• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Trigonometric function (Mechanics-Landau)

<Moderator's note: Moved from a technical forum and thus no template.>

Mechanics by Lev D. Landau & E. M. Lifshitz
Chapter 4 Collisions between particles
§16. Disintegration of particles
Problem 3
§16.1.png

The angle θ = θ1 + θ2
It is simplest to calculate the tangent of θ.
A consideration of the extrema of the resulting expression gives the following ranges of θ, depending on the relative magnitudes of V, v10 and v20 (for definiteness, we assume v20 > v10):
0 < θ < π if v10 < V < v20,
π-θ10 < θ < π if V < v10,
0 < θ < θ10 if V > v20.

So I calculated the tangent of θ and derivative of it
캡처.PNG

But I can not calculate the extrema, thus I can not solve it...

How to do calculate the extrema?

In addition,
How to do calculate the following?
sinθ10 = V(v10 + v20)/(V2 + v10v20)
 

Attachments

Last edited:

BvU

Science Advisor
Homework Helper
11,979
2,632
Last edited:
Please state the full problem statement. Not everybody has all books that have exercises in tehm.
The problem is
Determine the range of possible values of the angle θ following the figure.
 

BvU

Science Advisor
Homework Helper
11,979
2,632
Can't be the full problem statement. And: there is no ##\theta## in the fgure.

1. The problem statement, all variables and given/known data


2. Relevant equations


3. The attempt at a solution
 
Can't be the full problem statement. And: there is no ##\theta## in the fgure.
in description θ = θ1 + θ2
and θ10 + θ20 = π
 

BvU

Science Advisor
Homework Helper
11,979
2,632
What particle are we looking at ? What's its energy before decay ? What are the decay products ? etc etc etc
 
What particle are we looking at ? What's its energy before decay ? What are the decay products ? etc etc etc
In C system, total momentum is 0
it is just m1v10+m2v20 = 0

In L system, v1=v10+V, v2=v20+V, and total momentum is not 0

The particles we looking at are in L system.
 
Last edited:

Ray Vickson

Science Advisor
Homework Helper
Dearly Missed
10,705
1,710
<Moderator's note: Moved from a technical forum and thus no template.>

Mechanics by Lev D. Landau & E. M. Lifshitz
Chapter 4 Collisions between particles
§16. Disintegration of particles
Problem 3
View attachment 236072
The angle θ = θ1 + θ2
It is simplest to calculate the tangent of θ.
A consideration of the extrema of the resulting expression gives the following ranges of θ, depending on the relative magnitudes of V, v10 and v20 (for definiteness, we assume v20 > v10):
0 < θ < π if v10 < V < v20,
π-θ10 < θ < π if V < v10,
0 < θ < θ10 if V > v20.

So I calculated the tangent of θ and derivative of it
View attachment 236073
But I can not calculate the extrema, thus I can not solve it...

How to do calculate the extrema?

In addition,
How to do calculate the following?
sinθ10 = V(v10 + v20)/(V2 + v10v20)
To get the extrema you need to solve for ##\theta_{10}## from ##d \tan(\theta)/ d \theta_{10} = 0##. I have not checked your algebra, but assuming your final expression is correct, this condition simplifies to
$$ (V^2 - v_{10} v_{20}) \cos \theta_{10} +V(v_{10} - v_{20}) = 0$$ You can express the answer in terms of ##\arccos \theta_{10}##.

Note added in edit: When I do the question (using a computer algebra program) I get an expression for ##\tan \theta## much different from yours, and my derivative is much more complicated than yours. My equation for the max/min is a quadratic equation in ##\cos \theta_{10}##, so will have two roots, one for the minimum and the other for the maximum.

BTW: please try to avoid typesetting using the "[S UB]" and "[/S UB]" commands; it is much easier, faster and prettier to just use LaTeX, which comes pre-loaded in PF, and which I used for my answer above.
 
Last edited:
To get the extrema you need to solve for ##\theta_{10}## from ##d \tan(\theta)/ d \theta_{10} = 0##. I have not checked your algebra, but assuming your final expression is correct, this condition simplifies to
$$ (V^2 - v_{10} v_{20}) \cos \theta_{10} +V(v_{10} - v_{20}) = 0$$ You can express the answer in terms of ##\arccos \theta_{10}##.

Note added in edit: When I do the question (using a computer algebra program) I get an expression for ##\tan \theta## much different from yours, and my derivative is much more complicated than yours. My equation for the max/min is a quadratic equation in ##\cos \theta_{10}##, so will have two roots, one for the minimum and the other for the maximum.

BTW: please try to avoid typesetting using the "[S UB]" and "[/S UB]" commands; it is much easier, faster and prettier to just use LaTeX, which comes pre-loaded in PF, and which I used for my answer above.
What's your tanθ?
I also used algebra program but the result is same as my calculation
 

Ray Vickson

Science Advisor
Homework Helper
Dearly Missed
10,705
1,710
What's your tanθ?
I also used algebra program but the result is same as my calculation
Sorry: you are correct! I changed the notation but neglected to change the equation for the angle ##\theta##. When I correct that error, I end up getting exactly the same result as you.

Therefore, there is a single root for ##\cos(\theta_{10})##, but two values for ##\theta_{10}## in the interval ##(0,2 \pi)##. These correspond to the two extrema of ##\tan (\theta)##, the max and the min.
 
Last edited:
Sorry: you are correct! I changed the notation but neglected to change the equation for the angle ##\theta##. When I correct that error, I end up getting exactly the same result as you.

Therefore, there is a single root for ##\cos(\theta_{10})##, but two values for ##\theta_{10}## in the interval ##(0,2 \pi)##. These correspond to the two extrema of ##\tan (\theta)##, the max and the min.
How to calculate the extrema?
for example, When v10 < V < v20
캡처.PNG

From d(tanθ)/dθ10 < 0, how can get the solution 0 < θ < π ?
I can get it only when V2 = v10v20
 

Attachments

I have just solved it completely thanks everyone
 
  • Like
Reactions: BvU
5
2
5
2
Hi, can I ask you how you solved the problem? I got stuck just the way you did in the posts above.
I solved it! I'll copy and paste my solution, which could be useful in case someone comes across this thread.

It would have been much easier if the book had given a different name to the angle ##\theta_{0}##. In fact, it is not to be intended as the same angle as the one introduced by the book in prior graphical constructions and problems. Hence, even if the problem is conceptually very simple, this might create confusion.
For the sake of completion, to bring this post to an end, I shall add some details for the solution (also to try and redeem myself from some careless mistakes :) ).
The function we have to analyze is:
$$ \begin{align}\tan{\theta} = \frac {(v_{10}+v_{20})V\sin{\theta_{0}}}{V^2+\left(v_{10}-v_{20} \right)V\cos{\theta_0 - v_{10} v_{20} }} \nonumber\end{align}$$
or equivalently:
$$ \begin{align}\cot{\theta} = \frac {V^2+\left(v_{10}-v_{20} \right)V\cos{\theta_0 - v_{10} v_{20} }}{(v_{10}+v_{20})V\sin{\theta_{0}}} \nonumber\end{align}$$ where ##\theta_{0} \in \left[0,\pi \right]##, ##V \geq 0##.

In order to find the extrema, I took the derivative with respect to ##\theta_{0}## and found the expression:
$$ \begin{align} \frac {\partial\cot{\theta}} {\partial\theta_{0}} = \frac {(-v_{10}+v_{20})V^2-V\cos\theta_{0}\left[V^2-v_{10}v_{20}\right] }{(v_{10}+v_{20})V^2\sin^2\theta_{0}} \nonumber\end{align}$$
By setting the result to 0 one gets: $$ \begin{align} \cos\theta_{0}=\frac{(v_{20}-v_{10})V}{V^2-v_{10}v_{20}} \nonumber\end{align}$$

One should notice that the formula ## \begin{align} \cos\theta_{0}=\frac{(v_{20}-v_{10})V}{V^2-v_{10}v_{20}} \nonumber\end{align} ## holds only when the cosine obviously lies in the interval ##\left[-1,1\right]##; after some calculations, one finds that this happens when ##V \geq v_{20} ~\text{or}~ V\leq v_{10}##, meaning that when ##v_{10}<V<v_{20}## the derivative of ##\cot{\theta}## does not change sign.
Moreover, one could calculate the derivative of the cotangent with respect to ##V## and find the expression:
$$ \begin{align} \frac {\partial\cot{\theta}} {\partial V} = \frac { V^2+v_{10}v_{20} } { (v_{10}+v_{20})V^2\sin\theta_{0} } \nonumber\end{align}$$
which is positive for all values of ##V##.
This is why ##\cot{\theta}## may assume all the values between ##0## and ##\pi## when ##v_{10}<V<v_{20}##, because ## \frac {\partial\cot{\theta}} {\partial\theta_{0}}\neq 0## and ## \frac {\partial\cot{\theta}} {\partial V} \neq 0## ##\forall ~V, \theta_{0}## in that interval and it is easy to see that if ##V## is kept constant and ##\theta_{0}## is varied between 0 and ##\pi##, ##\cot\theta## spans the interval ##\left]-\infty,+\infty \right[##; similarly, if ##\theta_{0}## is kept constant and ##V## is varied between 0 and +##\infty##, ##\cot \theta## spans the interval ##\left]-\infty,+\infty \right[##.
When ##V \leq v_{10}##, it is easy to see that ##\cot{\theta}## has a maximum value. In fact, ## \lim_{\theta_{0} \rightarrow 0^+} {\cot(x)} = -\infty##, ##\lim_{\theta_{0} \rightarrow \pi^-} {\cot(x)} = -\infty## and the maximum value of the cotangent is reached when:
$$ \begin{align} \cos\theta_{0}=\frac{(v_{20}-v_{10})V}{V^2-v_{10}v_{20}} \nonumber\end{align}$$
The calculation of the maximum value leads to:
$$ \begin{align} \cot{\theta} &= \frac {V^2+\left(v_{10}-v_{20} \right)V\cos{\theta_0 - v_{10} v_{20} }}{(v_{10}+v_{20})V\sin{\theta_{0}}}= \nonumber \\
\dots &= \frac{v_{20}-v_{10}}{v_{20}+v_{10}} \frac{ \sin{\theta_{0}} }{\cos{\theta_{0}}}= \dots \nonumber \\
&=-\frac{\sqrt{\left( V^2-v_{10}v_{20}\right)^2-\left(v_{20}-v_{10}\right)^2 V^2}}{\left( v_{10}+v_{20}\right)V } \nonumber \end{align}$$
Then, after transforming the cotangent into a sine, one gets:$$ \begin{align} \sin\theta=\frac{(v_{20}+v_{10})V}{V^2+v_{10}v_{20}} \nonumber\end{align}$$ If we put: $$ \begin{align} \alpha=\sin^{-1}{\left(\frac{(v_{20}+v_{10})V}{V^2+v_{10}v_{20}}\right)} \nonumber\end{align}$$
then, in the interval where ##V\leq v_{10}##, we have the following ranges for ##\theta##: ##\pi-\alpha \leq \theta \leq \pi##.
Similarly, when ##V \geq v_{20}##, it is easy to see that ##\cot{\theta}## has a minimum value. In fact, ## \lim_{\theta_{0} \rightarrow 0^+} {\cot(x)} = +\infty##, ##\lim_{\theta_{0} \rightarrow \pi^-} {\cot(x)} = +\infty## and the minimum value of the cotangent is reached when ##\theta=\alpha##. In this case the ranges for ##\theta## are: ##0 \leq \theta \leq \alpha## (the equality ##\theta=0## is satisfied only when ##V \rightarrow +\infty##).
To summarize:

$$ \begin{align} &\text{if}~~ v_{10}<V<v_{20} \rightarrow 0<\theta<\pi \nonumber \\
\nonumber \\
&\text{if}~~ V\leq v_{10} \rightarrow \pi-\alpha \leq \theta \leq \pi \nonumber \\
\nonumber \\
&\text{if}~~ V\geq v_{20} \rightarrow 0\leq\theta\leq\alpha \nonumber\end{align} $$
where $$ \begin{align} \alpha=\sin^{-1}{\left(\frac{(v_{20}+v_{10})V}{V^2+v_{10}v_{20}}\right)} \nonumber\end{align}$$
 
Last edited:

Want to reply to this thread?

"Trigonometric function (Mechanics-Landau)" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top