Today, we have the opportunity to learn another fallacy. Previously, I shared in a comment in Najib's blog about another fallacy called "Poisoning the well". It involves discrediting someone's character in hopes of discrediting that person's argument WITHOUT actually debating the argument itself. For example, if I tell you that "Every one has the right to be stupid", then you say, "The Chess Ninja is a liar and you should never believe anything he says". This is what poisoning the well is. You did not debate the argument about how every one has the right to be stupid or not. Basically, you just claim I am a liar and hence, everything I say must be untrue. Hence, you would be able to claim that NOT every one has the right to be stupid.
So today, we have another opportunity, as I said, to learn a new fallacy. It is called, "False Analogies". While it is not a formal logical fallacy, it is one that is very commonly made by many people, including myself. We often drag in unrelated examples to prove a point. For example, I can argue that "Engineers, researchers, lawyers, accountants are allowed to refer to manuals, researches, law statutes, and accounting standards in their work. Therefore students should be allowed to refer to their books during exams." This is a form of false analogy because the purpose of a student taking an exam is to measure the student's aptitude and the amount he/she has learnt. This is quite different from the objectives of what an engineer, researcher, lawyer or accountant does.
To examine another example, take this statement, "In Physics, work is measured by effort followed by movement. If there is no movement, no work was done." Then we proceed to argue that we need milestones to show that there is effort and so on and so forth. Let us break it down in parts.
For starters, I am not disagreeing with the statement in any way. I am just trying to show how sometimes, we use fake examples to justify our statements. While they can appear related, they are actually distracting you to make an argument stronger than it actually is.
So the first sentence:
In Physics, work is measured by effort followed by movement.
This is easy enough. Every one knows work done ON an object is the NET FORCE (effort) applied on the object over a distance. Or for those of you who are more mathematically inclined, W = F x d (very simplified version, but it is actually the integral of F.d (since both force and displacement are both vectors)). But let us not complicate things. So the second sentence clearly follows:
If d=0, then W MUST = 0, i.e. if there is no movement, then there is no work done. Like I said, I am not trying to argue against the statement, but showing why this is a poor analogy.
But is the above relationship REALLY true?
WARNING: This part is for those of you who have some further training in Physics, as opposed to some people who try to use Physics without fully understanding it.
Now, you should have caught on that I used the term NET FORCE. The reason is this. If there are two equal forces pushing against each other, the net force is zero. When the net force is zero, it actually implies there is no acceleration. It does not mean that there is no movement. For example, an object moving at constant velocity (no acceleration), has no NET FORCE applied on it, but it is still moving. But because there is NO NET FORCE, i.e. F = 0, the work done is zero (W = F x d = 0 x d = 0).
So, if we use this analogy in "real life", we can claim that, if we are progressing on our own at constant velocity, and our own effort (FORCE), but moving against an equal but opposing resistance (opposing FORCE), NO WORK IS DONE!! This shows that even if we have milestones and meeting them, it is possible that no work is done. Yes, I have managed to twist this into something that is totally destructive, but I guess I can't help myself for having critical thinking. Perhaps next time, when you want to choose an example, use a real one. Or else, you will just sound stupid, but then again, you have the right to do so.
My head hurts just reading all that :)
ReplyDeleteI never really paid much attention in school especially in Science class, so I can't understand all those technical terms :)
Finally what I understood from this is that someone used a poor analogy to for his arguments.
1. chess players do not talk about chess
ReplyDelete2. non-chess 'player' talk so much about chess
3. ... this is Malaysia...
seng
Actually, work is done by the person. If I push against you with the same magnitude of force that you push against me, in the opposite direction, we're still applying a force. In other words, while net work is zero, the person is doing positive work while the resistive forces do negative work, both equal in magnitude. Conservation of energy :)
ReplyDeleteBut point taken anyway. And what YCS says is too true.
Chin Seng,
ReplyDeleteHaha, chess players can have other hobbies too :)
Anonymous,
You are incorrect. Work done is NOT a vector, and hence has no direction. It is a scalar quantity. Your Physics teacher might have taught you that scalar quantities "cannot" be negative. So, you can only use NET FORCE to multiply the distance. There is no such thing as "net work done". As I said, work done is actually the integral of (net force) dot (displacement). "Dot" is a vector operation that gives a scalar quantity.
The conservation of energy applies "in the system". When 2 opposing forces push against one another without movement, there is just a conversion of chemical energy to heat energy but NO kinetic energy. People exert energy by converting their "food energy" into heat. Heat is just dissipated. Thus, energy is still conserved. That is why you will feel hot if you just push against the wall. It is not because work done is cancelled out. The work done is zero because the net force is zero.
There is a case of "negative work done", but this applies in the case of thermodynamics. This is due to the frame of reference issue, i.e. are you looking at work done BY the system, or work done ON the system? I will leave this for your own research.
Another thing to add, an object that moves at a constant velocity has kinetic energy, yet it can have NO WORK DONE ON IT.
ReplyDeleteMovement without effort?!!
Work done can be negative if the direction of movement is against the direction of the force. Look up potential energy.
ReplyDeleteWhat your physics teacher might have not told you is that a scalar does not mean it has to be positive. Scalars just don't have direction, i.e. they are dimensionless constants. A number that can be negative does not automatically make it a vector. Temperature in Celsius is a scalar. Electric charge in Coulombs is a scalar. I will leave this for your own research.
Or simply look at the formula for work done, i.e. line integral over some trajectory C in the a vector field F. The dot product F.dr CAN be negative if the angle between the vectors is between 90 and 270 degrees.
Even simpler, look at it this way: The "person" is applying an effort. Yes? He's also moving. F is not zero, and the fact that he moves means that r is not zero. Hence F.dr can't be zero, and he's doing work. But then one might wonder, "This is impossible! The net work has to be zero!". Yes. Calculate the integral of F.dr for the resistive forces, and it's negative.
Your example of pushing against the wall is irrelevant; really no work is done there, because the wall does not move. Mathematically, r is zero, hence dr is zero and the integral of F.dr is zero.
Anyway, the whole point of an analogy is to simplify. So think about this: You're driving a car at a constant speed. Net work done is zero, hence you're not consuming energy, i.e. petrol to move the car?
Another thing to add, an object that moves at a constant velocity has kinetic energy, yet it can have NO WORK DONE ON IT is true. But if it moves against a resistance, then the driving force IS DOING WORK, equal to the WORK DONE AGAINST IT by the resistive forces. Net work is zero. But do not think that the force driving the object through its trajectory is not doing any work.
ReplyDeletehttp://en.wikipedia.org/wiki/Work_(physics)
Ctrl+F "negative".
I give you a dollar, you give me a dollar. Net flow of money is zero. It does not change the fact that I did give you a dollar.
More math:
ReplyDeleteLet the vector F1 be the driving force, and F2 be the resistive force. Since net force is zero, I claim that
F1-F2=0
Hence the integral over C of (F1-F2).dr=0
But since the dot product is linear, and a definite integral is a linear operator, over C,
{int}[(F1-F2).dr]={int}[F1.dr] + (- {int}[F2.dr])
But this is still zero!
so if {int}[F1.dr] is positive, it follows that you have to subtract {int}[F2.dr] for the expression to be zero. But what physical interpretation is there to subtracting work? As far as I know, it's equivalent to adding negative work.
Long story short, work done can be negative.
Q.E.D.
OK, my last post does not drive home the point, so let me add this:
ReplyDeleteAssume that only positive work can be done, and subtracting work is not equivalent to adding negative work. Then it follows that
{int}[F1.dr] = {int}[F2.dr]
since both forces do equal work. Now, if work done can't be negative, then the TOTAL work done within the system is {int}[F1.dr] + {int}[F2.dr] and hence, non-zero, as both terms are non-negative. And that's a contradiction to the premise that no net work has been done.
But reading again, I was wrong at the term "net work". It was the wrong choice of words, and does not exist due to the fact that work is indeed a scalar, and hence the proper terminology would be "total work". But what does exist is the work done by the driving force, and that done by the resistance which I can reassure you, is negative, if the total work done is zero.
ReplyDeleteAnyway, back onto the main topic:
Please do not talk condescendingly onto others (you are incorrect, "I will leave this for your own research" etc.), especially when you are uncertain on the correctness of your own statements (which I assure you, is invalid this one time). I am sure you come from a credible physics/mathematics background. But your credentials play no role when you make a false statement. Spending 3 paragraphs talking about the conservation of energy, thermodynamics and kinetic energy but only 1 about the dot product and scalars, does not change the fact that some scalars, work included, can be negative.
You've already made your point that there are bad analogies. The fact that your example is wrong does not mean that your whole post is invalid. There are many very true examples of lousy analogies to life(off the top of my head:quantum mechanics).
A shitty blogger is one who cannot accept that he can be wrong. There is already at least one in the Malaysia chess society, so please do not make it two.
1. I never said that work done CANNOT be negative. I said that your teacher might have said that. I even qualified cannot with " ". I didn't want to get into the details of negative work done. But thanks for the proof anyway.
ReplyDelete2. "Scalars just don't have direction, i.e. they are dimensionless constants".
They are not dimensionless. For example, distance is a scalar quantity and it has dimensions. But the first sentence is exactly what I just said. Your logic seems to have failed you as I also did not claim that negative numbers MUST have direction.
3. Actually, we are both talking about the same thing. You are just using "net work done" as work done. I am refraining from using the term "net" as it typically refers to vector quantities in physics. The "net" you are using indicates a arithmetic sum, whereas the term "net" that I am using is for a vector addition. Arguing on the terminology here is pointless.
4. I did not deny that there is work done by the force. I said work done ON the object is zero. Work done by an external system and work done on an object are completely different things.
And my point was that the object, assuming it was a person, was indeed doing work. So which part of my comment was wrong?
ReplyDeleteThanks for your advice. It is not my intention to be condescending. The internet does not convey tone very well. My apologies if it sounded as such.
ReplyDeleteMy intention for saying I will leave this for your own research was sincere. I just didn't want to get into the details of negative work done, which would totally bring this off tangent like you said. Hence, encouraging you to pursue it on your own.
Anyhow, thanks for the enlightening mathematical experience.
No part actually. Just to add, as you said, assuming the object itself is doing work, it is work done by the object. The resistive force is the external force applied on the object. From what you have shown, the work done ON the object is zero. Just pointing out the difference between work done on the object and by the object.
ReplyDeleteNow my head truly hurts...
ReplyDeletein chess, we also have W=Fd, where
ReplyDeleteW denotes war for chess is war game,
F (mating) forces,
d number of legal moves
W,F,d here are boolean ok, not scalar/vector
so we know when d=0 (is false) the game ends (W=0) as mate/stalemate, regardless of F
when F=0 (insufficient material, e.g. K vs. K) the game will eventually end (W=0) as draw but both sides may insist to play on, it's completely 'legal', until 50-move draw is reached,
...........LOL
Guys, please stop this. I cannot understand anything.
ReplyDeleteShould have been paying attention in school :)
Vomit blood!
ReplyDeleteMaybe this is what non chess playing feel whenever we start talking/ arguing chess e4 e5 castling, No fianchetto is better. Fritz don't agree...Rybka agreed, even give 1.25....
Ninja, nice argument and counter argument (though it all Greeks to me :) Keep it up!
That's how anyone can become a theoretical physicist! Raymond Siew included. Move over FGM. Here comes FTGM (First theoretical GM)
ReplyDeleteRaymond Siew can win Nobel Prize 2012 for being the FTGM to unify the theories of Quantum Chess and Theoretical Physics. He postulated that the primary driving force in the universe is simply WORK, WORK AND WORK!
ReplyDeleteNo, Raymond Siew has a chance of unifying the theory of everything with psychology. Well, at least he's trying.
ReplyDelete