| Topic: |
Science > Physics |
| User: |
"Don1" |
| Date: |
25 Feb 2006 06:45:39 AM |
| Object: |
Instantaneous acceleration |
Let it be known from the outset here: There is NO instantaneous
(infinite) acceleration. All motion is a change in position DURING
TIME.
If at least a couple of youse guys had paid attention to your algebra
lessons, you could have learned the equation for _average_
acceleration: a=vi-vt/(2t); _not_ a=vi-vt/t; which itself is called
"instantaneous acceleration"; of which it is not.
Then you could teach each other how to solve these sorts of problems:
To move a body from one place to another, and leave it stopped there,
it must first be accelerated, and then allowed to coast to where it is
stopped by it's own friction.
There are only two parts to such a problem: An accelerating part, to
start it moving, and a coasting to a stop part:
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| User: "Mr Clarke" |
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| Title: Re: Instantaneous acceleration |
25 Feb 2006 04:25:53 PM |
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"Don1" <dcshead@charter.net> wrote in message
news:1140871539.537518.95250@i39g2000cwa.googlegroups.com...
Let it be known from the outset here: There is NO instantaneous
(infinite) acceleration. All motion is a change in position DURING
TIME.
This may be so for particles, but what about constant forces which don`t use
particles
of matter to be able to create a motion? [Gravity]... magnetism even.
-----------------------------------------------------------------------
Ashley Clarke
-------------------------------------------------------
*** Free account sponsored by SecureIX.com ***
*** Encrypt your Internet usage with a free VPN account from http://www.SecureIX.com ***
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| User: "JEMebius" |
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| Title: Re: Instantaneous acceleration- |
26 Feb 2006 09:59:03 AM |
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AFAIK "instantaneous value of a quantity" means the value of that
quantity at a specific instant (point in time).
Instataneous acceleration is =by definition= the derivative of the
velocity as a function of time, like the instantaneous velocity is by
definition the derivative of the position as a function of time.
This is anyhow what I learned in elementary physics class.
Average acceleration, the average being taken over a certain time
interval [t1, t2],
is simply the increase of velocity divided by that time interval, i.e.
the difference quotient (v2 - v1) / (t2 - t1).
Assuming that vi and vt denote velocities, your formula a=vi-vt/(2t) is
dimensionally wrong, and so must be the result of a typing mistake.
Perhaps you meant a=(vi-vt)/(2t); and then I ask "from where does the
factor 2 come?".
Even the formula without the factor 2: a=(vi-vt)/(t), is no good,
assuming that vi, vt mean initial and terminal velocity for a time
interval of duration t. One would rather put a-average = (vt - vi) / t.
Tips:
(1) Always study math and physics =with a dictionary= and lots of paper,
pencils and erasers on your desk.
(2) Do not think of using computers and going public until you have
really mastered the elements of scientific work, including the art of
spelling.
(3) Classical textbooks, like Max Planck's and Arnold Sommerfeld's, may
also be of some help.
Happy studies: Johan E. Mebius
Don1 wrote:
Let it be known from the outset here: There is NO instantaneous
(infinite) acceleration. All motion is a change in position DURING
TIME.
If at least a couple of youse guys had paid attention to your algebra
lessons, you could have learned the equation for _average_
acceleration: a=vi-vt/(2t); _not_ a=vi-vt/t; which itself is called
"instantaneous acceleration"; of which it is not.
Then you could teach each other how to solve these sorts of problems:
To move a body from one place to another, and leave it stopped there,
it must first be accelerated, and then allowed to coast to where it is
stopped by it's own friction.
its
There are only two parts to such a problem: An accelerating part, to
start it moving, and a coasting to a stop part:
.
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| User: "Richard Henry" |
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| Title: Re: Instantaneous acceleration- |
26 Feb 2006 10:31:58 AM |
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"JEMebius" <jemebius@xs4all.nl> wrote in message
news:4401D047.1070705@xs4all.nl...
AFAIK "instantaneous value of a quantity" means the value of that
quantity at a specific instant (point in time).
Instataneous acceleration is =by definition= the derivative of the
velocity as a function of time, like the instantaneous velocity is by
definition the derivative of the position as a function of time.
This is anyhow what I learned in elementary physics class.
Average acceleration, the average being taken over a certain time
interval [t1, t2],
is simply the increase of velocity divided by that time interval, i.e.
the difference quotient (v2 - v1) / (t2 - t1).
Assuming that vi and vt denote velocities, your formula a=vi-vt/(2t) is
dimensionally wrong, and so must be the result of a typing mistake.
Perhaps you meant a=(vi-vt)/(2t); and then I ask "from where does the
factor 2 come?".
Even the formula without the factor 2: a=(vi-vt)/(t), is no good,
assuming that vi, vt mean initial and terminal velocity for a time
interval of duration t. One would rather put a-average = (vt - vi) / t.
Tips:
(1) Always study math and physics =with a dictionary= and lots of paper,
pencils and erasers on your desk.
(2) Do not think of using computers and going public until you have
really mastered the elements of scientific work, including the art of
spelling.
(3) Classical textbooks, like Max Planck's and Arnold Sommerfeld's, may
also be of some help.
Happy studies: Johan E. Mebius
I don't know which is more amusing: Donnie's attempts at physics or the
attempts of others to educate him.
Don1 wrote:
Let it be known from the outset here: There is NO instantaneous
(infinite) acceleration. All motion is a change in position DURING
TIME.
If at least a couple of youse guys had paid attention to your algebra
lessons, you could have learned the equation for _average_
acceleration: a=vi-vt/(2t); _not_ a=vi-vt/t; which itself is called
"instantaneous acceleration"; of which it is not.
Then you could teach each other how to solve these sorts of problems:
To move a body from one place to another, and leave it stopped there,
it must first be accelerated, and then allowed to coast to where it is
stopped by it's own friction.
its
There are only two parts to such a problem: An accelerating part, to
start it moving, and a coasting to a stop part:
.
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| User: "tadchem" |
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| Title: Re: Instantaneous acceleration- |
26 Feb 2006 03:00:07 PM |
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The efforts to 'educate' Don1 remind me of an old expression from my
youth out West: "You can't polish a turd."
Tom Davidson
Richmond, VA
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| User: "" |
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| Title: Re: Instantaneous acceleration |
25 Feb 2006 06:51:18 AM |
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Any other fun facts to know and tell????????
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| User: "G=EMC^2 Glazier" |
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| Title: Re: Instantaneous acceleration |
25 Feb 2006 11:37:20 AM |
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Don 1 Right you are Don Not only is no instant acceleration,but
reality is before motion takes place there is a time lapse. I put
together in my lab a fast picture that proves this. email me and I will
show it to you. TreBert
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| User: "Sam Wormley" |
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| Title: Re: Instantaneous acceleration |
25 Feb 2006 11:55:11 AM |
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G=EMC^2 Glazier wrote:
Don 1 Right you are Don Not only is no instant acceleration,but
reality is before motion takes place there is a time lapse. I put
together in my lab a fast picture that proves this. email me and I will
show it to you. TreBert
Herb used to sell used buggy whips.
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| User: "Robert Low" |
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| Title: Re: Instantaneous acceleration |
25 Feb 2006 08:33:33 AM |
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Don1 wrote:
If at least a couple of youse guys had paid attention to your algebra
lessons, you could have learned the equation for _average_
acceleration: a=vi-vt/(2t); _not_ a=vi-vt/t; which itself is called
"instantaneous acceleration"; of which it is not.
It's a searing indictment on the educational system that
only one person in the history of the world ever understood
this.
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| User: "tadchem" |
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| Title: Re: Instantaneous acceleration |
26 Feb 2006 07:41:38 AM |
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"Robert Low" <mtx014@coventry.ac.uk> wrote in message
news:46b85tFa12tfU2@individual.net...
Don1 wrote:
If at least a couple of youse guys had paid attention to your algebra
lessons, you could have learned the equation for _average_
acceleration: a=vi-vt/(2t); _not_ a=vi-vt/t; which itself is called
"instantaneous acceleration"; of which it is not.
It's a searing indictment on the educational system that
only one person in the history of the world ever understood
this.
It's a searing indictment on the mental health system that Don1 sHead is
still permitted a keyboard.
Tom Davidson
Richmond, VA
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| User: "Don1" |
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| Title: Re: Instantaneous acceleration |
26 Feb 2006 08:23:59 AM |
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tadchem wrote:
"Robert Low" <mtx014@coventry.ac.uk> wrote in message
news:46b85tFa12tfU2@individual.net...
Don1 wrote:
If at least a couple of youse guys had paid attention to your algebra
lessons, you could have learned the equation for _average_
acceleration: a=vi-vt/(2t); _not_ a=vi-vt/t; which itself is called
"instantaneous acceleration"; of which it is not.
It's a searing indictment on the educational system that
only one person in the history of the world ever understood
this.
It's a searing indictment on the mental health system that Don1 sHead is
still permitted a keyboard.
Tom Davidson
Richmond, VA
Like we say: It's a free country. AFAIK, no bill of indictment has been
served against me; searing, sad or otherwise. How's your mental health?
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| User: "Paul Cardinale" |
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| Title: Re: Instantaneous acceleration |
25 Feb 2006 08:20:04 AM |
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Don1 wrote:
Let it be known from the outset here: There is NO instantaneous
(infinite) acceleration. All motion is a change in position DURING
TIME.
If at least a couple of youse guys had paid attention to your algebra
lessons, you could have learned the equation for _average_
acceleration: a=vi-vt/(2t); _not_ a=vi-vt/t; which itself is called
"instantaneous acceleration"; of which it is not.
Wow. Not only are you 100% wrong, you have achieved a new low level of
stupitude.
Paul Cardinale
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| User: "Spaceman" |
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| Title: Re: Instantaneous acceleration |
25 Feb 2006 08:54:51 AM |
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"Paul Cardinale" <pcardinale@volcanomail.com> wrote in message
news:1140877204.043074.221710@u72g2000cwu.googlegroups.com...
Don1 wrote:
Let it be known from the outset here: There is NO instantaneous
(infinite) acceleration. All motion is a change in position DURING
TIME.
If at least a couple of youse guys had paid attention to your algebra
lessons, you could have learned the equation for _average_
acceleration: a=vi-vt/(2t); _not_ a=vi-vt/t; which itself is called
"instantaneous acceleration"; of which it is not.
Wow. Not only are you 100% wrong, you have achieved a new low level of
stupitude.
Paul Cardinale
Sam! Sam! Quick!
Paul says you and Don1 are wrong!
LOL
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| User: "PD" |
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| Title: Re: Instantaneous acceleration |
27 Feb 2006 04:11:31 PM |
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Don1 wrote:
Let it be known from the outset here: There is NO instantaneous
(infinite) acceleration. All motion is a change in position DURING
TIME.
If at least a couple of youse guys had paid attention to your algebra
lessons, you could have learned the equation for _average_
acceleration: a=vi-vt/(2t); _not_ a=vi-vt/t; which itself is called
"instantaneous acceleration"; of which it is not.
Then you could teach each other how to solve these sorts of problems:
To move a body from one place to another, and leave it stopped there,
it must first be accelerated, and then allowed to coast to where it is
stopped by it's own friction.
There are only two parts to such a problem: An accelerating part, to
start it moving, and a coasting to a stop part:
Don is likely confused by a calculus term.
Velocity is also something that describes the rate of displacement over
a certain interval. Since data are typically compiled in tabular form,
difference calculations over these intervals is what we customarily get
from data.
However, we can also recognize by looking at something that it has a
velocity at each and every moment for that whole continuous time, and
that it makes sense to ask what velocity it had at a particular
*instant*. Though we can't get that from the tabular data, we *can* get
that by fitting the data with a curve and then asking what the value of
that curve is at a particular instant.
In the same way, the "instantaneous acceleration" is a calculus term
with a specific meaning: it means the limit of the acceleration over an
interval, as that interval is made smaller and smaller and smaller.
Though we may not be able to actually *measure* the acceleration when
the interval is zero, we can certainly figure out what the limit is. In
this context, "instantaneous acceleration" does not mean "jump from one
velocity to another different velocity over a zero time period," which
is what I think Shead thought it meant, being largely ignorant of
calculus and what those terms mean in calculus.
PD
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| User: "Traveler" |
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| Title: Re: Instantaneous acceleration |
27 Feb 2006 04:58:45 PM |
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On 27 Feb 2006 14:11:31 -0800, "PD" <TheDraperFamily@gmail.com> wrote:
In the same way, the "instantaneous acceleration" is a calculus term
with a specific meaning: it means the limit of the acceleration over an
interval, as that interval is made smaller and smaller and smaller.
Though we may not be able to actually *measure* the acceleration when
the interval is zero, we can certainly figure out what the limit is. In
this context, "instantaneous acceleration" does not mean "jump from one
velocity to another different velocity over a zero time period," which
is what I think Shead thought it meant, being largely ignorant of
calculus and what those terms mean in calculus.
This is actually a perfect argument against continuity. Continuity
(infinite divisibility) makes cause and effect impossible. But I'm
sure this aspect of it went over your head, PD. Pfftttt! ahahaha...
Making phun of "physicists" is so much phucking phun. ahahaha...
Louis Savain
Why Software Is Bad and What We Can Do to Fix It:
http://www.rebelscience.org/Cosas/Reliability.htm
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| User: "Sam Wormley" |
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| Title: Re: Instantaneous acceleration |
25 Feb 2006 07:46:25 AM |
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Don1 wrote:
Let it be known from the outset here: There is NO instantaneous
(infinite) acceleration. All motion is a change in position DURING
TIME.
You got one right Shead... ticker tape parade in New York!
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| User: "ma1ibu" |
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| Title: Re: Instantaneous acceleration |
25 Feb 2006 08:07:42 AM |
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So Sam,
(Is your wife's name Ella?- Sam'n'Ella :-)
When a photon is given off, *how long*
does it take to go from zero
to c?
John
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| User: "Gregory L. Hansen" |
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| Title: Re: Instantaneous acceleration |
27 Feb 2006 11:10:47 AM |
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In article <1140876462.608518.246600@e56g2000cwe.googlegroups.com>,
ma1ibu <vegan16@accesscomm.ca> wrote:
So Sam,
(Is your wife's name Ella?- Sam'n'Ella :-)
When a photon is given off, *how long*
does it take to go from zero
to c?
John
When you throw a rock into a lake, how long does it take the ripples in
the water to go from zero to their final speed?
--
"The probability of anything happening is in inverse ratio to its
desirability." -- Gumperson's Law
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| User: "Tony Orlow" |
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| Title: Re: Instantaneous acceleration |
27 Feb 2006 12:13:59 PM |
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Gregory L. Hansen said:
In article <1140876462.608518.246600@e56g2000cwe.googlegroups.com>,
ma1ibu <vegan16@accesscomm.ca> wrote:
So Sam,
(Is your wife's name Ella?- Sam'n'Ella :-)
When a photon is given off, *how long*
does it take to go from zero
to c?
John
When you throw a rock into a lake, how long does it take the ripples in
the water to go from zero to their final speed?
Depends how fast the rock is falling.
When you throw two stones in a pond, and they each emit a ripple, and those two
ripples expand until they meet at a point, which instantaneously splits into
two points of intersection, at that moment of contact, how fast are the two
points moving away from each other?
Do those two points actually represent objects?
;)
--
Smiles,
Tony
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| User: "PD" |
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| Title: Re: Instantaneous acceleration |
27 Feb 2006 12:22:05 PM |
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Tony Orlow wrote:
Gregory L. Hansen said:
In article <1140876462.608518.246600@e56g2000cwe.googlegroups.com>,
ma1ibu <vegan16@accesscomm.ca> wrote:
So Sam,
(Is your wife's name Ella?- Sam'n'Ella :-)
When a photon is given off, *how long*
does it take to go from zero
to c?
John
When you throw a rock into a lake, how long does it take the ripples in
the water to go from zero to their final speed?
Depends how fast the rock is falling.
Don't think so. If you smack a board on the end, does the onset of the
propagation of the disturbance depend on how hard you smacked it? (No.)
When you throw two stones in a pond, and they each emit a ripple, and those two
ripples expand until they meet at a point, which instantaneously splits into
two points of intersection, at that moment of contact, how fast are the two
points moving away from each other?
Do those two points actually represent objects?
;)
--
Smiles,
Tony
.
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| User: "Tony Orlow" |
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| Title: Re: Instantaneous acceleration |
27 Feb 2006 01:01:32 PM |
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PD said:
Tony Orlow wrote:
Gregory L. Hansen said:
In article <1140876462.608518.246600@e56g2000cwe.googlegroups.com>,
ma1ibu <vegan16@accesscomm.ca> wrote:
So Sam,
(Is your wife's name Ella?- Sam'n'Ella :-)
When a photon is given off, *how long*
does it take to go from zero
to c?
John
When you throw a rock into a lake, how long does it take the ripples in
the water to go from zero to their final speed?
Depends how fast the rock is falling.
Don't think so. If you smack a board on the end, does the onset of the
propagation of the disturbance depend on how hard you smacked it? (No.)
No, but can the impulse be said to have reached its final speed of propagation
before it is even done being started as a wave? It seems to me that if the
stone is still in the process of passing through the surface of the water, the
wave is not yet complete. I suppose the front of the wave has begun its
propagation before the back is fully defined, but would you say there is a wave
propagated in real life in the very first moment the rock touches the first
molecule of water? I'd say it takes a little finite time to create the wave to
begin with.
Now, what did you think of the question below? It wasn't really an idle one.
When you throw two stones in a pond, and they each emit a ripple, and those two
ripples expand until they meet at a point, which instantaneously splits into
two points of intersection, at that moment of contact, how fast are the two
points moving away from each other?
Do those two points actually represent objects?
;)
--
Smiles,
Tony
--
Smiles,
Tony
.
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| User: "Gregory L. Hansen" |
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| Title: Re: Instantaneous acceleration |
27 Feb 2006 01:56:50 PM |
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In article <MPG.1e6d169599817ded98aa7b@newsstand.cit.cornell.edu>,
Tony Orlow <aeo6@cornell.edu> wrote:
PD said:
Tony Orlow wrote:
Gregory L. Hansen said:
In article <1140876462.608518.246600@e56g2000cwe.googlegroups.com>,
ma1ibu <vegan16@accesscomm.ca> wrote:
So Sam,
(Is your wife's name Ella?- Sam'n'Ella :-)
When a photon is given off, *how long*
does it take to go from zero
to c?
John
When you throw a rock into a lake, how long does it take the ripples in
the water to go from zero to their final speed?
Depends how fast the rock is falling.
Don't think so. If you smack a board on the end, does the onset of the
propagation of the disturbance depend on how hard you smacked it? (No.)
No, but can the impulse be said to have reached its final speed of propagation
before it is even done being started as a wave? It seems to me that if the
stone is still in the process of passing through the surface of the water, the
wave is not yet complete. I suppose the front of the wave has begun its
propagation before the back is fully defined, but would you say there is a wave
propagated in real life in the very first moment the rock touches the first
molecule of water? I'd say it takes a little finite time to create the wave to
begin with.
Define "wave". Does it have to go up and down at least once? Sinusoids
are not the only solutions to wave equations. Any disturbance in the
water will propagate at a characteristic speed, or else there just isn't a
disturbance. When the rock hits the water the leading edge of the
disturbance will travel away at that characteristic speed, and for as long
as the rock (or log or diver or whatever) continues to fall into the water
it will continue to generate waves which will continue to propagate at
that speed.
Maxwell's equations can be combined to form a wave equation, and that
tells you how fast electromagnetic radiation goes. It either goes that
fast, or there is no radiation to assign a speed to.
Quantum electrodynamics still have Maxwell's equations in them. The speed
of the photon wavefunction is still c. When you go to the probabilistic
mechanics you can't always say exactly at what time energy is transferred
to the field, or when energy is transferred from field to detector, or
even that energy is transferred within a given interval of time. But
that's all tied up with the usual interpretation of the wavefunction that
you'll run into in the undergrad non-electrodynamic regime. But the
photon propagators and other theoretical machinery all describe things
going at c.
Now, what did you think of the question below? It wasn't really an idle one.
A wave carries energy. The points where two waves meet do not represent a
transmission of energy.
When you throw two stones in a pond, and they each emit a ripple,
and those two
ripples expand until they meet at a point, which instantaneously splits into
two points of intersection, at that moment of contact, how fast are the two
points moving away from each other?
Do those two points actually represent objects?
;)
--
Smiles,
Tony
--
Smiles,
Tony
--
"No one need be surprised that the subject of contagion was not clear to
our ancestors."-- Heironymus Fracastorius, 1546
.
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| User: "QCD Apprentice" |
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| Title: Re: Instantaneous acceleration |
27 Feb 2006 01:09:32 PM |
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Tony Orlow wrote:
PD said:
Tony Orlow wrote:
Gregory L. Hansen said:
In article <1140876462.608518.246600@e56g2000cwe.googlegroups.com>,
ma1ibu <vegan16@accesscomm.ca> wrote:
So Sam,
(Is your wife's name Ella?- Sam'n'Ella :-)
When a photon is given off, *how long*
does it take to go from zero
to c?
John
When you throw a rock into a lake, how long does it take the ripples in
the water to go from zero to their final speed?
Depends how fast the rock is falling.
Don't think so. If you smack a board on the end, does the onset of the
propagation of the disturbance depend on how hard you smacked it? (No.)
No, but can the impulse be said to have reached its final speed of propagation
before it is even done being started as a wave? It seems to me that if the
stone is still in the process of passing through the surface of the water, the
wave is not yet complete. I suppose the front of the wave has begun its
propagation before the back is fully defined, but would you say there is a wave
propagated in real life in the very first moment the rock touches the first
molecule of water? I'd say it takes a little finite time to create the wave to
begin with.
It seems like this point is just going to be really
pedantic. I think PD was talking about the disturbance
being created the instant the rock began to have real
contact with the water.
Now, what did you think of the question below? It wasn't really an idle one.
When you throw two stones in a pond, and they each emit a ripple, and those two
ripples expand until they meet at a point, which instantaneously splits into
two points of intersection, at that moment of contact, how fast are the two
points moving away from each other?
Do those two points actually represent objects?
I'm trying to picture what you mean. The two points you're
talking about are just where the wave fronts are
intersecting with the other waves? So if we're talking
about wave A and wave B, the two points are the point on
wave B closest to the origin of A and visa versa?
.
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| User: "Tony Orlow" |
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| Title: Re: Instantaneous acceleration |
28 Feb 2006 09:15:40 AM |
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QCD Apprentice said:
Tony Orlow wrote:
PD said:
Tony Orlow wrote:
Gregory L. Hansen said:
In article <1140876462.608518.246600@e56g2000cwe.googlegroups.com>,
ma1ibu <vegan16@accesscomm.ca> wrote:
So Sam,
(Is your wife's name Ella?- Sam'n'Ella :-)
When a photon is given off, *how long*
does it take to go from zero
to c?
John
When you throw a rock into a lake, how long does it take the ripples in
the water to go from zero to their final speed?
Depends how fast the rock is falling.
Don't think so. If you smack a board on the end, does the onset of the
propagation of the disturbance depend on how hard you smacked it? (No.)
No, but can the impulse be said to have reached its final speed of propagation
before it is even done being started as a wave? It seems to me that if the
stone is still in the process of passing through the surface of the water, the
wave is not yet complete. I suppose the front of the wave has begun its
propagation before the back is fully defined, but would you say there is a wave
propagated in real life in the very first moment the rock touches the first
molecule of water? I'd say it takes a little finite time to create the wave to
begin with.
It seems like this point is just going to be really
pedantic. I think PD was talking about the disturbance
being created the instant the rock began to have real
contact with the water.
Yeah, I guess I was being pedantic, too. Like, it seems to me it takes more
than an instant for the rock to come into contact with the water. But,
anyway...
Now, what did you think of the question below? It wasn't really an idle one.
When you throw two stones in a pond, and they each emit a ripple, and those two
ripples expand until they meet at a point, which instantaneously splits into
two points of intersection, at that moment of contact, how fast are the two
points moving away from each other?
Do those two points actually represent objects?
I'm trying to picture what you mean. The two points you're
talking about are just where the wave fronts are
intersecting with the other waves? So if we're talking
about wave A and wave B, the two points are the point on
wave B closest to the origin of A and visa versa?
No, that's not what I mean.
Two stones thrown simultaneously into a pond make two ripples...
(.) (.)
....which expand towards each other...
( . ) ( . )
....and meet at a point halfway between the two stones...
( . ).( . )
....and pass through each other, intersecting at two points from that moment on.
.
( . (|) . )
.
Hope those crappy pictures helped. The two points are the points of
intersection between the two circular point-source waves. So, given the rate of
propagation for the two waves of X, what is the initial rate at which those two
points are moving apart?
:D
--
Smiles,
Tony
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| User: "SCW" |
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| Title: Re: Instantaneous acceleration |
28 Feb 2006 09:57:04 AM |
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Tony Orlow wrote:
<snip>
Two stones thrown simultaneously into a pond make two ripples...
(.) (.)
...which expand towards each other...
( . ) ( . )
...and meet at a point halfway between the two stones...
( . ).( . )
...and pass through each other, intersecting at two points from that moment on.
.
( . (|) . )
.
Looks more like an emoticon movie of breast enlargement.
<snip>
:D
--
Smiles,
Tony
Just kiddin'
SCW
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| User: "Randy Poe" |
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| Title: Re: Instantaneous acceleration |
27 Feb 2006 01:49:03 PM |
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Tony Orlow wrote:
PD said:
Tony Orlow wrote:
Gregory L. Hansen said:
In article <1140876462.608518.246600@e56g2000cwe.googlegroups.com>,
ma1ibu <vegan16@accesscomm.ca> wrote:
So Sam,
(Is your wife's name Ella?- Sam'n'Ella :-)
When a photon is given off, *how long*
does it take to go from zero
to c?
John
When you throw a rock into a lake, how long does it take the ripples in
the water to go from zero to their final speed?
Depends how fast the rock is falling.
Don't think so. If you smack a board on the end, does the onset of the
propagation of the disturbance depend on how hard you smacked it? (No.)
No, but can the impulse be said to have reached its final speed of propagation
before it is even done being started as a wave?
Yes.
It seems to me that if the
stone is still in the process of passing through the surface of the water, the
wave is not yet complete.
What if the source is a speaker transmitting Beethoven's
Ninth Symphony and the wave isn't complete for 45 minutes
or however long it takes? Do you think it's 45 minutes before
the wave begins propagating?
I suppose the front of the wave has begun its
propagation before the back is fully defined,
At the molecular level, as soon as you disturb the
first molecule, it will disturb it's neighbors, those in turn
will disturb their neighbors, etc. This process occurs
at the speed of sound. The clock begins as soon as
the first molecule has changed position.
but would you say there is a wave
propagated in real life in the very first moment the rock touches the first
molecule of water?
Yes.
I'd say it takes a little finite time to create the wave to
begin with.
Define that time in molecular terms.
Now, what did you think of the question below? It wasn't really an idle one.
When you throw two stones in a pond, and they each emit a ripple, and those two
ripples expand until they meet at a point, which instantaneously splits into
two points of intersection, at that moment of contact, how fast are the two
points moving away from each other?
Consider two sources a distance D apart. Two
wavefronts expand, one from each source, with radius
r = vt. You are asking about the rate of motion of the two
points which are horizontal distance D/2 from each
source.
The y coordinate of these points is given by +-sqrt(r^2 - (D/2)^2).
Their separation is thus 2 sqrt(r^2 - D^2/4) and the velocity
of the points is the time derivative of 2 sqrt(v^2* t^2 - D^2/4)
I get 2v^2*t/ sqrt[v^2*t^2 - D^2/4) for that derivative, and that
at the time where vt = D/2, the speed is infinite.
Do those two points actually represent objects?
No.
- Randy
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| User: "Tony Orlow" |
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| Title: Re: Instantaneous acceleration |
28 Feb 2006 09:31:54 AM |
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Randy Poe said:
Tony Orlow wrote:
PD said:
Tony Orlow wrote:
Gregory L. Hansen said:
In article <1140876462.608518.246600@e56g2000cwe.googlegroups.com>,
ma1ibu <vegan16@accesscomm.ca> wrote:
So Sam,
(Is your wife's name Ella?- Sam'n'Ella :-)
When a photon is given off, *how long*
does it take to go from zero
to c?
John
When you throw a rock into a lake, how long does it take the ripples in
the water to go from zero to their final speed?
Depends how fast the rock is falling.
Don't think so. If you smack a board on the end, does the onset of the
propagation of the disturbance depend on how hard you smacked it? (No.)
No, but can the impulse be said to have reached its final speed of propagation
before it is even done being started as a wave?
Yes.
It seems to me that if the
stone is still in the process of passing through the surface of the water, the
wave is not yet complete.
What if the source is a speaker transmitting Beethoven's
Ninth Symphony and the wave isn't complete for 45 minutes
or however long it takes? Do you think it's 45 minutes before
the wave begins propagating?
I suppose the front of the wave has begun its
propagation before the back is fully defined,
At the molecular level, as soon as you disturb the
first molecule, it will disturb it's neighbors, those in turn
will disturb their neighbors, etc. This process occurs
at the speed of sound. The clock begins as soon as
the first molecule has changed position.
but would you say there is a wave
propagated in real life in the very first moment the rock touches the first
molecule of water?
Yes.
Doesn't it take some finite time to transmit the first energy to the next
meolecule? If not, then why isn;t the speed of sound infinite?
I'd say it takes a little finite time to create the wave to
begin with.
Define that time in molecular terms.
For the point source of the wave to create the entire wave takes an amount of
time equal to the inverse of the frequency. That's independent of the molecular
nature, but a property of waves in general, I'd say.
Now, what did you think of the question below? It wasn't really an idle one.
When you throw two stones in a pond, and they each emit a ripple, and those two
ripples expand until they meet at a point, which instantaneously splits into
two points of intersection, at that moment of contact, how fast are the two
points moving away from each other?
Consider two sources a distance D apart. Two
wavefronts expand, one from each source, with radius
r = vt. You are asking about the rate of motion of the two
points which are horizontal distance D/2 from each
source.
Yes, the two points which move apart perpendicular to the line which connects
the two sources.
The y coordinate of these points is given by +-sqrt(r^2 - (D/2)^2).
Their separation is thus 2 sqrt(r^2 - D^2/4) and the velocity
of the points is the time derivative of 2 sqrt(v^2* t^2 - D^2/4)
I get 2v^2*t/ sqrt[v^2*t^2 - D^2/4) for that derivative, and that
at the time where vt = D/2, the speed is infinite.
Very good. That's what I get, too. Does it mean anything? Hmmm....
Do those two points actually represent objects?
No.
Hmmm, perhaps not, but they would appear to be like 0-dimensional waves within
each of the 1-dimensional spaces defined by the two intersecting waves. They
would start with an initially infinite speed of separation, slowing down to a
speed equal to pi times the speed of the waves, if measured within the waves.
So, I wonder, if one were to start making waves in an infinite dimensional
space, which expanded at some infinitesimal rate, and intersected after an
eternity, could their intersection expand at an initially finite rate? Hmmmm...
Given such a scenario, where oo-1 dimensional waves intersect to make oo-2
dimensional waves, etc, all the way down to finite dimensional waves, how would
the finite waves tend to expand? This seems to be a pretty hard problem and not
one I know how to solve, but I suspect that finite-dimensional waves produced
this way would have a speed of expansion inversely proportional to their age.
Maybe that's way off. Any clues?
- Randy
--
Smiles,
Tony
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| User: "Virgil" |
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| Title: Re: Instantaneous acceleration |
28 Feb 2006 02:00:29 PM |
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In article <MPG.1e6e36efc820134498aa7e@newsstand.cit.cornell.edu>,
Tony Orlow <aeo6@cornell.edu> wrote:
Randy Poe said:
Tony Orlow wrote:
PD said:
Tony Orlow wrote:
Gregory L. Hansen said:
In article <1140876462.608518.246600@e56g2000cwe.googlegroups.com>,
ma1ibu <vegan16@accesscomm.ca> wrote:
So Sam,
(Is your wife's name Ella?- Sam'n'Ella :-)
When a photon is given off, *how long*
does it take to go from zero
to c?
John
When you throw a rock into a lake, how long does it take the
ripples in
the water to go from zero to their final speed?
Depends how fast the rock is falling.
Don't think so. If you smack a board on the end, does the onset of the
propagation of the disturbance depend on how hard you smacked it? (No.)
No, but can the impulse be said to have reached its final speed of
propagation
before it is even done being started as a wave?
Yes.
It seems to me that if the
stone is still in the process of passing through the surface of the
water, the
wave is not yet complete.
What if the source is a speaker transmitting Beethoven's
Ninth Symphony and the wave isn't complete for 45 minutes
or however long it takes? Do you think it's 45 minutes before
the wave begins propagating?
I suppose the front of the wave has begun its
propagation before the back is fully defined,
At the molecular level, as soon as you disturb the
first molecule, it will disturb it's neighbors, those in turn
will disturb their neighbors, etc. This process occurs
at the speed of sound. The clock begins as soon as
the first molecule has changed position.
but would you say there is a wave
propagated in real life in the very first moment the rock touches the
first
molecule of water?
Yes.
Doesn't it take some finite time to transmit the first energy to the next
meolecule? If not, then why isn;t the speed of sound infinite?
I'd say it takes a little finite time to create the wave to
begin with.
Define that time in molecular terms.
For the point source of the wave to create the entire wave takes an amount of
time equal to the inverse of the frequency. That's independent of the
molecular
nature, but a property of waves in general, I'd say.
Now, what did you think of the question below? It wasn't really an idle
one.
When you throw two stones in a pond, and they each emit a ripple, and
those two
ripples expand until they meet at a point, which instantaneously
splits into
two points of intersection, at that moment of contact, how fast are
the two
points moving away from each other?
Consider two sources a distance D apart. Two
wavefronts expand, one from each source, with radius
r = vt. You are asking about the rate of motion of the two
points which are horizontal distance D/2 from each
source.
Yes, the two points which move apart perpendicular to the line which connects
the two sources.
The y coordinate of these points is given by +-sqrt(r^2 - (D/2)^2).
Their separation is thus 2 sqrt(r^2 - D^2/4) and the velocity
of the points is the time derivative of 2 sqrt(v^2* t^2 - D^2/4)
I get 2v^2*t/ sqrt[v^2*t^2 - D^2/4) for that derivative, and that
at the time where vt = D/2, the speed is infinite.
Very good. That's what I get, too. Does it mean anything? Hmmm....
Points, being non-physical and without mass, have no inertia.
As those points are not attached to any physical objects. as are the
molecules of water which form the waves, there is no constraint on their
velocities or accelerations such as would occur on the velocities and
accelerations points attached to physical objects.
Under suitable conditions, for example, a shadow can undergo momentarily
infinite acceleration so as to have a discontinuous change in velocity:
A horizontal beam of light strikes the side of a horizontally
aligned circular cylinder.
A ball is tossed up vertically in that beam casting a shadow on the
cylinder, so that at the ball's highest point the top of its shadow
is just at the top of the cylinder.
Describe the velocity of the top of the ball's shadow!
One finds that the velocity of the top point of the shadow jumps from
strictly non-zero value in one direction to an equal non-zero value in
the opposite direction along the surface of the cylinder.
.
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| User: "Tony Orlow" |
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| Title: Re: Instantaneous acceleration |
01 Mar 2006 10:21:42 AM |
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Virgil said:
In article <MPG.1e6e36efc820134498aa7e@newsstand.cit.cornell.edu>,
Tony Orlow <aeo6@cornell.edu> wrote:
Randy Poe said:
Tony Orlow wrote:
PD said:
Tony Orlow wrote:
Gregory L. Hansen said:
In article <1140876462.608518.246600@e56g2000cwe.googlegroups.com>,
ma1ibu <vegan16@accesscomm.ca> wrote:
So Sam,
(Is your wife's name Ella?- Sam'n'Ella :-)
When a photon is given off, *how long*
does it take to go from zero
to c?
John
When you throw a rock into a lake, how long does it take the
ripples in
the water to go from zero to their final speed?
Depends how fast the rock is falling.
Don't think so. If you smack a board on the end, does the onset of the
propagation of the disturbance depend on how hard you smacked it? (No.)
No, but can the impulse be said to have reached its final speed of
propagation
before it is even done being started as a wave?
Yes.
It seems to me that if the
stone is still in the process of passing through the surface of the
water, the
wave is not yet complete.
What if the source is a speaker transmitting Beethoven's
Ninth Symphony and the wave isn't complete for 45 minutes
or however long it takes? Do you think it's 45 minutes before
the wave begins propagating?
I suppose the front of the wave has begun its
propagation before the back is fully defined,
At the molecular level, as soon as you disturb the
first molecule, it will disturb it's neighbors, those in turn
will disturb their neighbors, etc. This process occurs
at the speed of sound. The clock begins as soon as
the first molecule has changed position.
but would you say there is a wave
propagated in real life in the very first moment the rock touches the
first
molecule of water?
Yes.
Doesn't it take some finite time to transmit the first energy to the next
meolecule? If not, then why isn;t the speed of sound infinite?
I'd say it takes a little finite time to create the wave to
begin with.
Define that time in molecular terms.
For the point source of the wave to create the entire wave takes an amount of
time equal to the inverse of the frequency. That's independent of the
molecular
nature, but a property of waves in general, I'd say.
Now, what did you think of the question below? It wasn't really an idle
one.
When you throw two stones in a pond, and they each emit a ripple, and
those two
ripples expand until they meet at a point, which instantaneously
splits into
two points of intersection, at that moment of contact, how fast are
the two
points moving away from each other?
Consider two sources a distance D apart. Two
wavefronts expand, one from each source, with radius
r = vt. You are asking about the rate of motion of the two
points which are horizontal distance D/2 from each
source.
Yes, the two points which move apart perpendicular to the line which connects
the two sources.
The y coordinate of these points is given by +-sqrt(r^2 - (D/2)^2).
Their separation is thus 2 sqrt(r^2 - D^2/4) and the velocity
of the points is the time derivative of 2 sqrt(v^2* t^2 - D^2/4)
I get 2v^2*t/ sqrt[v^2*t^2 - D^2/4) for that derivative, and that
at the time where vt = D/2, the speed is infinite.
Very good. That's what I get, too. Does it mean anything? Hmmm....
Points, being non-physical and without mass, have no inertia.
As those points are not attached to any physical objects. as are the
molecules of water which form the waves, there is no constraint on their
velocities or accelerations such as would occur on the velocities and
accelerations points attached to physical objects.
Under suitable conditions, for example, a shadow can undergo momentarily
infinite acceleration so as to have a discontinuous change in velocity:
A horizontal beam of light strikes the side of a horizontally
aligned circular cylinder.
A ball is tossed up vertically in that beam casting a shadow on the
cylinder, so that at the ball's highest point the top of its shadow
is just at the top of the cylinder.
Describe the velocity of the top of the ball's shadow!
One finds that the velocity of the top point of the shadow jumps from
strictly non-zero value in one direction to an equal non-zero value in
the opposite direction along the surface of the cylinder.
Yes, as long as there's no mass involved, instantaneous acceleration is not
forbidden, but mathematically possible. It's too bad you didn't have an answer
to the questions of mine that you snipped. I guess they didn't make enough
sense to you, or something. Too bad. I have been wondering if the result of an
infinite number of sucessive intersections between waves from infinite
dimensions on down would result in an expansion rate inversely proportional to
the time since initial intersection for a finite dimensional wave. This has to
do with a possible model of the universe where our space bubble is the
intersection between two expanding 4D space bubbles in time. So, I was just
wondering....
--
Smiles,
Tony
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| User: "Virgil" |
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| Title: Re: Instantaneous acceleration |
01 Mar 2006 12:37:15 PM |
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In article <MPG.1e6f941ca80de7f198aa83@newsstand.cit.cornell.edu>,
Tony Orlow <aeo6@cornell.edu> wrote:
Yes, as long as there's no mass involved, instantaneous acceleration
is not forbidden, but mathematically possible. It's too bad you
didn't have an answer to the questions of mine that you snipped.
As TO generally chooses either to ignore or to misrepresent what I say,
I did not think him to be that interested in what I say.
.
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| User: "Randy Poe" |
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| Title: Re: Instantaneous acceleration |
28 Feb 2006 10:12:46 AM |
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Tony Orlow wrote:
Randy Poe said:
Tony Orlow wrote:
I suppose the front of the wave has begun its
propagation before the back is fully defined,
At the molecular level, as soon as you disturb the
first molecule, it will disturb it's neighbors, those in turn
will disturb their neighbors, etc. This process occurs
at the speed of sound. The clock begins as soon as
the first molecule has changed position.
but would you say there is a wave
propagated in real life in the very first moment the rock touches the first
molecule of water?
Yes.
Doesn't it take some finite time to transmit the first energy to the next
meolecule?
Yes. And that transmission has begun as soon as the rock
begins disturbing the surface, while it is still in the
process of falling.
If not, then why isn;t the speed of sound infinite?
The speed of sound is finite. It takes time for a disturbance
to move. This does not allow you to conclude that the
disturbance won't begin to propagate until it "knows" that
it's done making a complete period.
Imagine you are wiggling one end of a string. Do you
think that as you pull up your end, the part of the
string one millimeter away from your hand stays
back where it started?
No, when you pull up your hand, you disturb a
big chunk of string, and by the time you complete
one whole oscillation, a disturbance one wavelength
long is already moving down the string. It doesn't wait
for you to finish your first period.
I'd say it takes a little finite time to create the wave to
begin with.
Define that time in molecular terms.
For the point source of the wave to create the entire wave takes an amount of
time equal to the inverse of the frequency. That's independent of the molecular
nature, but a property of waves in general, I'd say.
Yes, but for a point source to create a propagating disturbance
does not take an entire period. When you've completed 0.001
of a period, there's a propagating disturbance 0.001 of a
wavelength long. That is true of waves in general.
When the waves are coming from a rock hitting water, your
description of what is happening should be explainable
in molecular terms.
So in molecular terms, as the first molecules are being disturbed,
what do you think defines the finite time until the disturbance
starts propagating outward?
- Randy
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