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Dimensional analysis

  1. Nov 20, 2014 #1
    Once an equation is well put and there is dimensional homogeneity like:

    ##mx''+\beta x' + kx=f(x)## ; ##Lq''+Rq'+q/c=E(t)## (mass-spring; Kirchoff's diff eq.)

    One proceeds with the math as if there were no actual units involved and just solved a problem dealing with only numbers.

    Is there anyway to prove that there will always be dimensional validity without me having to go and see? I'm looking for some general argument.

    This thing is causing me a huge headache and any help will be greatly appreciated.
     
  2. jcsd
  3. Nov 20, 2014 #2
    If you don't make a mistake, the units will always work out. A correct physics law will never cause mismatched units, so this is one way of checking if a law or equation is not invalid.
     
  4. Nov 20, 2014 #3
    I guess I was looking for some proof of "a correct law of physics will never cause mismatched units"
     
  5. Nov 20, 2014 #4
    You can reduce the equations to dimensionless form, and then solve them in terms of the dimensionless groups.

    Chet
     
  6. Nov 20, 2014 #5
    Chet. I thought about this, but then another question came to my mind: When one redistributes the corresponding units to all quantities involved how do you prove that nothing "artificial" was done? Where artificial means something according to your own convenience.
     
  7. Nov 20, 2014 #6
    I know that in every case I've tried no contradiction will be produced, but I want some sort of general argument. Thanks.
     
  8. Nov 20, 2014 #7
    Well, that's why textbooks and papers generally have derivations for the less familiar equations they use, or references to where these derivations are to be found.
     
  9. Nov 20, 2014 #8
    Khashishi: I don't know if I got you correctly, so correct me if I'm wrong, but are saying one always has to go check and see, and theres no general argument that says: "If an equation is well put from the beginning, then you can treat as if it were dimensionless". ?
     
  10. Nov 20, 2014 #9
    Actually, I'm not exactly sure what you are asking. Are you asking if you can simply drop all the units during calculation and add them back in at the end, throwing in the appropriate unit? If that's what you are asking, then the answer is that that will often work, but it is dangerous. If you stick with one consistent set of units, such as SI, it should work out. For example, F=ma. If mass is 10 kg, and acceleration is 10 m/s^2, then the force is 100 N, where I simply took 10*10 = 100, and I knew the answer had to be in newtons without explicitly checking that kg*m/s^2 = N. If you mix multiple unit systems, or you use scaled units such as cm instead of m, then you are asking for trouble though.

    Certain unit systems, such as atomic units, make atomic calculations very convenient because most of the units drop out, and you can add them back in at the end. But it's dangerous to do unless you are experienced, and even experts make mistakes.
     
  11. Nov 20, 2014 #10
    suppose

    ##x''+5x'+6x=0## models a physical system.

    I say, my solution will be of the form

    ##x=C_1 e^{\lambda t} + C_2 e^{\lambda_2 t}##

    To know this I use the characteristic equation

    ##\lambda ^2 + 5\lambda+6=0## ##\lambda_1=-3 , \lambda_2=-2##

    Throughout this whole process I've thought of this problem as if it had no dimensions. Supposing it had dimensions, how could I guarantee that this is an actual solution and my equation always preserves dimensional homogeneity?.

    Suppose that I'm not really sure if my "characteristic equation" has dimensional homogeneity. I know it should ( which is crucial because this would make my solution correct), but how could I know it **has** to without actually checking mechanically.

    Thanks
     
  12. Nov 20, 2014 #11
    The original differential equation has problems with units. I assume x to be some field, and the prime marks to indicate a derivative with respect to space. Well, then, there's already a problem because space is measured in some unit like meters, so a derivative will have units of x/m. Real physical equations, like the wave equation, have coefficients with dimensions which will make everything match up. Your example doesn't.
     
  13. Nov 20, 2014 #12
    Oh, sorry, I was assuming those units were implicitly in there. Suppose I jotted that equation down when studying a mass-spring system. That's usually the way I would've proceeded to solve it. How did I know there was always going to be dimensional homogeneity?

    I've thought of one argument though: an equation tells you how variables are numerically dependent on one another. You can do all sorts of manipulations here because you do not care about units when you want to know about relationships. Once you have arrived at an expression, only then you'll use units (and the correct ones) to see what's up with your equation. I think this interpretation works, but sounds to shallow IMHO.

    Thanks.
     
  14. Nov 20, 2014 #13
    You seem to describe it as if you are doing something arbitrary. When you reduce the equations to dimensionless form, you are using only the variables that are present in the original equations, and none others. The mathematics takes care of the rest, to guarantee that the solution to the dimensionless equations is exactly the same as the solution of the original equations. Start getting used to the idea that the real world is governed by dimensionless groups. They make the job of describing the behavior of a physical system much more concise and simpler.

    Chet
     
  15. Nov 20, 2014 #14
    Chet: Thanks, what you said about dimensionless groups seems interesting and definitely relieves my current headache a bit. But I kind of don't like saying "the mathematics will take care of the rest", I want to know why this is true! Sorry if I'm being irritating by now, haha.
     
  16. Nov 20, 2014 #15
    Are you saying that you don't have confidence in the ability of the original formulation of the mathematical equations to describe the behavior of the physical system? Reducing the equations to dimensionless form can't change anything, since it just involves multiplying and dividing by constants. The solution to the dimensionless equations is exactly the same as the solution to the original equations, and you can reconvert to dimensional parameters afterwards.

    Chet
     
  17. Nov 20, 2014 #16
    I think that nails it! Thanks! Whatever I can say about the dimensionless form is also true for the one with dimensions since all the intermediate steps (cancelling) were valid. Is that right?
     
  18. Nov 20, 2014 #17
    Sure.
     
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