Science > Physics > Cold start problem reading Morrison's Understanding QM
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Science > Physics |
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"" |
| Date: |
27 Oct 2003 09:07:20 AM |
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Cold start problem reading Morrison's Understanding QM |
On page 4, the author uses an unsidedown triangle:
(m)d^2/dt^2 r(t) = -*V(r,t)
Since I'm reading this to myself, I don't have any idea the
English incantation for the upsidedown triangle (* in the eq.).
All I want is the spoken word.
This is not a book where I can read the ending first to see how
the story ends :-). One of the integrals in Appenix I
looks wrong to me---but don't tell me--I want to do it myself).
/BAH
Subtract a hundred and four for e-mail.
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| User: "Greg Neill" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
27 Oct 2003 09:57:19 AM |
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<jmfbahciv@aol.com> wrote in message news:bnjchv$cv8$3@bob.news.rcn.net...
On page 4, the author uses an unsidedown triangle:
(m)d^2/dt^2 r(t) = -*V(r,t)
Since I'm reading this to myself, I don't have any idea the
English incantation for the upsidedown triangle (* in the eq.).
All I want is the spoken word.
That would be the Dell operator. It implies taking
the partial derivatives w.r.t. each of the i,j,k
vector components. So if V(r,t) is a vector potential
DelV = (dV/dx)*i + (dV/dy)*j + (dV/dz)*k
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| User: "Edward Green" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
27 Oct 2003 05:05:33 PM |
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wrote in message news:<bnjchv$cv8$3@bob.news.rcn.net>...
On page 4, the author uses an unsidedown triangle:
(m)d^2/dt^2 r(t) = -*V(r,t)
Since I'm reading this to myself, I don't have any idea the
English incantation for the upsidedown triangle (* in the eq.).
All I want is the spoken word.
You know, it's not obvious to me a priori which way a triangle would
be "upside down"! In context, since this is a gradient ... as you've
since learned ... apparently an upside down triangle is standing on
its pointy end. Ok ... I guess that makes sense: they don't call one
of the flat sides "the base" for nothing. :-)
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| User: "" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
28 Oct 2003 06:03:03 AM |
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In article <2a0cceff.0310271505.3151d4d3@posting.google.com>,
(Edward Green) wrote:
jmfbahciv@aol.com wrote in message news:<bnjchv$cv8$3@bob.news.rcn.net>...
On page 4, the author uses an unsidedown triangle:
(m)d^2/dt^2 r(t) = -*V(r,t)
Since I'm reading this to myself, I don't have any idea the
English incantation for the upsidedown triangle (* in the eq.).
All I want is the spoken word.
You know, it's not obvious to me a priori which way a triangle would
be "upside down"! In context, since this is a gradient ... as you've
since learned ... apparently an upside down triangle is standing on
its pointy end. Ok ... I guess that makes sense: they don't call one
of the flat sides "the base" for nothing. :-)
Especially if you have to measure it. Balancing something on its
pointy end and measuring does not assure accuracy.
/BAH
Subtract a hundred and four for e-mail.
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| User: "Edward Green" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
29 Oct 2003 05:44:53 PM |
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wrote in message news:<bnlph3$rp7$1@bob.news.rcn.net>...
In article <2a0cceff.0310271505.3151d4d3@posting.google.com>,
nulldev00@aol.com (Edward Green) wrote:
...it's not obvious to me a priori which way a triangle would
be "upside down"! In context, since this is a gradient ... as you've
since learned ... apparently an upside down triangle is standing on
its pointy end. Ok ... I guess that makes sense: they don't call one
of the flat sides "the base" for nothing. :-)
Especially if you have to measure it. Balancing something on its
pointy end and measuring does not assure accuracy.
But reading a historical page Uncle Al gave the link to, I see the
original symbol was even balanced on the pointy end, but cantilevered
off to one side! At least the modern one is in unstable equilibrium.
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| User: "Uncle Al" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
27 Oct 2003 10:14:02 AM |
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wrote:
On page 4, the author uses an unsidedown triangle:
(m)d^2/dt^2 r(t) = -*V(r,t)
Since I'm reading this to myself, I don't have any idea the
English incantation for the upsidedown triangle (* in the eq.).
All I want is the spoken word.
This is not a book where I can read the ending first to see how
the story ends :-). One of the integrals in Appenix I
looks wrong to me---but don't tell me--I want to do it myself).
The "upsidedown triangle" is a del or nabla.
http://mathworld.wolfram.com/Del.html
http://en.wikipedia.org/wiki/Del
http://www.hypercomplex.com/education/intro_tutorial/nabla.html
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
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| User: "" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
28 Oct 2003 06:24:28 AM |
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In article <3F9D444A.5F7F53E0@hate.spam.net>,
Uncle Al <UncleAl0@hate.spam.net> wrote:
jmfbahciv@aol.com wrote:
On page 4, the author uses an unsidedown triangle:
(m)d^2/dt^2 r(t) = -*V(r,t)
Since I'm reading this to myself, I don't have any idea the
English incantation for the upsidedown triangle (* in the eq.).
All I want is the spoken word.
This is not a book where I can read the ending first to see how
the story ends :-). One of the integrals in Appenix I
looks wrong to me---but don't tell me--I want to do it myself).
The "upsidedown triangle" is a del or nabla.
http://mathworld.wolfram.com/Del.html
http://en.wikipedia.org/wiki/Del
http://www.hypercomplex.com/education/intro_tutorial/nabla.html
Thanks for doing the lookup work. I'm not going to be able
to get to the library until January. But a tuorial sounds
so tempting rather than working through page 875 of my calc text.
I just scanned the chapter; there are a lot of terms in there
that I see here in the newsgroup. I suspect that skipping
the word problems in calc class was not a Good Thing.
Having physics texts that only used algebra was a Bad Thing.
/BAH
Subtract a hundred and four for e-mail.
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
27 Oct 2003 09:20:41 AM |
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wrote:
On page 4, the author uses an unsidedown triangle:
(m)d^2/dt^2 r(t) = -*V(r,t)
Since I'm reading this to myself, I don't have any idea the
English incantation for the upsidedown triangle (* in the eq.).
All I want is the spoken word.
Try "nabla operator" or "gradient". (the equation above is Newton's law,
right? I don't know the book...)
[snip rest]
Bye,
Bjoern
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| User: "" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
28 Oct 2003 06:16:40 AM |
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In article <3F9D37C9.B6BB0D9D@ix.urz.uni-heidelberg.de>,
Bjoern Feuerbacher <bfeuerba@ix.urz.uni-heidelberg.de> wrote:
jmfbahciv@aol.com wrote:
On page 4, the author uses an unsidedown triangle:
(m)d^2/dt^2 r(t) = -*V(r,t)
Since I'm reading this to myself, I don't have any idea the
English incantation for the upsidedown triangle (* in the eq.).
All I want is the spoken word.
Try "nabla operator" or "gradient". (the equation above is Newton's law,
right? I don't know the book...)
Right. I didn't want to create more thread drift than necessary
by trying to describe the equation. I just needed a word that
teachers speak when they write it on the blackboard. It also
helps when I try to use an index.
Thank you all for answering. Each added to my learning what
I don't know. I also found gradient (but not nabla) in my
calc text. Now I know what needs work.
I don't think (but my memory's seems to be pretty broken) I've
heard anybody say nabla.
/BAH
/BAH
Subtract a hundred and four for e-mail.
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| User: "Jon Bell" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
28 Oct 2003 09:55:37 AM |
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In article <bnlqak$rp7$2@bob.news.rcn.net>, <jmfbahciv@aol.com> wrote:
In article <3F9D37C9.B6BB0D9D@ix.urz.uni-heidelberg.de>,
Thank you all for answering. Each added to my learning what
I don't know. I also found gradient (but not nabla) in my
calc text. Now I know what needs work.
I don't think (but my memory's seems to be pretty broken) I've
heard anybody say nabla.
"Nabla" is the technical name for the actual symbol (the inverted
triangle), whereas "gradient" is the mathematical operation that it
represents in your example. The nabla is also used in writing a couple of
mathematical operations involving vector fields, specifically the
"divergence" and the "curl."
But I've never heard anybody actually use the word "nabla" when reading an
equation out loud (as I often do when writing an equation on the board in
lecture). Everybody actually says "del."
For a scalar field V:
Gradient of V --> "del V"
For a vector field E:
Divergence of E --> "del dot E" (written to look like a vector dot
product)
Curl of E --> "del cross E" (written to look like a vector cross product)
You probably won't encounter divergence and curl in studying QM at
Morrison's level. You find them most often in electricity and magnetism,
where they appear in Maxwell's Equations.
--
Jon Bell <jtbellap8@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA
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| User: "" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
12 Apr 2005 06:05:31 AM |
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In article <bnm3hp$g4f$1@jtbell.presby.edu>,
(Jon Bell) wrote:
In article <bnlqak$rp7$2@bob.news.rcn.net>, <jmfbahciv@aol.com> wrote:
In article <3F9D37C9.B6BB0D9D@ix.urz.uni-heidelberg.de>,
Thank you all for answering. Each added to my learning what
I don't know. I also found gradient (but not nabla) in my
calc text. Now I know what needs work.
I don't think (but my memory's seems to be pretty broken) I've
heard anybody say nabla.
"Nabla" is the technical name for the actual symbol (the inverted
triangle), whereas "gradient" is the mathematical operation that it
represents in your example. The nabla is also used in writing a couple of
mathematical operations involving vector fields, specifically the
"divergence" and the "curl."
But I've never heard anybody actually use the word "nabla" when reading an
equation out loud (as I often do when writing an equation on the board in
lecture). Everybody actually says "del."
For a scalar field V:
Gradient of V --> "del V"
For a vector field E:
Divergence of E --> "del dot E" (written to look like a vector dot
product)
Curl of E --> "del cross E" (written to look like a vector cross product)
You probably won't encounter divergence and curl in studying QM at
Morrison's level. You find them most often in electricity and magnetism,
where they appear in Maxwell's Equations.
Subtract a hundred and four for e-mail.
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| User: "" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
12 Apr 2005 05:41:11 AM |
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In article <SKadnb15xu6jJMbfRVn-1g@rcn.net>, wrote:
Please ignore this. Something fucked up on my computer...
which I have yet to figure out. This post is on my todo-list
and demonstrates how much I'm getting done these days [wry
emoticon here].
/BAH
Subtract a hundred and four for e-mail.
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| User: "Bruce Scott TOK" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
27 Oct 2003 12:25:30 PM |
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BAH wrote:
|> On page 4, the author uses an unsidedown triangle:
|>
|> (m)d^2/dt^2 r(t) = -*V(r,t)
|>
|> Since I'm reading this to myself, I don't have any idea the
|> English incantation for the upsidedown triangle (* in the eq.).
|>
|> All I want is the spoken word.
mass times acceleration equals force,
where force equals minus gradient of force potential
note force is a vector (like r) and potential is a scalar...
--
cu,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
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| User: "" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
28 Oct 2003 06:28:11 AM |
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In article <200310271825.h9RIPUYx008525@ipp.mpg.de>,
Bruce Scott TOK <Use-Author-Supplied-Address-Header@[127.1]> wrote:
BAH wrote:
|> On page 4, the author uses an unsidedown triangle:
|>
|> (m)d^2/dt^2 r(t) = -*V(r,t)
|>
|> Since I'm reading this to myself, I don't have any idea the
|> English incantation for the upsidedown triangle (* in the eq.).
|>
|> All I want is the spoken word.
mass times acceleration equals force,
where force equals minus gradient of force potential
note force is a vector (like r) and potential is a scalar...
I think one of the places I get confused is not knowing the
difference between scalar and magnitude.
/BAH
Subtract a hundred and four for e-mail.
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| User: "Bruce Scott TOK" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
29 Oct 2003 01:00:26 PM |
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BAH wrote:
|> In article <200310271825.h9RIPUYx008525@ipp.mpg.de>,
|> Bruce Scott TOK <Use-Author-Supplied-Address-Header@[127.1]> wrote:
|> >BAH wrote:
|> >
|> >|> On page 4, the author uses an unsidedown triangle:
|> >|>
|> >|> (m)d^2/dt^2 r(t) = -*V(r,t)
|> >|>
|> >|> Since I'm reading this to myself, I don't have any idea the
|> >|> English incantation for the upsidedown triangle (* in the eq.).
|> >|>
|> >|> All I want is the spoken word.
|> >
|> >mass times acceleration equals force,
|> >where force equals minus gradient of force potential
|> >
|> >note force is a vector (like r) and potential is a scalar...
|>
|> I think one of the places I get confused is not knowing the
|> difference between scalar and magnitude.
Ah, that's important. A scalar _field_ is an object with a single
number associated with every point on the manifold (i.e., the space
time). In physics we expect these to be continuous, so that you can
take space or time derivatives without worrying about special math
pitfalls. The gradient of a scalar field is itself a _vector_ field. A
vector is an object with a set of numbers at every point, that
transforms between coordinate systems according to some specific rules.
The magnitude of a vector is its L2-norm, which means (in simple xyz
Cartesian geometry) you add the squares of the compontents and take the
square root, to get "how strong" the vector is. So for a velocity
vector, which has a direction, the speed is the magnitude.
Basically, the gradient of a scalar gives you the direction of the
"uphill". The magnitude of that vector gives you how fast the value of
the scalar changes.
What you need to get this is the last quarter of freshman calculus (or
something equivalent to "div grad curl and all that")... calculus in
more than one variable.
--
cu,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
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| User: "Gregory L. Hansen" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
27 Oct 2003 01:10:46 PM |
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In article <bnjchv$cv8$3@bob.news.rcn.net>, <jmfbahciv@aol.com> wrote:
On page 4, the author uses an unsidedown triangle:
(m)d^2/dt^2 r(t) = -*V(r,t)
Since I'm reading this to myself, I don't have any idea the
English incantation for the upsidedown triangle (* in the eq.).
All I want is the spoken word.
This is not a book where I can read the ending first to see how
the story ends :-). One of the integrals in Appenix I
looks wrong to me---but don't tell me--I want to do it myself).
/BAH
Subtract a hundred and four for e-mail.
If the triangle indicates the difference in two quantities, it's a delta.
If the triangle is the gradient operator say gradient, divergence, or
curl, as appropriate. Or del or nabla.
If the triangle is that which indicates the difference in two quantites
but is used for del^2, curse the author quietly for being such a putz,
imagine scenarios where you'd write him a letter where del^2 will mingle
with differences of quantities, and call it del-squared or nabla-squared.
--
"Let us learn to dream, gentlemen, then perhaps we shall find the
truth... But let us beware of publishing our dreams before they have been
put to the proof by the waking understanding." -- Friedrich August Kekulé
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| User: "" |
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| Title: Re: Cold start problem reading Morrison's Understanding QM |
27 Oct 2003 01:56:44 PM |
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In article <bnjchv$cv8$3@bob.news.rcn.net>, writes:
On page 4, the author uses an unsidedown triangle:
(m)d^2/dt^2 r(t) = -*V(r,t)
Since I'm reading this to myself, I don't have any idea the
English incantation for the upsidedown triangle (* in the eq.).
All I want is the spoken word.
Gradient.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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