On 5/1/2025 7:25 AM, Alan Grayson wrote:


On Wednesday, April 30, 2025 at 11:06:07 PM UTC-6 Brent Meeker wrote:



    On 4/30/2025 5:54 PM, Alan Grayson wrote:


    On Wednesday, April 30, 2025 at 2:41:43 PM UTC-6 Brent Meeker wrote:



        On 4/30/2025 4:29 AM, John Clark wrote:
        On Tue, Apr 29, 2025 at 8:09 PM Brent Meeker
        <[email protected]> wrote:

                *>> If you place two macroscopic conductive plates
                close to each otherthe Casimir Effect will cause the
                two plates to attract each other; this occurs
                regardless of if you make any measurements or not.
                It happens because there are fewer virtual particles
                between the two plates than there are outside the
                plates. And virtual particles exist because it's
                impossible for the energy in the electromagnetic
                field to be exactly zero for any arbitrary length of
                time; and the shorter the time the greater the
                deviation from zero it's likely to be. *


            />That's why the qualification about measure like
            interactions.  The two conductive plates exclude longer
            wavelengths. /


        *Yes.*
        *
        *

            /> I don't recall that the effect depended on duration. /


        *Heisenberg's uncertainty principle is not just about the
        relationship between momentum and position, it also insists
        there is a similar relationship between energy and time; the
        shorter amount of time the greater the random variation from
        a zero value there is. *

        In quantum mechanics /*energy*/ and the /*time per unit
        change of a variable*/ are conjugate variables. So they
        satisfy an Heisenberg uncertainty relation, often written
        $\Delta E \Delta t \geq \hbar$ . This is sloppy though and
        not quite right. What is right is given any operator $A$ and
        the Hamiltonian $H$ defining the time evolution of $A$, then
        $\Delta A \Delta H \geq \frac{1}{2} \hbar [d<A>/dt]$ . In
        this case I don't see what is the time per unit change in the
        expected value of the energy density between the plates?  The
        plates are assumed stationary.

        Brent


    In the time-energy form of the HUP, what is the role of time as
    an operator? What does *time per unit change of a variable* mean?
    Which variable is referenced? About virtual particles; aren't
    they elements of a perturbation expansion and thus not to be
    considered real since those terms violate conservation of energy?
    TY, AG
    That's why I include the equations (although I see they didn't get
    converted to display).  It can be any variable whose change is
    encoded by the Hamiltonian, A and H respectively in the equation. 
    It doesn't have anything to do with how you might solve the
    equations; which is where perturbation expansions and virtual
    particles enter.

    Brent


Can you give some examples of what A could be, and mustn't A be an /operator,/ not a variable, that commutes with H?
Yes, A is an operator, but it doesn't commute with H.  That would imply the variable measured by A is constant in time.  The time per unit change in the expected value of the variable is the inverse of [d<A>/dt].

If your claim is correct, ISTM that Clark cannot apply the Time-Enegy form of the HUP to make his claim about the Casmir Effect. Do you agree?
I don't know.  I don't understand the proposed role of time.  It doesn't seem to have anything to do with conducting plates.  Perhaps he means the period of EM fields filling the gap between the plates.  Those of long period being excluded from between the plates would thereby remove their repulsive pressure and leave an unbalanced compressive pressure.

Brent
TY, AG
--
You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/fb1df801-e5b4-47ee-b7c0-1c364a565ec3n%40googlegroups.com <https://groups.google.com/d/msgid/everything-list/fb1df801-e5b4-47ee-b7c0-1c364a565ec3n%40googlegroups.com?utm_medium=email&utm_source=footer>.

--
You received this message because you are subscribed to the Google Groups 
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email 
to [email protected].
To view this discussion visit 
https://groups.google.com/d/msgid/everything-list/53274ba5-b4b2-42f1-97dc-c394a3163369%40gmail.com.

Reply via email to