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Unit cancellation in integration with natural logarithm?

  1. May 7, 2008 #1
    I was recently doing some problems with Entropy, and came to the startling conclusion of incorrect units for most of my answers!
    One problem, for example--was to calculate the change in entropy of a given sample as it gained heat per temperature.

    [tex]\Delta S=\int^{f}_{i}\frac{dQ_{r}}{T}[/tex]

    The dQr is assumed to be relatively reversible for the problem, and you end up solving the problem by recognizing Q as mc(delta)t.

    [tex]\Delta S=mc\int^{f}_{i}\frac{dT}{T}[/tex]

    Integrating yields mc(ln(Tf)-ln(Ti)), and then you recognize the properties of the natural logarithm and divide Tf by Ti, take the natural log, and finish the problem.

    The point where I get screwed up logically is that the units don't come out correctly if you don't treat the natural logarithm this way, and simply compute through subtraction. I know its a property of the natural logarithm to divide, and I've gotten stuck on problems like this in the past (simple rocket fuel propulsion examples, etc). My main question is--does this happen every time a natural logarithm is involved after integration? Are there cases where the units don't cancel like this? I know its a silly question, but it has been bugging me all night.
  2. jcsd
  3. May 7, 2008 #2


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    Hi Cvan! :smile:

    Yeah … never really occurred to me before …

    when we write:

    the dT and T have the same units, so dT/T is dimensionless (a scalar),

    but the logT looks as if it has the same dimensions as T.

    I suppose the answer is that log (to any base) can only act on dimensionless numbers, so technically one ought to write log(T/c), where c is a constant with the same dimensions as T! :smile:

    For sanity-threatening conundrums like this, I only have two pieces of advice …
    a) panic!
    b) don't panic!​

    … they both work … ! :smile:
  4. May 7, 2008 #3


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  5. May 7, 2008 #4


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    It may help to write the difference-in-logs as

    \ln(\frac{T_f}{1 K})-ln(\frac{T_i}{1 K})

    This gives pure-number arguments for the logs. Also, it does not matter what units you use in the denominators, as long as they're the same, since

    \ln(\frac{T_f}{1 \ Anything})-ln(\frac{T_i}{1 \ Anything})
    = \ln\left[(\frac{T_f}{1 \ Anything}) \cdot (\frac{1 \ Anything}{T_i})\right]
    = \ln(\frac{T_f}{T_i})

    independent of what "1 Anything" actually is.
  6. May 9, 2008 #5
    Crazier idea: treat the dimension as a scalar multiplier (which is what we do when we cancel them in a division), then you end up with the log of the dimensioned quantity equal to the log of the quantity plus the log of the dimension (whatever that means). You end up with the log of the dimensions cancelling by subtraction after just like they cancel by division in the ratio.

    I've gotta get a simple equation editor going here .........
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