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Wrong Solution/Right Answer and Logrithms

  1. Jun 22, 2011 #1
    The topic of this post is about the applicability of mathematical formalism in applied physics, and especially how physical units as well as physical variables might have proper attention.

    I've come upon an engineering problem requiring integration of 1/R. R is in centimeters. It has associated units.

    To simplify, I have a radially symmetric problem in thermal resistance, that can be stated as

    [tex]R_{Th} = K \int^{R_b}_{R_a}dR \frac{1}{R}[/tex]

    R is the radial dimension.
    So I went to a table of indefinite integrals and found this

    [tex]\int \frac{dx}{x} = ln(x)[/tex]
    Applying this template function in the usual way I get

    [tex]R_{Th} = K [ln(R_{b}) - ln(R_a)][/tex]
    Now, I'm stuck. I can find ln(R), but I don't know how many centimeters are in one log centimeter.

    How many cm are in one ln(cm)?

    I've rewritten my log tables to say

    [tex]\int \frac{dx}{x} = ln(x/c)\ , c>0[/tex]
    Anything wrong with this? (Please don't nit-pick over x vs. |x|.)
    Last edited: Jun 22, 2011
  2. jcsd
  3. Jun 22, 2011 #2


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    Isn't it dimensionless?

    Like you showed above, you can simply reformulate the expression to be

    [tex]K\ln R_b - K\ln R_a = K\ln \left(\frac{R_b}{R_a}\right) [/tex]
    The dimensions of the result is just the dimensions of your constant K.
  4. Jun 22, 2011 #3
    Thank you for responding, B2bw.

    Of course, I know the answer. But I also know the equation you have given is meaningless with dimensionful units of R.

    The issue, which I addressed, in ammendment to my post (sorry), is about the formal treatment of physical units in mathematics and where it fails.
  5. Jun 22, 2011 #4


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    Bomb2bwire is right. If you aren't familiar with basic facts about logs such as log x - log y = log (x/y), you probably shouldn't be trying to "correct" your formula sheet just yet.

    That is a meaningless question - it's like asking "how many cm are in one cm2", etc.
  6. Jun 22, 2011 #5

    Vanadium 50

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    The logarithm of a product is a sum. log(5 cm) is therefore log(5) + log(cm). While log(cm) is, by itself, meaningless, in your equation it is subtracted off by another log(cm) so it never appears in the final answer. It's a silly way to write it, to be sure, but there is no problem, and at no point are you ever (as you suggest) adding cm to log(cm).
  7. Jun 24, 2011 #6
    My already poorly developed communicative skills seem to have taken a turn for the worst--again.

    Could you show what you mean at each step with attached units? For instance, a value like 12 centimeters could be written as 12[cm]. A pure scalar might be written 3.

    If I could back up for a moment, rather than making a bland assurtion, I should ask

    "Are all formal statements of scalar equations within simple algebra equally true when unit modifiers are bound to scalar values?" I'm not so sure this is so.​

    For instance, given ln x/y = ln x - ln y as a true algebraic statement,

    then ln x[L]/y[L] = ln x[L] - ln x[L].

    Each side is unitless.

    What are the units of k in the expression k=ln x[M] ?
    Last edited: Jun 24, 2011
  8. Jun 24, 2011 #7

    Claude Bile

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    Expressions in logs and exponentials are inherently dimensionless.

    Examine a few common formulae with exponents and logs and see this for yourself.

  9. Jun 24, 2011 #8
    Yeah, well, I guess I'll try this in the math folder.
  10. Jun 25, 2011 #9


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    Phrak, I think I understand your question, as I had wondered about it myself years ago.

    The value of an expression like [itex]\ln{R_b} - \ln{R_a}[/itex] is independent of the units used because, as has been pointed out, it is equivalent to [itex]\ln{\frac {R_b}{R_a} }[/itex], for which the units obviously cancel out.

    It can be confusing that the value of [itex] \ln{R_b} [/itex] , by itself, does depend on the units used for Rb. However, in physics such an expression (where the argument is not dimensionless) always appears as the difference between two logarithms, and that difference is independent of the units. As it must, to be physically meaningful.

    Hope that helps.

    EDIT / p.s.:

    Strictly speaking, it is not allowed to take the logarithm of a dimensioned quantity such as "10 cm". But as long as the two logarithm arguments are in the same units, it works out mathematically:

    [tex]\ln(x/y) \ = \ \ln{\frac{x / \text{(1 cm)}}{y / \text{(1 cm)}}}
    \ = \ \ln{\frac{x}{\text{(1 cm)}}} - \ln{\frac{y}{\text{(1 cm)}}}[/tex]

    In the final expression, each logarithm argument is a dimensionless quantity, numerically equal to x (or y) in cm. But we could easily have used inches or meters instead of cm; the result is the same no matter what the units are.
    Last edited: Jun 25, 2011
  11. Jun 26, 2011 #10
    Redbelly, many years ago when solving exam or homework questions the same thing crossed my mind. I didn't pursue it then. Now it bothers me.

    The failure is mine; not to construct a well posed question. But this should be expected. Students that come to this forum do this all the time when it doesn't come straight out of a textbook. The hardest part is forming a well constructed question. Maybe I'll figure out what it is.
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