Q.A famous relation in physics relates 'moving mass' $m$ to the 'rest mass' $m_0$ of a particle in terms of its speed $v$ and the speed of light, $c$. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant $c$. He writes: $m = \dfrac{m_0}{\left(1 - v^2\right)^{1/2}}$. Guess where to put the missing $c$.

The key idea is that the denominator must be dimensionless, so the term $v^2$ must be divided by $c^2$ to make it a pure number.

Reasoning:

1. The expression $1 - v^2$ is dimensionally inconsistent — $v^2$ has dimensions $[L^2 T^{-2}]$, while 1 is dimensionless.
2. To fix this, $v^2$ must be divided by $c^2$, giving the dimensionless ratio $v^2/c^2$. …

The boy’s guess — $m = \frac{m_0}{\sqrt{1 - v^2}}$ — is almost right, but it has a serious problem: the term $v^2$ inside the square root is subtracted from 1. Since $v$ has dimensions of velocity (say, m/s), $v^2$ has dimensions of (velocity)$^2$. You cannot subtract a dimensional quantity from the pure number 1. That’s like asking “what is 1 metre minus 3 seconds?” — it’s meaningless. The expression must be dimensionally homogeneous: every term inside the square root must be dimensionless.

The fix is to divide $v^2$ by $c^2$, because $c$ is also a speed. Then $v^2/c^2$ is a pure number, and $1 - v^2/c^2$ makes perfect sense. So the correct relativistic mass formula is:

Let’s walk through the reasoning step by step.

1. Identify the dimensions.

   Mass $m$ and rest mass $m_0$ both have dimension $[M]$. Speed $v$ has dimension $[L T^{-1}]$, and $c$ has the same dimension. The number 1 is dimensionless. For the fraction $\frac{m_0}{\sqrt{1 - \text{something}}}$ to give a mass, the denominator must be dimensionless — that means the “something” must also be dimensionless.

2. Check the boy’s version.

   He wrote $m = \frac{m_0}{\sqrt{1 - v^2}}$. Here $v^2$ has dimension $[L^2 T^{-2}]$, which is not dimensionless. So the expression is dimensionally illegal. The only way to fix it is to introduce $c$ in such a way that the subtracted term becomes $v^2/c^2$, a pure number.

3. Where does $c$ go?

   The boy already has $c$ nowhere — he forgot it entirely. We need to insert $c$ so that $v^2$ is divided by $c^2$. That means the correct form is:

Could $c$ go elsewhere? For instance, $m = \frac{m_0}{\sqrt{1 - (v/c)^2}}$ is the same thing. Could it be $m = \frac{m_0}{\sqrt{1 - v^2}}\,c$? That would give dimensions of $[M][L T^{-1}]$, not mass. Could it be $m = \frac{m_0 c}{\sqrt{1 - v^2}}$? Again, wrong dimensions. The only placement that makes the denominator dimensionless and preserves mass dimension is dividing $v^2$ by $c^2$ inside the square root. …

Method: Dimensional Analysis

Why this method?

The boy’s equation is dimensionally wrong — $v^2$ is subtracted from $1$, but $v^2$ has dimensions of $(\text{speed})^2$, while $1$ is dimensionless. In physics, you can only add/subtract quantities with the same dimensions. The speed of light $c$ must be inserted to make the term dimensionless.

Steps

1. Write the boy’s equation

2. Identify the problem

   Inside the square root: $1$ is dimensionless, but $v^2$ has dimensions $[L^2 T^{-2}]$.

   → They cannot be subtracted.

3. Insert $c$ to make $v^2$ dimensionless

   The only way is to divide $v^2$ by $c^2$, giving $\frac{v^2}{c^2}$, which is dimensionless.

4. Write the corrected relation

5. Verify dimensions …

The Core Mistake

The boy writes:

The problem: $v^2$ is subtracted from a pure number $1$. But $v$ has dimensions of speed (e.g., m/s), so $v^2$ has dimensions of (speed)$^2$. You cannot subtract a dimensional quantity from a dimensionless number.

Common Mistakes & How to Avoid Them

1. Forgetting that $v^2$ must be dimensionless inside the square root

• Mistake: Writing $1 - v^2$ as if $v$ is a pure number.
• Why it's wrong: In physics, you can only add/subtract quantities with the same dimensions. Here, $1$ is dimensionless, so $v^2$ must also be dimensionless.
• Fix: The speed $v$ must be divided by another speed to cancel its dimensions. The only universal speed in relativity is $c$, the speed of light. So the correct term is:

Thus the correct formula is:

2. Placing $c$ in the numerator instead of the denominator

• Mistake: Writing $m = \frac{m_0}{(1 - c^2 v^2)^{1/2}}$ or similar.
• Why it's wrong: $c^2 v^2$ has dimensions of (speed)$^4$, which still doesn't match the dimensionless $1$.
• Fix: Always put $c$ in the denominator with $v$: $\frac{v}{c}$ or $\frac{v^2}{c^2}$.

3. Forgetting that $c$ appears squared

• Mistake: Writing $m = \frac{m_0}{(1 - v/c)^{1/2}}$.
• Why it's wrong: The correct Lorentz factor is $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$, not $\frac{1}{\sqrt{1 - v/c}}$. The square on $v$ and $c$ comes from the geometry of spacetime (the invariant interval).
• Fix: Always use $v^2/c^2$ inside the square root.

4. Misplacing $c$ outside the square root

• Mistake: Writing $m = \frac{m_0}{c(1 - v^2)^{1/2}}$ or $m = \frac{m_0}{(1 - v^2)^{1/2}} \cdot c$.
• Why it's wrong: The Lorentz factor $\gamma$ is dimensionless. Adding $c$ outside changes the dimensions of $m$ (mass would no longer have units of mass). …