Definite product of zero and infinity?

[Antonio Lao]

Can the product of zero and infinity be defined as a finite quantity?

The differential calculus makes the limit approach to zero possible (existence of a derivative).

The integral calculus makes the limit approach to infinity possible (convergence of infinite series).

 

[matt grime]

The product of something that tends to zero, say a function f as x tends to 0, and something that tends to infinity also at zero may have a limit or it may not.

x and 1/x say, the product is 1, and the limit is 1. 2x and 1/x has 2 as the limit of the product. x and 1/x^2, the product has no limit as x tends to zero.

If you just want to declare that 0*infinity is something you'll need to justify why there is number 'infinity' in your number system, or indeed what your number system is, as it ain't the reals.

 

[Antonio Lao]

Thanks for your reply.

The cases you gave look like product of inverse functions. I am going to give an example of what is it I have in mind.

In the theory of gases, Boyle's law says at constant temperature, the product of pressure and volume is a constant.

If the pressure goes to infinity then the volume goes to zero to keep the product constant. Can a definition be made when the values are exactly at infinity and zero and still keep the constant?

 

[matt grime]

What makes you think the law is still valid there? What even makes you think that isn't a nonsensical question to ask physically? Mathematically if y = 1/x for x not zero, then $$\lim_{x\to 0} xy = 1$$

 

[Chen]

Why should the law still apply when talking about something as unphysical as "infinite pressure"?

 

[Antonio Lao]

Thanks for all your replies. But physically, we can say such thing as zero pressure which is the same thing as there is no force. Now the volume is infinite.

 

[Michael D. Sewell]

Zero pressure is in no way indicative of infinite volume.

 

[matt grime]

No, definitely not - 'zero pressure' would correspond to absolute zero, there is still a volume there. The ideal gas law is not valid at these extremes.

Note, this answer was directed at Antonio, not Michael

 

[Michael D. Sewell]

Extra credit question on test:

Does a vacuum with a volume of 3 meters^3 contain three times as much Nothing as a vacuum with a volume of 1 meter^3?

 

[Antonio Lao]

Thanks. Now to give another case where this is more paradoxical is to show that photon has zero mass.

$$m(v) = \frac {m_0}{\sqrt {1 - \frac {v^2}{C^2}}}$$

 

[Zurtex]

Extra credit question on test:

Does a vacuum with a volume of 3 meters^3 contain three times as much Nothing as a vacuum with a volume of 1 meter^3?

I'm not a physicist but don't all vacuums contain at least something even if this be a very small amount of energy?

If not then by the way you are measuring them they must contain dimensions (at least 3 of them) and therefore are in fact not nothing but rather just space.

 

[Antonio Lao]

Vacuum is known to be the site of infinite amount of energy. The phenomenon of vacuum fluctuation attests to this fact. The period of time that this infinite energy exists is almost zero. These are conjugate variables in quantum mechanics.

$$\delta E \delta t \geq h$$

E is energy approaches infinity. t is time approaches zero. h is Planck constant.

 

[matt grime]

"infinite" amount of energy? Wow, stand well back from that or you might lose your eyebrows.

 

[Antonio Lao]

This is again a case where the product of near infinity and near zero is finite.

 

[matt grime]

Doesn't what you've written say the product "near zero and infinity" is not zero; it doesn't put an upper bound on it at all.

 

[Antonio Lao]

The product cannot be less than Planck constant (lower bound). The upper bound is beyond quantum mechanics into classical mechanics.

 

[HallsofIvy]

The product cannot be less than Planck constant (lower bound). The upper bound is beyond quantum mechanics into classical mechanics.

No, mathematics is not bound by physical constraints. The moment you talk about "infinity" you are no longer talking about any mathematics that can be applied to physics. If you are talking about numbers "near infinity" and "near zero" then you cannot have a specific numerical answer because you do not have specific numerical input.

 

[Antonio Lao]

Still I don't understand how the product of "infinity" and "zero" is finite? This happens very often in physics. To understand is the purpose of my thread.

 

[matt grime]

The product of infinity and zero is not defined. You seem to have misinterpreted a statement about limits. It is perfectly possible for the limit as x tends to zero of f(x)g(x) to be any real number (or infinity) if f tends to zero as x tends to zero and g tends to infinity as x tends to zero. eg f(x)=kx g(x) = 1/x then f(x)g(x) tends to k as x tends to zero. Whether or not there is any meaning "at infinity" is debatable.

 

[Antonio Lao]

I encountered two situations when I tried to find the energy of the universe whether it is zero or it is infinite.

If I am allowed to play with just two physical constants of nature, the speed of light c in vacuum and Planck's constant h, and further I assume two more variables that of acceleration a and length r, I can give two energy formulations as the following:

$$E_0 = \frac {ah}{c}$$

and

$$E_{\infty} = \frac {hc}{r}$$

 

[Antonio Lao]

Matt, can I do the following limiting processes?

$$E_0 \rightarrow h \nu$$ as $$a \rightarrow 0$$

and

$$E_{\infty} \rightarrow mc^2$$ as $$r \rightarrow 0$$

 

[matt grime]

Neither of those is correct. h and c are constant. thus the limits are 0 and infinity resp.

 

[Antonio Lao]

When do constants become variable? It is crucial to the next step which is that after combining the formulations:

$$\overrightarrow{a} \cdot \overrightarrow{r} = c^2$$

 

[matt grime]

It's perfectly possible for the product of two variables to be constant. Nothing wrong there.

 

[Antonio Lao]

Thank you vergy much. I don't really need to show how I derived the product of two variables. Is this what is called deduction instead of induction?

 

[Antonio Lao]

Can the limit of the following exist?

$$\frac {1}{dt}$$

 

[matt grime]

limit of what as it tends to where, and what's t and what's d?

 

[Antonio Lao]

dt is the differential of time. I did the following manipulations (not mathematically allowed?).

acceleration a, constant c changed to a variable velocity v, r is length.

$$\frac {a}{c}$$

changed to

$$\frac {a}{v}$$

$$a = \frac {dv}{dt}$$

$$v = \frac {dr}{dt}$$

after manipulations

$$\frac {a}{c} = \frac {1}{dt}$$

 

[Zurtex]

O.K this has been something getting to me for some time now while starting to get into this and other maths forums. In our maths class we are taught something like:

$$\frac{dy}{dx} = x^2$$

Can not be split up into:

$$dy = x^2 dx$$

Although it appears you are doing just that when doing integration by substitution, I have gone through exactly why you are not doing that and how it is just an extension of the chain rule. So is just accepted notation where as long as people are careful about how they use it, it works. If so do statements like this:

$$\frac {a}{c} = \frac {1}{dt}$$

make any sense, if so what does it mean?

 

[Antonio Lao]

The inverse of time is a period for some period functions. Under theory development site of this forum, I started a thread for the discussion of the physical meaning of time inverse.

 

[matt grime]

These do have uses in applied mathematics and are called infinitesimals, rigorously dt is a 1 form, and rearranging antonios equation yields a 1 form = a function which is not permitted, you've canceled things that can't be canelled, and your manipulations omit many equalities that need to be satisfied, you should always write out in full.

a/c is (1/c)dv/dt

I don't see how you took any of the steps above without implicitly assuming some things that you've not told us, such as somehow deciding that dv/c = 1, which is amazing as c is a constant.

 

[Antonio Lao]

Is the following limits acceptable in mathematics?

$$\lim_{dt\rightarrow 0} \frac {1}{dt} = \infty$$

$$\lim_{dt\rightarrow \infty} \frac{1}{dt} = 0$$

$$\lim_{dt\rightarrow a} \frac {1}{dt} = \frac {1}{a}$$

 

[Muzza]

The last one is invalid if a = 0...

 

[Antonio Lao]

Thanks. But if a=0, it is just the first one again. Physically speaking, can absolute time (absolute spacetime) ever be zero? Spacetime is not defined at the singularity.

 

[matt grime]

Try reading Segal's original papers to see the formalization of zero time.