• Support PF! Buy your school textbooks, materials and every day products Here!

Very simple question on negative powers

  • #1
167
0

Homework Statement



The problem is stated in the attempt as a solution

Homework Equations



There are none

The Attempt at a Solution



Suppose you have

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

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
No. Why would you think that? n^(-1)=1/n, n^(-2)=1/n^2.
 
  • #3
167
0
Because in Dimensional analysis, we tend to use the notation ie. [tex]MTl^{-2}[/tex] which would mean that it is mass times time divided by length squared...


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

[tex]\epsilon \mu = c^{-2}[/tex]

such that

[tex]\sqrt{\epsilon \mu} = \frac{1}{c}[/tex]

I could be totally off, mind my ignorance.
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
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?
 
  • #5
167
0
Let's try this again. Let us say, E is energy.

What would be E^{-2} be?
 
  • #6
Dick
Science Advisor
Homework Helper
26,258
618
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?
 
  • #7
167
0
In fact let's do this another way. I was reading about the dirac operator

[tex]D(\psi) = \hbar^2 R^{-2} \psi(x)[/tex]

How does R^{-2} read to you? Just 1/R^2 like you said before?
 
  • #8
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
In fact let's do this another way. I was reading about the dirac operator

[tex]D(\psi) = \hbar^2 R^{-2} \psi(x)[/tex]

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
 
  • #9
167
0
R is taken to be a radius or a length.
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
41,833
955
Because in Dimensional analysis, we tend to use the notation ie. [tex]MTl^{-2}[/tex] 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

[tex]\epsilon \mu = c^{-2}[/tex]

such that

[tex]\sqrt{\epsilon \mu} = \frac{1}{c}[/tex]

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

Related Threads on Very simple question on negative powers

  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
3
Views
2K
Replies
12
Views
691
Replies
4
Views
4K
  • Last Post
Replies
3
Views
926
  • Last Post
2
Replies
30
Views
3K
Replies
3
Views
3K
Replies
6
Views
1K
Top