Philosophy of Physics Philosophy of Physics Forum  Philosophical questions about our universe  12Likes  2 Post By HallsofIvy
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Dec 27th 2016, 12:23 PM

#1  Junior Member
Join Date: Dec 2016
Posts: 10
 0/0 = 1
Hello there.
My thoughts on the multiverse, and the question whether it is infinite or not, guided me to some searches on the net and articles which are related to the subject.
I have found an article which claims 0/0 = 1 according to Relativity. Here is the link: Anti Aristotle?The Division of Zero by Zero
Unfortunately my math and physics level don't allow me to qualify the article, but the conclusion 0/0 = 1 has a deep meaning for me, having a philosophical taste.
So my main question is whether the article is valid, and also I'd be very interested to read your thoughts and debates on the subject.
Thanking you in anticipation.
EDIT: I found some other article which addresses the issue in some other way which makes it realky interesting for me. Here is the link: http://www.friesian.com/zero.htm
In the article it is mentioned:
"The logical problem of 0/0 has a real world application. Photons, subatomic particles that are the field quanta, the bosons, of electromagnetic radiation, have zero rest mass. However, they do have momentum, which ordinarily is mass times velocity. Since their mass is zero, and zero times anything is zero, one might think that, whatever their velocity, their momentum would be zero. Not so. And we can see why. Photons instantaneously travel at the velocity of light. Indeed, they are light. According to Einstein's theory of Special Relativity, any moving object approaching the velocity of light has its mass multiplied to the extent that its mass would be infinite at the velocity of light. This is the physical reason why things can't go faster than light, since it would take an infinite force to accelerate something that has an infinite mass. With light itself, however, it's already going the velocity of light. So in effect, we have zero mass times infinity. Zero times anything is zero, but then infinity times anything is infinite. So what happens to the photon? Well, its zero rest mass times the infinite multiplication of Relativity results in... a finite momentum, as though it had a nonzero rest mass times a velocity less than light. Since the reciprocal of zero (1/0) is infinite (or that "undefined" thing again), zero times infinity is equivalent to our problem of 0/0."
Is zero really multiplied by infinite in the equations of Relativity?!
Are massless particles essentially 0*infinite from a mathematical point of view?!
Thanks again
Last edited by Koohyar; Dec 27th 2016 at 05:23 PM.
Reason: adding the link to some other article...

 
Dec 27th 2016, 07:12 PM

#2  Senior Member
Join Date: Aug 2010
Posts: 126

"Anti Aristotle?The Division of Zero by Zero" is a very weird paper but what it really reduces to is that the limit of something that would evaluate as 0/0 if you just the put the limit value in can be taken so that the limit is 1. There is nothing strange about that. The paper itself includes a lot that would make me look askance. For example, the write goes on at length about basic Calculus facts and the history of Calculus, none of which is relevant. He talks about what a proof is, the rules of logic, etc. which are very strange for such a paper.

 
Dec 28th 2016, 10:23 AM

#3  Junior Member
Join Date: Dec 2016
Posts: 10

Unfortunately I'm not good at math and can't think about the subject mathematically.
What I got from the paper is that the writer is trying to show that L’Hôpital’s rule (whatever it is) is wrong and must be checked for validity, otherwise Reativity is wrong.I'm not sure if I have a correct clue about the paper. But I think what it is getting at is that accourding to Relativity zero divided by zero has a definition. For me division by zero and infinity are philosophically important. Considering the paper and the other link I mentioned in the first post are directly related to the subject, I have the same concern the writers of the articles have.
I think division by zero has a deep philosophical meaning and the "undefined" thing is a matter of definition. After searching more and more I see some other people apperantly have the same idea. What I can't be sure of is whether the links I sent here are scientifically valid or not.

 
Dec 28th 2016, 02:58 PM

#4  Senior Member
Join Date: Aug 2010
Posts: 126

I hadn't looked that far into the paper. But, for the record, "L'Hopital's rule" certainly is valid a simple proof is given in every Calculus text.
One interesting thing he says is
"Clearly, it is incorrect that 1 = 2. Multiplying this equation by 0 it is 1 0 = 2 0 and we obtain 0 = 0 which is correct. Dividing by zero, it is (0/0) = (0/0) and due to our finding (0/0) = 1 we obtain 1 = 1 which is of course correct. Thus far, we started with something incorrect by claiming that 1 = 2. After the division of zero by zero we obtained 1 = 1, i.e. something correct. This is a contradiction. "
First, this depends on the incorrect statement that "0/0= 1". Second is the logic error that "false implies true" is a "contradiction". In fact, from a false statement, any statement, true of false, follows. "Proof by contradiction" works the other way if you can show that statement A implies a false statement, that is a contradiction and statement A must be false.
Last edited by HallsofIvy; Dec 28th 2016 at 03:11 PM.

 
Dec 29th 2016, 04:24 AM

#5  Senior Member
Join Date: Jun 2016 Location: England
Posts: 148

0/0 on it's own tells you nothing about the answer.
However if the 0/0 is the result of a larger equation then the correct answer can sometimes be deduced.
For example x/x is 1 for all other values of x,
so we can deduce it is 1 when x=0.
However (3x(x1))/(x1) gives 0/0 when x=1 but gives 3x for all other values,
we can deduce that for this equation 0/0=3
Getting 0/0 from your equation generally implies either you have arranged your equation improperly
(easy to see with the simple examples above, but not so easy with Quantum equations...)
or you have missed some (small but vital) feature of the physical world from the mathematical description.

 
Dec 29th 2016, 02:26 PM

#6  Junior Member
Join Date: Dec 2016
Posts: 10

Thanks for your replies.
Considering this article: Zero Divided by Zero
Is what the article concludes a case of improper equation arrangement? Or is there something profound about the subject (something missing)?

 
Dec 29th 2016, 03:50 PM

#7  Senior Member
Join Date: Jun 2016 Location: England
Posts: 148

The article you reference makes the (correct) observation that 0/0 can have any value, but it then goes astray.
What is meant by 0/0 being undefined is not that it does not have a value, but that the value it should have cannot be determined.
For most equations that give rise to a 0/0 result there are mathematical methods available to deduce what actual value should be used (for example the values of the equation just a sort distance away from the 0/0 point).
Alternatively, when the equation is modelling a real world situation, the physical constraints of the situation will often clearly indicate the correct value.
With regard to the photon momentum mentioned in the article, this is an example of incorrectly understanding the mathematics.
The momentum should properly be expressed as a four dimensional vector (3 space dimensions and one time dimension).
The space dimensions have mass times velocity, which (since mass=0 are zero)
the time dimension has energy divided by the speed of light.
The energy is not zero, so the momentum is not zero, when considered in the wider four dimensional context.
This is not easy maths, so the mistake is understandable.
Last edited by Woody; Dec 29th 2016 at 03:55 PM.

 
Dec 30th 2016, 06:02 PM

#8  Senior Member
Join Date: Aug 2010
Posts: 126

Originally Posted by Woody The article you reference makes the (correct) observation that 0/0 can have any value, but it then goes astray.
What is meant by 0/0 being undefined is not that it does not have a value, but that the value it should have cannot be determined. 
Actually, at least in mathematics, "undefined" does mean that it does not have a value. Every math text book that I have seen, that mentions the point, distinguishes between "undefined" and "undetermined" with "undetermined" referring to things like "0/0" that can be given different values in the sense of limits.
For most equations that give rise to a 0/0 result there are mathematical methods available to deduce what actual value should be used (for example the values of the equation just a sort distance away from the 0/0 point).
Alternatively, when the equation is modelling a real world situation, the physical constraints of the situation will often clearly indicate the correct value.
With regard to the photon momentum mentioned in the article, this is an example of incorrectly understanding the mathematics.
The momentum should properly be expressed as a four dimensional vector (3 space dimensions and one time dimension).
The space dimensions have mass times velocity, which (since mass=0 are zero)
the time dimension has energy divided by the speed of light.
The energy is not zero, so the momentum is not zero, when considered in the wider four dimensional context.
This is not easy maths, so the mistake is understandable.
 
 
Jan 2nd 2017, 04:13 AM

#9  Senior Member
Join Date: Jun 2016 Location: England
Posts: 148

Ok, I stand corrected
(actually I'm sitting, but the phrase "I sit corrected" doesn't sound right).
However I suggest (with no more evidence than this is the way I would like things to be) that while equations in pure mathematics can be devised to give an undefined result, any applied mathematical model of the "real" world which results in 0/0 would be of the undetermined type.
(I bet someone out there can find an example to prove me wrong) 
 
Jan 9th 2017, 07:12 AM

#10  Forum Admin
Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 2,070

Originally Posted by Koohyar "The logical problem of 0/0 has a real world application. Photons, subatomic particles that are the field quanta, the bosons, of electromagnetic radiation, have zero rest mass. However, they do have momentum, which ordinarily is mass times velocity. Since their mass is zero, and zero times anything is zero, one might think that, whatever their velocity, their momentum would be zero. Not so. And we can see why. Photons instantaneously travel at the velocity of light. Indeed, they are light. According to Einstein's theory of Special Relativity, any moving object approaching the velocity of light has its mass multiplied to the extent that its mass would be infinite at the velocity of light. This is the physical reason why things can't go faster than light, since it would take an infinite force to accelerate something that has an infinite mass... 
This thread has been pretty well discussed so I'm not going to pick up where it left off. However I do want to put in my two cents for the first post.
What I am concerned with are the Relativistic parts.
There is little wrong with the analysis of a moving particle being discussed in terms of mass, but most modern sources use the concept of a variable momentum and not a variable mass. But the argument goes through correctly, so no big deal here.
It's the other part that bothers me.
...With light itself, however, it's already going the velocity of light. So in effect, we have zero mass times infinity. Zero times anything is zero, but then infinity times anything is infinite. So what happens to the photon? Well, its zero rest mass times the infinite multiplication of Relativity results in... a finite momentum, as though it had a nonzero rest mass times a velocity less than light. Since the reciprocal of zero (1/0) is infinite (or that "undefined" thing again), zero times infinity is equivalent to our problem of 0/0."
Is zero really multiplied by infinite in the equations of Relativity?!
Are massless particles essentially 0*infinite from a mathematical point of view?!

The source is incorrect. If we have a speed large enough to compare with the speed of light then the problem is relativistic and thus the equation p = mv is incorrect: It should be $\displaystyle p = \gamma mv$. You could argue that this still makes the momentum "effectively infinite" but we still cannot use an infinity argument here...Any object with mass can't have a speed equal to c, it must have a speed less that c. The correct equation to use (and I'm not going to derive it unless you specifically want it) is $\displaystyle E = \frac{hc}{\lambda}$ which contains no mention of p at all!
Dan
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