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V0ODO0CH1LD
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I am actually going to post the problem that sparked the question I am about to ask, but I don't need help with the answer and that's why I didn't start this thread in the homework and coursework section. Actually, I just realized I read the problem statement in the wrong way, which makes my question pointless with regards to the problem. Anyway, I am still wondering about it independently of weather it helps with the problem or not. So here it it:
The problem stated that an acceleration has magnitude [itex] \mu(r+\frac{a^3}{r^2}) [/itex], where [itex] a [/itex] is the initial displacement and [itex] r [/itex] is the distance from the origin. What the problem asked was to check what were the dimensions of [itex] \mu [/itex]. When you read it right it's an easy problem, right?
But in the way I had originally read it, it stated: an acceleration has magnitude [itex] \mu(\frac{r + a^3}{r^2}) [/itex]. Which makes the thing inside the parenthesis look like [itex] \mu(\frac{1}{r}+\frac{a^3}{r^2}) [/itex].
Well, [itex]\frac{a^3}{r^2}[/itex] would just be something like [itex]\frac{(am)^3}{(rm)^2}[/itex], where [itex] m [/itex] is some unit of distance and that would simplify to [itex]\frac{a^3}{r^2}m[/itex]. Which is fine.
But what would [itex]\frac{1}{rm}[/itex] mean? Like, a dimensionless something per unit of distance? How should I think about [itex]\frac{1}{r}m^{-1}[/itex]? Or does it not even exist on the account that I read the problem wrong? At first I was like: well; I have something like 1/r inverse meters, so that must mean I have r regular meters.. But that makes no sense. Is there a correct way to view this?
The problem stated that an acceleration has magnitude [itex] \mu(r+\frac{a^3}{r^2}) [/itex], where [itex] a [/itex] is the initial displacement and [itex] r [/itex] is the distance from the origin. What the problem asked was to check what were the dimensions of [itex] \mu [/itex]. When you read it right it's an easy problem, right?
But in the way I had originally read it, it stated: an acceleration has magnitude [itex] \mu(\frac{r + a^3}{r^2}) [/itex]. Which makes the thing inside the parenthesis look like [itex] \mu(\frac{1}{r}+\frac{a^3}{r^2}) [/itex].
Well, [itex]\frac{a^3}{r^2}[/itex] would just be something like [itex]\frac{(am)^3}{(rm)^2}[/itex], where [itex] m [/itex] is some unit of distance and that would simplify to [itex]\frac{a^3}{r^2}m[/itex]. Which is fine.
But what would [itex]\frac{1}{rm}[/itex] mean? Like, a dimensionless something per unit of distance? How should I think about [itex]\frac{1}{r}m^{-1}[/itex]? Or does it not even exist on the account that I read the problem wrong? At first I was like: well; I have something like 1/r inverse meters, so that must mean I have r regular meters.. But that makes no sense. Is there a correct way to view this?