Why do we have red shift and blue shift, and not only one?

AntigenX

I hardly understand anything about relativity (as would be evident from my threads and posts), and thus, this dumb question.

If there is only length contraction, and not length expansion, why do we have two shifts, namely redshift and blue shift? should not only there be only blue shift due to length contraction? Red shift would imply length expansion, I suppose!

Doc Al

Before you start another wild goose chase (), do you understand the ordinary Doppler shift for sound? After all, there's no length contraction or time dilation at all there, yet there are "two" shifts.

Study this: Doppler Effect

AntigenX

Served me better .

My guess was (in this case), that the red/blue shift was due to apparent (wave)length contraction/expansion when an observer is in relative motion with the source of light.

DaleSpam

The relativistic Doppler is nothing more than the classical Doppler (e.g. for sound waves) with time dilation included. The time dilation part only goes one way (red) but the classical Doppler goes both ways (red and blue). The reason that you still get red and blue shifts overall is that the classical Doppler effect is HUGE compared to the time dilation effect, even for relativistic velocities.

calis

Nice this have been broaght up...

I agree, that light shifting is not a case of relativity whatsoever... but what I cannot get... If a visible light is just a small part of all wavelength... then how come that ultraviolet rays spred by an apparting star doesn't become visible as a violet light (after all ultraviolet ray frequency also get longer and reaches the visible light value...??? and why red light doesnot go out of the visible light area and doesn't became a FM radio wave... red whift shouldnt accure at all. just we see an apparting star in different colour

DaleSpam

This does happen. In fact, the redshift is the way that the velocity of distant stars or galaxies is measured.

gggarnet

When mosquito flies near you,you can hear sharp voice and when mosquito flies away,the sound of mosquito becomes not sharp as before.

calis

To DaleSpam
you helped me on the other thread. I hope you will help on this one too

you dind't get me right.
I know about the dopler effect - actually it is called hubble law - that by determining red shift we measure the speed of departuring thus we can measure current distance

what concernes me is that red shift shouldnt exist

since the visible light ends with violet (high freqency) and red(low frequency) than I undersnatd... if a low frequency light becomes lower it gets out of the visible area.... and if a too high frequence gets a little lower than it enters the visible light area.

So I understand it should be balanced equaly

Doc Al

If the source of "light" were completely featureless, then you wouldn't be able to observe any shift. But it's not. To measure a shift, one looks for specific spectral features (for example, hydrogen absorption lines) and measures the shift in wavelength of those features. This might help: Wiki: Redshift

DaleSpam

I don't understand your comments. If the frequency of the electromagnetic waves goes down it is called redshift regardless of whether or not it goes in or out of the visible range.

yuiop

Hi Calis,

Doc Al is right. All elements give off a unique signiture spectrum that is like a bar code. You can see the emission spectrums for iron and hydrogen here http://en.wikipedia.org/wiki/Emission_spectrum

It is easy to see that the signiture for iron is very distinct from hydrogen. Most stars are mostly hydrogen (at the surface anyway) and you would expect to see the barcode for hydrogen. If the basic signiture code for hydrogen is there, but shifted to the left or the right, you can calculate the degree of redshift. This is probably a gross over simplification, but I hope you get the general idea.

dancing queen

hi,

where can i find a list of the stars
--------------------------------
1-shifting to red (getting away from us)
2-shifting to blue (getting closer to us)

I'm particularly interested in
--------------------------
1-Orion constellation
2-Pleiades's star system

Passionflower

I am not sure how you wish to support this statement DaleSpam.

The ratio between the relativistic Doppler shift of an approaching and retreating object with the same speed is exactly 1. So how could you argue that the relativistic Doppler shift includes a time dilation part that only goes one way but is small compared to the classical Doppler shift?

That relativistic time dilation plays no role in the relativistic Doppler factor is clear after we consider the relationship between the square of the relativistic Doppler factor and the change in radar roundtrip time for increasing or decreasing distances, the clock speed of the ship containing the mirror plays absolutely no role in this.

DaleSpam

The relativistic Doppler shift formula is:

[tex]\frac{f_r}{f_e}=\sqrt{\frac{c-v}{c+v}}=\frac{c}{c+v} \sqrt{1-\frac{v^2}{c^2}}=\frac{k}{\gamma }[/tex]

Where γ is the relativistic time dilation factor and k is the classical Doppler shift. In general γ is much closer to 1 than k is, especially for v<<c.

JesseM

Suppose something is moving away from you at 0.6c, and sending out light signals once every 20 seconds according to its own clock. Then if there is no time dilation, then it sends out signals once every 20 seconds in your frame too, and during this time it has moved a distance of 0.6*20 = 12 light-seconds further away from you, so if we still assume the light signals travel at c in your frame, you'll only receive signals from the object once every 20+12=32 seconds. On the other hand, if it's moving towards you at 0.6c and sending out signals once every 20 seconds with no time dilation, then it will be 12 light-seconds closer to you each time it sends a signal, so the you'll receive signals from the object once every 20-12=8 seconds. And 1/32 and 1/8 are the numbers you get if you use the classical Doppler shift formula [tex]f = \frac{v + v_r}{v + v_s} f_0[/tex], assuming that the receiver is in the frame where light waves move at c (the rest frame of the luminiferous aether) so v_r=0, the velocity of waves is v=c, the frequency f₀ at which the waves are being sent is 1/20, and the velocity of the source is v_s=-0.6c or v_s=0.6c.

Now see what happens if we include relativistic time dilation. In this case, if the object is moving away at 0.6c and sending out signals once every 20 seconds according to its own clock, then its clock is slowed down by a factor of 1.25 in your frame, so in your frame it only sends out signals once every 20*1.25 = 25 seconds. So, if it's moving at 0.6c it will have moved further away from you by a distance of 25*0.6 = 15 light-seconds each time it sends a signal. So, you will only receive signals once every 25+15=40 seconds. On the other hand, if it's moving towards you at 0.6c, it'll be 15 light-seconds closer each time it sends a signal, so you'll receive a signal once every 25-15=10 seconds. 1/40 and 1/10 are the numbers you get if you use the relativistic Doppler shift formula [tex]f_{received} = f_{source}\sqrt{\frac{1 + v/c}{1 - v/c}}[/tex] with v=-0.6c or v=0.6c respectively, with f_source = 1/20. So, you can see that time dilation does figure into the relativistic Doppler shift formula, and that the relativistic blueshift of 1/10 is equal to 8/10 times the non-relativistic blueshift of 1/8, and likewise the relativistic redshift of 1/40 is equal to 8/10 times the non-relativistic redshift of 1/32.

Passionflower

We all seem to agree on the formula, how to calculate the effect and on the results.

[tex]f_{received} = f_{source}\sqrt{\frac{1 + v}{1 - v}}[/tex]

Approaching and retreating objects with speed 0.6 get a resp. shift factor of 2 and .5

Completely symmetrical!

Then I do not see how one can sensibly argue this case:

"The time dilation part only goes one way (red) but the classical Doppler goes both ways (red and blue). The reason that you still get red and blue shifts overall is that the classical Doppler effect is HUGE compared to the time dilation effect, even for relativistic velocities."

There is absolutely nothing in the formula that refers to time dilation!

DaleSpam

Right, it is only when you express it as I did in post 14 that you can separate it out into a "time dilation" part and a "classical Doppler" part. The time dilation part is a function of v², so it is always red, but the classical part is a function of v, so it can be red or blue depending on the sign of v.