Spectrum Shift

[YT2095]

a Bit like Red-shift but wider.

 

a space craft is on a heading to pass in front of you from the left to the right mega fast.

as it approaches it will be shifted to the blue/violet, when directly in front of you it will be it`s real color and as it passes it will shift to the Red end of the spectrum.

at least that`s how I understand it.

 

so will you get all the colors of the "rainbow" from UV to IR, or does it snap from one to the other the second it passes you?

[Klaynos]

I can't remember the maths to work it out, and it depends on the velocity of the ship.

 

But I doubt you will get all the spectrum. When we talk about "red shift" we mean that it is more 'red' than it would other wise be, not that it is actually a true red colour. It is just a way of noting which way the frequency has shifted.

[YT2095]

well it`s all to do with the rate that the peaks and troughs of the waveform hit your eye.

 

and yes it can be true Red if moving fast enough, unlike Sound waves where if you go too fast you get a sonic boom, I`ve used Light for this instead to preclude that annoyance.

 

also, forget it`s a space ship and those limitations as well, it`s a Lightbulb moving at half c (or near c if you like, I don`t care).

[Klaynos]

You just need to find a medium with a higher speed of sound

 

Anyway, if we pick a light bulb emitting only one wavelength, 550nm which is near the middle of visible spectrum.

 

[math]\Delta f = \frac {v}{\lambda}[/math]

 

so we'd want a change of 150nm

 

and I've gtg now, if no one else finishes this bbl...

[YT2095]

I hope they don`t finish it, I would the reply in PLAIN SPEAK please

[Gypsy Cake]

What are you trying to work out? What colours you get.

[Klaynos]

Gypsy Cake how fast it'd have to be going....

 

anyway where was I...

 

a green light source (as this is the middle of the spectrum so easiest to get all the colours from) we want it's approch to be violet. So that's 150nm or 2000000GHz (10^9Hz)

 

[math]

\Delta f * \lambda= v

[/math]

 

[math]

2000000 * 10^9 * 550 * 10^{-9}= v = 1.1*10^9m/s

[/math]

 

Which I think is greater than c, so, in summary no you can't get all of the visible spectrum to be shifted passed you in one go.

 

Although I think I recall something about a relativistic doppler shift.... let me look into that....

[Klaynos]

Ah yes, here we go (http://en.wikipedia.org/wiki/Relativistic_Doppler_effect):

 

[math]f_o = \frac{1}{t_o} = \gamma (1-v/c) f_e = \sqrt{\frac{1-v/c}{1+v/c}}\,f_e [/math]

 

Some rearangement:

 

[math] c \frac {\frac {f_0^2} { f_e^2} - 1} {1 - \frac {f_0^2} { f_e^2}} = v

[/math]

 

f_e = c / 400nm

f₀ = c / 550nm

 

Which actually seems to give c, how interesting, so again, no but the close to c you get the closer to being able to sweet 300nm you get....

[swansont]

Ah yes, here we go (http://en.wikipedia.org/wiki/Relativistic_Doppler_effect):

 

[math]f_o = \frac{1}{t_o} = \gamma (1-v/c) f_e = \sqrt{\frac{1-v/c}{1+v/c}}\,f_e [/math]

 

Some rearangement:

 

[math] c \frac {\frac {f_0^2} { f_e^2} - 1} {1 - \frac {f_0^2} { f_e^2}} = v

[/math]

 

f_e = c / 400nm

f₀ = c / 550nm

 

Which actually seems to give c, how interesting, so again, no but the close to c you get the closer to being able to sweet 300nm you get....

 

I think you have a sign error. I believe it should be

 

[math] c \frac {\frac {f_0^2} { f_e^2} - 1} {1 + \frac {f_0^2} { f_e^2}} = v

[/math]

 

 

If you want to double (or halve) the frequency, go 3/5 c

[Klaynos]

I thought my result was a bit strange but when I put in different (similar) numbers I didn't get c out so ignored it.

 

Think I found my error

 

 

[math]

\frac {f_0^2} { f_e^2} + \frac {f_0^2} { f_e^2} \frac {v}{c} = 1 - \frac {v}{c}

[/math]

 

got the sign on the rhs wrong I think

 

[math]

\frac {f_0^2} { f_e^2} - 1 = - \frac {f_0^2} { f_e^2} \frac {v}{c} - \frac {v}{c}

[/math]

 

[math]

\frac {f_0^2} { f_e^2} - 1 = - ( \frac {f_0^2} { f_e^2} + 1 ) \frac {v}{c}

[/math]

 

 

[math]

\frac {\frac {f_0^2} { f_e^2} - 1} {1 + \frac {f_0^2} { f_e^2}} = -\frac {v}{c}

[/math]

 

 

[math]

c \frac {\frac {f_0^2} { f_e^2} - 1} {1 + \frac {f_0^2} { f_e^2}} = -v

[/math]

 

and v +/- is just a definition issue... Unless I've made another sign error, this is why equation rearranging when tired is not a good idea....

[YT2095]

so it does snap from the violet to the red then, with no in-between?

 

so this Prismatic/rainbow effect you see in Star Trek when they go to "warp" whatever, wouldn`t happen in reality?

[Klaynos]

It would scan, but the time taken to scan would depend on how quickly it's velocity changed relative to you. So the closer it passes you the faster the scan...

[swansont]

It would scan, but the time taken to scan would depend on how quickly it's velocity changed relative to you. So the closer it passes you the faster the scan...

 

Right. If the source is passing by at some distance, rather than coming right at you, there's going to be a cosine(theta) term in there somewhere, and that angle will change in time.