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I've found this website (click here for the link), the link leads you to a problem about the velocity Andromeda is going towards the Milky way galaxy, and you can try three approaches. I like to challenge myself, so I click on the second one, about spectra. I already know what the Electromagnetic Spectrum is, but this one is a tad different. It tells you about the redshift equation, which I already know about, and I've used it a few times in examples I made for myself, but I am confused about the next page (to get there, click the button at the bottom that says "Use This Technique to Solve for M31's velocity"), and the next page is worded in a way that I can't really understand it, and using the clues I get from the Spectrum Graph, I can't figure out what the answer is. What is Relative intensity of light, and what does the Calcium have to do with it? Also, isn't their value for the speed of light inaccurate, as far as I know, it's $3\times 10^8$, not $3\times 10^5$. Even further on, they want you to calculate the radial velocity from the graph, which I have no idea how to do.

If an answer about this here is acceptable, please give me a feasibly worded explanation of all this madness, and thanks in advance!

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  • $\begingroup$ The speed of light is correct if the units are km/sec -- very normal for astronomy -- rather than m/sec (as you are assuming). $\endgroup$
    – Peter Erwin
    Commented Apr 30, 2021 at 14:51
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    $\begingroup$ Okay, thanks @PeterErwin, I already found that out from the answer. $\endgroup$
    – Questioner
    Commented Apr 30, 2021 at 14:52
  • $\begingroup$ Also, if you think my other post on Aviation Stack Exchange is worthy of reopening by the community (I've just made some edits to it), please vote to undelete it; here it is, if any of you are interested: Aviation, What are its Elements? $\endgroup$
    – Questioner
    Commented Apr 30, 2021 at 14:57

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Okay, well, calcium is just an easily recognizable spectral feature, so I guess that's why they used it. The y-axis shows the relative intensity. I am assuming this is a spectrum that is normalized or scaled down so that the highest value of the intensity is 1, but this is not so important. The laboratory wavelengths of the absorption lines are shown in red in the graph. If you look carefully, you can see that the minimum of each spectral line does not exactly line up with these red tick marks. That is because the lines are Doppler shifted. M31 is moving and thus the light that we see has a different color (or frequency) from what we would observe in the lab. Let's focus on the left absorption line. The minimum is a bit to the left of the red mark. I conceive this to be approximately 4 angstrom off. The wavelength that you measure here on Earth will thus be 3630 angstrom and not 3934. If I take the formula from the previous page and rewrite I get

$$\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}$$

where $\Delta \lambda$ is the difference between the observed and the lab wavelength of the spectral line. In this case $\Delta \lambda = -4$ angstrom and $\lambda_0 = 3934$ angstrom. The speed of light is given in km/s, not in m/s, so the value is indeed correct. Now you can get the speed of M31 from $$v = c\frac{\Delta \lambda}{\lambda_0} = 3 \times 10^5 \frac{4}{3934} = -305km/s$$ I entered this value and it was accepted.

enter image description here

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  • $\begingroup$ What is in the image? $\endgroup$
    – Questioner
    Commented Apr 30, 2021 at 14:03
  • $\begingroup$ By the way, why are we considering the wavelength of Calcium K, and not Calcium H? $\endgroup$
    – Questioner
    Commented Apr 30, 2021 at 14:09
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    $\begingroup$ @Questioner the image I just took from the link you provided, to give other people an idea what the question is about. You can do the same with Calcium H, it should give you the same answer, but I found it harder to guess the wavelength difference from that line. $\endgroup$
    – theWrongAlice
    Commented Apr 30, 2021 at 14:16
  • $\begingroup$ Okay. Thanks a lot @theWrongAlice, +1 on your post(s), I added a vote to your comment, too. $\endgroup$
    – Questioner
    Commented Apr 30, 2021 at 14:22

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