver. That's what Doppler is. Each equation establishes a relationship between these two different frequencies under different well-defined conditions of motion. Some of your equations include both frequencies but others include just one on both sides of the equation which doesn't make any sense. Not only that, but you aren't defining your terms or your conditions and you leave out steps so I don't know what you are doing. It would also help if you would explain what your purpose is in going through this exercise. It seems like you are trying to derive the Relativistic Doppler formula starting with the Classical Doppler formula by applying some logic but I'm not sure. It really helps to explain all the details.