Principia

Here is the analysis provided in the BE thread.
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But, for the lurkers, let us check the fundamental principle of dimensional consistency (from the MIT textbook above):

This principle is extremely important in order for a physical quantity to have meaning. If one measures, say, the side of a square to be 5 meters and another to be 2 meters, then dimensional consistency requires that the ratio of the areas be equivalent in any other units. In this case the function to calculate area of a square is f(x) = x^2 and (5[units])^2/(2[units])^2 is invariant under any change of units.

However, in Douglas's case, the following ratio, ln(x [units]) / ln(y [units]), is not invariant under change of units. Try it with [units] = [B years] and convert it to [T years]. As a result, dimensional consistency cannot be assured based on the above principles.

On the other hand ln(x [units]/1 [units]) / ln(y [units]/1 [units]) is dimensionally consistent. Converting x [units] to another system of units in this case requires a conversion back to [units] so that the division by 1 [units] is possible. This is the physical reason why the arguments of all transcendental functions must be dimensionless with respect to a reference unit.

The fact of the matter is that all definitions of ln(x) must be consistent and equivalent. Douglas is the one who is creating inconsistencies with your assertions. His admission that I was right from the previous posts is sufficient proof that he is mistaken. But, out of mercy, I offer him:

Hint: {The integral of 1 [unit] to x [unit] 1/t dt} = {The integral of 1 to x 1/u du}, where I substitute u = t/(1 [unit]). Notice that the RHS of the equation has limits that are unitless. Consequently, the equation says ln(x [unit]) = ln(x)! Thus, ln(x [unit1]) = ln(x [unit2]) = ln(x [unit3]) = ... = ln(x) for any unit [unitX]! Clearly such a result is nonsensical and physically meaningless (see the principle defined above). In other words, the integral definition in of itself provides no reasons why the argument may be dimensional. In fact, the result suggests otherwise.

Still, let me give another example by analogy. Douglas has claimed that sin(x) requires a dimesionless argument and returns a dimensionless quantity. But, let us look at the Taylor series definition of sin(x):

sin(x) = sin(0) + x sin'(0) + 1/2 x^2 sin''(0) + ... =
sum from n = 0 to infinity of x^n / n! sin[n](0), where sin[n](0) means the nth derivative of sin(x) evaluated at x = 0.

Notice that if x has units [unit], then x^n has units [unit]^n. Also notice that sin[n](0) has units of 1/([unit]^n) (since each derivative divides by [unit]). Thus, the units cancel! Does that mean sin(x) can take dimensional arguments? Absolutely not! The Taylor series per se provides no reasons why sin(x) may be dimensional (or dimensionless). Similarly, the integral definition of ln(x) per se gives no reason why x may be dimensional.

[ June 04, 2002: Message edited by: Scientiae ]</p>

Principia

And I completely agree. The problem of course is constructing an explanation that is acceptable to cranks. If one is willing to accept an argument from authority, then I am more than willing to provide it (and in this case, one was indeed requested).

Perchance, do you have any reasons of your own to add to the discussion about the dimensional analysis of ln(x)? It seems that all you have done so far is taken a stance on the issue at hand.

[ May 26, 2002: Message edited by: Scientiae ]</p>

Principia

And there we have it, DJB's <a href="http://iidb.org/ubb/ultimatebb.php?ubb=get_topic&f=47&t=000386&p=9" target="_blank">best argument</a>: if there are inconsistent definitions of ln(x), ignore them or better yet pretend they don't exist. LOL

I win.

[ May 26, 2002: Message edited by: Scientiae ]</p>

AdamWho

Scientiae:
No I really can't add anymore facts to the discussion, who can, the facts are out there, one can except them or not. Maybe we could talk about more pertinent things like: Do you know how to make tamales? We are looking at recipes but they seem to differ quite a bit, which one to choose from…? Or are you still getting it?

Principia

You're right, quite right. I think I'll claim this a victory and move on.

Tamales? How about, your best chili recipe?

Ragnarok

I hear that taking the natural log of tamales improves the flavor.

Coragyps

Tamales! Mi amiga Martina puede hacer los mejores del mundo! Or, hell, just go to Tia Juanita's in Post, Texas! And they're all unitless!

Principia

And if you make 36.5 of them, your odds of getting some that night improves significantly...

[ May 31, 2002: Message edited by: Scientiae ]</p>

AdamWho

Tamales are in a log shape. I like to make vege chill with lintels and tofu.
But this is NOT a proclaimation vegatarianism.

[ May 26, 2002: Message edited by: AdamWho ]</p>