Hi all,I am reading a paper related to astrophysics and I am stuck

[llama123]

Hi all,

I am reading a paper related to astrophysics and I am stuck in a step in a calculation. It is about orbiting gas in a spiral galaxy and it calculates the errors in the fitted rotation velocity if one of the viewing angles is incorrectly estimated. My problem is that I don't understand one of the steps which has to do with either inserting a trigonometric identity or performing a Taylor expansion. The following equations are given:
[itex] x' = \hat{R} \cos(\hat{\psi})[/itex], [itex] y' = (q+\delta q) \hat{R} \sin(\hat{\psi})[/itex] as well as: [itex] \cos(\psi) =\frac{x'}{\sqrt{x'^2 + y'^2/q^2}}[/itex]

Combining these, one should arrive at the following:

[itex]\cos(\psi) = (1-\frac{\delta q}{4q}) \cos(\hat{\psi})+\frac{\delta q}{4q} \cos(3 \hat{\psi})[\itex]

I tried inserting the expressions for x' and y' into the equation for psi and then tried to apply some trigonometric identities but with no luck. In the paper it doesn't say that it is a Taylor approximation, but I can't really exclude this. I get stuck with the following expression:
[itex]\cos(\psi) = \frac{1}{\sqrt{1+(\frac{\delta q}{q} \tan(\hat{\psi}))^2}}[\itex]

Thanks very much!

 

[mathman]

Fix the latex expressions.

 

[llama123]

Sorry for the mistake, now it should be OK.

Hi all,

I am reading a paper related to astrophysics and I am stuck in a step in a calculation. It is about orbiting gas in a spiral galaxy and it calculates the errors in the fitted rotation velocity if one of the viewing angles is incorrectly estimated. My problem is that I don't understand one of the steps which has to do with either inserting a trigonometric identity or performing a Taylor expansion. The following equations are given:
[itex]x′=\hat{R} \cos(\hat{\psi})[/itex], [itex]y′=(q+\delta q) \hat{R} \sin(\hat{\psi}) [/itex] as well as: [itex]\cos(\psi)=\frac{x'}{\sqrt{x′^2+y′^2/q^2}} [/itex]

Combining these, one should arrive at the following:

[itex]cos(\psi) = (1-\frac{\delta q}{4q}) \cos(\hat{\psi})+\frac{\delta q}{4q} \cos(3 \hat{\psi})[/itex]

I tried inserting the expressions for x' and y' into the equation for psi and then tried to apply some trigonometric identities but with no luck. In the paper it doesn't say that it is a Taylor approximation, but I can't really exclude this. I get stuck with the following expression:
[itex]\cos(\psi) = \frac{1}{\sqrt{1+(\frac{\delta q}{q} \tan(\hat{\psi}))^2}}[/itex]

Thanks very much!

 

[mathman]

I get for the coefficient of tan to be 1 + δq/q not δq/q.

 

[llama123]

Thanks for your reply. Yes you are right, excuse me for the error, it was a mistake in copying it from my notes. So now the problem is how to go from this:

[itex]\cos(\psi) = \frac{1}{\sqrt{1+(\frac{q + \delta q}{q} \tan(\hat{\psi}))^2}}[/itex]

To the result:

[itex]\cos(\psi) = (1-\frac{\delta q}{4q}) \cos(\hat{\psi})+\frac{\delta q}{4q} \cos(3 \hat{\psi})[/itex]

My colleague believed that it could be that there is information missing to solve this and that the information is located somewhere in the rest of the text. However, it is a calculation in the appendix and it explicitly says that from x', y', cos([itex]\psi[/itex])=... follows this result...so I believe there is no information missing.

 

[mathman]

[I haven't got the hang of latex, so I'll omit the angle on the right side]

It looks like an expansion in first order in δq/q. The expression under the square root is then:

1 + (1 + 2 δq/q) tan² = {1 + 2 δq/q sin²}/cos²

Multiply numerator and denominator by cos and get:

cos/(1 + 2 δq/q sin²)^1/2

Expand (binomial) the denominator and keep the first term and get:

cos{1 - δq/q(1- cos²)}

This is fairly close to what you have. I suggest you work on it to fill in the details.

 

[llama123]

Wonderful! Many thanks!
It works perfectly, all I needed to use was:
[itex] \cos^3(\phi) = (3\cos(\phi)+\cos(3\phi))/4 [/itex] to finish the job.
Thanks again.