Re: Infinities in Physics.

Arnold Neumaier wrote

> Michael C Price wrote:

[note: I'm leaving the previous quoted material since
the earlier messages in the thread have probably expired
on most newsreaders]

>> Mark <[EMAIL PROTECTED]> wrote
>>
>>> "michaelprice" <[EMAIL PROTECTED]> writes:
>>>> Rich wrote
>>>>> Can you give me an example of where infinity has a
>>>>> definite physical meaning?
>>>>Yes, an infinite number of photons are emitted by an
>>>> accelerated electric charge (the infrared [divergence] in
>>>> quantum field theory).
>>>
>>> Only a finite number of photons are observed;
>>
>> Correct.  But an infinite number are emitted.  The number we
>> observe is obviously limited by the detector's sensitivity.  I
>> thought this too obvious to mention in my earlier post :-)
>>
>>> the energy and wavelength of what can be observed is strictly
>>> limited by the size and duration of the spacetime region in
>>> question;
>>
>> I assume this is some heuristic appeal to the uncertainty principle.
>> Is it backed up by precise QED calculations?   Perhaps the size
>> of the detector limits of its sensitivity?  Even so, this doesn't
>> [] alter the emission processes themselves.  In Bremsstrahlung
>> processes the probability of any finite number of photons being
>> emitted is precisely zero.  Ergo there is no lower limit on the
>> energy of the photons emitted.
>>
>> > and as a general rule in quantum theory, nothing exists until
>> > it's observed.
>>
>> No.  Nothing has an operational existence unless it is
>> *capable* of being detected.  Which brings us back to the
>> detector's sensitivity - a hard physical variable - not some
>> vague philosophy
>>>
>>> So, no.  Only a finite number of photons are actually emitted.
>>
>> Again you are confusing what we detect with what occurs.
>>
>>> The
>>> infinity is a artifact of the formalism, because the initial and
>>> final Cauchy surfaces you're taking the states over enclose a
>>> non-compact spacetime region -- which is unphysical.
>>
>> Although many QFT calculations assume a non-compact,
>> infinite volume of space-time this is not required.  The same
>> QFT calculations can be re-worked on finite volumes,
>> atlhough the calculations are more intricate.
>>
>>> To get a realistic state transition, you have to take the transition
>>> between two Cauchy surfaces that enclose a compact region.
>>
>> True.  Can you give me a QFT text book quote that states that
>> the volume of the region of interest determines the infrared
>> cut-off?  You seem to be saying that the infrared divergences
>> won't occur in any finite, compact region of space time.m
>> Refs?
>
> Already in ordinary QM, finite volume implies a discrete set
> of momentum eigenvalues,

Correct.

> and hence a smallest nonzero momentum.

No, the non-zero "zero-point" momentum/energy is due to the
boundary conditions we impose, not the finite, compact volume.
A compact volume need not have boundary conditions (e.g. a
hypersphere or any realistic closed universe) and hence may
have infinite wavelength photons that wrap around the entire
universe, with zero momentum and zero energy.  Ergo in such
a closed universe the zero-point energy really is zero.

Same for the open universe, of course.

In either case the absence of any infrared cut-off permits the
divergence of photon number during Bremsstrahlung, i.e.
an infinite number of real photons are emitted, even though
they only sum to a finite energy.

Cheers,
Michael C Price
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