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Dimensional Analysis Question

  1. Jan 8, 2008 #1
    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?
     
  2. jcsd
  3. Jan 8, 2008 #2

    rbj

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    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: Jan 8, 2008
  4. Jan 8, 2008 #3

    Dale

    Staff: Mentor

    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!
     
  5. Jan 8, 2008 #4
    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!
     
  6. Jan 8, 2008 #5

    Ben Niehoff

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    Science Advisor
    Gold Member

    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]
     
  7. Jan 8, 2008 #6

    rbj

    User Avatar

    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.
     
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