I How can I identify the appropriate spectral line for Doppler calculation?

[jagadeeshr]

Hi,

I was looking into Doppler shift calculations, and I came across this blog post. It gives a very simple and straight forward account of calculating the velocity of a star.

Following is the summary the blog post:

Spectrum of Alpha Centauri (A. Cen.) is obtained through Lhires III spectroscope. Spectrum of the sky/sun is used as a reference for calculations since A. Cen. and Sun are G2V type stars.

When the A. Cen.'s spectrum is superimposed on the Sun's spectrum, A. Cen's lines have shifted to the left (blue). Shift is 4 pixels.

From the spectrum, two Iron (Fe) lines are identified at 5371.5 Å and 5424.1 Å. They are 52.58 Å apart, and number of pixels between them is 258 pixels. Therefore 1 pixel = 0.2 Å.

Shift between A. Cen. and Sun is is 4 pixels. Therefore, shift in terms of wavelength is 0.8 Å (-0.8 Å, because of blue shift).

Radial Velocity is calculated using the formula: V = C * (Δλ / λ). Where: C ( speed of light) is 3*10^5 km/s, Δλ is 0.8 Å and λ is 5424.1 Å. The equation gives velocity as -44.25 km/s. Taking into account Earth's heliocentricity of 20 km/s, the final velocity of A. Cen. is -44.25 + 20 = -24.25 km/s. This is very close to the astronomical database value of -22.3 km/s.

Here is what I'm confused about: If we use the other Fe line (at 5371.5 Å) as the rest wavelength, the velocity will be -44.68 + 20 = 24.68 km/s (0.4 km/s increase). If we consider spectral lines towards the left (blue region) of the spectrum, the velocity will be higher.

For example, consider an object that emits lines at 4000 Å, 5500 Å and 7000 Å. The shift in wavelength due to Doppler is 1 Å. The Doppler calculations will indicate velocity of 75 km/s at 4000 Å; 54.54 km/s at 5500 Å and 42.85 km/s at 7000 Å.

Therefore, how to identify the appropriate spectral line for Doppler calculation?

 

[BvU]

Hello there,

Do I understand correctly that you conclude a velocity of 24.25 km/s from a shift of 4 pixels ? What is your estimate for the accuracy of this calculated velocity ?

Re your last paragraph: what is the formula for doppler shift as a function of velocity ?

 

[jagadeeshr]

Hello there,

Do I understand correctly that you conclude a velocity of 24.25 km/s from a shift of 4 pixels ? What is your estimate for the accuracy of this calculated velocity ?

Re your last paragraph: what is the formula for doppler shift as a function of velocity ?

Hello there,

Do I understand correctly that you conclude a velocity of 24.25 km/s from a shift of 4 pixels ? What is your estimate for the accuracy of this calculated velocity ?

Re your last paragraph: what is the formula for doppler shift as a function of velocity ?

24.25 km/s from a shift of 4 pixels in that particular spectral image. 1 pixel is 0.2 Å, shift is 4 pixels or 0.8 Å.

3*10^5*(-0.8/5424.1) = -44.25 km/s+ 20 km/s = -24.25 km/s

Formula of Doppler shift as function of velocity: Δλ / λ = V/ C (Calculations)

 

[BvU]

24.25 km/s from a shift of 4 pixels in that particular spectral image. 1 pixel is 0.2 Å, shift is 4 pixels or 0.8 Å.

3*10^5*(-0.8/5424.1) = -44.25 km/s+ 20 km/s = -24.25 km/s

So what if the actual shift is 4.43 pixels ? Or 3.85 ?

Formula of Doppler shift as function of velocity: Δλ / λ = V/ C (Calculations)

So not all frequencies will shift the same 1 Angstrom

 

[DrClaude]

Formula of Doppler shift as function of velocity: Δλ / λ = V/ C (Calculations)

To make it clearer:
$$
\Delta \lambda = \frac{v \lambda}{c}
$$
i.e., the shift is not the same for all wavelengths, since it depends on $\lambda$.

 

[BvU]

So what if the actual shift is 4.43 pixels ? Or 3.85 ?

My point is that from the given 4 pixels (which can mean anything between 3.5 and 4.5 pixels, I must assume), you can not calculate a velocity in 4 digits accuracy. You get $44\pm 5$ km/s. And the same absolute error in the final $24\pm 5$ km/s (provided the 20 km/s correction has a considerably smaller error)

 

[jagadeeshr]

So what if the actual shift is 4.43 pixels ? Or 3.85 ?

So not all frequencies will shift the same 1 Angstrom

To make it clearer:
$$
\Delta \lambda = \frac{v \lambda}{c}
$$
i.e., the shift is not the same for all wavelengths, since it depends on $\lambda$.

Ah. Based on the image in the blog, I had assumed that the shift will be uniform across the entire spectrum.

Thank you.