Question

If energy $\left(\mathrm{E}\right)$, velocity $\left(\mathrm{V}\right)$ and force $\left(\mathrm{F}\right)$ be taken as fundamental quantities, then what are the dimensions of mass?

Answer 1

Step 1: Given

The energy $\left(\mathrm{E}\right)$, velocity $\left(\mathrm{V}\right)$ and force $\left(\mathrm{F}\right)$ are taken as a fundamental quantity.

Step 2. Concept to be used

The Einstein mass-energy relation states that mass and energy are related to each other and every substance has energy because of its mass also.

The energy that exists due to the motion of an object is known as kinetic energy.

The formula of kinetic energy is,

$\mathrm{KE}=\frac{1}{2}{\mathrm{mV}}^2$

Where, $\mathrm{KE}$ is the kinetic energy, $\mathrm{m}$ is the mass of the object and $\mathrm{V}$ is the velocity of the object.

Since $\mathrm{KE}$ is another form of $\mathrm{E}$, they have the same units.

The velocity is the ratio of displacement by time.
The formula of velocity is,

$\mathrm{V}=\frac{\mathrm{d}}{\mathrm{t}}$
where $\mathrm{d}$ is the displacement, and $\mathrm{t}$ is the time taken.

The formula for force is,
$\mathrm{F}=\mathrm{ma}$

where $\mathrm{a}$ is the acceleration of mass $\mathrm{m}$ due to force $\mathrm{F}$.

Step 3: Find the dimensions of force, velocity and energy

The unit of the velocity $\left(\mathrm{V}\right)$ is $\mathrm{m}/\mathrm{s}$, as the velocity is the ratio of displacement by time given as,
$\mathrm{V}=\frac{\mathrm{d}}{\mathrm{t}}$
The dimensional formula of this above unit is,
$\mathrm{V}=\left[{\mathrm{LT}}^{-1}\right]$

The unit of the force is $\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2$.
As the force $\left(\mathrm{F}\right)$ is the product of the mass and acceleration.
So, the dimensional formula of this above unit is,
$\mathrm{F}=\left[{\mathrm{MLT}}^{-2}\right]$

The unit of the energy is $\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2$ ,
As the energy $\left(\mathrm{E}\right)$ is the product of mass and the velocity of light in the air, that is,
$\mathrm{E}={\mathrm{mc}}^2$
The dimensional formula of this unit is, $\mathrm{E}=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]$

Therefore, the dimensional formulae,

For velocity is $\mathrm{V}=\left[{\mathrm{LT}}^{-1}\right]$.

For force is $\mathrm{F}=\left[{\mathrm{MLT}}^{-2}\right]$.

For energy is $\mathrm{E}=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]$

Step 4: Express dimension of mass in terms of new units

The SI unit of the mass is $\mathrm{kg}$.
So, the dimensional formula of mass is,

$\text{M}=[\text{M}{\text{L}}^0{\text{T}}^0]$ --------- $\left(1\right)$

Let the dimensional formula of the mass in terms of the energy, force, and velocity be,

$\mathrm{M}=\left[{\mathrm{V}}^{\mathrm{a}}{\mathrm{F}}^{\mathrm{b}}{\mathrm{E}}^{\mathrm{c}}\right]$ --------- $\left(2\right)$

Putting the dimensional formulae of the energy, force, and velocity in the above equation, we get,
$\Rightarrow \text{M}=[[\text{L}{\text{T}}^{-1}{]}^{\text{a}}[\text{ML}{\text{T}}^{-2}{]}^{\text{b}}[\text{M}{\text{L}}^2{\text{T}}^{-2}{]}^c]$

Removing the brackets and adding the like terms, we get,

$\mathrm{M}=\left[{\mathrm{M}}^{\mathrm{b}+\mathrm{c}}{\mathrm{L}}^{\mathrm{a}-\mathrm{b}-2\mathrm{c}}{\mathrm{T}}^{-\mathrm{a}-2\mathrm{b}-2\mathrm{c}}\right]$ --------- $\left(3\right)$

Comparing the equations $\left(2\right)$ and $\left(3\right)$, so we can write,

$\left[{\mathrm{ML}}^0{\mathrm{T}}^0\right]=\left[{\mathrm{M}}^{\mathrm{b}+\mathrm{c}}{\mathrm{L}}^{\mathrm{a}+\mathrm{b}+2\mathrm{c}}{\mathrm{T}}^{-\mathrm{a}-2\mathrm{b}-2\mathrm{c}}\right]$
Now, compare and equate the components accordingly.

So, we can write,

$1=\mathrm{b}+\mathrm{c}$ ----- $\left(4\right)$

$0=\mathrm{a}+\mathrm{b}+2\mathrm{c}$-------- $\left(5\right)$

$0=-\mathrm{a}-2\mathrm{b}-2\mathrm{c}$ ----- $\left(6\right)$

Step 5: Solve the equations

Now let us consider the equation $\left(6\right)$ again,

So,

$0=-\mathrm{a}-2\mathrm{b}-2\mathrm{c}$

$0=-\mathrm{a}-2\left(\mathrm{b}+\mathrm{c}\right)$

Substituting equation $\left(4\right)$ in the above equation, we get,
$0=-\mathrm{a}-2\left(1\right)$

$\mathrm{a}=-2$
Now let us consider the equation $\left(5\right)$

$0=\mathrm{a}+\mathrm{b}+2\mathrm{c}$
We can break the $\mathrm{c}$ as,

$0=\mathrm{a}+\left(\mathrm{b}+\mathrm{c}\right)+\mathrm{c}$
Substitute the values of $\mathrm{a}$ and $(\mathrm{b}+\mathrm{c})$ in the above equation, we get,

$0=-2+\left(1\right)+\mathrm{c}$
$c=1$

Now considering the equation $\left(4\right)$ again, we can substitute the value of $\mathrm{c}$ and get,

$1=\mathrm{b}+\mathrm{c}$

$1=\mathrm{b}+1$

$\mathrm{b}=1-1$

$\mathrm{b}=0$

Now put the values of $\mathrm{a},\mathrm{b}$ and $\mathrm{c}$ in the equation $\left(2\right)$, we get,

$\mathrm{M}=\left[{\mathrm{V}}^{-2}{\mathrm{F}}^0{\mathrm{E}}^1\right]$
Hence, the dimension of mass when energy, velocity, and force are taken as fundamental units is $\left[{\mathrm{V}}^{-2}{\mathrm{F}}^0{\mathrm{E}}^1\right]$.