Understanding dimensional analysis in a system with only one independent dimension

Question

Suppose it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity X as follow: [position]=[$X^\alpha$], [speed]=[$X^\beta$],[acceleration]=[$X^p$], [linear momentum]=[$X^q$], [force]=[$X^r$]. Then:

A) $\alpha + p=2 \beta$.
B)$p+q-r=\beta$
C) $p-q+r=\alpha$.
D) $p+q+r=\beta$

I tried writing all of them in their basic dimensions $M,L,T$ as [position]=$L$, [speed]=$LT^{-1}$ etc. , but I could not proceed further. How are they different exponents of the same dimension $X$?

Answer 1

When your question writes

Suppose it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity

this is not a hypothetical $-$ this is how 'natural units' work. This is the system used in particle physics: there you set your units such that $c=1$ and $\hbar=1$, so that you only have one dimension left (as described e.g. here), normally taken to be mass.

If you have a system like that, where you've set

[position]=[$X^\alpha$], [speed]=[$X^\beta$],[acceleration]=[$X^p$], [linear momentum]=[$X^q$], [force]=[$X^r$]

then saying $p+q+r=\beta$ is equivalent to saying that $$[\mathrm{acceleration} \times \mathrm{(linear\ momentum)} \times \mathrm{force}] = [\mathrm{speed}],$$ with similar equivalences for the other equations in your set-piece.

Answer 2

@EmilioPisanty answered your question, but you appear shaky on scaling nondimensionalization, so I'll make it more concrete for you. You say you already wrote down the five relevant equations, $$ [X^\alpha]=L \tag{1} $$ $$[X^\beta]=L T^{-1} \tag{2} $$ $$ [X^p]=L T^{-2} \tag{3} $$ $$ [X^q]=M L T^{-1}\tag{4} $$ $$ [X^r]=ML T^{-2} \tag{5}.$$ These five equations must be dependent if they are to span a three-dimensional space (of dimensions), so two of the vector variables α,β,p,q,r must be superfluous and need be replaced as soon as possible here and in your checking or rejecting the relations of your homework. If the relations are identities, they are valid. But if they are not, they are wrong, since the remaining variables are independent--they are nontrivial exponents that make the natural units' magic work!

Squaring (2) and dividing by (1) nets you (3), so p=2β-α, superfluous, and is never heard from again, including its avatar (3). Likewise, (5) divided by (4) gives you (2) divided by (1), so that r=q+β-α, superfluous, and disappears together with (5).

You now only have $$ [X^\alpha]=L \tag{1} $$ $$[X^\beta]=L T^{-1} \tag{2} $$ $$ [X^q]=M L T^{-1}\tag{4} .$$ Equivalently, $$ [X^\alpha]=L \tag{1} $$ $$[X^{\alpha-\beta}]= T \tag{6} $$ $$ [X^{q-\beta}]=M \tag{7} .$$

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In particle physics, the common unit is typically the MeV, so $[c]=[X^\beta]$; $ [\hbar]=[X^{\alpha +q}]$; $[MeV]=[X^{\beta+q}] $, that is, all quantities are measured along the β+q direction, with the other two trivialized out, to be reinstated in the end, if desired. (This means that all speeds are measured in units of c and actions in units of ℏ. So $\beta=0$, and $ \alpha=-q$.)

You may then take q=1 for X an energy, measured in MeV. In (1,6,7) above, you see L and T measured in $MeV^{-1}$, while M is measured in MeV.

Now, if I understand your conceptual question, you wonder if there is a problem using the same unit, MeV, to measure masses, energies, inverse lengths, and inverse times. No, not really; it's actually convenient, since you automatically relate lifetimes with inverse energy widths, etc. But you cannot tell the quantity merely from its units.

Once you do know what you are talking about, you may uniquely reinstate c and ℏ with the appropriate powers for dimensional consistency. So, if you are told your MeV represents a mass, you may immediately reconstruct its engineering units to be $MeV/c^2$. If, instead, you are told it represents an inverse length, then its engineering units are $MeV/c\hbar $. Countless engineer readers of popular HEP articles are confused by the convention.