Very simple question on negative powers

Homework Statement

The problem is stated in the attempt as a solution

There are none

The Attempt at a Solution

Suppose you have

$$n^{-1}$$ am I right in thinking this is just $$\frac{n}{1}$$ so for something like $$n^{-2}$$ would it be simply $$\frac{n}{2}$$?

Dick
Homework Helper
No. Why would you think that? n^(-1)=1/n, n^(-2)=1/n^2.

Because in Dimensional analysis, we tend to use the notation ie. $$MTl^{-2}$$ which would mean that it is mass times time divided by length squared...

mmm... has it got something to do with things like

$$\epsilon \mu = c^{-2}$$

such that

$$\sqrt{\epsilon \mu} = \frac{1}{c}$$

I could be totally off, mind my ignorance.

Dick
Homework Helper
I have no idea what you are confused about. The definition of n^(-k) where k>=0 is pretty simple. It's 1/n^k. Could you try and explain again?

Let's try this again. Let us say, E is energy.

What would be E^{-2} be?

Dick
Homework Helper
Let's try this again. Let us say, E is energy.

What would be E^{-2} be?

It would be 1/E^2. Since energy has units M*L^2/T^2, E^2 has units M^2*L^4/T^4. So 1/E^2 would have units T^4/(M^2*L^4). Wouldn't it?

In fact let's do this another way. I was reading about the dirac operator

$$D(\psi) = \hbar^2 R^{-2} \psi(x)$$

How does R^{-2} read to you? Just 1/R^2 like you said before?

Ray Vickson
Homework Helper
Dearly Missed
In fact let's do this another way. I was reading about the dirac operator

$$D(\psi) = \hbar^2 R^{-2} \psi(x)$$

How does R^{-2} read to you? Just 1/R^2 like you said before?

If R is a number, then YES, it is just 1/R2. If R is an operator, then NO, it means something different.

RGV

R is taken to be a radius or a length.

HallsofIvy
Homework Helper
Because in Dimensional analysis, we tend to use the notation ie. $$MTl^{-2}$$ which would mean that it is mass times time divided by length squared...
Yes, "divided by length squared" which contradicts what you wrote! MTl2= MT/l2, Not "MTl/2".

mmm... has it got something to do with things like

$$\epsilon \mu = c^{-2}$$

such that

$$\sqrt{\epsilon \mu} = \frac{1}{c}$$

I could be totally off, mind my ignorance.
You are making this much too difficult- $\epsilon\mu= \frac{1}{c^2}$ which again contradicts what you said originally.