Exploring the Relativistic Doppler Shift

In summary, the conversation discusses the equation for relativistic doppler shift and how it reduces to -v/c for v<<c. The conversation includes steps for expanding the equation and dropping higher order terms, as well as different approaches to prove the equation. Ultimately, the conversation concludes that the equation can be simplified to -v/c by dropping second order terms and using the approximation 1/(1+x) ~ 1-x for x close to zero.
  • #1
gnome
1,041
1
relativistic doppler shift

I'm trying to show that this equation for the doppler shift for light:
f' = [&radic;(1+(v/c))/&radic;(1-(v/c))] * f
reduces to
&Delta;f/f = -v/c for v<<c

So I expanded (1+v/c)^(1/2) = 1 + v/(2c)
and (1-v/c)^(1/2) = 1 - v/(2c)
dropping higher order terms on the assumption that they are negligible.
Now I have
f' = [(1+v/(2c))/(1-(v/(2c))] * f
&Delta;f = f' - f = f * [(1+v/(2c))/(1-(v/(2c)) - 1]
&Delta;f/f = (2c + v)/(2c - v)

That doesn't look like -v/c.

On the other hand, if I multiply the original equation by &radic;(1-(v/c))/&radic;(1-(v/c)) I get
f' = [&radic;(1-(v^2/c^2))/(1-(v/c))] * f
and then expand &radic;(1-(v^2/c^2)) = 1-v^2/(2c^2)
then
f' = [(1-v^2/(2c^2))/(1-(v/c))] * f
&Delta;f/f = (1-v^2/(2c^2))/(1-(v/c)) - 1
= (2c^2 - v^2)/(2c^2 - 2cv) - 1
= (2cv - v^2)/(2c^2 - 2cv)
= v(2c - v)/(c(2c - 2v))

Close but no cigar.

Or, can I just claim that (2c-v)/(2c-2v) &sim; 1?

(The minus sign can be attributed to how we define &Delta;f.)
 
Last edited:
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  • #2


Starting here:

Originally posted by gnome

&Delta;f = f' - f = f * [(1+v/(2c))/(1-(v/(2c)) - 1]
&Delta;f/f = (2c + v)/(2c - v)

Can't you say:

[tex] \frac{\Delta f}{f} = \frac{1+\frac{v}{2c}}{1-\frac{v}{2c}} - 1 [/tex]
[tex] \frac{\Delta f}{f} = \frac{2\frac{v}{2c}}{1-\frac{v}{2c}} [/tex]

and since v<<c, 1>>v/2c --> 1-v/2c ~ 1, and then you've proven it.

Or can you expand again at that point, so that: 1/(1+x) ~ 1 - x (for x close to zero), then:


[tex] \frac{\Delta f}{f} = \frac{v}{c}(1-\frac{v}{2c}) [/tex]

drop the second order terms and you've got the answer.

I'm not sure if that's all right, but that'd be my guess; hope it helps.
 
  • #3
I guess that's it. I'm always uncertain as to which terms I can legitimately "drop", but I don't see any other argument that works here.

Thanks, James.
 

1. What is the Relativistic Doppler Shift?

The Relativistic Doppler Shift is a phenomenon in which the frequency and wavelength of electromagnetic waves, such as light, appear to change depending on the relative motion between the source of the waves and the observer.

2. How does the Relativistic Doppler Shift differ from the Classical Doppler Shift?

Unlike the Classical Doppler Shift, which only takes into account the relative velocities of the source and observer, the Relativistic Doppler Shift also considers the effects of time dilation and length contraction due to the high speeds involved in the movement of the source and observer.

3. What is the formula for calculating the Relativistic Doppler Shift?

The formula for calculating the Relativistic Doppler Shift is:
f' = f * √((1 ± v/c) / (1 ∓ v/c)),
where f' is the observed frequency, f is the emitted frequency, v is the relative velocity between the source and observer, and c is the speed of light.

4. Can the Relativistic Doppler Shift be observed in everyday life?

Yes, the Relativistic Doppler Shift can be observed in everyday life, particularly in situations involving high speeds, such as in space travel or when observing celestial objects. It is also used in technologies such as GPS systems and satellite communications.

5. How does the Relativistic Doppler Shift affect the color of light?

The Relativistic Doppler Shift can cause a change in the color of light, as the observed frequency and wavelength of the light may appear different to an observer due to their relative motion. This is known as the relativistic Doppler effect and can result in a shift towards the red or blue end of the electromagnetic spectrum.

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