Dimensional Analysis Question

  • Thread starter dougr81
  • Start date
  • #1
6
0
I understand the concept of dimensional analysis and what it's used for etc..My question refers to a dimensionally correct or incorrect equation being true or false. Here are the possiblities:

1. A dimensionally correct equation may be correct.
2. A dimensionally incorrect equation may be correct.
3. A dimensionally correct equation must be correct.
4. A dimensionally incorrect equation must be wrong.
5. A dimensionally correct equation may be wrong.

I think all but #3 and #4 are true, but I may be wrong. Any takers?
 

Answers and Replies

  • #2
rbj
2,226
9
1 and 5 do not conflict. 2 and 4 do conflict as do 3 and 5. i would disagree with you about 2 and 4. i think 2 is wrong and 4 is correct. so i think that 1, 4, and 5 are correct. 2 and 3 are incorrect.

physical quantities that are added, subtracted, equated, or compared need to be commensurable. if you discover that you are adding apples to oranges, then it's time to stop and look for a previous mistake.
 
Last edited:
  • #3
30,275
6,736
1. True
2. False
3. False
4. True
5. True

4 is true for the same reason that 2 is false. If you have a dimensionally incorrect equation then you are saying at some level that a physical unit equals a pure number. An expression like "1 m = 23.43" is never true.

EDIT: I agree with rbj who was faster on the post!
 
  • #4
6
0
Thanks for the explanation you two, it makes sense! I thought a dimensionally correct eqn had to be correct but I now see why that's not true. Thanks again!
 
  • #5
Ben Niehoff
Science Advisor
Gold Member
1,879
162
As an example, one can easily construct equations which are dimensionally correct, but not physically correct, such as

[tex]\vec{F} = 2m\vec{a}[/tex]
 
  • #6
rbj
2,226
9
As an example, one can easily construct equations which are dimensionally correct, but not physically correct, such as

[tex]\vec{F} = 2m\vec{a}[/tex]
well, if you define a Newton of force to be the force needed to accelerate 1/2 kg of mass by 1 m/s2, then it would be correct. but it's a dumb definition for the unit force.
 

Related Threads on Dimensional Analysis Question

  • Last Post
Replies
6
Views
5K
Replies
2
Views
2K
Replies
3
Views
1K
Replies
4
Views
1K
  • Last Post
Replies
1
Views
529
  • Last Post
Replies
16
Views
1K
  • Last Post
Replies
5
Views
927
  • Last Post
2
Replies
26
Views
5K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
17
Views
1K
Top