Doubt about a probability density

[Kaguro]

I was trying a problem from Griffith's Introduction to QM. The problem was:

The needle on a broken car speedometer is free to swing, and bounces perfectly off the pins at either end, so that if you give it a flick it is equally likely to come to rest at any angle between 0 to $\pi$.

a)Find probability density $\rho$($\theta$).
b)If x is the projection of needle on the horizontal line and r is length of needle, then find probability density $\rho$(x)
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The first one is simply:
$\rho$($\theta$)$d\theta$ = $d\theta$/$\pi$
so, $\rho$($\theta$) = 1/$\pi$

I couldn't do the b part. So I looked at the solution and it says:
$\rho$($\theta$)$d\theta$ = $\rho$(x)dx

Why? Can you please tell me the reasoning?

 

[PeroK]

I couldn't do the b part. So I looked at the solution and it says:
$\rho$($\theta$)$d\theta$ = $\rho$(x)dx

Why? Can you please tell me the reasoning?

That's the definition of probability density. If we take a small angular range $(\theta, \theta + d \theta)$, then the probability the needle lands in that range is $\rho(\theta)d\theta$.

And, if we project that range onto the x-axis, $(x, x+dx)$, then the probability it lands in that range is $\rho'(x) dx$

Note I've used $\rho'$ to indicate that it's actually a different function in each case.

 

[Kaguro]

Why does projecting this range conserves probability?
The intervals are different...

Instead of projection I could have taken a completely weird relation like
u = rcos($\theta$) + 3rsin($\theta$/2)

Now the total range is r to 2r.

But interval of d$\theta$ doesn't correspond to the interval of du... then there's no sense of "probability in this region of space" should be same...

 

[PeroK]

Why does projecting this range conserves probability?
The intervals are different...

Instead of projection I could have taken a completely weird relation like
u = rcos($\theta$) + 3rsin($\theta$/2)

Now the total range is r to 2r.

But interval of d$\theta$ doesn't correspond to the interval of du... then there's no sense of "probability in this region of space" should be same...

In this case $x$ and $dx$ can be expressed as the appropriate functions of $\theta$. Probability must be conserved.

E.g. if we had a line at an angle, then there would be a linear relationship between the length along the line and the $x$ coordinate. In this case the mapping is from the semi-circle to the diameter.

 

[Kaguro]

I think that, yes... it should be correct..

The interval is just a matter of system...

The physical reality of probability must not depend upon the system used...

If the interval is different then the density also should be different so as to conserve the probability..

 

[PeroK]

I think that, yes... it should be correct..

The interval is just a matter of system...

The physical reality of probability must not depend upon the system used...

If the interval is different then the density also should be different so as to conserve the probability..

Yes, which is why $\rho$ is a different function in each case.

 

[vanhees71]

It's really a nuissance with this book. Of course you must write $\tilde{\rho}(x)$ since it's NOT the same function. A density by definition transforms as a density! Who'd guessed this.

It's easy to understand. If you have a random number $\theta$ with a probability distribution $\rho(\theta)$ then the meaning is: When measuring $\theta$ the probability to find its value to be in an infinitesimal intervall $(\theta,\theta+\mathrm{d}\theta)$ is given by $\mathrm{d} P=\mathrm{d} \theta \rho(\theta)$.

Now we have a one-to-one map from $\theta \in [0,\pi]$ to $x \in[-L,L]$ given by $x=L \cos \theta$, where $L$ is the length of the needle. Since it's a one-to-one mapping the probability to find $\theta$ in the intervall $(\theta,\theta+\mathrm{d} \theta)$ we have
$$\rho(\theta) \mathrm{d} \theta = \tilde{\rho}(x) \mathrm{d} x,$$
from which we get
$$\tilde{\rho}(x)=\rho(\theta) \left |\frac{\mathrm{d} \theta}{\mathrm{d} x} \right|.$$
Now $\theta=\arccos(x/L)$ and thus $|\mathrm{d} \theta/\mathrm{d} x|=\frac{1}{\sqrt{L^2-x^2}}$. So we finally get
$$\tilde{\rho}(x)=\frac{1}{\pi \sqrt{L^2-x^2}}, \quad x \in [-L,L],$$
because obviously $\Theta(\theta)=1/\pi$.