# Pi

Discussion in 'Physics & Math' started by Pi-Sudoku, Aug 15, 2005.

1. ### James RJust this guy, you know?Staff Member

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Aer:

I'm consistent, but I don't always give the most complete possible answer. I tailor my answers to the audience, so as not to swamp people with unnecessary or confusing details. If they require clarification, I'm happy to give it. Sometimes it is better to sacrifice absolute accuracy for clarity.

I don't know what you're referring to here.

The right tools for the right job, Aer.

I've already explained that to you, above. Do you still have questions? If so, ask away.

1/0 is undefined for integers or real numbers. When v=c, γ is undefined, not infinite. v=c is not an approximation, either - it is exact.

No. I think you're arguing for the overthrow of relativity because you say relativity is BS, and you've said so.

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3. ### AerRegistered Senior Member

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Horseshit! You explained that 1/0 is infinity and then explained that it is not. Call that not always complete all you want, but that is just inconsistent no matter how you spin it.

I told you before, You need not give me dumbed down answers. In fact, if you do give me a dumbed down answer, I'm going to call you out on it everytime, ok?

You did not provide any clarification here. You provided a contradiction to your previous statement, then you recited the same BS that I had just told you about forgetting the definition of pc and just assuming pc=hf from the quantum physics result.

Definition of p in relativity: p=&gamma;mv

Therefore pc would be &gamma;mvc, BUT NO, we have to forget all about this for a photon and just redefine it as hf because that is what quantum physics gives us. Remember that relativity and quantum physics do not mingle very well! Interchanging results is what leads to the stupid deal of forgetting a definition just like I said before you offered your BS explanation which I know is not yours because I've heard it before and call it out for what it is - BS. Call it mainstream all you want, you cannot deny that you have to disregard the relativity definition of p to get pc=hf for a photon -from quantum mechanics- I might add. Just to be sure that you are not confused (your statements below seem to indicate that you are confused) I understand everything here, I am not seeking your knowledge of phyics for explanations. I know the explanations - I contend the stupidity inherient within. If you still don't know what stupidity I am talking about (I am sure you don't) Reread this paragraph. If you have any further questions, repeat the preceeding sentence until all clear. This should keep you busy for a couple hours.

Right, relativity can't explain it, so let's jump to quantum physics and force that back into relativity. Yeah... explain why exactly quantum physics and relativity cannot be unified again? How about you try to fix relativity instead of doing these ad hoc jobs.

I don't need any explanations from you. I explained everything you did before you explained it - and you did precisely as I predicted, recited the same BS argument of GR->QM->GR.

0 can only be an integer when 1/0 is undefined, including real numbers is redundant (all integers are real numbers).

You think wrong. Slight amendments to a theory is not the same as overthrowing it.

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5. ### James RJust this guy, you know?Staff Member

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Aer:

You seem to want to score points, so have this one for free. I've explained myself sufficiently for you to understand if you want to.

Actually, momentum is best defined in relativity using 4-vectors. We start with energy and 3-momentum and combine them to make a 4-vector, called the 4-momentum. The magnitude of the 4-momentum is then taken to be the rest mass of the particle.

In the case of photons, the magnitude of the 4-momentum is always zero, so photons have zero rest mass. For particles of non-zero rest mass, the 3-momentum, which is part of the 4-momentum vector, is given by the expression you have given. For photons, the 3-momentum is p=E/c, where E is the photon energy. This equation is also correct for massive particles in the limit of high speeds.

We don't forget it. We put the two theories together. We wouldn't know anything about h, Planck's constant, if not for quantum physics. If we didn't know any quantum mechanics, we could still use E=pc.

That's not quite true. In fact, one of our the most successful scientific theories, in terms of sheer accuracy, is quantum electrodynamics, is a combination of relativity and quantum physics.

You need quantum mechanics to get hf, of course, but not to get E=pc.

You tell me, since you claim to know all the "standard" BS explanations.

Which ad-hoc jobs?

I'm not going to argue with you that 0 is an integer. But you've changed your tune. Before, you were claiming 0 is just an "approximation" to something non-zero.

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7. ### AerRegistered Senior Member

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Sufficiently enough for me to point out your inconsistencies, yes.

The 4 vector momentum is just a contrived explanation of the E<sup>2</sup>=(pc)<sup>2</sup>+(mc<sup>2</sup>)<sup>2</sup> result. In fact, it is basically one in the same, just new terminology.

The result is circular. The 4 vector momentum does not explain the above energy equation, it is a result from it.

It is derived from E=&gamma;mc<sup>2</sup> so of course it is going to give the same result. The problem is, you must forget the definition of p=&gamma;mv which is also used in the derivation (so therefore should always be valid). This is the point you seem to be forgetting.

And E=pc=&gamma;mvc = &gamma;mc<sup>2</sup> for a photon moving at v=c (yes, I know they always move at v=c...). You don't get pc=hf anywhere from relativity and to say that pc=constant for a photon makes no sense unless you forget the definiton of p, because as we've seen p is undefined for anything moving at v=c in relativity. And just so we make clear that multiplying by c doesn't fix the issue:

undefined*1 = undefined
undefined*&infin; = undefined
undefined*0 = undefined

multiplication does nothing for an undefined quantity, no matter what you multiply it by.

Funny, I thought it was almost completely quantum theory. How is relativity involved?

I never argued that point. However, p is well defined in relativity. p=&gamma;mv.

Yours.

No. I claimed that for most instances in the real world, 0 is an approximation. The only exception is when dealing with counts and when dealing with theory when something can be exactly 0 by definition. I've never said anything other than this.

Last edited: Aug 26, 2005
8. ### James RJust this guy, you know?Staff Member

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30,358
Aer:

No. You're starting from the wrong end of the stick. You assume you know what m is, then you set about deriving the 4-momentum. A better approach, which is the one normally used in relativity, is to work out what 4-momentum should be, taking into account that we expect conservation of 4-momentum in collisions, and then derive what (rest) mass is from that. Of course, this is all a formal approach, and needs to be justified by matching predictions to experiments (e.g. is 4-momentum really conserved in collisions in the real world? If so, then our definition is correct and useful.)

4-momentum is not "just new terminology". It gives us fundamentally new insights into the nature of spacetime, because it is a tensor. It also unifies the treatment of massive and massless particles - something which could not be done before relativity.

Other way round.

As I explained, that equation is just a special case of three components of the 4-momentum.

Except that γmvc for a photon is a number which can't be calculated by working out γ and m and trying to multiply them - which brings us right back to the start of this conversation.

Not true. p=E/c for photons moving at v=c. That's well defined.

QED is a relativistic field theory. It's correct result for the gyromagnetic ratio of the electron, for example, depends on including relativistic effects.

Ok then. I have no argument with that. In many theories, though, things can be exactly zero, as I'm sure you'll agree. For instance, the rest mass of a photon is exactly zero.

9. ### AerRegistered Senior Member

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2,250
OK, derive for us rest mass in this way. The link in which Pete provided for deriving the quantity uses E=&gamma;mc<sup>2</sup> and p=&gamma;mv. In fact, it provided those as definitions and proceeded from there.

I keep telling you this but you keep not listening. Massless particles can only be analysed if the definition of p is forgotten so that massless particles can have momentum.

Any equation used in a derivation should be satisfied.

I am not arguing that it can't be used in the way you describe, only that you must forget the definition of p, because that will never be satisfied for a massless object.

Yes it does. I say that this means that &gamma; is only approximately correct, that is 1/0 is only an approximation and that there is another term that needs to be added onto the energy equation so that the m term becomes (m+b). Because the equation E=&gamma;mc<sup>2</sup> should be satisfied and p=&gamma;mv should be satisfied since both are used to derive the precious energy-momentum relation.

That comes from the energy-momentum relation and is not a definition. If we apply the relativistic definition of energy E=&gamma;mc<sup>2</sup> (I know it is undefined for a photon!) We get: p=&gamma;mc which is &gamma;mv for a photon since v=c and this is what one would expect since E=&gamma;mc<sup>2</sup> and p=&gamma;mv are used to derive the relation.

Now let me be clear, I never said that pc=hf is incorrect for a photon, I only said that you cannot get this result from relativity. It comes from quantum mechanics. I only said that you must disregard the definition of p=&gamma;mv in relativity to be able to import the relation pc=hf for a photon. That is it. That is all I am saying.

10. ### Pi-SudokuSlightly extremeRegistered Senior Member

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This getting realy complicated now

11. ### AerRegistered Senior Member

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2,250
What more would you like to know about pi?

We concluded: it is defined as the circumference divided by the radius of a circle, it is between 3.14 and 3.15, there is no known pattern to its digits, it is transcendental, what am I leaving out?

Oh yes, SL or someone said it is tasty.

12. ### James RJust this guy, you know?Staff Member

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30,358
Aer:

I'm not the link Pete provided. We're different.

My definition of p is the 3 space components of the 4-momentum. It works just fine for massless particles.

That's messy and unnecessary.

Both those equations are satisfied.

I agree. There's no h in relativity.

And all I am saying is that p=γmc for a photon is correct, but useless.

13. ### AerRegistered Senior Member

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Really?

Just when I had you all figured out you deliver this shocker.

Anyway, what exactly were you refering to when you said "That's exactly what I've been trying to explain to Aer in another thread." to Pete?

I assumed it was refering to this quote:
So we have: E=γmc<sup>2</sup> and p=γmv.

As David Morin said, we have two quantities E and p that are just given the label "Energy and momentum" and stated specifically that they were just labels.

Now, clearly these definitions are undefined for any object moving with v=c. I say "object" in the generic sense because we consider a photon an "object" with no implied meaning as to the attributes of this object.

Now, you must understand, the energy-momentum relation is derived from the definition of the above quantities, E and p. I know of no other way to derive them. For instance, let's take a look at the vector version of the derivation as adapted from the text of David Morin:

Start with the vector:

(cdt,dx,dy,dz)

Divide by proper time d&tau;=dt/&gamma; and define as V:

V &equiv &gamma;(c,dx/dt,dy/dt,dz/dt) = (&gamma;c,&gamma;v)

Multiply by m and define as P:

P &equiv (&gamma;mc,&gamma;mv)

Multiply by c:

Pc = (&gamma;mc<sup>2</sup>,&gamma;mvc) = (E,pc)

Take the inner product of Pc&middot;Pc:

Pc&middot;Pc = (&gamma;mc<sup>2</sup>)<sup>2</sup> - (&gamma;mvc)<sup>2</sup>

The value of the inner product is invariant and can be found most easily by choosing v=0:

(&gamma;mc<sup>2</sup>)<sup>2</sup> - (&gamma;mvc)<sup>2</sup> = (mc<sup>2</sup>)<sup>2</sup>

Rearrange and we get:

E<sup>2</sup> = (pc)<sup>2</sup> + (mc<sup>2</sup>)<sup>2</sup>

As you can see, the definitions of E and p are used in the same derivation considering the 4 vector energy-momentum that you've been yapping about. If you can provide a derivation of E<sup>2</sup> = (pc)<sup>2</sup> + (mc<sup>2</sup>)<sup>2</sup>, without using E=&gamma;mc<sup>2</sup> and p=&gamma;mv, then please do.

Does not. It is derived above, still uses p=&gamma;mv.

No, pc is undefined for a massless particle traveling at v=c in the definiton p=&gamma;mv.

So then you are changing your mind once again. You are saying that &gamma;m=constant since pc=hf for a photon and hf=constant, not "undefined".

14. ### shmoeRegistred UserRegistered Senior Member

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No kidding. In this case "Inf" is clearly being used to say "outside maximum range", and they assign gamma(1000) the value of "Inf" for convenience, not because it has a profound mathematical meaning, or that it's the definition of gamma(1000). My point was to not use matlab as your source for mathematics definitions as they are sometimes forced to make concessions for practical purposes.

I made no such argument. I said radians were the 'mathematically natural' choice to measure you angles in and I made it pretty clear that my own definition of 'natural' doesn't give a rats ass about making life more convenient for humans.

15. ### shmoeRegistred UserRegistered Senior Member

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Never have I even hinted at disregarding anything that is not a Real Number. You're wasting your time trying to convince me that the complex numbers are important, I'm well aware of them.

The text of mine you quoted was hopefully to get Aer (and anyone else) to at least think about what the things they called 'numbers' are before they go manipulating them. Before you can even talk about trying to divide N by 0, you have to know what N and 0 are and what sort of rules they obey. I kept mentioning the reals because they made sense given the context, that's all.

16. ### James RJust this guy, you know?Staff Member

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30,358
Aer:

Define the 4-momentum as P=(E/c, p), where p is the 3-momentum (which is not defined in advance) and E is the total energy.

The magnitude of P is (E/c)^2 - p^2 = (E<sub>0</sub>/c)^2

The value E<sub>0</sub> is frame-independent, and we can define a new quantity, called "rest mass" m, by E<sub>0</sub> = mc<sup>2</sup>.

For a photon, we find that m=0.

We can now go on to find an expression for p for massive quantities in terms of m, if we wish.

We've already covered this. Please re-read my previous post.

That's consistent with what I said before, not a change of mind.

17. ### AerRegistered Senior Member

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2,250
That is not how it works, the 4-momentum is derived, not defined, from E=&gamma;mc<sup>2</sup> and p=&gamma;mv as shown above in the way David Morin of Harvard shows. It is from QM that a photon is shown to have energy/momentum, so this is brought into relativitiy by using the 4-momentum P and disregarding the definitions E=&gamma;mc<sup>2</sup> and p=&gamma;mv because those equations are undefined for a massless object such as a photon.

Yes, this is from the invariance of the inner product, I already showed this above.

For a photon, we assume that m=0 becuase it is measured as 0 as far as we can tell from experiment.

That is not how it works. The 4 momentum was derived from the defintions of E and p. Now you are claiming that the 4 momentum is simply defined and that E and p are just results. No... that argument is BS.

OK OK - I finally get it!

&gamma;m = 1/0 * 0 = infinity * 0 "but [is not] the most complete possible answer" because 1/0 is undefined and &gamma;m is not defined. However, since &gamma;m=constant, 1/0 is defined after all, as... infinity?

Yeah.. uh huh, ok. :m:

Last edited: Aug 29, 2005
18. ### James RJust this guy, you know?Staff Member

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Aer:

That's one way to do it, but not the only way.

Yes. Now you're getting it! We can't use expressions which aren't defined, can we?

Yes, just as experiments ultimately support or refute any definitions we make and conclusions we draw from them.

E was previously defined, in my approach. p wasn't.

There are many ways to perform a formal derivation, Aer. Just because you've only seen it done one way doesn't mean that's the only way possible.

No, it's undefined.

Yes it is, for this particular expression. Once you discover it's undefined, you know you can't use it.

Right.

No. γ is undefined. m=0. You should avoid trying to use γm for photons, because it doesn't tell you anything.

19. ### AerRegistered Senior Member

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2,250
Well, since we are just defining things, I'll define P = (&gamma;<sup>4</sup>E&radic;c, 2p/c). Any idea why my definition is invalid? Because I defined it without any derivation which is precisely my point. There is no meaning in defining something if there is nowhere from whence it came. In the case of the 4 momentum vector, it comes from E=&gamma;mc<sup>2</sup> and p=&gamma;mv. Those equations must be satisfied for anything that uses the energy-momentum relation which is simply a result of the 4 vector momentum.

Technically I am not in the process of getting anything from you because I've known this all along, in fact I am not sure if it was you or I who brought it up - but one thing is for sure, I've known this for quite a long time.

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Good. We can now agree that experiments are the only and final say.

No, you just were not treating it in a mathematically rigorous manner and just chose to not show what p really is or how you can define a 4 vector momentum as such. Because if you did, you'd see that your p is very much well defined.

I've now show two ways. SL provided a third way which happens to be very similar to the derivation in the OP. Yours is very similiar to the second one I provided, just that you chose to not be thorough at all which any formal derivation requires.

Last edited: Aug 29, 2005
20. ### AerRegistered Senior Member

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2,250
Uh huh. Consistent you are not.

Curious... How do you define undefined*0? Undefined

21. ### James RJust this guy, you know?Staff Member

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Aer:

Ultimately, because it doesn't match experimental results, which we agree are the final arbiter, right?

Let's go back to Newton for a minute. Why did he define p=mv? Did he do it "without any derivation which is precisely my point"? What was his motivation?

Everything starts from an unproven assumption, Aer. Then you follow the consequences through and see whether the initial assumption was justified or not, with reference to experiments.

We're just starting at different ends of the derivation, Aer.

To give you an analogy: There are two possible ways to derive classical electromagnetism. One way is to start from observations such as Coulomb's law, and work your way through to Maxwell's equations. The other way is to assume Maxwell's equations, and show that you can derive things like Coulomb's law from them.

Exactly the same thing can be done in relativity.

Have another point for free, Aer. You're smart enough to understand my explanations, without me laboring them further, even though you pretend otherwise.

22. ### AerRegistered Senior Member

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Newton defined a quantity, p, known as momentum from the well defined and measurable mass, m, and the well defined and measurable velocity, v as the product of the two. As such, it is a mathematical definition that is well defined from known quanities.

In your case, you are defining p as what exactly? That's right, because of your arbitrary and not thorough ways, p could be just about anything and is merely a little tooting symbol. What is Q = (E/c<sup>3</sup>, <b>q</b>)? It's the same idea as your P = (E/c, <b>p</b>). If I accidentally find a neat result between E and p does that mean anything? Of course that definition for P was no accident, consequently, because it follows from E=&gamma;mc<sup>2</sup> and p=&gamma;mv.

p=mv was not an unproven assumption, it was simply a definition of momentum. As to where "momentum" means anything follows from the math. Since we know what a force is physically, we found that F = ma which happens to be F = m d(v)/dt; However this was always the case for an object with constant mass. With the concept of m varying with time comes F = d(mv)/dt and hence the defintion of p=mv.

Yes, you are starting with the result and claiming it leads to the fundamentals which is not a very good way of going about anything.

Would you care to share a result of Maxwell's equation that is undefined in Coulomb's law? Such that your analogy can be completed... Not saying that I agree with assuming equations and deriving fundamentals, but, still...

Exactly the same thing can be done with any equation, it does make it proper.

Last edited: Aug 29, 2005
23. ### James RJust this guy, you know?Staff Member

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Aer:

Sure. Take Gauss's law, for example. It introduces something called "the electric field", which it defines by the amount of charge enclosed by a surface in space. Using that concept, we can derive the Coulomb electrostatic force between two charged particles. If we measure that force in an experiment, we find that it is given correctly by the mathematics, which can be taken to retrospectively justify our arbitrary assumption of Gauss's law.

Similarly, if we start with a definition of the 4-momentum, we can derive the expected 3-momentum for a massive particle, and compare its behaviour with experimental evidence. Then, the initial definition is retrospectively justified, if we find that the definition makes accurate predictions.