3 Q?s
LawDog
Recently, the "ask the the weather bee" item in theSacramento Bee implied
that sunlight has weight. I don't buy it.
I also heard on the radio that an uncoiled spring weighs less than a coiled
spring proving that energy has mass. Again, I have trouble believing this.
And, this one has had me for years, is "Zero" a discovery, or an invention?
I think it's an invention, but we can probably go on and on about it.
In earnest
jsdesquire
Dr. Michael Albert
> Recently, the "ask the the weather bee" item in theSacramento Bee implied
> that sunlight has weight. I don't buy it.
>
> I also heard on the radio that an uncoiled spring weighs less than a coiled
> spring proving that energy has mass. Again, I have trouble believing this.
In both cases, the answer is yes, but negligibly so.
Taking the second question first, yes, if you can store energy
in an object, you have increased it's mass, according to the
theory of special relativity. The mass *increase* is
change in mass = (added energy) / (speed of light squared)
and if you get the units right you will find that, because the
speed of light is so large, the change is mass is negligibly
small. In the case of the spring, the change in mass due
to relativity would probably be smaller than the change in
air bouancy due to some slight change in the volume of the
spring as it is compressed--the point being, a negligibly
small amount. Even in nuclear reactions, the mass change is
on the order of a 1%.
In the first case, again, in a sense light has weight. If
you had a box whose walls were perfect mirrors, then the mass
of the box would increase if you "filled" it with light.
Again, this is a negligible and irrelevant for all purposes
of which I can conceive.
I'd like to add, by the way, that as usual I dislike agreeing
with the newspaper report because, while technically correct,
it almost certaintly was so misleading that one should really
consider it incorrect. The "spring" example is correct in
principle, but it must be emphasized that the effect predicted
by special relativity is extremely small, because otherwise
every-day experience would falsify special relativity. Of
course, the reason while special relativity is interesting
is because there are other cases, like nuclear reactions, where
the effect is large enough to measure and the result is
in agreement with special relativity. But he point is
that special relativity is also consistent with every-day
experience, it's just that is the case of everyday
experience the difference between the predictions of
special relativity and classical physics are negligibly
small, as they must be or classical physics would never
have been clasical physics :-).
> And, this one has had me for years, is "Zero" a discovery, or an invention?
> I think it's an invention, but we can probably go on and on about it.
In this case, your second sentence is abosolutely right. Whether
mathematical "objects" really exist in some "Platonic universe"
or whether mathematics should be seen as a formal game in which
proofs are "valid" in some formal sense without necessarily
being true is a philosophical question more than a real one.
Uncle Al
>
> O.K., I am disturbed.
>
> Recently, the "ask the the weather bee" item in theSacramento Bee implied
> that sunlight has weight. I don't buy it.
Energy has mass-equivalent. Divide by c^2.
> I also heard on the radio that an uncoiled spring weighs less than a coiled
> spring proving that energy has mass. Again, I have trouble believing this.
A compressed spring weighs more. Energy (force times distance) has
mass-equivalent. Divide by c^2. If coiling causes strain, teh
strained spring weighs more than its unstrained precursor.
> And, this one has had me for years, is "Zero" a discovery, or an invention?
> I think it's an invention, but we can probably go on and on about it.
"Zero" is a concept.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
Gregory L. Hansen
LawDog <jsdes...@eartDeletethiShlink.net> wrote:
>O.K., I am disturbed.
>
>Recently, the "ask the the weather bee" item in theSacramento Bee implied
>that sunlight has weight. I don't buy it.
>
>I also heard on the radio that an uncoiled spring weighs less than a coiled
>spring proving that energy has mass. Again, I have trouble believing this.
E=mc^2, my good man. This is easily seen in nuclear reactions where, for
instance, the mass of a helium nucleus is less than the mass of two
protons plus two neutrons. But a helium nucleus is made from two protons
and two neutrons, no particles are missing. The mass difference is due to
the arrangement of the particles, the "binding energy".
This has also been measured in chemical reactions, where the atoms that
make up the molecules don't change, they're all there, but there's a mass
difference associated with their arrangement.
The increased weight of a compressed spring is, at the moment, strictly a
theoretical conclusion. But it's a theory that agrees well with all
experiments that have been done, and there's no reason to think it will
fail here.
>
>And, this one has had me for years, is "Zero" a discovery, or an invention?
>I think it's an invention, but we can probably go on and on about it.
Are numbers themselves a discovery or an invention? Some say discovery,
but I say numbers don't even make sense until someone decides to count
something, and sets out criteria for what will be counted and what will
not. In the classic example that Asimov as a student embarrassed a
professor with, if you break a peice of chalk do you have a peice of chalk
broken in half, or do you have two peices of chalk? Do you only count the
peices big enough to write with, or do you also count the fragments and
dust? There isn't even such a thing as counting until you define what
you're interested in.
"Zero" as we know it started out as a place marker in positional numbering
systems going back to Babylonia. It was not thought of as a number, it
just indicated the value of a number. I believe it was first used as a
number and given a circular shape in India, where it harmonized with
Eastern philosophy. But scholars throughout the learned world considered
it an abomination, "the nothing that is". It was even banned for a time
by the Turks. It also raised some ambiguities, like if x^2=x*x, what is
x^0? x multiplied zero times? And we know that if a*b=c then a=c/b, but
what if b=0? a*0=0, but 0/0 is undefined. These are the sorts of
conundrums that enraged some mathematicians. It took a few centuries to
win universal acceptance.
--
"For every problem there is a solution which is simple, clean and wrong. "
-- Henry Louis Mencken
LawDog
I "mostly" understood them and failed "physics for football players" at
Berkeley.
best
jsdesquire
"LawDog" <jsdes...@eartDeletethiShlink.net> wrote in message
news:vkLV8.5187$Yx5...@newsread1.prod.itd.earthlink.net...
Dr Arm®
>
> O.K., I am disturbed.
>
> Recently, the "ask the the weather bee" item in theSacramento Bee implied
> that sunlight has weight. I don't buy it.
If they ment weight is rest mass they are wrong. No mass, but it is
influenced by a gravitational field and it does have momemtum. Though a
lightbeam will be "attracted" towards a gravitational mass, it wil not
in turn attract that mass.
>
> I also heard on the radio that an uncoiled spring weighs less than a coiled
> spring proving that energy has mass. Again, I have trouble believing this.
Yes it does, as a charged battery has mass that an uncharged battery
doesn't, but not that you would ever notice.
>
> And, this one has had me for years, is "Zero" a discovery, or an invention?
> I think it's an invention, but we can probably go on and on about it.
It's neither. It's a quantity, not a contrivance.
da
>
> In earnest
> jsdesquire
Old Man
news:vkLV8.5187$Yx5...@newsread1.prod.itd.earthlink.net...
>
> Recently, the "ask the the weather bee" item in theSacramento Bee implied
> that sunlight has weight. I don't buy it.
Your imprecise statement leaves much to be improved upon,
but in a tit for tat manner, you ought to buy it. Does light
carry energy? Energy gravitates. Euclidean geometry was a
great achievement for mankind, but if jsdesquire insists on
believing that, "the shortest distance between two points is a
straight line", then he will never understand how light and
energy can gravitate.
> I also heard on the radio that an uncoiled spring weighs less than a
coiled
> spring proving that energy has mass. Again, I have trouble believing
this.
You had better believe it. How else is energy to be conserved?
Springs, the bomb, and E = mc^2. It's all great physics.
> And, this one has had me for years, is "Zero" a discovery, or an
invention?
> I think it's an invention, but we can probably go on and on about it.
Earnestly, who cares. Mathematics is abstract. There is nothing
inherently natural about it. Applied mathematics is a brand new
ball game. Add the scientific method for "discovery". Now,
unlike the Greeks, one can move from one side of a room to
another in a finite amount of time. [Old Man]
> In earnest
> jsdesquire
>
>
Maleki
<Uncl...@hate.spam.net> wrote in
<3D27A4BC...@hate.spam.net> that:
>A compressed spring weighs more.
Bullshit.
Franz Heymann
"Dr Arm®" <Genuin...@Yahoo.com> wrote in message
news:3D27CD...@Yahoo.com...
> LawDog wrote:
> >
> > O.K., I am disturbed.
> >
> > Recently, the "ask the the weather bee" item in theSacramento Bee
implied
> > that sunlight has weight. I don't buy it.
>
> If they ment weight is rest mass they are wrong. No mass, but it is
> influenced by a gravitational field and it does have momemtum. Though
a
> lightbeam will be "attracted" towards a gravitational mass, it wil not
> in turn attract that mass.
Two or more photons may have rest mass.
[snip]
Franz Heymann
Sam Wormley
>
>
> Bullshit.
Measure it! A Measurement (well done, of course) is worth a thousand
expert opinions! Unfortunately, for you, the opinions expressed in this
thread are not opinions at all, but scientific fact. Measure it!
Maleki
<swor...@mchsi.com> wrote in
<3D284210...@mchsi.com> that:
You mean, "weigh it", no? Well, it is also scientific
fact beyond what nonsense Uncle Al may say, that if you
weigh protons and neutrons as free particles and
compare the result to what you get by weighing the same
number of them inside nucleus, you will see that they
weigh more as free particles exactly by the mass
equivalent of their binding energies inside nucleus.
Binding energy doesn't show itself to weighing. Energy
equivalent of mass is not mass, and weighing means
measuring of the mass.
Franz Heymann
"Maleki" <male...@hotmail.com> wrote in message
news:e6tfiu0nikr5nkqi5...@4ax.com...
A compressed spring has a mass which is increased by the amount E/c^2,
where E is the stored energy. Mass has weight. Ergo a compressed
spring weighs more than the same spring uncompressed.
The effect is immeasurably small.
Franz Heymann
Spaceman
>A compressed spring has a mass which is increased by the amount E/c^2,
>where E is the stored energy. Mass has weight. Ergo a compressed
>spring weighs more than the same spring uncompressed.
>
<ROFLOL>
The parrot wants a cracker!
Franz,
That is so full of it..
it's sad.
stored energy has no mass unless it is being released.
a compressed spring weighs and has the same mass
as an uncompressed spring.
It has more potential for energy.
not more mass at rest.
Stop twisting it to such magic.
If it were true
we would ship them compressed
for even an ounce over a thousands springs
could save millions of dollars in shipping costs.
you are full of it!
and have no clue about REALITY
I suggest you drop your books.
and try making something with them for REAL.
James M Driscoll Jr
Spaceman
http://www.realspaceman.com
Jan C. Bernauer
wrote:
>>From: "Franz Heymann" Franz....@btopenworld.com
>
>>A compressed spring has a mass which is increased by the amount E/c^2,
>>where E is the stored energy. Mass has weight. Ergo a compressed
>>spring weighs more than the same spring uncompressed.
>>
>
><ROFLOL>
>
>The parrot wants a cracker!
>
>Franz,
>
>That is so full of it..
>it's sad.
>
>stored energy has no mass unless it is being released.
>
>a compressed spring weighs and has the same mass
>as an uncompressed spring.
>
>It has more potential for energy.
>not more mass at rest.
>Stop twisting it to such magic.
>
>If it were true
>we would ship them compressed
>for even an ounce over a thousands springs
>could save millions of dollars in shipping costs.
You are dumb. If space is no concern, we would ship them uncompressed.
They would have less mass and you don“t need to supply energy to
compress them.
The changing of mass can be measurable. A Deuterium, that is, a
hydrogen isotop with a neutron in the core doesn“t have the same mass
as a hydrogen with just a proton in the core plus the mass of a single
neutron.
That“s caused by the energy involved in the nuclear bonds.
----
Jan C. Bernauer
Sam Wormley
>
> >From: "Franz Heymann" Franz....@btopenworld.com
>
> >A compressed spring has a mass which is increased by the amount E/c^2,
> >where E is the stored energy. Mass has weight. Ergo a compressed
> >spring weighs more than the same spring uncompressed.
> >
>
Driscoll, when posting in a physics newsgroup, you should at least
look up the well established and widely known physics.
See: http://www.google.com/search?q=mass+of+compressed+spring
Marco Nelissen
> and have no clue about REALITY
Are you talking about the reality that *you* live in, or the reality
that the rest of the world lives in? As we all know, those two realities
are vastly different. In your reality, you're a brilliant scientist running
a succesful tire company. In our reality, you're an uneducated crackpot troll,
who should never have been allowed to procreate, let alone be allowed to
have custody over the offspring.
Maleki
Heymann" <Franz....@btopenworld.com> wrote in
<ag9lub$bgk$5...@helle.btinternet.com> that:
But the effect of stored energy in nucleus is not
immeasurably small. It is easy to weigh atoms and look
for that effect.
A nucleus is a good example for a compressed spring,
no? All those positive charges are bunched up in a very
small space. We can use the nuclear force ("strong"
nuclear force? Or "weak"? I don't remember) that holds
the protons together as equivalent to that invisible
massless string that keeps our spring in question
compressed.
Now if you are right, then nucleus must be heavier than
sum of the masses of same particles in their free
states, but you can easily weigh atoms and see that
this is not the case. For instance carbon, C12, by
definition has exactly a mass of 12 amu. Now add the
masses of the same number of _free_ protons and
neutrons and electrons using any table of elementary
particles' masses, and you'll come up with a mass
slightly above 12 amu, not below it. If you
theoretically convert the difference to energy you get
exactly the binding energy inside C12 nucleus.
Therefore a portion of mass in C12 nucleus is consumed
to provide the energy necessary to compress and keep 12
positrons together. And when you weigh the atom it
shows! The weight is less than same particles when
free.
I'm not saying that I'm a hundred percent right in this
argument :) But if I'm wrong, can you clearly explain
why?
Spaceman
>You are dumb.
No,
I just twisted the wron way on the shipping issue
You are an ass.
>The changing of mass can be measurable.
It has never been measured.
you are full of it anbd have no clue abotu springs
nevermind anything mechanical in nature.
>?A Deuterium, that is, a
>hydrogen isotop with a neutron in the core doesn´t have the same mass
>as a hydrogen with just a proton in the core plus the mass of a single
>neutron.
Does not matter,
all bologna and nothing to do with springs.
>That´s caused by the energy involved in the nuclear bonds.
twsity twisty as usual..
seems I need to ignore you as usual again too.
Tell me why steel is heavier when bent under pressure.
don't bring this other crap about Deuterium in.
Why do you twist so much
I will tell the others.
It's so you can twist away from being the wrong one.
Go away Clock God worshipper.
you bother the REAL.
and still ignore it and call time real.
sad case or morons.
can't figure out how a clock MUST work
and needs to make up so much crap they can lose
most people becasue simply they don't want to read it all.
You are a joke.
Go say a prayer to your atomic clock.
I'm smashing it's Godly powers slowly but surely.
and physically.
Spaceman
>Driscoll, when posting in a physics newsgroup, you should at least
>look up the well established and widely known physics.
>
Wormley,
when posting to my posts.
have REAL stuff instead of just THEORY CRAP
all the time.
your theories are still WRONG
and even 100 yrs after they were made up
they will still be wrong
as long as they keep worshipping a clock God as such.
you are a follower and parrot only now and it's sad.,
not a physicist nor a scientist.
Go away Clock God worshipper
try getting a REAL spring and scale yourself !
It DOES NOT CHANGE MASS just from compression of the spring alone
Physics is a joke to any REAL tech.
I bet you have not even ever worked on a REAL machine
with a REAL spring ever in your life.
for if you had.
you would know that is bologna.
and all theories are stupid and WRONG.
and that is why they are still theory
almost 100 yrs later.
no facts
facts are backed up by Planet rotation and actual things
that have force .
not magical time and magical curved space and Zero point energy crap.
IT's a friggen joke to a REAL mind that works with the REAL
and not just thought "theory"
Your Clock God is falling and you are not going to be able to catch
it.
I suggest you move out of the way.
for it is falling hard.
and I have stepped far away from the false God you worship.
Spaceman
>Are you talking about the reality that *you* live in, or the reality
>that the rest of the world lives in? As we all know, those two realities
>are vastly different. In your reality, you're a brilliant scientist running
>a succesful tire company. In our reality, you're an uneducated crackpot
>troll,
>who should never have been allowed to procreate, let alone be allowed to
>have custody over the offspring.
and the insultation physics is piling up as usual from the clock
worshippers..
BTW folks.,
I killed thier God so they really hate me now.
each and every insult is proof of thier hate.
<LOL>
You will notice some great peopel on these groups.
and not one of them would insult me to prove me wrong.
and You can rule Marco from ever giving any proof of anything.
Spaceman
>A nucleus is a good example for a compressed spring,
>no? All those positive charges are bunched up in a very
>small space. We can use the nuclear force ("strong"
>nuclear force? Or "weak"? I don't remember) that holds
>the protons together as equivalent to that invisible
>massless string that keeps our spring in question
>compressed.
you forgot to measure the stuff that hold it together
first you need the "clamps" measured for thier mass.
and then the spring.
add the 2 masses
and you will have the compressed spring and binding energy
mases.
If you did not measure the bindign energies seperate and add
it to the spring,
fo course the spring and clamps will be heavier.
I hope you get it.
I made it as simple as possible.
basically they missed measureing the clamps themsleves
for the before (the clamps outside..)
all the stuff that holds the spring compressed.
PS..
even air pressure can be part of that mass depending on the
springs composition and denisity and such.
compressed or not..
the "spring alone" always has the same mass.
but can have more energy potential when compressed..
the mass itself is not different,
the energy it can produce is.
just like a bowling ball in motion.
It's mass does not change.
it's kinetic energy with the "same mass" changes.
Dr Arm®
>
> "Dr ArmŽ" <Genuin...@Yahoo.com> wrote in message
I don't get that. Why "two or more" photons? I'm missing somethig here.
da
Dr Arm®
>
> >From: "Franz Heymann" Franz....@btopenworld.com
>
> >A compressed spring has a mass which is increased by the amount E/c^2,
> >where E is the stored energy. Mass has weight. Ergo a compressed
> >spring weighs more than the same spring uncompressed.
> >
>
> <ROFLOL>
>
> The parrot wants a cracker!
>
> Franz,
>
> That is so full of it..
> it's sad.
>
> stored energy has no mass unless it is being released.
No, that's not correct. The mass of the spring increases by the amont of
energy added to it. Honest. Has nothing to do with releasing it, don't
even know where that comes from. Who says it needs to be "released" What
equations?
>
> a compressed spring weighs and has the same mass
> as an uncompressed spring.
No.
>
> It has more potential for energy.
> not more mass at rest.
Yes, it does.
> Stop twisting it to such magic.
Blame that "Albert" guy. It's only magic depending on which side of the
curtain you're on.
>
> If it were true
> we would ship them compressed
> for even an ounce over a thousands springs
> could save millions of dollars in shipping costs.
Your actual savings would be somewhat less than that and would occur
when the springs were unwound., the battery discharged and the pendulum
at its lowpoint.
>
> you are full of it!
> and have no clue about REALITY
Some day you are going to regret saying that!
>
> I suggest you drop your books.
> and try making something with them for REAL.
>
> James M Driscoll Jr
> Spaceman
> http://www.realspaceman.com
Like a spaceman? <bg>
da
Spaceman
>I don't get that. Why "two or more" photons? I'm missing somethig here.
>
>da
Franz won't help you get it either.
he has no clue.
he believes the photon is massless and magical things
as such.
he also believes in a clock God.
and refuses to admit the atomic clock goofed up and not time itself.
None of them will even look at what clocks count,
nevermind any other REALities.
a photon is a vibration in the electron cloud.
electrons DO NOT have to be part of an atomic system
There are free ones all over the place.
and they even FILL space up to the brim to create
electronic pressure in space .
the center of the Universe is still smashing things up and releasing even more
of the things.
and that is the simple Universe .
It's so simple ..
physics of today should be flushed.
REALITY rules!
physics drools.
and should REMOVe thier clock God.
and find the real causes from now on.
The excess electrons.
without them
light would not travel at all.
light is not magical..
It needs a "road"
and it need a "push against the next to transfer the vibration"
like clogged highway that shoots off in every direction.
and every road is filled to the brim....
no real motion..
just back and forth bumping..
that is light.
It's a lot more simple than the physics bologna makes it.
Spaceman
>No, that's not correct. The mass of the spring increases by the amont of
>energy added to it. Honest. Has nothing to do with releasing it, don't
>even know where that comes from. Who says it needs to be "released" What
>equations?
Honest ...it's not the mass changing.
It's the kinetic energy potential.
you have forced that potential into it.
like winding a rubber band
or pushing a bowling ball up and then releasing it at the top.
the mass has not changed at all.
it's energy potentail has.
same mass more potential energy.
does the rubber band gain mass too?
No..
It gains energy.
not mass.
the energy can move more mass but
the mass is the same as it was before it was compressed (charged)
I know ..
I work with springs ALL the time.
Do you?
>Blame that "Albert" guy. It's only magic depending on which side of the
>curtain you're on.
I have seen both sides.
I choose the side of real.
not the side of magical time bologna changes.
>Some day you are going to regret saying that!
the shipping thing.
yup..
I admitted it alread..
it was today.
:)
anyway.
the mass did not change.
the energy alone did.
When you throw a ball.
the mass stays the same.
the motion variable changes and that
changes the energy variable.
the mass variable ..
guess what..
remains constant.
I know my basic math and kinetic energy crap.
and ..
I went further and found out why these "theories"
are still theories.
I hope you want to learn why too.
otherwise.
there is nothing in physics you can teach me
for I have learned about both sides of the curtain.
and..
I have shot the atomic clock God many times.
of course...
it still lives..
but only in spirit and faith of the bad math
and ..
bad clocks.
Gregory L. Hansen
If the helium nucleus is like a spring, then it's like a spring that's
bunched up in its relaxed state. You put energy into it by pulling it
apart. Deuterium fusion releases energy. That tells you there's less
energy in the aggregate than there was in the free particles.
Fission of heavy elements is closer to the spring analogy you were looking
for. A U235 nucleus is heavier than the fission fragments that would be
released. That's why fission happens, because it lowers the potential
energy, and the nucleus was originally shoved together by stellar
processes.
Iron is sort of the cut-off. Below iron and you gain energy by fusion,
above iron and you gain energy by fission.
Dr Arm®
>
> >From: Dr Arm® Genuin...@Yahoo.com
>
> >I don't get that. Why "two or more" photons? I'm missing somethig here.
> >
> >da
>
> Franz won't help you get it either.
> he has no clue.
> he believes the photon is massless and magical things
> as such.
You speak for Franz yet you are not Franz. Who are you?
> that is light.
> It's a lot more simple than the physics bologna makes it.
So I take it you did _not_ get that A in physics?
da
Dr Arm®
>
>
> does the rubber band gain mass too?
Yes.
> I know ..
> I work with springs ALL the time.
> Do you?
Springs, yes; loose springs, starting just now.
da
>
Spaceman
>Springs, yes; loose springs, starting just now.
Good!
I hope you find out the springs don't gain "true" mass.
or loose any when un-compressed.
only energy potential is lost or gained.
the mass remains the same.
try placing the mass in a kinetic energy equation.
the mass does not change.
placing it any equation that works will lock it as
the measured mass and the other variables must change.
not the one you actually measured.
the mass variable does not vary if you had it to begin with.
that would just be bad math.
Spaceman
>You speak for Franz yet you are not Franz. Who are you?
I don't speak for Franz,
I am trying to give you advice.
too bad you won't take it I guess.
>So I take it you did _not_ get that A in physics?
You would be wrong.
and.
it's sad you now have to also resort to insultatoin physics.
Don't know how a clock works huh?
<LOL>
poor guy.
try some reality once in while.
look at what a clock counts once.
just once..
if you don't get it.
You are the one that should have failed physics.
I learned both sides of the coin.
I see you were only brainwashed by one side.
(the magical force without form side)
<LOL>
Clock worshipper too huh?
too bad..
so many could do so much more.
yet ..
all stuck in time.
..
sad..
just sad.
I am sorry I killed your Godly clock.
Find a new God.
Sam Wormley
>
>
> Honest ...it's not the mass changing.
> It's the kinetic energy potential.
>
No spaceman--"kinetic energy potential"?? All of the energy used to
compress (wind) the spring is NOT dissipated as kinetic energy.
Spaceman
>No spaceman--"kinetic energy potential"?? All of the energy used to
>compress (wind) the spring is NOT dissipated as kinetic energy.
potential was added for extra explanation.
and I'm sorry,
you are wrong.
all the energy placed in.
will come out.
too bad you never worked with springs.
or you would know that.
dang desktop mechanics..
Go away Sam.
Go outside and learn some REAL for once.
chekc out how clock can goof up.
and time remains the same.
Find the real,
stop worshipping the magical clock God known
as Atomic.
Spaceman
>No spaceman--"kinetic energy potential"?? All of the energy used to
>compress (wind) the spring is NOT dissipated as kinetic energy.
>
I blame it on physics.
George is too brainwashed.
He knows relativities magic side like nobody else does.
It's sad.
all that brain waste going to a false God known as Atomic clock.
It seems you all need a nice ride outside with an hourglass and a
fast car.
watch me change time!
I'm amazing!
<LOL>
Steve Harris
>On Sun, 07 Jul 2002 02:18:07 GMT, Uncle Al
><Uncl...@hate.spam.net> wrote in
><3D27A4BC...@hate.spam.net> that:
>
>>A compressed spring weighs more.
>
>Bullshit.
Well, in theory it does. Tension in the metal pushes atoms together and the
energy is stored as increased electric fields between atoms. These fields
have mass and weight, although it's quite small: m = E/c^2.
This is true of metals, but not generally true of solids. Many elastomers
would not weigh more on tension, at least if you let them cool off. For
example in rubber, to a first approximation, all work done on it in tension
comes off as heat, and the what is being stored after that is not energy,
but rather negentropy: you have a system with lowered entropy, which is
capable of delivering usable energy later on demand, by extracting heat from
the environment to do it. So a stretched rubber band is much like the
concentration cell we've been discussing. Or like a compressed ideal gas
which has been allowed to cool to ambient temp (which would also store some
potential energy as negentropy, without weighing more).
SBH
--
I welcome email from any being clever enough to fix my address. It's open
book. A prize to the first spambot that passes my Turing test.
Bruce Bowen
> O.K., I am disturbed.
>
> I also heard on the radio that an uncoiled spring weighs less than a coiled
> spring proving that energy has mass. Again, I have trouble believing this.
If the spring is pushing with one pound of force (~4 Newtons) and you
compress it 2 inches (0.05 m) you have increased its mass by 0.2
Joules or 0.2/9*10^16 or 2.222*10^-18 kilograms.
-Bruce
Spaceman
>If the spring is pushing with one pound of force (~4 Newtons) and you
>compress it 2 inches (0.05 m) you have increased its mass by 0.2
>Joules or 0.2/9*10^16 or 2.222*10^-18 kilograms.
NO,
you have increased it's energy potential.
the mass of the spring alone has not changed.
Why do you twist like that.
It's WRONG
please learn how kinetic energy works
the mass stays the same
the OTHER variables VARY.
Wake up.
you math is wrong and so is your mass
The energy is higher.
the mass is the same..
Do I need to hit you in the head with a baseball
to prove it?
Franz Heymann
"Spaceman" <agents...@aol.combination> wrote in message
news:20020707165456...@mb-mu.aol.com...
[snip]
Go and sell a retreaded tyre.
Franz Heymann
Franz Heymann
"Spaceman" <agents...@aol.combination> wrote in message
Repaint a stolen bicycle before selling it
Franz Heymann
Franz Heymann
"Spaceman" <agents...@aol.combination> wrote in message
Go mend a puncture.
Franz Heymann
Franz Heymann
"Spaceman" <agents...@aol.combination> wrote in message
[snip]
Go and emulate the kuru kuru bird.
Franz Heymann
Spaceman
>Repaint a stolen bicycle before selling it
Get lost.
oh..
nevermind,
you are already.
lost in time.
<LOL>
Franz Heymann
"Spaceman" <agents...@aol.combination> wrote in message
[snip]
Go and fix a brake pad.
Franz Heymann
Spaceman
>[snip]
>
>Go and sell a retreaded tyre.
>
Go and shove a clock up your ass and time travel.
Franzy boy.
want to see me stop time..
a simple hammer can stop the time being measured by
the atomic clock
you wuld say time stopped huh?
<LOL>
you are such a moron that you can't even
bring one dang tiny bit of evidence to
prove me wrong
you are also an ashole that deserves each and every insult you recieve.
Fuh off dipwad.
you have no clue about Reality.
you live in fanatasy world where you are king of the Universe
and time is your power force.
you arean idiot in true form.
Britanica told me they will now accept your picture
for the definition of "stupidtimetravelmoron.
You are part of a new class of morons.
the magical morons!
<LOL>
go away magic man.
you have no magic in this Universe of mine.
Maleki
<sbha...@ix.RETICULATEDOBJECTcom.com> wrote in
<agao8k$2kp$1...@slb4.atl.mindspring.net> that:
>Maleki wrote in message ...
>>On Sun, 07 Jul 2002 02:18:07 GMT, Uncle Al
>><Uncl...@hate.spam.net> wrote in
>><3D27A4BC...@hate.spam.net> that:
>>
>>>A compressed spring weighs more.
>>
>>Bullshit.
>
>
>
>Well, in theory it does.
? What's that supposed to mean.
>Tension in the metal pushes atoms together and the
>energy is stored as increased electric fields between atoms. These fields
>have mass and weight, although it's quite small: m = E/c^2.
>
But I'm taking the E/c^2 as the "mass equivalent of the
added field energy", not a portion of total mass that
you can weigh. You can't weigh energy. Uncle Al is
saying a compressed spring "weighs" more. He is not
saying "a compressed spring has a total mass equivalent
higher than the mass of the uncompressed spring."
Spaceman
>But I'm taking the E/c^2 as the "mass equivalent of the
>added field energy", not a portion of total mass that
>you can weigh. You can't weigh energy. Uncle Al is
>saying a compressed spring "weighs" more. He is not
>saying "a compressed spring has a total mass equivalent
>higher than the mass of the uncompressed spring."
Eeither way put,
the total mass of the spring does not change at all.
The energy it can produce is gained (kinetic energy under pressure)
but the mass is the same mass.
equations don't allow the measured variable to change.
the variables you want to find ..
change.
not the one you feed as a "real"
C,mon guys!
Why are you adding a mass that is not there?
why are you saying the mass changes when in every equation
you placed the mass in so of course it won't change.
the mass,
remains the same.
C,mon all wake up to the REAL.
not the non changes
the m in the equation always stays the same if you enter it first.
the base ball is the m.
the motion is the v
the ke is gaining with the motion.
the mass reamins the same
stop twisting the correct physics stuff to be wrong.
:)
the mass does not change
the energy does.
spin, squish, roll up a hill, throw.
all kinetic.
the kinetic energy changes,.
the mass does not.
It's how the equation works.
I'm not making it up.
:)
the m
does not change if you actually got that measurement first.
It' that simple.
the force it can produce changes.
but the baseball still only has one baseballs mass
just as the spring still only has one springs mass.
even with the math,
the mass stays the same.
anyone that wants can show thier math,
and I will gladly and nicely show you where you went wrong if the mass
changes at all.
Gregory L. Hansen
Maleki <male...@hotmail.com> wrote:
>On Sun, 7 Jul 2002 18:00:15 -0700, "Steve Harris"
><sbha...@ix.RETICULATEDOBJECTcom.com> wrote in
><agao8k$2kp$1...@slb4.atl.mindspring.net> that:
>
>>Maleki wrote in message ...
>>>On Sun, 07 Jul 2002 02:18:07 GMT, Uncle Al
>>><Uncl...@hate.spam.net> wrote in
>>><3D27A4BC...@hate.spam.net> that:
>>>
>>>>A compressed spring weighs more.
>>>
>>>Bullshit.
>>
>>
>>
>>Well, in theory it does.
>
>? What's that supposed to mean.
In theory, there's no difference between theory and practice. But in
practice there is.
Compress a spring and say you add ten joules of energy to something that
weights .25 kg. The differential increase in mass would be (10/c^2)/.25kg
= 4e-16. No weight scale on Earth has 4e-16 precision. So don't expect
to be able to measure it.
But it's from the same theory that predicts with amazing precision mass
differences that can be measured, like in nuclei and even chemical
reactions, and there's no reason why we think it should go wrong with a
spring, even if we can't measure it.
>But I'm taking the E/c^2 as the "mass equivalent of the
>added field energy", not a portion of total mass that
>you can weigh. You can't weigh energy. Uncle Al is
>saying a compressed spring "weighs" more. He is not
>saying "a compressed spring has a total mass equivalent
>higher than the mass of the uncompressed spring."
What's the difference? In theory, you put it in a scale and read the
dial. That's what the m in E=mc^2 means, just plain ordinary mass.
Maleki
glha...@steel.ucs.indiana.edu (Gregory L. Hansen)
wrote in <agajfp$flf$2...@wilson.uits.indiana.edu> that:
>
>If the helium nucleus is like a spring, then it's like a spring that's
>bunched up in its relaxed state. You put energy into it by pulling it
>apart.
You might be right, I don't know enough to know off
hand if this is the correct picture or not. But the
energy that you say you put into it to pull the nucleus
apart, in my understanding, is the energy we are using
to cancel out the effect of the nuclear force holding
two positrons close together. In my analogy for the
compressed spring, the nuclear force itself is out of
the picture, as it is just a force equivalent to
"massless invisible" imaginary string that are wrapped
around our compressed spring to hold it in that state.
So the energy that you say you put into the nucleus to
pull it apart would be like the energy we use to
stretch that invisible string far enough to allow the
spring to relax. So this energy you're talking about
doesn't apply to what we're discussing.
>Deuterium fusion releases energy. That tells you there's less
>energy in the aggregate than there was in the free particles.
>
It tells that there is less of **mass + mass equivalent
of field energy** in the aggregate. This still says
nothing about what helium would weigh, because weighing
applies only to mass part of **mass + mass equivalent
of field energy**; i.e., it applies only to the first
term within the asterisks. But Uncle Al says the second
term in there also affects the result of weighing, and
this doesn't sound right.
I did not use a good analogy. It presupposes additional
knowledge that I might not have. But as digression from
the compressed spring thread I'm still interested to
continue the discussion of mass and energies in
nucleus.
>Fission of heavy elements is closer to the spring analogy you were looking
>for.
Actually, for what I had in mind for that analogy,
fission and fusion both work the same.
>A U235 nucleus is heavier than the fission fragments that would be
>released.
Aren't two deuterium atoms heavier than a helium atom?
I don't know the answer, but I'm guessing that they're
heavier exactly by an amount of mass equivalent to the
released energy of fusion.
>That's why fission happens, because it lowers the potential
>energy, and the nucleus was originally shoved together by stellar
>processes.
>
But this applies to fusion as well, the way I see it.
In fission as well as in fusion, the process results a
lower **mass + mass equivalent of field energy** in the
products.
>Iron is sort of the cut-off. Below iron and you gain energy by fusion,
>above iron and you gain energy by fission.
Yet another reason my analogy of nucleus was not a good
one. It just presupposes a number of things before it
even begin to elucidate anything.
Maleki
glha...@steel.ucs.indiana.edu (Gregory L. Hansen)
wrote in <agcdr6$k86$2...@wilson.uits.indiana.edu> that:
>>>
>>>Well, in theory it does.
>>
>>? What's that supposed to mean.
>
>In theory, there's no difference between theory and practice. But in
>practice there is.
>
If in practice there is a difference, then the "theory"
has never been a theory to begin with.
>Compress a spring and say you add ten joules of energy to something that
>weights .25 kg. The differential increase in mass would be (10/c^2)/.25kg
>= 4e-16. No weight scale on Earth has 4e-16 precision. So don't expect
>to be able to measure it.
>
But my point is that if we could weigh it to any
precision, still we wouldn't see any difference in the
results of weighing them in the two states. Energy
cannot be weighed. Mass can. Photon can have a lot of
energy but has no mass (rest mass). But all of a sudden
you can weigh its mass when it is converted to an
electron and positron in pair production process. You
can only weigh mass, not energy.
>But it's from the same theory that predicts with amazing precision mass
>differences that can be measured, like in nuclei and even chemical
>reactions, and there's no reason why we think it should go wrong with a
>spring, even if we can't measure it.
>
I'm not saying the theory is wrong. I'm saying theory
doesn't say what you assert.
>>But I'm taking the E/c^2 as the "mass equivalent of the
>>added field energy", not a portion of total mass that
>>you can weigh. You can't weigh energy. Uncle Al is
>>saying a compressed spring "weighs" more. He is not
>>saying "a compressed spring has a total mass equivalent
>>higher than the mass of the uncompressed spring."
>
>What's the difference? In theory, you put it in a scale and read the
>dial. That's what the m in E=mc^2 means, just plain ordinary mass.
The difference is between mass and energy. They are two
different physical quantities. The fact that they're
mutually convertible doesn't negate the existence of
this difference.
Maleki
(Bruce Bowen) wrote in
<b824a8a0.02070...@posting.google.com> that:
>you have increased its mass by 0.2
>Joules
This is not physics. This looks like economy :)
Gregory L. Hansen
Maleki <male...@hotmail.com> wrote:
>On Sun, 7 Jul 2002 23:39:37 +0000 (UTC),
>glha...@steel.ucs.indiana.edu (Gregory L. Hansen)
>wrote in <agajfp$flf$2...@wilson.uits.indiana.edu> that:
>
>>
>>If the helium nucleus is like a spring, then it's like a spring that's
>>bunched up in its relaxed state. You put energy into it by pulling it
>>apart.
>
>You might be right, I don't know enough to know off
>hand if this is the correct picture or not. But the
>energy that you say you put into it to pull the nucleus
>apart, in my understanding, is the energy we are using
>to cancel out the effect of the nuclear force holding
>two positrons close together. In my analogy for the
>compressed spring, the nuclear force itself is out of
>the picture, as it is just a force equivalent to
>"massless invisible" imaginary string that are wrapped
>around our compressed spring to hold it in that state.
Change that to a "massless invisible" stick tied to the spring to keep it
stretched. A helium atom has a lower energy than four free particles, a
relaxed spring has a lower energy than a stretched spring.
The direction really doesn't matter.
>>Deuterium fusion releases energy. That tells you there's less
>>energy in the aggregate than there was in the free particles.
>>
>
>It tells that there is less of **mass + mass equivalent
>of field energy** in the aggregate. This still says
>nothing about what helium would weigh, because weighing
>applies only to mass part of **mass + mass equivalent
>of field energy**; i.e., it applies only to the first
>term within the asterisks. But Uncle Al says the second
>term in there also affects the result of weighing, and
>this doesn't sound right.
How do you think the weight is figured in the first place? F=ma. Ionize
it, give it kinetic energy, spin it around in a magnetic field.
It should show up on a static scale, too, except that requires precision
on the order of 1/1000 and a temperature near absolute zero, or else try
to weigh it as a gas. I don't know if people have done that, but it
seems doable.
>>A U235 nucleus is heavier than the fission fragments that would be
>>released.
>
>Aren't two deuterium atoms heavier than a helium atom?
Yes, by around 24 MeV/c^2.
>I don't know the answer, but I'm guessing that they're
>heavier exactly by an amount of mass equivalent to the
>released energy of fusion.
Well, masses of subatomic particles are typically given in MeV, the energy
equivalent. It works in well with the units that nuclear and particle
physicists use in their work.
>
>>That's why fission happens, because it lowers the potential
>>energy, and the nucleus was originally shoved together by stellar
>>processes.
>>
>
>But this applies to fusion as well, the way I see it.
>In fission as well as in fusion, the process results a
>lower **mass + mass equivalent of field energy** in the
>products.
>
>>Iron is sort of the cut-off. Below iron and you gain energy by fusion,
>>above iron and you gain energy by fission.
The strong nuclear force is short range, the electromagnetic force is
long-range. Two deuteriums need to be shoved together to overcome the
electrostatic repulsion, but then the strong force latches on and pulls
them together into a stable nucleus.
In U235 it's somewhat the same picture but it's metastable. The partons
are not at the lowest possible energy, but the strong force makes a
potential barrier that, for a time, keeps them together. But you know
that U235 will spontaneously fission. It's like a compressed spring that
breaks the massless, invisible spring that's keeping it compressed. And
helium is like a spring that is compressed in its normal, unforced state.
Michael Mcneil
news:vkLV8.5187$Yx5...@newsread1.prod.itd.earthlink.net
<O.K., I am disturbed.
Recently, the "ask the the weather bee" item in theSacramento Bee
implied
that sunlight has weight. I don't buy it.
I also heard on the radio that an uncoiled spring weighs less than a
coiled
spring proving that energy has mass. Again, I have trouble believing
this.
And, this one has had me for years, is "Zero" a discovery, or an
invention?
I think it's an invention, but we can probably go on and on about it.
In earnest
jsdesquire>
Sometimes I think the spam is preferable to twaddle like this.
--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG
Steve Harris
>
>? What's that supposed to mean.
The difference is there in theory but in practice is too small to measure.
>
>>Tension in the metal pushes atoms together and the
>>energy is stored as increased electric fields between atoms. These fields
>>have mass and weight, although it's quite small: m = E/c^2.
>>
>
>But I'm taking the E/c^2 as the "mass equivalent of the
>added field energy", not a portion of total mass that
>you can weigh. You can't weigh energy.
Sure you can. If you fission uranium into lighter elements, they weigh less
than the uranium. And that's enough that we CAN measure it with ordinary
scales. The difference is the energy you get when you fission uranium. It's
the weight of the energy. Which in the uranium is various fields before it
is split, and kinetic energy (heat) afterwards. Heat has weight also. A
good big H-bomb releases POUNDS of heat.
>Uncle Al is
>saying a compressed spring "weighs" more. He is not
>saying "a compressed spring has a total mass equivalent
>higher than the mass of the uncompressed spring."
If he said either one he'd be right. You have some idea that the total mass
of an object is composed of mass you can weigh and mass you can't? Duh.
Pay no attention to Spaceman. He's a wonderful weathervane on sci.physics,
inasmuch as he's ALWAYS wrong. It's uncanny.
Bruce Bowen
> NO,
> you have increased it's energy potential.
>
> the mass of the spring alone has not changed.
>
> Why do you twist like that.
> It's WRONG
>
> please learn how kinetic energy works
>
> the mass stays the same
> the OTHER variables VARY.
>
> Wake up.
> you math is wrong and so is your mass
>
> The energy is higher.
> the mass is the same..
>
> Do I need to hit you in the head with a baseball
> to prove it?
Have you fogotten to take your Prozac today Spaceman?
-Bruce
Maleki
<sbha...@ix.RETICULATEDOBJECTcom.com> wrote in
<agd1ja$gkt$1...@slb4.atl.mindspring.net> that:
>> You can't weigh energy.
>
>Sure you can. If you fission uranium into lighter elements, they weigh less
>than the uranium. And that's enough that we CAN measure it with ordinary
>scales. The difference is the energy you get when you fission uranium. It's
>the weight of the energy.
Hehe :) I have two problems with this. First is that
measuring a mass difference by weighing two different
masses and subtracting the results is not quite
weighing the absent mass in the lighter one. An absent
mass cannot be weighed, it can only be measured
indirectly. The second is that even if you could weigh
the absent mass, you'd still be weighing mass not
energy. Energy cannot be weighed (but following what I
just read and posted in another thread about light
deflection around massive bodies I'm not so sure of
this anymore).
>Which in the uranium is various fields before it
>is split, and kinetic energy (heat) afterwards. Heat has weight also. A
>good big H-bomb releases POUNDS of heat.
>
But such use of terms are mere conventions. Jargons
that develop in any trade.
>[...]
> You have some idea that the total mass
>of an object is composed of mass you can weigh and mass you can't? Duh.
>
No, mass I can weigh and another mass equivalent to the
energy that I cannot weigh.
Steve Harris
><sbha...@ix.RETICULATEDOBJECTcom.com> wrote in
><agd1ja$gkt$1...@slb4.atl.mindspring.net> that:
>
>>> You can't weigh energy.
>>
>>Sure you can. If you fission uranium into lighter elements, they weigh
less
>>than the uranium. And that's enough that we CAN measure it with ordinary
>>scales. The difference is the energy you get when you fission uranium.
It's
>>the weight of the energy.
>
>Hehe :) I have two problems with this. First is that
>measuring a mass difference by weighing two different
>masses and subtracting the results is not quite
>weighing the absent mass in the lighter one.
It is if you believe in the conservation of mass. If a kg of U or Pu
fissions into 999 grams of products, do you think that missing gram escapes
into the 5th dimention? And if the energy released is exactly 1 gram times
c^2 do you think that's this is some kind of giant frigging coincidence?
>An absent
>mass cannot be weighed, it can only be measured
>indirectly.
So also with your brain while you're alive. Be careful. What do you wish me
to *infer* about your brain with this argument?
>The second is that even if you could weigh
>the absent mass, you'd still be weighing mass not
>energy.
Again, one supposes that it's just a giant frigging coincidence that the
mass deficit happens to be exactly the energy release in nuclear processes,
according to Einstein's equation.
>Energy cannot be weighed (but following what I
>just read and posted in another thread about light
>deflection around massive bodies I'm not so sure of
>this anymore).
Good, because you're wrong.
>Which in the uranium is various fields before it
>>is split, and kinetic energy (heat) afterwards. Heat has weight also. A
>>good big H-bomb releases POUNDS of heat.
>>
>
>But such use of terms are mere conventions. Jargons
>that develop in any trade.
No. It releases pounds of heat and loses the same number of pounds of mass.
That's it.
>> You have some idea that the total mass
>>of an object is composed of mass you can weigh and mass you can't? Duh.
>>
>
>No, mass I can weigh and another mass equivalent to the
>energy that I cannot weigh.
Well, you're wrong.
Dr Arm®
>
> In article <u2bjiu4a82sjbk715...@4ax.com>,
> >>Well, in theory it does.
> >
> >? What's that supposed to mean.
>
> In theory, there's no difference between theory and practice. But in
> practice there is.
lol! I love that! Hahaha! ROFLMAO!
da
Maleki
<sbha...@ix.RETICULATEDOBJECTcom.com> wrote in
<agdd3k$akg$1...@slb4.atl.mindspring.net> that:
>Maleki wrote in message ...
>>On Mon, 8 Jul 2002 14:51:47 -0700, "Steve Harris"
>><sbha...@ix.RETICULATEDOBJECTcom.com> wrote in
>><agd1ja$gkt$1...@slb4.atl.mindspring.net> that:
>>
>>>> You can't weigh energy.
>>>
>>>Sure you can. If you fission uranium into lighter elements, they weigh
>less
>>>than the uranium. And that's enough that we CAN measure it with ordinary
>>>scales. The difference is the energy you get when you fission uranium.
>It's
>>>the weight of the energy.
>>
>>Hehe :) I have two problems with this. First is that
>>measuring a mass difference by weighing two different
>>masses and subtracting the results is not quite
>>weighing the absent mass in the lighter one.
>
>
>It is if you believe in the conservation of mass.
But in measuring the nuclei before and after fission we
are measuring their rest masses. The rest masses are
not conserved in fission.
>If a kg of U or Pu
>fissions into 999 grams of products, do you think that missing gram escapes
>into the 5th dimention? And if the energy released is exactly 1 gram times
>c^2 do you think that's this is some kind of giant frigging coincidence?
>
No, but this would only demonstrate that one gram of
mass has converted to its equivalent energy and is not
present anymore. The one gram isn't the weight of the
released energy, it is the weight of the mass
equivalent of that energy. Energy doesn't have weight.
Mass does.
>
>
>>An absent
>>mass cannot be weighed, it can only be measured
>>indirectly.
>
>So also with your brain while you're alive. Be careful. What do you wish me
>to *infer* about your brain with this argument?
>
Ok, one more time. To find out the amount of an absent
mass, one needs to _both_ weigh what's left _and_ know
the total amount before that absence occurred. So
compared with simply weighing something one more piece
if information is required. Does it make sense?
>
>
>>The second is that even if you could weigh
>>the absent mass, you'd still be weighing mass not
>>energy.
>
>Again, one supposes that it's just a giant frigging coincidence that the
>mass deficit happens to be exactly the energy release in nuclear processes,
>according to Einstein's equation.
>
But of course I accept that some mass has converted to
energy. And that we can calculate that missing mass by
measuring masses of nuclei before and after fission.
Question is whether energy has mass equivalent or has
mass! It does not have mass, so we cannot weigh energy.
It has mass equivalent, and if and only if it has
already converted into mass, then we can weigh that
mass. While it is in the form of energy we cannot weigh
it. We can just measure the energy, and this is
different from weighing its equivalent mass.
>
>
>>Energy cannot be weighed (but following what I
>>just read and posted in another thread about light
>>deflection around massive bodies I'm not so sure of
>>this anymore).
>
>
>Good, because you're wrong.
>
>
I will know that as soon as I find out about a question
I had in that thread.
>>
>>But such use of terms are mere conventions. Jargons
>>that develop in any trade.
>
>
>No. It releases pounds of heat and loses the same number of pounds of mass.
>That's it.
>
You're being mule-headed. "Pounds of heat" is not
physics. It's jargon, and not that of scientists
either.
>
>>> You have some idea that the total mass
>>>of an object is composed of mass you can weigh and mass you can't? Duh.
>>>
>>
>>No, mass I can weigh and another mass equivalent to the
>>energy that I cannot weigh.
>
>Well, you're wrong.
>
Show me why. Or read my question in the light's index
of refraction thread and answer it for me. That would
help me see I'm wrong.
Herman Trivilino
> The difference is between mass and energy. They are two
> different physical quantities. The fact that they're
> mutually convertible doesn't negate the existence of
> this difference.
Mass is just one of many different forms of energy. Consider any
collection of particles. Find the frame of reference in which their
momenta sums to zero. This frame is, by definition, the rest frame of this
collection. Now, measure the total energy of the collection, in this rest
frame. That total energy is, by definition, the mass of that collection.
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Herman Trivilino
> It tells that there is less of **mass + mass equivalent
> of field energy** in the aggregate. This still says
> nothing about what helium would weigh, because weighing
> applies only to mass part of **mass + mass equivalent
> of field energy**; i.e., it applies only to the first
> term within the asterisks. But Uncle Al says the second
> term in there also affects the result of weighing, and
> this doesn't sound right.
But it is right. Mass and "mass equivalent of field energy" are both mass.
When you weigh yourself most of the weight is the "mass equivalent of field
energy". If you added up the masses of all the quarks and leptons in your
body it would account for but a fraction of your total weight.
> Aren't two deuterium atoms heavier than a helium atom?
> I don't know the answer, but I'm guessing that they're
> heavier exactly by an amount of mass equivalent to the
> released energy of fusion.
>
> >That's why fission happens, because it lowers the potential
> >energy, and the nucleus was originally shoved together by stellar
> >processes.
> >
>
> But this applies to fusion as well, the way I see it.
> In fission as well as in fusion, the process results a
> lower **mass + mass equivalent of field energy** in the
> products.
>
> >Iron is sort of the cut-off. Below iron and you gain energy by fusion,
> >above iron and you gain energy by fission.
>
> Yet another reason my analogy of nucleus was not a good
> one. It just presupposes a number of things before it
> even begin to elucidate anything.
>
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James B. Glattfelder
> And, this one has had me for years, is "Zero" a discovery, or an invention?
> I think it's an invention, but we can probably go on and on about it.
Well, as somebody already mentioned, the issue concerns the reality of
mathematics. Is mathematics invented or discovered? (AFAIK
zero is an ancient concept first appearing in Hindu metaphysics.)
Back to your question: is the concept of nothingness real? Physically
the notion of an empty vacuum dissolves at small scales, i.e. when
quantum effects kick in.
If one observes that
infinity = 1 / zero
then the abstract notion of zero is seen in a different context: zero
is actually just as inconceivable to our minds as its cousin infinity,
we are just more used to it. (Interesting though, that the mind can
grapple with these brain teasers in a formal, mathematical fashion -
indeed, the mathematics of the infinitesimal can be traced back to
Keppler in 1615, trying to measure the volume of wine barrels.)
And somehow reality is really revealing itself as being of finite
nature. We've had a century to come to terms with the notion that
energy is not continuous but comes in discrete packets. Now
string/M-theory and loop quantum gravity is saying: space and *time*
are also discrete. (This is also the foundation for the algorithmic
approach to reality, as proposed by Wolfram and Fredkin - check
sci.physics.research for various posts).
So, now, if the universe has a slightly (infinitesimally) positive
curvature it will also have the right topology to not be infinite;-)
Spaceman
>Have you fogotten to take your Prozac today Spaceman?
Have you forgotten to post physics in your insult?
Get lost moron.
The paradox twins are the same revs of Earth WRT the Sun..
you have no clue about relativity for if you did.
you would know hwat kills it too.
Itself!
Steve Harris
>>It is if you believe in the conservation of mass.
>
>But in measuring the nuclei before and after fission we
>are measuring their rest masses. The rest masses are
>not conserved in fission.
Comment: They are if you don't let the ensemble cool, and you define "rest
mass" of the ensemble as you yourself did in another message, as the mass of
the collection in the COM frame.
For example, we can make this easy: if you put a fission bomb in a
superstrong ideal box and weight it, then detonate it so the box is sitting
there full of hot plasma at tremendous pressure, the weight won't change.
Because the total mass of the system hasn't changed. All the particles are
lighter, but now you're weighing a collection of lighter particles which are
all in rapid motion. So now you're weighing the particles PLUS (in effect)
weighing their kinetic energy. Weighing kinetic energy in a plasma of idea
gas is the same as weighing heat, because that's what heat IS in an ideal
gas: Just kinetic energy. In this case part of what you're weighing before
the bomb blows is energy in electric fields. That gets converted to kinetic
energy of motion (heat), but the weight doesn't change, and neither does the
mass. Only if you let the stuff cool down and extract the heat that way,
does the mass change. That's because heat has mass, and you let it leak out
and its mass with it.
If you merely heat a box of ideal gas with a bunsen burner, all the
particles all go faster and their total mass (as a system) increases, and as
you see in special relativity, you're weighing kinetic energy (heat) there
also. Energy in systems which can be weighed DOES have mass and weight.
Thus, if you add energy (of any kind) and you add weight.
>>If a kg of U or Pu
>>fissions into 999 grams of products, do you think that missing gram
escapes
>>into the 5th dimention? And if the energy released is exactly 1 gram times
>>c^2 do you think that's this is some kind of giant frigging coincidence?
>>
>
>No, but this would only demonstrate that one gram of
>mass has converted to its equivalent energy and is not
>present anymore. The one gram isn't the weight of the
>released energy, it is the weight of the mass
>equivalent of that energy. Energy doesn't have weight.
>Mass does.
Nope. You're making it way too complicated. See the box example above. I can
simplify even further: this is the same system as the one where two masses
at rest in empty space are joined by a compressed spring. Now, release the
spring and let the masses fly apart. If you believe that the mass of this
system increases after the spring potential has been converted to kinetic
energy of the two masses flying apart, you're going to have some difficulty
explaining where that extra mass came from. But if you simply accept the
idea that the mass of this system doesn't change when the spring does its
work, all those problems go away.
>Ok, one more time. To find out the amount of an absent
>mass, one needs to _both_ weigh what's left _and_ know
>the total amount before that absence occurred. So
>compared with simply weighing something one more piece
>if information is required. Does it make sense?
Yes, and we do know weights of the isotopes involved before and after-- both
uranium and the isotopes it fissions into.
>But of course I accept that some mass has converted to
>energy. And that we can calculate that missing mass by
>measuring masses of nuclei before and after fission.
>Question is whether energy has mass equivalent or has
>mass!
"Mass equivalent" is meaningless. All energy present in such resting and
confined systems which you can define and weigh has both mass and weight.
Period.
Franz Heymann
"Dr Arm®" <Genuin...@Yahoo.com> wrote in message
news:3D28BF...@Yahoo.com...
> Franz Heymann wrote:
> >
> > "Dr Arm®" <Genuin...@Yahoo.com> wrote in message
> > news:3D27CD...@Yahoo.com...
> > > LawDog wrote:
> > > >
> > > > O.K., I am disturbed.
> > > >
> > > > Recently, the "ask the the weather bee" item in theSacramento
Bee
> > implied
> > > > that sunlight has weight. I don't buy it.
> > >
is
> > > influenced by a gravitational field and it does have momemtum.
Though
> > a
> > > lightbeam will be "attracted" towards a gravitational mass, it wil
not
> > > in turn attract that mass.
> >
> > Two or more photons may have rest mass.
> >
> > [snip]
> >
> > Franz Heymann
>
>
> I don't get that. Why "two or more" photons? I'm missing somethig
here.
The rest mass of a system is the energy in the frame of reference in
which the CM is at rest. Two or more photons may move in such a way
that their CM is at rest. In this frame, their resultant energy is their
mass. This situation is not possible for a single photon.
Franz Heymann
Franz Heymann
"Maleki" <male...@hotmail.com> wrote in message
news:u2bjiu4a82sjbk715...@4ax.com...
[snip]
> But I'm taking the E/c^2 as the "mass equivalent of the
> added field energy", not a portion of total mass that
> you can weigh. You can't weigh energy.
You can. Energy gravitates.
Franz Heymann
Franz Heymann
"Spaceman" <agents...@aol.combination> wrote in message
> >From: Dr Arm® Genuin...@Yahoo.com
>
> >I don't get that. Why "two or more" photons? I'm missing somethig
here.
> >
>
> Franz won't help you get it either.
> he has no clue.
Naturally you won't retract that ststement when you read my answer which
I have already posted. You do not have the intellectual capacity to
understand my answer.
Go sell a tyre.
Franz Heymann
Dr Arm®
>
> "Dr ArmŽ" <Genuin...@Yahoo.com> wrote in message
> news:3D28BF...@Yahoo.com...
> > Franz Heymann wrote:
> > >
> > > news:3D27CD...@Yahoo.com...
> > > > LawDog wrote:
> > > > >
> > > > > O.K., I am disturbed.
> > > > >
> > > > > Recently, the "ask the the weather bee" item in theSacramento
> Bee
> > > implied
> > > > > that sunlight has weight. I don't buy it.
> > > >
> > > > If they ment weight is rest mass they are wrong. No mass, but it
> is
> > > > influenced by a gravitational field and it does have momemtum.
> Though
> > > a
> > > > lightbeam will be "attracted" towards a gravitational mass, it wil
> not
> > > > in turn attract that mass.
> > >
> > > Two or more photons may have rest mass.
> > >
> > > [snip]
> > >
> > > Franz Heymann
> >
> >
> > I don't get that. Why "two or more" photons? I'm missing somethig
> here.
>
> The rest mass of a system is the energy in the frame of reference in
> which the CM is at rest. Two or more photons may move in such a way
> that their CM is at rest. In this frame, their resultant energy is their
> mass. This situation is not possible for a single photon.
So, if two photons of equal energy are moving directly away from each
other then CM remains stationary.
Q: Is there some process that emits pairs of photons of equal energy in
exactly opposite directions? It is even possible to arrange this
circumstance?
da
>
> Franz Heymann
Jan C. Bernauer
wrote:
>Franz Heymann wrote:
>>
>> "Dr Arm®" <Genuin...@Yahoo.com> wrote in message
>> news:3D28BF...@Yahoo.com...
>> > Franz Heymann wrote:
>> > >
>> > > "Dr Arm®" <Genuin...@Yahoo.com> wrote in message
If I remember right, Pi^0 decay should yield this result.
----
Jan C. Bernauer
Dr Arm®
So even if the two photons each have no rest mass it is possible to
arrange them under specific conditions such that the system of photons
has a rest mass. Will that rest mass then have all the other
characteristics of a mass such as the ability to bend space and curve a
passing lightbeam? What other kind of mass is there?
So a pair of particles with a stationary Cm will bend a lightbeam while
an adjacent beam of light will not.
da
me...@cars3.uchicago.edu
>Franz Heymann wrote:
>>
>> The rest mass of a system is the energy in the frame of reference in
>> which the CM is at rest. Two or more photons may move in such a way
>> that their CM is at rest. In this frame, their resultant energy is their
>> mass. This situation is not possible for a single photon.
>
>So, if two photons of equal energy are moving directly away from each
>other then CM remains stationary.
Yep.
>
>Q: Is there some process that emits pairs of photons of equal energy in
>exactly opposite directions? It is even possible to arrange this
>circumstance?
>
Particle-antiparticle annihilation at rest.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
Herman Trivilino
> So even if the two photons each have no rest mass it is possible to
> arrange them under specific conditions such that the system of photons
> has a rest mass. Will that rest mass then have all the other
> characteristics of a mass such as the ability to bend space and curve a
> passing lightbeam? What other kind of mass is there?
Lots of other kinds have been invented. But more and more we are seeing
the introductory textbooks adopting a rational terminology where there is
only one kind of mass, the same mass as is used in Newtonian physics. As
soon as you adopt this rational terminology there is no longer any need to
refer to it as the rest mass, it is simply the mass.
Einstein's great discovery was rest energy. He discovered that rest energy
is equivalent to mass, that they are one and the same thing. If we use a
system of units where c=1, then we can write
m^2 = E^2 - p^2
where m is the mass, E the total energy, and p the momentum. If we have a
collection of particles we define their rest frame to be the frame in which
their total momentum sums to zero. In such a frame we have m=E, or in
other words, the mass equals the total energy. This is the essence of the
Einstein mass-energy equivalence.
> So a pair of particles with a stationary Cm will bend a lightbeam while
> an adjacent beam of light will not.
Anything with energy-momentum will act as a source of gravity. Mass seems
to be, by far, the greatest contributor to the energy-momentum. It is,
however, but one component. In Einstein's General Theory, there is no need
to have mass to have a source of gravity. That is a Newtonian
approximation that seems to be very very valid in our neck of the universe.
Herman Trivilino
> So, if two photons of equal energy are moving directly away from each
> other then CM remains stationary.
Unless the two photons are moving in the same direction, it is always
possible to find a frame of reference in which their CM is stationary.
> Q: Is there some process that emits pairs of photons of equal energy in
> exactly opposite directions?
Again, unless the two photons are co-moving, there will always be some
frame of reference in which they move in exactly opposite directions.
> It is even possible to arrange this
> circumstance?
It happens more often than not.
Dr Arm®
>
> "Dr Arm®" <Genuin...@Yahoo.com> wrote ...
>
> > So, if two photons of equal energy are moving directly away from each
> > other then CM remains stationary.
>
> Unless the two photons are moving in the same direction, it is always
> possible to find a frame of reference in which their CM is stationary.
>
> > Q: Is there some process that emits pairs of photons of equal energy in
> > exactly opposite directions?
>
> Again, unless the two photons are co-moving, there will always be some
> frame of reference in which they move in exactly opposite directions.
OK, I'll buy that. Does the vector cmoponent count?
da
Dr Arm®
You avoided the real issue: Will a beam of light bend an adjacent
parallel beam of light towards it. Accorfding to your discussion it
seems you are implying that it wi;ll. I have it on other authority that
it will not.
da
Dr Arm®
> >
> Particle-antiparticle annihilation at rest.
Which reminds me of another issue I have: Hawkins radiation.
If I understand it, just beyond the event horizon a
particle-antiparticle pair is spontaneously created. And the anti
particle falls into the black hole, dimishing the black hole while the
particle goes off into the universe with the net result that the
universe gained while the black hole lost.
If this is correct, here is the issue I have: As far as I know, it is
just as likely for the particle to fall into the event horizon as the
antiparticle, thereby the long term average is that the universe obtains
an equal number of antiparticles as particles, as also does the black
hole.
Is there something else operating here that tilts the bias of particle
distribution such that the anti particles are more likely to fall into
the BH?
da
me...@cars3.uchicago.edu
Herman Trivilino
> > Again, unless the two photons are co-moving, there will always be some
> > frame of reference in which they move in exactly opposite directions.
>
> OK, I'll buy that. Does the vector cmoponent count?
The vector component of what? Momentum is a vector. The rest frame of a
collection of particles is the frame in which the vector sum of the momenta
of the particles is zero. In this frame the total energy of the collection
is, by definition, its mass.
Dr Arm®
Then how does there result a net particle flow of Hawking radiation?
da
me...@cars3.uchicago.edu
>me...@cars3.uchicago.edu wrote:
>>
>> In article <3D30D8...@Yahoo.com>, Dr Arm® <Genuin...@Yahoo.com> writes:
>> >me...@cars3.uchicago.edu wrote:
>> >> >
>> >> Particle-antiparticle annihilation at rest.
>> >
>> >
>> >Which reminds me of another issue I have: Hawkins radiation.
>> >
>> >If I understand it, just beyond the event horizon a
>> >particle-antiparticle pair is spontaneously created. And the anti
>> >particle falls into the black hole, dimishing the black hole while the
>> >particle goes off into the universe with the net result that the
>> >universe gained while the black hole lost.
>> >
>> >If this is correct, here is the issue I have: As far as I know, it is
>> >just as likely for the particle to fall into the event horizon as the
>> >antiparticle, thereby the long term average is that the universe obtains
>> >an equal number of antiparticles as particles, as also does the black
>> >hole.
>> >
>> >Is there something else operating here that tilts the bias of particle
>> >distribution such that the anti particles are more likely to fall into
>> >the BH?
>> >
>> If there is, it is not something we know about, currently.
>>
>
>
would you think there is more of one than of another?
Dr Arm®
I'm asking the same question a third time: Isn't the reason you get a
net radiation is because more particles flow outward than antiparticles.
If they were even, the net result would be zero radiation as emergent
antiparticles anilated with particles.
da
me...@cars3.uchicago.edu
>net radiation is because more particles flow outward than antiparticles.
No.
>If they were even, the net result would be zero radiation as emergent
>antiparticles anilated with particles.
>
And what happens when particles annihilate with antiparticles?
The delivery of equal numbers of particles and antiparticles is *not*
zero radiation. It is a hell of a lot of radiation.
Dr Arm®
Ah! I think I see a lightbulb. So whether you get particles or
antiparticles progressing away you wind up with a net energy or particle
contribution to the universe.
But wait, that would also mean that you are adding energy to the black
hole too, at the same rate, via the same processes.
See, somemthing doesn't add up if you consider both directions occuring
equally likely. The only way HR makes sense is if particles are more
likely to drift away fromthe BH than antiparticles.
And what about when it is an energy-pair fluctuation rather than a
massive particle pair that gets created? See, the accounting seems all
wrong.
da
me...@cars3.uchicago.edu
>antiparticles progressing away you wind up with a net energy or particle
>contribution to the universe.
>
>But wait, that would also mean that you are adding energy to the black
>hole too, at the same rate, via the same processes.
>
Ahh, but you forget that it was the energy of the black hole that was
used to generate the particle-antiparticle pair in the first place.
In other words, the blackhole paid for both but got only one delivered
back, the other went to outside. So, the net result is that the black
hole lost energy and the outside universe gained same amount of
energy. Everything adds up.
>See, somemthing doesn't add up if you consider both directions occuring
>equally likely. The only way HR makes sense is if particles are more
>likely to drift away fromthe BH than antiparticles.
>
It is absolutely irrelevant to the process whether what escapes is a
particle or an antiparticle. Both particles and antiparticles have a
positive mass and carry positive energy.
>And what about when it is an energy-pair fluctuation rather than a
>massive particle pair that gets created? See, the accounting seems all
>wrong.
If something escapes, it is no longer a fluctuation. No problems with
the accounting.
Dr Arm®
No, this part is wrong, on it's face. It has to be wrong, by definition.
And it just doesn't make sense. It's not the BH energy that generates
these pairs, it's the quantum fluctuations around the EH that causes the
generation.
Look, I'm obviously missing something about HR, but this isn't it. I'm
going to start a new thread about it, you're welcomeof course, but this
part is just wrong, I'm pretty sure.
> In other words, the blackhole paid for both but got only one delivered
> back, the other went to outside. So, the net result is that the black
> hole lost energy and the outside universe gained same amount of
> energy. Everything adds up.
No, not so. If what you're sying were true, then the black hole is not
black. The reason is, because the particle-pair or energy-pair is
created just beyond the event horizon is due to the Uncertainty
Principle in operation: you can't have absolutely zero energy even
though classical physics predics exactly that at the EH.
The pair-generation is spontaneous and does not, CAN NOT, draw energy or
matter from the black hole, like you are saying it does.
No, it cannot. You're as confused about HR as I am, maybe more so. Let's
see if someone else knows.
>
> >See, somemthing doesn't add up if you consider both directions occuring
> >equally likely. The only way HR makes sense is if particles are more
> >likely to drift away fromthe BH than antiparticles.
> >
> It is absolutely irrelevant to the process whether what escapes is a
> particle or an antiparticle. Both particles and antiparticles have a
> positive mass and carry positive energy.
>
> >And what about when it is an energy-pair fluctuation rather than a
> >massive particle pair that gets created? See, the accounting seems all
> >wrong.
>
> If something escapes, it is no longer a fluctuation. No problems with
> the accounting.
No, the problem still exists, you're overlooking it.
I'm pretty sure.
da
me...@cars3.uchicago.edu
What definition? Your definition? Well, you're wrong.
>And it just doesn't make sense. It's not the BH energy that generates
>these pairs, it's the quantum fluctuations around the EH that causes the
>generation.
Quantum fluctuations generate *nothing*. They just create the
possibility of generation. In order for anything to be actually
generated, the energy bill has to be paid. No exceptions other than
de*dt > hbar.
>
>Look, I'm obviously missing something about HR, but this isn't it. I'm
>going to start a new thread about it, you're welcomeof course, but this
>part is just wrong, I'm pretty sure.
>
Feel free to be sure, but you'll better educate yourself.
>> In other words, the blackhole paid for both but got only one delivered
>> back, the other went to outside. So, the net result is that the black
>> hole lost energy and the outside universe gained same amount of
>> energy. Everything adds up.
>
>
>No, not so. If what you're sying were true, then the black hole is not
>black.
Well, that's *exactly* the point of Hawking radiation. That when you
combine GR with QM then black holes end up being not quite black.
What did you think is the point?
The reason is, because the particle-pair or energy-pair is
>created just beyond the event horizon is due to the Uncertainty
>Principle in operation: you can't have absolutely zero energy even
>though classical physics predics exactly that at the EH.
>
See above.
>The pair-generation is spontaneous and does not, CAN NOT, draw energy or
>matter from the black hole, like you are saying it does.
>
It *does* draw energy from the black hole. The uncertainty principle
does not violate conservation of energy. It only allows for short
term credit which needs to be repaid if full.
>No, it cannot. You're as confused about HR as I am, maybe more so.
Feel free to speak for yourself. I won't add more.
Dr Arm®
I know I just made another post on this.
This is the part that's still giving me trouble. It looks as if you're
saying that the energy of the BH is creating particles just beyond the
the EH. But that's not the way I understand the process.
It's supposedly the Uncertainty Principle that causes the pair
generation, because the BH removes all energy at the EH, removes it
below the point allowed by the UP, and I don't see where that requires
an energy input from the BH that you're suggesting.
Anyway, thanks for being patient with me on this.
da
me...@cars3.uchicago.edu
>
>This is the part that's still giving me trouble. It looks as if you're
>saying that the energy of the BH is creating particles just beyond the
>the EH. But that's not the way I understand the process.
>
>It's supposedly the Uncertainty Principle that causes the pair
>generation, because the BH removes all energy at the EH, removes it
>below the point allowed by the UP, and I don't see where that requires
>an energy input from the BH that you're suggesting.
>
allows you to "borrow energy on credit" but the credit needs to be
repaid in full. Energy conservation is maintained and no process
requiring net energy input can occur without the energy being provided
from somewhere. No miracles.
>Anyway, thanks for being patient with me on this.
>
You ask good questions and you behave well, so no undue amounts of
patience are needed.
Dr Arm®
If a pair, +e and -e, whose energy sum is zero, are spontaneously
created at the EH, there is no borrowed energy to repay.
Maybe it's the nature of the pair that's giving me problem. While I see
that if an matter and antimatter particle pair are created, that results
in a 2*m creation with a 2m mass deficit to be repaid.
What about if a photon and anti photon are created? Is there such a
beast as an antiphoton of energy minus hv? Because that's the way I'm
looking at this, that there is no local accounting problem to be
resolved at the EH where the pair are created, and that doesn't hold for
two particles with mass.
da
Jan C. Bernauer
wrote:
>me...@cars3.uchicago.edu wrote:
A pair of e+ and e- has the energy of 2*0,511MeV + kinetic energy.
>
>Maybe it's the nature of the pair that's giving me problem. While I see
>that if an matter and antimatter particle pair are created, that results
>in a 2*m creation with a 2m mass deficit to be repaid.
And since mass is energy, this is the enerfy deficit the BH has to
balance when it sucks in just one of it. You, see, the particel
falling in the BH gains the energy neede out of it´s potential energy
in the gravitational field.
----
Jan C. Bernauer
Gregory L. Hansen
The stipulation is that particles with negative total energy fall into the
event horizon. That's not necessarily the particle or the antiparticle,
and both varieties will come out with equal probability.
--
"For every problem there is a solution which is simple, clean and wrong. "
-- Henry Louis Mencken
Franz Heymann
"Jan C. Bernauer" <taggedfo...@web.de> wrote in message
news:l691juc6t7rsqccge...@4ax.com...
Yes
Franz Heymann
Franz Heymann
"Dr Arm®" <Genuin...@Yahoo.com> wrote in message
The word "rest" is redundant when discussing "mass".
Will that rest mass then have all the other
> characteristics of a mass such as the ability to bend space and curve
a
> passing lightbeam? What other kind of mass is there?
There is only one kind of mass. And yes, it will gravitate.
>
> So a pair of particles with a stationary Cm will bend a lightbeam
while
> an adjacent beam of light will not.
It is not mass which gravitates, but energy.
Franz Heymann
Herman Trivilino
> If a pair, +e and -e, whose energy sum is zero, are spontaneously
> created at the EH, there is no borrowed energy to repay.
Mati has already told you, and you seemed to agree that you understood,
that the total energy of an electron-positron pair is NOT zero.
> Maybe it's the nature of the pair that's giving me problem. While I see
> that if an matter and antimatter particle pair are created, that results
> in a 2*m creation with a 2m mass deficit to be repaid.
See? You seem to have understood it.
> What about if a photon and anti photon are created? Is there such a
> beast as an antiphoton of energy minus hv?
The antiparticle does not have negative energy!
There you go again, seeming to have not understood it.
Franz Heymann
"Dr Arm®" <Genuin...@Yahoo.com> wrote in message
The pair as originally generated has a real particle moving away from
the black hole and a virtual particle moving into the black hole. The
black hole loses energy to this virtual particle in order to make it
real.
>
> Look, I'm obviously missing something about HR, but this isn't it. I'm
> going to start a new thread about it, you're welcomeof course, but
this
> part is just wrong, I'm pretty sure.
>
>
> > In other words, the blackhole paid for both but got only one
delivered
> > back, the other went to outside. So, the net result is that the
black
> > hole lost energy and the outside universe gained same amount of
> > energy. Everything adds up.
>
>
> No, not so.
Yes, yes so.
If what you're sying were true, then the black hole is not
> black. The reason is, because the particle-pair or energy-pair is
> created just beyond the event horizon is due to the Uncertainty
> Principle in operation: you can't have absolutely zero energy even
> though classical physics predics exactly that at the EH.
Virtual pairs are continuously created and annihilated everywhere.
Being "virtual" means that a particle is not on its mass shell. ("Does
not have the mass of its free mates"). Not both of the particles in
such a transient pair need to be off their mass shells. Hawking
radiation occurs when such a virtual pair is created near the event
horizon of a black hole, specifically with the "outward bound" particle
being on its mass shell and the inward bound one being virtual. During
a subsequent interaction with the gravitational field of the black hole,
the inward going virtual particle is jerked on to its mass shell. Nett
effect: The black hole loses some energy in the process of "realising"
the virtual particle, and the outward bound particle brings some energy
away into our part of the universe. If you do your sums, you will find
that the energy lost by the black hole is exactly balanced by the energy
of the newly created particle outside the black hole.
>
> The pair-generation is spontaneous and does not, CAN NOT, draw energy
or
> matter from the black hole, like you are saying it does.
Reread what I said above here.
>
> No, it cannot. You're as confused about HR as I am, maybe more so.
Let's
> see if someone else knows.
>
Reread what I said above here.
Mati is not confused. Only you are.
> >
> > >See, somemthing doesn't add up if you consider both directions
occuring
> > >equally likely. The only way HR makes sense is if particles are
more
> > >likely to drift away fromthe BH than antiparticles.
> > >
> > It is absolutely irrelevant to the process whether what escapes is a
> > particle or an antiparticle. Both particles and antiparticles have
a
> > positive mass and carry positive energy.
> >
> > >And what about when it is an energy-pair fluctuation rather than a
> > >massive particle pair that gets created? See, the accounting seems
all
> > >wrong.
> >
> > If something escapes, it is no longer a fluctuation. No problems
with
> > the accounting.
>
>
> No, the problem still exists, you're overlooking it.
> I'm pretty sure.
There is no problem of any description in the story of Hawking
radiation.
Franz Heymann
Franz Heymann
"Dr Arm®" <Genuin...@Yahoo.com> wrote in message
That paragraph does not make sense with me. I bet that on rereading it,
you will say the same.
>
> Anyway, thanks for being patient with me on this.
Franz Heymann
Franz Heymann
"Dr Arm®" <Genuin...@Yahoo.com> wrote in message
Impossible. Both particles, if real, have masses. Where does that
energy come from?
>
> Maybe it's the nature of the pair that's giving me problem. While I
see
> that if an matter and antimatter particle pair are created, that
results
> in a 2*m creation with a 2m mass deficit to be repaid.
Yes, ecxept that we ought to consult a GR wallah about the details of
the particle proceeding into the black hole.
>
> What about if a photon and anti photon are created? Is there such a
> beast as an antiphoton of energy minus hv?
An antiphoton does not have an energy -hv.
A photon is its own antiparticle. Photon pairs may well be generated in
a Hawking process, as far as I am aware.
> Because that's the way I'm
> looking at this, that there is no local accounting problem to be
> resolved at the EH where the pair are created, and that doesn't hold
for
> two particles with mass.
There is. The photon which escapes, escapes with positive energy. The
one going inward has to be put on its mass shell. That requires energy.
That has to be supplied by the black hole.
>
Franz Heymann
me...@cars3.uchicago.edu
>me...@cars3.uchicago.edu wrote:
>>
>> In article <3D3143...@Yahoo.com>, Dr Arm® <Genuin...@Yahoo.com> writes:
>> >me...@cars3.uchicago.edu wrote:
>> >>
>> >> In article <3D3107...@Yahoo.com>, Dr Arm® <Genuin...@Yahoo.com> writes:
...
>> >>
>> >> >But wait, that would also mean that you are adding energy to the black
>> >> >hole too, at the same rate, via the same processes.
>> >> >
>> >> Ahh, but you forget that it was the energy of the black hole that was
>> >> used to generate the particle-antiparticle pair in the first place.
>> >> In other words, the blackhole paid for both but got only one delivered
>> >> back, the other went to outside. So, the net result is that the black
>> >> hole lost energy and the outside universe gained same amount of
>> >> energy. Everything adds up.
>> >
>> >I know I just made another post on this.
>> >
>> >This is the part that's still giving me trouble. It looks as if you're
>> >saying that the energy of the BH is creating particles just beyond the
>> >the EH. But that's not the way I understand the process.
>> >
>> >It's supposedly the Uncertainty Principle that causes the pair
>> >generation, because the BH removes all energy at the EH, removes it
>> >below the point allowed by the UP, and I don't see where that requires
>> >an energy input from the BH that you're suggesting.
>> >
>> I can only repeat what I said before. The uncertainty principle
>> allows you to "borrow energy on credit" but the credit needs to be
>> repaid in full. Energy conservation is maintained and no process
>> requiring net energy input can occur without the energy being provided
>> from somewhere.
>
>
>If a pair, +e and -e, whose energy sum is zero, are spontaneously
>created at the EH, there is no borrowed energy to repay.
>
The energy sum of e+ and e- is not zero. It is 2mc^2
>Maybe it's the nature of the pair that's giving me problem. While I see
>that if an matter and antimatter particle pair are created, that results
>in a 2*m creation with a 2m mass deficit to be repaid.
What gives you problem is thinking that the energy sum is zero.
>
>What about if a photon and anti photon are created?
Photon is its own antiparticle, meaning an antiphoton is a photon.
> Is there such a
>beast as an antiphoton of energy minus hv?
Again, the energy of an antiparticle is *same* as this of the
corresponding particle, *not* its negative.
Dr Arm®
What is "mass shells"? I have never heard that before. Can't even
imagine what that might mean.
Hawking
> radiation occurs when such a virtual pair is created near the event
> horizon of a black hole, specifically with the "outward bound" particle
> being on its mass shell and the inward bound one being virtual. During
> a subsequent interaction with the gravitational field of the black hole,
> the inward going virtual particle is jerked on to its mass shell. Nett
> effect: The black hole loses some energy in the process of "realising"
> the virtual particle, and the outward bound particle brings some energy
> away into our part of the universe.
It seems you are saying that
1. two massless particles are spontaneously generated.
2. one particle obtains mass by moving toward the universe.
3. the other particle obtains mass by moving into the black hole.
Can you see my skepticism about that?
> If you do your sums, you will find
> that the energy lost by the black hole is exactly balanced by the energy
> of the newly created particle outside the black hole.
That part is easy if you take for granted that only one type of particle
falls back into the black hole or that particles gain mass by moving
toward or away from black holes.
da
Dr Arm®
but then you say...
>
> There is. The photon which escapes, escapes with positive energy. The
> one going inward has to be put on its mass shell. That requires energy.
> That has to be supplied by the black hole.
> >
> Franz Heymann
It appears that you are saying the same thing. What am I missing?
da
Dr Arm®
Right, that's the way I understand it. So, if roughly the same number of
positive and negative total energy particles fall into the BH, there
should be no net change in the BH due to this process.
da
Dr Arm®
>
> "Dr Arm®" <Genuin...@Yahoo.com> wrote ...
>
> > If a pair, +e and -e, whose energy sum is zero, are spontaneously
> > created at the EH, there is no borrowed energy to repay.
>
> Mati has already told you, and you seemed to agree that you understood,
> that the total energy of an electron-positron pair is NOT zero.
>
> > Maybe it's the nature of the pair that's giving me problem. While I see
> > that if an matter and antimatter particle pair are created, that results
> > in a 2*m creation with a 2m mass deficit to be repaid.
>
> See? You seem to have understood it.
>
> > What about if a photon and anti photon are created? Is there such a
> > beast as an antiphoton of energy minus hv?
>
> The antiparticle does not have negative energy!
>
> There you go again, seeming to have not understood it.
I don't understand the spontaneous generation of two particles with
mass. If there is spontaneous particle generation, then their local sums
of energy and mass has to be zero. THis is a conservation principle.
Generating particles now and paying for it later has not been adequately
explained to me.
da