The Physics Professor

Task

A physics professor says: "Of course, it is easy to see that $$ L_0 \sqrt{1 - \frac{v^2}{c^2}} = 0 $$ when $v = c$."

1. Give a possible explanation in terms of the structure of the expression on the left why the professor might say that.
2. Assuming that $L_0$ and $c$ are positive, what is the greatest possible value of the expression on the left? Explain your answer in terms of the structure of the expression.

IM Commentary

The purpose of this task is to provide students practice in drawing conclusions about expressions they might encounter in classes outside mathematics, by parsing them in terms of their algebraic structure. Teachers might stress the subtle difference between "seeing" why the expression must be zero and the more mechanical process of algebraically simplifying the expression upon substituting $v=c$. Although part (b) might initially be though of as an optimization problem in a calculus course, some elementary reasoning with the structure of the expression gives the same answer in a more fluid and conceptual fashion.

For reference, this formula is for the length contraction of an object travelling near the speed of light, denoted $c$. While the constant $c$ in the problem is indeed a positive constant, its positivity is irrelevant to the task at hand since only its square, $c^2$, appears in the formula.

Solution

1. When $v=c$, the fraction $\frac{v^2}{c^2}$ is 1 and so $\left(1-\frac{v^2}{c^2}\right)$ is 0. Since the square root of $0$ is $0$, the entire expression is $0$ when $v=c.$
2. The greatest possible value of the expression is $L_0$. To see this, we observe that the term $1-\frac{v^2}{c^2}$ is always at most 1, since the quantity $\frac{v^2}{c^2}$ is always at least zero. Now the square root of a quantity which is at most 1 is again at most 1, and so when multiplied by $L_0$, we find that the entire expression is at most $L_0$. Note this value does indeed occur, when (and only when) $v=0$.