Understanding 0/0: The Undefined Concept in Energy Momentum Relation

[repugno]

As far as I understand 0/0 = undefined. What does this actually mean? is it somekind of constant? I came across this in the Energy momentum relation where p = 0/0 for a photon, but it can still be stated that E = pc. Thanks in advance

 

[TD]

I can't tell you anything about the physical interpretation in your example, but mathematically it's nonsense, if I may say so. It's a meaningless expression. You cannot divide by zero (this is undefined) and 0/0 doesn't change that.

However, when we're dealing with functions, and more specifically limits of functions, we can encounter this form and the outcome may still be defined. We call this (and some others, usually involving 0 or infinity in some way) indeterminate forms because the actual outcome is unclear and will depend on the functions in question.

A trivial example

\mathop {\lim }\limits_{x \to 0} \frac{{x^2 }}{x}

The function itself is undefined at x = 0 and when filling in this value, we get the indeterminate form 0/0. The limit exists though, and is equal to 0 (which can be easily found after algebraic simplification, x²/x -> x).

 

[Nimz]

I thought for a photon, the relationship is E=hc/lambda. Thus Planck's constant divided by wavelength is the momentum for a photon, which is not 0/0.

 

[tmc]

p is not 0/0 for a photon...

0/0 is simply undefined because division by 0 is undefined. Dividing by a is multiplying by a^{-1}, where a^{-1} is defined as being the number such that a * a^{-1} = 1 = a^{-1}*a.

Taking a=0, you can see that there is no such a^{-1}, since for any b, 0 * b = 0 != 1.

Hence, 0 has no inverse, and multiplying by the inverse of zero makes absolutely no sense. Thus, division by zero is utter nonsense, and as such is left as "undefined".

 

[The Guru Kid]

Think of the Energy equation:

E^2 = (mc^2)^2 + (pc)^2

when m = 0, E=pc.

The equation by which you get 0/0 only applies for m>0.

 

[Hootenanny]

The momentum of a photon can never be 0/0! The momentum of a photon is given by;

p = \frac{h}{\lambda}

h is Plank's constant (which is obviously not zero) and \lambda is the wavelength (which is also not zero), so I can't see how you've found the momentum of a photon to be zero.

-Hoot

 

[repugno]

I got it from the relativistic definition for momentum.

p = mv/sqrt(1 - v^2/c^2) and when you put v = c it should become clear where I got it from. I was curious since they use this definition for momentum when they derive the energy momentum relation.

 

[Hootenanny]

That formula is not valid when considering the momentum of a particle which has zero rest mass for precisly the reason you stated, it approaches 0/0. Therefore, we must use the equation;

p = \frac{h}{\lambda}

which is a derivation of the formula you quoted (put the rest mass equal to zero, them apply E = hv).

Regards
-Hoot

 

[Nimz]

p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}
is only valid when m>0, as guru kid pointed out.

 

[Nimz]

Another way you could look at that formula is a proof that particles moving at the speed of light have zero rest mass, since energy (and hence momentum) for a single particle is less than infinity. The sea of virtual particles is another story, but that's what normalization is for :P

In order to get a meaningful expression for the momentum of a particle moving at the speed of light (or equivalently, for a massless particle) you have to use a formula that doesn't give you 0/0. And that formula has now been posted several times in this thread.

 

[repugno]

thanks, things are clearing up now but I'm still not completely satisfied. The derivation of the energy momentum relation was done (according to my uni notes) by originally defining p = mv/sqrt( 1 - v^2/c^2) it follows that E = pc for massless particles and that you can equate this with E = hf and find that p = h/lambda. But originally p = mv/sqrt( 1 - v^2/c^2) so how can you dismiss this and then just say p = h/lambda. So it's possible to just ignore the original definition because it doesn't make sense?

 

[Hurkyl]

As far as I understand 0/0 = undefined. What does this actually mean?

It means that it's gibberish. Nonsense. Gobbledegook. It's the formal equivalent of being gramatically incorrect. In the formal language of real number arithmetic, it says "Thou shalt put a real number into the numerator, and thou shalt put a nonzero real number into the denominator, and the combination shall henceforth be known as a real number."

Normally, when you get 0/0 in the course of doing arithmetic, it's because you've made the incorrect assumption that some number was nonzero. This has already been pointed out, but I just wanted to harp on it again.

by originally defining p = mv/sqrt( 1 - v^2/c^2)

I suspect they didn't actually define it -- the word "define" is often used inappropraitely, but let's assume you're right.

What was defined was some thing I'll call "tardyonic momentum" -- it is a momentum that is only defined for tardyons. (Particles that travel slower than c)

Then, an algebraic relation was derived: probably

E² = (mc²)² + (pc)²

so this formula suggests a new, better thing I'll call "tardyonic-luxonic momentum": this thing is applicable both to tardyons and to luxons (particles that travel at c).

So it's not that the original definition was ignored -- it's that a new and better concept was devised, and we're using that one now.

 

[HallsofIvy]

It shouldn't too difficult to see that "undefined" means: undefined.

It's not defined- there is no such thing! It's not a "constant", it's not number- it's just not anything!

 

[nrqed]

That formula is not valid when considering the momentum of a particle which has zero rest mass for precisly the reason you stated, it approaches 0/0. Therefore, we must use the equation;

p = \frac{h}{\lambda}

which is a derivation of the formula you quoted (put the rest mass equal to zero, them apply E = hv).

Regards
-Hoot

well, there is no need to get into quantum physics here! It seems that it will only confuse even more the OP.


To the OP:

The key point is that there are several properties:

Energy, momentum, mass, and velocity , for massive particles

OR

Energy and momentum, for massless particles (in which case the speed is automatically c so no need to list it as a variable).


For *massive particles) there is an equation relating momentum, mass and speed, an equation relating energy, mass and speed and, if you plug one into the other, an equation relating energy, mass and momentum.


For a massless particle, there is NO equation of the first two forms. Only one euqation is left: the one between momentum and energy, and that is E=cp. That's it.

 

[Mattara]

1/1 = 1
1/ 0.1 = 10
1/ 0.01 = 100
1/ 0.001 = 1000
1 / 0.0001 = 10000
1/ 0.0000.= 100000
/.../
1/0.00000000... -> \infty

Works the same with 0, only that it won't work at all as it is against almost every principle in math.

 

[HallsofIvy]

1/1 = 1
1/ 0.1 = 10
1/ 0.01 = 100
1/ 0.001 = 1000
1 / 0.0001 = 10000
1/ 0.0000.= 100000
/.../
1/0.00000000... -> \infty

Works the same with 0, only that it won't work at all as it is against almost every principle in math.

This is not only irrelevant to the original question, it is wrong.
The lim_{x\rightarrow 0}\frac{1}{x} does not exist.

 

[Doodle Bob]

Works the same with 0...

Um, no, it definitely does not.

0/1 = 0
0/ 0.1 = 0
0/ 0.01 = 0
0/ 0.001 = 0
0 / 0.0001 = 0
0/ 0.0000.= 0
...

Quite a different situation than yours.

 

[Mattara]

This is not only irrelevant to the original question, it is wrong.
The lim_{x\rightarrow 0}\frac{1}{x} does not exist.

Does this mean that when the number in the denominator goes closer and the closer to zero, the answer does not go towards \infty? Or did I forget the equal sign?

I was only trying to give a general background to what division with smaller and smaller numbers (and then ulitmately 0 leads to) so I do not agree that it is irrelevant even if the example had another numerator.

 

[Your.Master]

The limit as x approaches 0 from above is infinity as you described.

The limit as x approaches 0 from below is NEGATIVE infinity in a manner analogous to how you describe.

Since the left-hand and right-hand limits do not coincide, the limit as x approaches 0 period does not exist.

-------------------------------------------

The original question was about 0/0, which Doodle Bob has already pointed out does not lead to infinity as in your example because of the differing numerator. Furthermore, it was not stated as a limit question. I don't see how smaller and smaller limits are relevant to it, although I can see the relevance to later discussion in the thread.

(In high school they taught me that X/0 was undefined for all X but 0/0 was also indeterminate, citing that

0/0 ?= X

Multiplying both sides by 0:

0 ?= (X)(0) = 0

Which is so far consistent for all X, but becomes inconsistent when X1 and X2 are compared and found unequal despite the fact that 0/0 is consistent with being equal to them both.

Then in University that idea never really came up again.)