In 2D elasticity having axisymmetry condition we employ Airy's stress function (phi) in polar coordinate system.Here phi becomes a function only of radial distance (r). Integrating the subsequent bi harmonic equation with respect to ''r'' we obtain an expression for phi which contains terms like '''logr'. Clearly ''r'' is not non dimensional.
Can anyone please help me understand this fallacy
For reference one can kindly refer to ch-4, 2D problems in polar coordinates ,article 28,Theory of Elasticity (3rd Edition);Timoshenko ,Goodier
The argument cannot have dimensions. It can be a complex number. It is possible to express the logarithm in terms of a power series; clearly this does not work if r has dimensions.
Perhaps the equation was written in a dimensionless form which is very useful for many sitruations
@William... see my comment. There are two cases when you find arguments with an "apparent" dimension:
- when the formula is written as a difference of logs (i.e. log(r)-log(r0))
- when there is a division by a unitary dimensional quantity (i.e. log (r/1))
So the argument CAN have dimension, provided the overall formula is dimensionally correct
You are referring to equation (41) on page 69 of that book, right?
Note that there is the constant D. This constant term contains the -log(r_0) that Matteo referred to. So, indeed the equation is written correctly in terms of dimensions. Note that the same is true for all following equations containing log(r). They all contain a constant term that absorbs the normalisation term log(r_0), as required.
@ Jörg
thank you sir
I am indeed referring to eqn 41.I am still uncomfortable with the arguments since in eqn 42 a term 'B(1+2logr)' is appearing.In this term according to ur argument 'B' should have dimension of stress . I am fine with that, but there is no way 'r' can be non dimensionalised since its added to '1' that appears in the parentheses.
And what Matteo said if that were to be the case then log(r) ,log(r0) are individually dimensionally ambiguous quantities thereby contradicting a more fundamental defn of logarithm (which William was referring to).
We can actually add or subtract quantities which have the same dimension or dimensionless..this is what I think.So I feel ()
log(a)-log(b) = log(a/b)..is true only when all the three quantities here are dimensionless as well as 'a' and 'b'.
Further discussions are welcome
The notation log(5s)-log(1s) is certainly not nice, as each term individually is undefined. However, as a formal notation for log(5s/1s)=log(5), it is perfectly acceptable. You must not take the parentheses in B(1+log r) as an actual rule for in which sequence to calculate the answer, but only as a help to understand the contributions. For actual calculations, the implicit term B log r0 that is contained in C must be taken into account. Since this term only takes the unit from r, some authors find it acceptable to hide it in the general constant. In a formally more correct notation, this term would be explicit within the parentheses.
Many functions such as the trigonometric, hyperbolic, logaritmic functions can be defined in terms of power series and so their arguments cannot have units.
guys, I still don't see the problem with the formula... the answer is in my cryptic first message "Logarithms can be tricky.. "
Log[r]-Log[r0] is as valid as Log[r/r0] and the quantities don't need to be dimensionless.
There are two ways of demonstrating it and it is just a matter of elementary calculus (adding and subtracting the same quantity and using the property of logarithms that I have already written in that first message). In fact, if we define x(m) the quantity x expressed in the unit m, we can write:
Log[r(m)/r0(m)]
=
Log[r(m)]-Log[r0(m)]
=
Log[r(m)]-Log[r0(m)]+Log[1(m)]-Log[1(m)]
=
(Log[r(m)]-Log[1(m)])-(Log[r0(m)]-Log[1(m)])
=
Log[r(m)/1(m)]-Log[r0(m)/1(m)]
=
Log[r]-Log[r0]
You can observe that the quantity 1(m) i.e. a unit length, has been added and subtracted (i.e. it is not changing the result). The trick "eliminates" the units of measurement from the logarithm for those who do not like them there.
In the same way, the identity operation allows to write x(m) = x 1(m) and the same reasoning follows:
Log[r 1(m)]-Log[r0 1(m)]=Log[r]+Log[1(m)]-Log[r0]-Log[1(m)]=Log[r]-Log[r0]
In case of problems involving problems of real no.s, the argument of logarithm must be positive which is evident from its graph(i.e. graph of f(x) = log(x)) . This is available in any Mathematics text of Higher Secondary level. According to graph, the graph lies in 1st quadrant and 4th quadrant. Therefore value of log(x) can be negative; but its argument x must be positive. In problems involving complex analysis, as no ordering is present among complex numbers, hence question regarding +ive or -ive of argument of logarithm is meaningless.
@ Matteo
Thank you sir for your effort
If you check the derivation of the formula: log(a)-log(b) = log(a/b)
from a more fundamental defn of logarithm then it might help why we are having this reservation!
I will give it a try in my way ..correct me if I am wrong
ab=(a)*(b) % at this stage a and b can have any dimension for them not necessarily same or even dimension less also.
But when I write
exp(log(ab))=exp(log(a))*exp(log(b))...I definitely mean a and b are dimensionless!!
=exp(log(a)+log(b))
implies:log(ab)=log(a)+log(b)
@Debojoti.. please read my last answer a bit more carefully.. I have corrected a typo and added a further more explanation. If you still don't get it ,take
Log[3 m] -Log[2 m]
where 3 m is 3 meters and 2 m is 2 meters. Now this is EQUIVALENT to:
Log [3m/1m]-Log[2m/1m] = Log[3] -Log[2]
where I divided each of the lengths by 1 m (3m/1m = 3 and there is no measurement units). It can be easily seen that :
Log[3m/1m]= Log[3m]-Log[1m]
-Log[2m/1m]= -Log[2m]+Log[1m]
now sum the two and Log[1m]-Log[1m] cancel independently of the fact that you might consider this "correct" or not, as they are equal and opposite. This proves that
Log[3 m] -Log[2 m] = Log[3] -Log[2]
I hope this concludes the debate ...
Matteo,
I'm sorry to say that you are doing a very bad job. By no means Log[3 m] is equal to Log[3 m/1m]. Equivalence relation you mention is different from equality and it *does not* apply here, as during calculations we put *equality* sign between successive steps.
Unfortunately, you are not the only person doing such terrible things. I see repeatedly (even in best journals!) the formula:
M(t) = M(t=0) - S Log(t),
describing the magnetic viscosity, with 't' being time, of course (in seconds? days? years?). The only sense of such a formula is for presentation that its graph is a straight line, when time axis is taken in logarithmic scale. The same goes for lognormal distribution, only rarely presented in a correct, dimensionless form.
Dear Marek,
I accept your criticism, but I am also going to accept your apologies after you will take the time to read carefully what I wrote in my comment, before saying I do "such terrible things".
Nothing more terrible than doing a strong or a harsh but wrong comment, right? This is something common on the site: people tend to answer, no matter if thy are 100% sure about the answer
If you read carefully my comments you will see that I wrote ALWAYS ONLY differences of logarithms or log of ratios of positive quantities (which is mathematically equivalent). So, unfortunately for you, what I wrote is correct from a mathematical point of view. I did not write that Log[3m] = Log[3m/1m] as this is clear nonsense. What I wrote, instead is Log[3m/1m] = Log[3m]-Log[1m] and it was clear that I did not wrongly assume that Log[1m] = 0 as this is an assumption that cannot be made, exactly for the same reason that Log[3m] alone cannot be calculated.
Of course everything works because measurements are real and positive quantities. Have fun. Mathematics can be !
Dear Matteo,
All the things like our RG scores, number of publications, perhaps number of citations, and who knows what else, keep telling me: shut up and apologize. Yes, it is my fault to respond to briefly, up to the point of skipping the "important" part of an intermediate expression. But let' s analyze your way of thinking again. You wrote:
Log[3m/1m]= Log[3m]-Log[1m]
-Log[2m/1m]= -Log[2m]+Log[1m]
If the equality character is meaningful, i.e. shows the equality of both sides, then the above "equations" are no equations at all. Their left hand sides are perfect, well defined, while their right hand sides mean *nothing*. Consequently, the next step, where the terms Log[1m], carrying no sense at all, cancel out, is doubtful at least - to say it politely. I AM 100% sure about that.
You also wrote:
Of course everything works because measurements are real and positive quantities.
Wrong again. The equality Log[a/b] = Log[a] - Log[b] is true even for complex a and b, provided both sides of this equality have sense.
All I can do for you is not to down-vote your previous contribution to this thread. But I will be forced to do so, if you continue to proliferate such pitiful practices. Otherwise some day you may want to convince us that Log[ 1 hour] = Log[1] + Log[hour] = Log[hour]. Are you sure this is 'correct from a mathematical point of view' as you wrote?
Dear Marek,
no problem having apparently contrasting ideas. This is the engine of science and I have no problem in accepting criticism, negative comments or anything else. And when people tell me they would be forced to down-vote me for trying to make my point (this is usually a polite way of saying: stop I am right by definition), I definitely have no problem in replying :) Nothing personal, don't worry.
What it seems, is that you don't like symbolic calculations.
There are still a few issues with your comment: yes I still believe that
Log[3m/1m]= Log[3m]-Log[1m]
is correct from a mathematical and a physical point of view. You would not say it is a scandal if I write:
a = 3 m
b = 1 m
Log[a/b]= Log[a]-Log[b]
or
Log[3 a/1 a]= Log[3a]-Log[1a]
right? It's a matter of symbols and my Mathematica digests all of this symbolically. What my equation tells is that something you can't calculate (i.e. Log[3 m] or Log[1 m] as separate entities), can be calculated if you take a difference. The = sign tells you that you can substitute the rsh for the lhs. It does not tell you anything about the validity of each terms. Now if you tell me that the right side means nothing to you, this implicitly means the left side means nothing as well. So, you can't tell anything about one of the terms on the left or right side unless you have properly defined it. The fact you can't calculate each individual terms (or you are not give n the definition), does not mean they are meaningless, right? If I tell you
Log[a/b] = f[a] +g[b]
Log[c/b] = f[c] + g[b]
then, is it correct for you to write:
Log[a/b]-Log[c/b]=f[a]-f[c]
or not? According to what you wrote above it is not. I did not give you the definition of the functions and I can easily tell you a posteriori that g[b] = 4*b + m+ s + kg^1/3. No meaning at all, but it works in any case.
In any case when you find two equal symbols (as in my case), can't you use again the rule of logarithms that we defined?
Log[1m] -Log[1m]= Log[1 m/1 m] = Log[1]= 0
If it is a rule, it is a rule.
What I do not understand is why you keep insisting in working with single logs like your Log[1 hour] alone and you insist I give you a definition of this. Is like getting back to defining Log[1 m]. I would not do it and I do not intend to do it as it has no meaning to me as an entity alone. I can give you a definition e.g. of
Log[1 hour]-Log[3 hour] = -Log[3]
if you like, or any such differences. But do not ask me or pretend I gave the definition of the single entity Log[ number unit] where number is a real and unit is a measurement unit.
In any case, I like challenges and it is easy to demonstrate that your definition in any case works . So let's take it:
Log[ 1 hour] = Log[1] + Log[hour] = Log[hour]
This simply tells you that you can write something known as the difference of two unknowns, as done previously (symbolic maths):
0 = Log[1 hour]-Log[hour]
we can also write then
Log[ 3 hour] = Log[3] + Log[hour]
I insist I don't need to know the meaning of what's on the rhs of Log[ n hour]. What can certainly do is:
Log[1 hour/3hour]=Log[ 1 hour]-Log[3hour]=Log[hour]- Log[hour]-Log[3]=-Log[3]
as it should be.
So yes, from my opinion this is correct from a mathematical point of view. I definitely hope I can find some mathematician here showing me it's not. I am using exactly the same idea (using equivalent symbols or combinations of symbols for parts of an expression I don't even know how to calculate), to help Mathematica simplifying some hundred lines formulas and so far it always worked. Just luck?
Dear Matteo,
It is nice to see your answer after all. You are brave to present your point of view, but the way you do it may only horrify mathematicians. The formula Log[a/b] = Log[a] - Log[b] is nothing else but theorem. As such it is valid only when its prerequisites are satisfied. There are many mathematical theorems looking as equality relations but ending with phrase "when both sides exist", see for example de l"Hospital's "rules" to calculate limit of a/b when a and b go to zero. The point that "Mathematica says so" is in fact a very poor argument. Unbelievable? You certainly know that solving a difficult equation you obtain nothing else but solution candidate(s). You should check which one of them are indeed solutions, if any. Mathematica (or any other symbolic package) never performs such checks on its own. It is your task to ask it to perform such a check. Thus mechanically applying well known "rules", without necessary care leads in discussed case to
Log[1/2] = Log[(-1)/(-2)] - true, but following carelessly your way it is also
= Log[-1] - Log[-2] - right?
Of course, it may happen that some of your intermediate results are extremely difficult to evaluate. This is not the same as the "results" having no sense, or non-existing in other words, not being the valid objects.
Are you lucky? Yes and no. Truth always implies truth but anything may follow from false, truth (your luck) or false. I'm sure, you know that.
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