0
$\begingroup$

I want to show that acceleration, pressure and area are independent dimensions meaning that there is no relationship between either pair of values. One cannot be written in terms of the others.

Here is my reasoning.

$$[acceleration]=[a]=LT^{-2}$$ $$[pressure]=[p]=ML^{-1}T^{-2}$$ $$[area]=[A]=L^{-2}$$

Now if there was any kind of relationship such as for example: $$a=pA$$ then it would be possible to multiply all terms to one side as follows: $$1=\frac{pA}{a}$$ This i just an example but for any relationship this is always true because of the equation must remain dimensionally homogeneous.

So in general we have: $$1=[a]^e[p]^{f}[A]^g$$ were $e,f,g$ are quotients that might be part of the equation. Now we do dimensional analysis: $$1=L^{a-b-2c}M^{b}T^{-2a-2b}$$

$$L: 0 = a-b-2c$$ $$M: 0 = b$$ $$T: 0 = -2a-2b$$

There the solutions are $a=b=c=0$. This, indicates that the only way in which a relationship can be achieved using these three dimensions is if each of them has a power $0$ i.e. there is no dimension. If by contrast we got some other type of solution, that would mean that there is a relationship between the dimensions are they are therefore not independent of each other.

Now, my question. Is is a correct proof? Does it make sense?

$\endgroup$
1
$\begingroup$

Your solution is correct (you were led to a 3 by 3 system with only the trivial solution, so the units are independent), but you need to make one more assumption clear. You have justified the equation that you got by dimensional homogeneity of any physically meaningful relation, but there is one more assumption that you need to state: that length, time and mass themselves are independent!

Suppose there was a fundamental quantity, let's call it lentimass, denoted by $[Q]$, so that $[L]=[Q]^2$, $[T]=[Q]^3$ and $[M]=[Q]^4$. Then your three expressions would collapse to one and you have infinitely many solutions. The fact that lentimass does not exist (as far as we know) is one of those empirical assumptions that we should do well not to brush off in our deductions of results critically dependent on them!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.