The curve between `log_(e)` L and `log_(e)` P is (L is the angular momentum and P is the linear momentum).

Options

A.
B.
C.
D.

Explanation

To solve the problem regarding the relationship between the logarithm of angular momentum (L) and the logarithm of linear momentum (P), we can follow these steps:

### Step-by-Step Solution

1. **Understand the Relationship**:
We start with the relationship between angular momentum (L) and linear momentum (P). The angular momentum \( L \) is given by the equation:
\[ L = R \times P \]
where \( R \) is the position vector and \( P \) is the linear momentum.

2. **Express Angular Momentum**:
The magnitude of angular momentum can also be expressed as:
\[ L = R P \sin(\theta) \]
where \( \theta \) is the angle between the position vector \( R \) and the linear momentum vector \( P \).

3. **Take the Natural Logarithm**:
We take the natural logarithm of both sides:
\[ \log_e(L) = \log_e(R P \sin(\theta)) \]

4. **Apply Logarithmic Properties**:
Using the properties of logarithms, we can expand this:
\[ \log_e(L) = \log_e(R) + \log_e(P) + \log_e(\sin(\theta)) \]

5. **Identify the Linear Relationship**:
Rearranging the equation gives:
\[ \log_e(L) = \log_e(P) + \log_e(R) + \log_e(\sin(\theta)) \]
This can be written in the form of a linear equation:
\[ y = mx + c \]
where:
- \( y = \log_e(L) \)
- \( x = \log_e(P) \)
- \( m = 1 \) (the coefficient of \( x \))
- \( c = \log_e(R) + \log_e(\sin(\theta)) \) (a constant)

6. **Conclusion**:
Since the relationship between \( \log_e(L) \) and \( \log_e(P) \) is linear, the graph of \( \log_e(L) \) versus \( \log_e(P) \) will be a straight line with a slope of 1 and a y-intercept equal to the constant \( c \).

### Final Answer
The curve between \( \log_e(L) \) and \( \log_e(P) \) is linear.