9.6. ** Here is an example of an experiment in which we would expect a nega- tiv

Question

9.6. ** Here is an example of an experiment in which we would expect a nega- tive correlation between two measured quantities (high values of one correlated with low values of the other). Figure 9.2 represents a photograph taken in a bubble chamber, where charged subatomic particles leave clearly visible tracks. A positive particle called the K+ has entered the chamber at the bottom of the picture. At point A, it has collided with an invisible neutron (n) and has undergone the reaction The proton's track (p) is clearly visible, going off to the right, but the path of the K° (shown dotted) is really invisible because the K° is uncharged. At point B, the Ko decays into two charged pions, whose tracks are again clearly visible. To investigate the conservation of momentum in the second process, the experimenter needs to measure the angles ? and ? be- tween the paths of the two pions and the invisible path of the K°, and this measure- ment requires drawing in the dotted line that joins A and B. The main source of error in finding a and B is in deciding on the direction of this line, because A and B are often close together (less than a cm), and the tracks that define A and B are rather wide. For the purpose of this problem, let us suppose that this is the only source of error.

Explanation / Answer

Answer:

Applying law of conservation of energy we can infer that the angle between ?- and ?+ is equal for each decay due to them having same mass and originating at the same instant with equal momentum. So the sum of ? and ? should be same for every observation, at least theoretically but in practice both ? and ? may have errors ?? and ?? respectively. Following a mathematical approach, if ? + ? = ?, differentiating we get ??/?? + ??/?? = 1, and hence ?? + ?? = ?? = 0. Here, ?? is zero since sum of ? and ? should be same for every observation, so no change. Now we can infer that from  ?? + ?? = 0, if ? is overestimated (?? > 0) then ? is being underestimated (?? < 0).

We have an important relation here as ?? = - ??.

Let we have N observations for both ? and ? with mean <?> and <?> (here <a> represent mean of a) then,

Variance for ? is ?? = [ ? (?i - <?>)2 ]/ N = [ ? (?i)2 / N ] - <?>2( this is also a property of variance).

Even from above it is evident that for all obeservations we have ?? = [ ? (?i - <?>)2 ]/ N = [ ? (??)2 ]/ N.

Similarly for ?, ?? = [ ? (?i - <?>)2 ]/ N = [ ? (??)2 ]/ N.

Putting ?? = - ?? in any one of ?? or ?? formula we can see that ?? = ??.

Now, covariance between ? and ? is given as

??? = (1/N) [? (?i - <?>)(?i - <?>)] = (1/N) [? (??*??)] = -(1/N) [? (??2)] = -(1/N) [? (??2)] = some -ve value, since ?? = - ??.

and we can also see that it equals to the largest value allowed by ??? = ??*??

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