How long must physics put up w/f=ma?
Donald G. Shead
interpreted his Second Law of motion; when the inverse of mass [m] entered
physics texts; written as 1/m!
Now I ask you: Who would think that the inverse of mass [m], is [1/m]?
Any intellegent pros or cons are welcome; from anyone.
Gene Nygaard
Sam Wormley
F = ma is probably the most important equation in the history of
physics, Shead.
http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html
Try to get it right for a change!
Double-A
>interpreted his Second Law of motion; when the inverse of mass [m] entered
>physics texts; written as 1/m!
>Now I ask you: Who would think that the inverse of mass [m], is [1/m]?
>Any intellegent pros or cons are welcome; from anyone.
Because the inverse of anything is 1/anything.
2
The number 2 = ---
1
1
The inverse (standing it on its head) of 2 = ---
2
m
The quantity of mass m = ---
1
1
The inverse of m is therefore ---
m
O
Donald Shead = /|\
|
/ \
\ /
The inverse of Donald Shead = |
\|/
O
Get it?
Double-A
David Murdock
> The equation: f = ma has been attributed to Newton; ever since someone first
> interpreted his Second Law of motion; when the inverse of mass [m] entered
> physics texts; written as 1/m!
>
> Now I ask you: Who would think that the inverse of mass [m], is [1/m]?
Not me, that's fer sure. Inverting [m] don't give either 1/m or
[1/m]. Once
you've got them brackets stuck to the m, they's stuck on GOOD and
ain't nothin
gonna get em off. Screwdrivers, chisels, I've tried em all, and nothin
gets
those damn brackets off. At best I was able to pry one a little bit
off, kinda like: [m ]. Trust me on this.
When you invert [m], you gotta invert the m **with the brackets
attached** and
so you get 1/[m]. Yep.
Steven Douglas
"David Murdock" <mur...@tntech.edu> wrote in message
news:bef937fb.03092...@posting.google.com...
> "Donald G. Shead" <u10...@snet.net> wrote in message
news:<6c3cb.7915$nS2....@newssvr31.news.prodigy.com>...
<snippage>
> Not me, that's fer sure. Inverting [m] don't give either 1/m or
> [1/m]. Once
> you've got them brackets stuck to the m, they's stuck on GOOD and
> ain't nothin
> gonna get em off. Screwdrivers, chisels, I've tried em all, and nothin
> gets
> those damn brackets off. At best I was able to pry one a little bit
> off, kinda like: [m ]. Trust me on this.
>
> When you invert [m], you gotta invert the m **with the brackets
> attached** and
> so you get 1/[m]. Yep.
>
> > Any intellegent pros or cons are welcome; from anyone.
LMFAO Murdock!
Minor Crank
news:<6c3cb.7915$nS2....@newssvr31.news.prodigy.com>...
> > Now I ask you: Who would think that the inverse of mass [m], is [1/m]?
>
> Not me, that's fer sure. Inverting [m] don't give either 1/m or
> [1/m]. Once
> you've got them brackets stuck to the m, they's stuck on GOOD and
> ain't nothin
> gonna get em off. Screwdrivers, chisels, I've tried em all, and nothin
> gets
> those damn brackets off. At best I was able to pry one a little bit
> off, kinda like: [m ]. Trust me on this.
>
> When you invert [m], you gotta invert the m **with the brackets
> attached** and
> so you get 1/[m]. Yep.
I find that soaking in hot SuperGlue remover works, except you have to be
careful not to let it soak too long or the "m" tends to fall apart into its
component curves, and you can get nonsense statements like
"inverting [m] gives you 1/nn"
Minor Crank
Rusty Shackleford
news:pHbcb.564355$o%2.251063@sccrnsc02...
I don't know why but this discussion is so inane and silly and yet I am
ROTFL'ing to the point of GooFawing. At which point I settle down and
re-read the discussion and try to analyze why it is funny but I start
ROTFL'ing and GooFawing again. I must be in a silly mood and maybe tomorrow
I will look at it find it totally inane and not worth a whimper.
--
Rusty Shackleford
'What ever happens, happens necessarily'
msha...@NOSPAMrglobal.net
Remove NOSPAM from E-mail address to reply.
Mark
>"Donald G. Shead" <u10...@snet.net> wrote:
>> Now I ask you: Who would think that the inverse of mass [m], is [1/m]?
>Not me, that's fer sure. Inverting [m] don't give either 1/m or
>brackets attached** and so you get 1/[m]. Yep.
Actually, I think those bracket-thingeys he made up might be an ASCII
for sampling mean taken on a set of N measurements. In that case, you
can find everything from the expectation of the generating function
ln([exp(Am)])
which, according to my calculations, will obviously be given by:
<ln([exp(Am)])> = (1 + A/N d/dB)^N ln(<exp(Bm)>) | B = 0.
However, the inverse of [m] (i.e., the generating function ln([exp(A/m)])
need not be well-defined, unless m is sufficiently distributed away from 0.
It's quita a complex issue, when you get down to it.
ghytrfvbnmju7654
Well I'll be darned, I reckoned that inverting [m] gets you a [w].
Mark Mallory
Donald G. sHead wrote:
>
> The equation: f = ma has been attributed to Newton; ever since someone first
> interpreted his Second Law of motion; when the inverse of mass [m] entered
> physics texts; written as 1/m!
>
> Now I ask you: Who would think that the inverse of mass [m], is [1/m]?
Learn algebra, stupid shitHead.
Henry Wilson
The only relationship that works is a=F/m.
A few idiots jumped on this little identity and twisted it around in the hope
that they would appear knowledgeable.
Randy Poe
>The only relationship that works is a=F/m.
For values of "works" that include "doesn't always apply".
David Murdock
> On Tue, 23 Sep 2003 21:47:46 GMT, "Donald G. Shead" <u10...@snet.net> wrote:
>
> >The equation: f = ma has been attributed to Newton; ever since someone first
> >interpreted his Second Law of motion; when the inverse of mass [m] entered
> >physics texts; written as 1/m!
> The only relationship that works is a=F/m.
> A few idiots jumped on this little identity and twisted it around in the hope
> that they would appear knowledgeable.
Damn right. So here's my advice to people who have to take college physics
courses from idiot professors who tell you to use F=ma:
DON'T GIVE IN!
If they give you "a" and "m" and ask you to find "F", just keep trying
different values of F until you can make a=F/m work out. Eventually, by
trial-and-error, you'll do it.
With some work you can automate the process with a simple computer program; do
a search on F so as to minimize |a-(F/m)|. (But only if you accept the validity
of subtraction, and I'm not saying that I do.)
Sure, it's tough, but you'll have the satisfaction of knowing that you did not
cave in and multiply.
---DPM
Brian Quincy Hutchings
One has to retain the units in one's equations!
mur...@tntech.edu (David Murdock) wrote in message news:<bef937fb.03092...@posting.google.com>...
> When you invert [m], you gotta invert the m **with the brackets
> attached** and
> so you get 1/[m]. Yep.
--les ducs d'Enron!
http://larouchepub.com/other/2003/3036meacher_cheney.html
http://members.tripod.com/~american_almanac/
Henry Wilson
I am sitting with two main forces on my body right now.
I am clearly not experiencing a positive 'dv/dt' in anyone's frame.
Force cannot be defined as 'ma'.
Bill Jones
If mass is defined as a ratio, say f/a or e/(c^2); then 1/m has meaning as in
a/f or (c^2)/e. But I hesitate to say that 1/m is the inverse of 'mass'
I don't understand the relation of the message text to the message subject?
Minor Crank
news:ctacnv4e46blmthoq...@4ax.com...
> I am sitting with two main forces on my body right now.
> I am clearly not experiencing a positive 'dv/dt' in anyone's frame.
>
> Force cannot be defined as 'ma'.
Bzzt!!!
There is a deep and important difference between FORCE and TENSION.
Force makes things accelerate. Which way is your body accelerating? If it
isn't accelerating, what your body is experiencing is not a force, but a
tension.
Tensions result from DIFFERENT forces acting on different parts of the body.
Tensions can break things. If you are tied to four horses to be drawn and
quartered, your body is experiencing tension.
Hey, you wanna try that out?
Pure forces can not break things.
Pure tensions do not move things.
Force and tension use the same units (Newtons) and because of that are very
often confused, but they are quite different physical quantities. Force is a
vector, and tension is a tensor.
Here's an analogy, in case the distinction is too much for you to
appreciate: Energy and torque use the same units kg*m^2/s^2, but despite
their being similar in that regards, I'm sure even -you- will agree that
energy and torque are different physical quantities.
Minor Crank
Gene Nygaard
What is their sum?
>I am clearly not experiencing a positive 'dv/dt' in anyone's frame.
Wrong.
>Force cannot be defined as 'ma'.
Wrong.
Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/
Henry Wilson
<blue_whal...@comcast.net> wrote:
>"Henry Wilson" <HW@..> wrote in message
>news:ctacnv4e46blmthoq...@4ax.com...
>
>> I am sitting with two main forces on my body right now.
>> I am clearly not experiencing a positive 'dv/dt' in anyone's frame.
>>
>> Force cannot be defined as 'ma'.
>
>Bzzt!!!
>
>There is a deep and important difference between FORCE and TENSION.
Crank, you are very confused. The difference is a '-'.
>
>Force makes things accelerate. Which way is your body accelerating? If it
>isn't accelerating, what your body is experiencing is not a force, but a
>tension.
God I hope not. You are extremely confused Crank.
>
>Tensions result from DIFFERENT forces acting on different parts of the body.
>Tensions can break things. If you are tied to four horses to be drawn and
>quartered, your body is experiencing tension.
>
>Hey, you wanna try that out?
I do, often just for kicks.
>
>Pure forces can not break things.
>Pure tensions do not move things.
You are completely confused Crank.
This is quite amusing.
You really haven't a clue about basic physics.
>
>Force and tension use the same units (Newtons) and because of that are very
>often confused, but they are quite different physical quantities. Force is a
>vector, and tension is a tensor.
>
>Here's an analogy, in case the distinction is too much for you to
>appreciate: Energy and torque use the same units kg*m^2/s^2, but despite
>their being similar in that regards, I'm sure even -you- will agree that
>energy and torque are different physical quantities.
Why don't you and Dirk both enrol in first year physics? You sure need it.
>
>Minor Crank
>
>
Henry Wilson
Zero. That is why I am not acelerating and that is why 'ma' cannot be used to
define force.
Accelerations result from forces. Forces don't result from accelerations.
>
>>I am clearly not experiencing a positive 'dv/dt' in anyone's frame.
>
>Wrong.
Go on! Tell me more.
>
>>Force cannot be defined as 'ma'.
>
>Wrong.
You'll have to do better than that.
>
>Gene Nygaard
>http://ourworld.compuserve.com/homepages/Gene_Nygaard/
Dirk Van de moortel
"Henry Wilson" <HW@..> wrote in message news:ctacnv4e46blmthoq...@4ax.com...
Ralph, you are brilliant:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Force.html
Dirk Vdm
S. Enterprize Company
Yes it can.
>
>Ralph, you are brilliant:
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Force.html
>
>Dirk Vdm
Smart's Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
Sam Wormley
Henry Wilson wrote:
>
> I am sitting with two main forces on my body right now.
> I am clearly not experiencing a positive 'dv/dt' in anyone's frame.
And those forces are opposite and equal... so the net force is
zero and you are not accelerating... your velocity with respect
to the chair or room is zero.
But with respect to the Sun, you are in free fall around it with
constantly changing acceleration and your orbital speed is also
changing as the orbit is elliptical.
>
> Force cannot be defined as 'ma'.
Newton's Second Law is an excellent definition of force in mechanics.
http://scienceworld.wolfram.com/physics/Force.html
Gene Nygaard
Velocity is a vector. Need more yet?
Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/
G00glyMinataur
Please let us not simplify this equation to just F=ma or a=F/m...
You all miss an extremely important part of the relationship between
force and acceleration... Essentially, force has nothing to do with
acceleration as it does with fricken momentum...
Force is simply the change in momentum over an interval of time...
F = d(momentum)/d(time) or... F = d(mv)/dt (1)
Try basic calculus before you try and limit this to only F=ma... You
will instead come to find...
F = [dm/dt]*v + m*[dv/dt] (2)
In "Intro level physics" you will only see the latter part of (2),
mass times the change in velocity per unit time... But for any sort of
propulsion problem or whatever, where mass is changing, this will also
induce and contribute to force... so please...
F = [dm/dt]*v + ma
Essentially, there is no logical reason to define or refer to this
equation as "a=F/m"
-G00glyMinataur-
---------------------------------------------
Oh yes, and to whoever got his/her panties in a bunch with the crap
about a=F/m and the inverse of "m"... F*** you, sissy... its a
mathematical model that can be interpreted either way...
Nonetheless, I have a solution for you... leave mathematics and
physics and go pursue a career in social work or business... You
aren't cut out for any of this...
ghytrfvbnmju7654
Therefore:
The equation f = nv must be written as: f = (f/v)v; where the v's cancel.
The equation v = f/n must be written as v = f/(f/v); where the f's cancel.
Surely you _all_ must understand how parentheses work with algebra.
Bob Pease
"ghytrfvbnmju7654" <ghytrfvb...@mail.com> wrote in message
news:cb623e6.03092...@posting.google.com...
Hasn't the full moon for September already passed?
Dr S.
S. Enterprize Company
>
>Please let us not simplify this equation to just F=ma or a=F/m...
>
>You all miss an extremely important part of the relationship between
>force and acceleration... Essentially, force has nothing to do with
>acceleration as it does with fricken momentum...
>
>Force is simply the change in momentum over an interval of time...
>
> F = d(momentum)/d(time) or... F = d(mv)/dt (1)
>
>Try basic calculus before you try and limit this to only F=ma... You
>will instead come to find...
>
> F = [dm/dt]*v + m*[dv/dt] (2)
>
>In "Intro level physics" you will only see the latter part of (2),
>mass times the change in velocity per unit time... But for any sort of
>propulsion problem or whatever, where mass is changing, this will also
>induce and contribute to force... so please...
>
> F = [dm/dt]*v + ma
>
>
>Essentially, there is no logical reason to define or refer to this
>equation as "a=F/m"
>
>
>-G00glyMinataur-
>
Sure there is.
1 + 1 = 2
1 = 2 - 1
You can equate things on both sides of the equal sign as long as the units are
correct.
Henry Wilson
What was all that stuff we learnt about 'dependent' and 'independent'
variables?
If 'f' dependent on 'a' or 'a dependent on 'f'?
ghytrfvbnmju7654
The point I was trying to make was that mass is more than just the
force divided by the acceleration; it is a property of an object that
is constant (at least to a good approximation). Otherwise one could
just as meaningfully talk about nass=force/velocity.
ghytrfvbnmju7654
> What was all that stuff we learnt about 'dependent' and 'independent'
> variables?
> If 'f' dependent on 'a' or 'a dependent on 'f'?
Yes, of course.
Henry Wilson
>:::aye, I cry tears of blood for thee, for this is sad:::
>
>Please let us not simplify this equation to just F=ma or a=F/m...
>
>You all miss an extremely important part of the relationship between
>force and acceleration... Essentially, force has nothing to do with
>acceleration as it does with fricken momentum...
>
>Force is simply the change in momentum over an interval of time...
>
> F = d(momentum)/d(time) or... F = d(mv)/dt (1)
>
>Try basic calculus before you try and limit this to only F=ma... You
>will instead come to find...
>
> F = [dm/dt]*v + m*[dv/dt] (2)
>
>In "Intro level physics" you will only see the latter part of (2),
>mass times the change in velocity per unit time... But for any sort of
>propulsion problem or whatever, where mass is changing, this will also
>induce and contribute to force... so please...
>
> F = [dm/dt]*v + ma
You are still defining F as the dependent variable.
This equation should not be use as a definition of force.
Henry Wilson
wrote:
>HW@..(Henry Wilson) wrote in message news:<28hinvcgnopjdkvlb...@4ax.com>...
>> What was all that stuff we learnt about 'dependent' and 'independent'
>> variables?
>> Is 'f' dependent on 'a' or 'a dependent on 'f'?
>
>Yes, of course.
Well? Is 'f' dependent on 'a' or is 'a' dependent on 'f'?
Dirk Van de moortel
"ghytrfvbnmju7654" <ghytrfvb...@mail.com> wrote in message news:cb623e6.03093...@posting.google.com...
:-))
Dirk Vdm
Laszlo Lemhenyi
> interpreted his Second Law of motion; when the inverse of mass [m] entered
> physics texts; written as 1/m!
>
>
> Any intellegent pros or cons are welcome; from anyone.
Depends on what you understand by "inverse" .
If it is just math, then the answer is YES.
Regarding f=ma ,I would mention that if we have let's say b=d,
this does not mean that b IS d.
b and d can be two totally different things.
Best wishes
Laszlo Lemhenyi
Randy Poe
wrote:
>
>"The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in
>message news:c1gs41-...@lexi2.athghost7038suus.net...
>Cut<
>>
>> F and a *are* vectors.
>
>That sounds like something from some hosses mouth(;^) Force is a thrust
>Goulie; between two or more masses of matter; a push or a pull.
That means it has a direction. That makes it a vector.
> Acceleration
>is the rate of change in the velocity of a mass of matter; caused by a
>force.
As velocity is a vector, so is it's rate of change.
>Force can be graphically represented as the component of the acceleration
>that it causes; but 'vectors' will only suffice where the force does not
>change direction.
Huh? Are you saying that there IS no direction in orbital motion?
- Randy
Jeff Relf
You say :
" But with respect to the Sun ,
you are in free fall around it
In an ideal orbit ,
there is no Net acceleration with respect to the center ,
because the gravitational acceleration
is counterbalanced by the centrifugal acceleration .
The orbiting object is just " Sitting " there ,
kind of like how I'm just sitting here now .
Jeff Relf
You correctly say :
" Force cannot be defined as ' ma ' . "
That's right ,
Only the _ Net _ force in the local frame
is Mass * Acceleration .
Relativity has a name for that : The 4-force .
The gravitational force is not a local Net force ,
so it's not a 4-force .
Pmb
"Jeff Relf" <__.Jef...@NCPlus.NET> wrote in message
news:ujm29if41z6r.dlg@__.Jeff.Relf...
> Hi Henry ,
> You correctly say :
> " Force cannot be defined as ' ma ' . "
Well you can define force to be a short stemmed rose. However in physics
this is not how force is defined
>
>
> That's right ,
> Only the _ Net _ force in the local frame
> is Mass * Acceleration .
No. Force is *defined* as f = dp/dt
> Relativity has a name for that : The 4-force .
No. 4-force and force are different. 4-force is a 4-vector whereas force is
a 3-vector. 4-vectors are just that, 4-vectors. It doesn't make them any
more 'real' then any other quantity. For example: "The velocity of Light" is
well defined. But "The 4-velocity of light" cannot be defined. 4-momentum is
P = (E/c, p) where E = free-particle energy and p = momemtum (3-momentum).
The rules for 3-vectors are not the same rules applied to 4-vectors. For
example: While momentum, p, can be zero, 4-momentum, P, can never be zero.
> The gravitational force is not a local Net force ,
> so it's not a 4-force
Why do you use the term "local" to refer to the gravitational force?
Force = dp/dt and that is not zero when a particle is in free-fall in a
gravitational field. Force is part of the 4-force 4-vector just as momentum
is part of the 4-momentum 4-vector.
Pmb
Gene Nygaard
wrote:
Henry Wilson's claim was that he has no acceleration in anybody's
frame of reference. The most you are showing is that there is some
frame in which he is not accelerating.
Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/
Jeff Relf
You say :
" 4-momentum , P , can never be zero . "
Maybe so ,
But in an ideal orbit , the centrifugal force is
counterbalanced by the gravitational force .
So the local forces do in fact add up to a zero 4-force ,
From Wiki
http://www.wikipedia.org/wiki/Four-force :
" In the special theory of relativity ,
four-force is a four-vector that
replaces the classical force ;
The four-force of the four-vector is defined as
the change in four-momentum
over the particle's own time . "
I wrote :
" The gravitational force is not a local Net force ,
so it's not a 4-force . "
You replied :
" Why do you use the term ' local '
to refer to the gravitational force ? "
I didn't , I said the opposite :
I said that 4-forces are net forces ... and local ,
quite unlike gravitational forces .
You say :
" Force = dp/dt and that is not zero
when a particle is in free-fall
in a gravitational field . "
Again , In an ideal orbit , the centrifugal force is
counterbalanced by the gravitational force .
With respect to the center ,
The satellite is not even moving closer or father away .
Despite this lack of net local 4-forces ,
The gravitational force still exists .
Would it them be proper to say that :
In the coordinates of the satellite's free fall frame ,
The satellite has no vertical movement ?
Jeff Relf
You say :
" The most you are showing is that there is some
frame in which he is not accelerating . "
No , In that ideal orbit , All local forces are balanced ,
the 4-force is equal to zero .
And when the 4-force vanishes in the local frame ,
it also vanishes in the very distant frame .
But the component forces ,
including the gravitational force ,
is still there , in both the local and distant frames .
YBM
[...]
>>>Is 'f' dependent on 'a' or 'a dependent on 'f'?
>>
>>Yes, of course.
>
>
> Well? Is 'f' dependent on 'a' or is 'a' dependent on 'f'?
You didn't get it Henri, did you ?
And you pretend to write "simulations" ?
Laszlo Lemhenyi
> "Donald G. Shead" <u10...@snet.net> wrote in message news:<6c3cb.7915$nS2....@newssvr31.news.prodigy.com>...
> > The equation: f = ma has been attributed to Newton; ever since someone first
> > interpreted his Second Law of motion; when the inverse of mass [m] entered
> > physics texts; written as 1/m!
> >
> > Now I ask you: Who would think that the inverse of mass [m], is [1/m]?
> >
> > Any intellegent pros or cons are welcome; from anyone.
>
>
> Depends on what you understand by "inverse" .
> If it is just math, then the answer is YES.
> Regarding f=ma ,I would mention that if we have let's say b=d,
> this does not mean that b IS d.
> b and d can be two totally different things.
> Best wishes
>
> Laszlo Lemhenyi
------------
I would like to make some more comments on the subject.
In F=ma , F it is not the force itself.
The force is an action like a push or a pull .
It could be exerted at 1 point of a body or at the
whole body . F=ma , does not say anything about this,
so it represents clearly less than the force.
But,
the measure of the force it is F = intensity of action.
It is unconfortable for us to say " the measure of the
intensity of action" etc. so we call it simply: force.
Thus the word "force" has actually 2 meanings!
We do not need ma either understand or measure a force.
Acceleration is only 1 of the effects produced by
a force.You can feel a force .
You can measure it using a spring, for example.
So, f=ma it is not the definition of a
force. (ma or dp/dt measures the force only)
On its turn, acceleration has a definition
independent from force or mass.
What about mass ? Well, this time m=F/a,
defines the mass.It IS the definition of mass!
And 1/m ? Yes, you might define a physical property
through a/F as well.
One could name it the "lightness" of a body....
but I think it would not be practical.
We better stick with the good old newtonian
physics!
All the best
Laszlo Lemhenyi,Toronto
Henry Wilson
You are the one who doesn't get it YB.
'ma' might mathematically equal the magnitude of a force but it is not a
definition of force
Henry Wilson
This raises an interesting point.
We are only aware of the existence of a force when it acts on a mass and
produces an acceleration.
Can force exist without matter or acceleration?
I know it is usually referred to as 'potential' - but what does it actually
mean. Does a force exist in, say, an electric field even without the presence
of a charge?
Edward Green
>
> This raises an interesting point.
> We are only aware of the existence of a force when it acts on a mass and
> produces an acceleration.
> Can force exist without matter or acceleration?
You know, that question's nutzo enough to round the corner and be a
good question -- or maybe it's something I've been drinking. ;-)
A possible glib answer is that we can measure force even at zero
acceleration: say with a strain guage.
ghytrfvbnmju7654
> We are only aware of the existence of a force when it acts on a mass and
> produces an acceleration.
> Can force exist without matter or acceleration?
>
> I know it is usually referred to as 'potential' - but what does it actually
> mean. Does a force exist in, say, an electric field even without the presence
> of a charge?
The answer depends on several things:
- When you would say a force exists
- What you consider to be matter
You can have stress without any mass accelerating. In that case,
although there are forces between objects, all the forces on any
particular object cancel out, so there is no net force on any
object. Thus, there will be no masses accelerating.
The electromagnetic field does contain stress where there are no
charges. Many people do not call the electromagnetic field
"matter," but the distinction between it and "matter" is an
artificial one.
If there is a charge, the electrical force on the charge is
equal to the electric field vector times the magnitude of the
charge. If there is no charge, there is no force on any
particle.
The electrostatic potential is different from the electric
field, but related to it mathematically. The electrostatic
potential multiplied by the charge of an object is equal
to its electrostatic potential energy.
Henry Wilson
Ah! But a strain guage only comes to equilibrium when two forces are balanced.
Henry Wilson
wrote:
>HW@..(Henry Wilson) wrote in message news:<3te2ov059qgdungbc...@4ax.com>...
>> This raises an interesting point.
>> We are only aware of the existence of a force when it acts on a mass and
>> produces an acceleration.
>> Can force exist without matter or acceleration?
>>
>> I know it is usually referred to as 'potential' - but what does it actually
>> mean. Does a force exist in, say, an electric field even without the presence
>> of a charge?
Yep. That's the question.
>
>The answer depends on several things:
> - When you would say a force exists
> - What you consider to be matter
>
>You can have stress without any mass accelerating. In that case,
>although there are forces between objects, all the forces on any
>particular object cancel out, so there is no net force on any
>object. Thus, there will be no masses accelerating.
>
>The electromagnetic field does contain stress where there are no
>charges. Many people do not call the electromagnetic field
>"matter," but the distinction between it and "matter" is an
>artificial one.
>
>If there is a charge, the electrical force on the charge is
>equal to the electric field vector times the magnitude of the
>charge. If there is no charge, there is no force on any
>particle.
>
>The electrostatic potential is different from the electric
>field, but related to it mathematically. The electrostatic
>potential multiplied by the charge of an object is equal
>to its electrostatic potential energy.
Hmm!
I suppose that since the dimensions of force are MLT-2, it is not possible to
have force without matter.
I retract my question.
tadchem
"Jeff Relf" <__.Jef...@NCPlus.NET> wrote in message
> Hi Pete ,
> You say :
> " 4-momentum , P , can never be zero . "
>
> Maybe so ,
> But in an ideal orbit , the centrifugal force is
> counterbalanced by the gravitational force .
>
> So the local forces do in fact add up to a zero 4-force ,
>
> From Wiki
> http://www.wikipedia.org/wiki/Four-force :
>
> " In the special theory of relativity ,
> four-force is a four-vector that
> replaces the classical force ;
> The four-force of the four-vector is defined as
> the change in four-momentum
> over the particle's own time . "
A constant 4-momentum has a zero time-derivative (force).
If p is not changing, F = dp/dt is 0.
Tom Davidson
Richmond, VA
tadchem
"Gene Nygaard" <gnyg...@nccray.com> wrote in message
news:908rnv8llrlb9idvr...@4ax.com...
<snip>
> Henry Wilson's claim was that he has no acceleration in anybody's
> frame of reference. The most you are showing is that there is some
> frame in which he is not accelerating.
He apparently can't run away from himself...
Tom Davidson
Richmond, VA
Edward Green
Huh? Are you suggesting that there is no "force" in the absense of
acceleration?
You know, that is provocative: in some sense of course you are right.
The total force on an object must be zero in the absense of
acceleration. Yet we are normally comfortable discussing component
forces as real forces. E.g., if I push on you and Tom pushes on you
in opposite directions with zero total force, you don't accelerate.
So you would buy it if I said, well, you are not accelerating,
therefore there is no force, so stop carping about the force I am
supposed to be exerting on your chest and he on your back -- you have
just proven it is zero? :-)
Of course purists sometimes confuse matters by insisting that force
must mean "total force", as you do, but without explicitly saying so.
Is there some problem with allowing real non-zero component forces
whose resultant is zero? It seems to me their "reality" comes down to
material deformation -- in effect, using a strain gauge.
Double-A
What about the very common case where a force pushes an object to
accelerate until friction matches the force, at which point the object
continues moving at a constant velocity. The force must still be
there since without it the object would stop moving. But there is no
further acceleration.
Double-A
Robert J. Kolker
Double-A wrote:
> What about the very common case where a force pushes an object to
> accelerate until friction matches the force, at which point the object
> continues moving at a constant velocity. The force must still be
> there since without it the object would stop moving. But there is no
> further acceleration.
A body with no force acting on it will either be at rest or move
uniformly in a straight line. Newton's first law. So if a body is moving
uniformly there is no net force acting on it. That means either there is
no force, periord, or all forces are cancelling out which is the same
thing.
Bob Kolker
Henry Wilson
wrote:
The 'thing' being that force cannot be defined in terms of 'ma'.
You just admitted it.
>
>Bob Kolker
Henry Wilson
>HW@..(Henry Wilson) wrote in message news:<31u6ov0asi3771nvn...@4ax.com>...
>> On 6 Oct 2003 18:03:00 -0700, null...@aol.com (Edward Green) wrote:
>>
>> >HW@..(Henry Wilson) wrote in message news:<3te2ov059qgdungbc...@4ax.com>...
>> >>
>> >> This raises an interesting point.
>> >> We are only aware of the existence of a force when it acts on a mass and
>> >> produces an acceleration.
>> >> Can force exist without matter or acceleration?
>> >
>> >You know, that question's nutzo enough to round the corner and be a
>> >good question -- or maybe it's something I've been drinking. ;-)
>> >
>> >A possible glib answer is that we can measure force even at zero
>> >acceleration: say with a strain guage.
>>
>> Ah! But a strain guage only comes to equilibrium when two forces are balanced.
>
>Huh? Are you suggesting that there is no "force" in the absense of
>acceleration?
>
>You know, that is provocative: in some sense of course you are right.
>The total force on an object must be zero in the absense of
>acceleration. Yet we are normally comfortable discussing component
>forces as real forces. E.g., if I push on you and Tom pushes on you
>in opposite directions with zero total force, you don't accelerate.
>So you would buy it if I said, well, you are not accelerating,
>therefore there is no force, so stop carping about the force I am
>supposed to be exerting on your chest and he on your back -- you have
>just proven it is zero? :-)
You are completely misquoting me.
I said ma cannot be used as a definition of force.
How could the forces you just mentioned be assessed in your example where there
is NO acceleration? Are you saying both forces are zero because ma is clearly
zero?
a=d2x/dt2, remember.
Mark
>The only relationship that works is a=F/m.
In general, for a system deriveable from a (non-singular) Lagrangian
L = L(q,v), if describeable by a finite number of coordinates
q=(q^1,q^2,..,q^N) you can write:
dq^A/dt = v^A, dv^A/dt = sum (W^AB F_B).
The W matrix is the inverse of the mass matrix W = m^{-1}, given by
the second derivatives of L with respect to v:
m_{AB} = @^2L/@v^A@v^B
and the force by:
F_B = @L/@q^B - (@/@t + sum v^C @/@q^C)@L/@v^B.
That m have an inverse, by the way, is precisely the condition that the
Lagrangian system have a Hamiltonian H = H(q,p), with the velocity-momentum
relation p_A = @L/@v^A being invertible to v^A = V^A(q,p) and
H(q,p) = sum (p_A V^A(q,p)) - L(q, V(q,p)).
So anyway, you then have either the matrix equation F = m d^q/dt^2 or
d^2v/dt^2 = WF and W is the (matrix) inverse of the mass.
All other things being equal, a given coordinate is going to vary more,
the larger W is. For instance, for a 1-D system, you have
dq/dt = v, dv/dt = W F(q,v) = a(q,v).
So, for initial values q(0) = 0, v(0) = 0, you get approximately for
short time intervals h:
q = 1/2 W F(0,0) h^2,
which is proportional to W.
So, the W matrix is best throught of as a "dispersion" matrix, and the
inverse mass best named the "dispersion".
This is exactly the same matrix that appears in the small-time-interval
commutator relation
[q^A(t), q^B(t + dt)] = i h-bar W^AB,
thus lending further justification to the name "dispersion matrix".
In fact, what's not widely-known is that the mere requirement that
a relation of this form exist in conjunction with a 2nd order system
of equations of motion
q''(t) = a(q(t),q'(t))
FORCES there to be a Lagrangian (in the classical limit) whose
dispersion matrix will be lim ([q,v]/(i h-bar)) as h-bar -> 0.
Mark
>The point I was trying to make was that mass is more than just the
>[sic] force divided by the acceleration
i.e. "m = (F_x, F_y, F_z)/(d^2x/dt^2, d^2y/dt^2, d^2z/dt^2)", when written
out in full -- which doesn't even make sense, since you can't divide
vectors by vectors.
HelpRaptor
As long a F = ma describes reality
tadchem
"HelpRaptor" <HelpR...@yahoo.com> wrote in message
news:1ae6862a.0310...@posting.google.com...
> The answer to the question "How long must physics put up w/f=ma?" is:
>
>
> As long a F = ma describes reality
And just what does that have to do with *w/f=ma*?
You don't read very carefully, do you?
Tom Davidson
Richmond, VA
The Ghost In The Machine
<tadche...@comcast.net>
wrote
on Sun, 12 Oct 2003 18:44:12 -0600
<wPudnY23pNe...@comcast.com>:
>
> "HelpRaptor" <HelpR...@yahoo.com> wrote in message
> news:1ae6862a.0310...@posting.google.com...
>> The answer to the question "How long must physics put up w/f=ma?" is:
>>
>>
>> As long a F = ma describes reality
>
> And just what does that have to do with *w/f=ma*?
I think the 'w/' in this case is an abbreviation for 'with',
not part of the formula proper. Of course writing it that
way isn't exactly helpful.
>
> You don't read very carefully, do you?
>
>
> Tom Davidson
> Richmond, VA
>
>
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
Gene Nygaard
<ew...@sirius.athghost7038suus.net> wrote:
>In sci.physics, tadchem
><tadche...@comcast.net>
> wrote
>on Sun, 12 Oct 2003 18:44:12 -0600
><wPudnY23pNe...@comcast.com>:
>>
>> "HelpRaptor" <HelpR...@yahoo.com> wrote in message
>> news:1ae6862a.0310...@posting.google.com...
>>> The answer to the question "How long must physics put up w/f=ma?" is:
>>>
>>>
>>> As long a F = ma describes reality
>>
>> And just what does that have to do with *w/f=ma*?
>
>I think the 'w/' in this case is an abbreviation for 'with',
>not part of the formula proper. Of course writing it that
>way isn't exactly helpful.
Of course it isn't helpful.
But if you haven't figured out yet, Shead is an idiot. He'll never
change. Yet we other idiots keep replying to a thread whose subject
asks about "w/f=ma?"
It might be understandable that after all these years, Shead still
hasn't the foggiest idea what a factorial is, or the symbol for this
operation, a symbol he keeps throwing into his formulas. But he damn
sure out to be held to a standard of knowing that a slash / is a
symbol for a mathematical operation. Don't be blaming this confusion
on anyone other than the culprit, Dense Donny Shead.
Donald Shead is also a crackpot who keeps using "ft" as a symbol for
the multiplication of force by time--and is too stupid to figure out
why people don't understand what he is saying.
Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/
The Ghost In The Machine
<gnyg...@nccray.com>
wrote
on Mon, 13 Oct 2003 13:15:49 GMT
<hr8lovo2f15vhuq6k...@4ax.com>:
I was under the impression that he was using 'ft' as an abbreviation
for 'foot' (since he highly prefers English units, for some
bizarre reason).
As for being an idiot -- I prefer not to make judgements
on the person, but certainly Shead's postings are
consistent with being an intractable quack, unable to
see that SI, for all of its warts (one of them being
the requirement of an existent of the mass-artifact of 1
kg; another being the liter/litre controversy) doesn't
have the conversion problems and/or other difficulties
of contemporary English/Imperial units: 1 km = 1,000
m = 1,000,000 mm and that's more or less that, all
throughout SI, regardless of the item being measured.
(Some artifacts, however, still exist: 1 cal = 4.180 J,
for instance. I'm not sure where 'cal' came from apart
from its relationship to water (1 cal heats 1 gm water 1
degree C) and it may not be, strictly speaking, part of SI.
There are also some issues on the electromagnetic side of
the issues, which I'm not that familiar with. Of course
this is a far cry from the grains/oz/lbs [pick a flavor!],
the mil/in/ft/rod/mile/furlough, or the dry vs wet quart
issues perambulating all over Imperial measurements.
Even pints are different.)
I base this primarily on his non-response to the elementary
observation that a mass artifact (the aforementioned 1 kg
item) is far easier to handle than an apparatus that can be
proven to generate a 1 Newton (or, for those so inclined,
1 lb-force) force. (Arguably, the simplest method to
reliably produce this force would rely on the ideal gas
law, and require precise temperature measurements.)
Nor is he treading with the crowd on his notions regarding
inertia, mass, and force. I don't see why it makes any
difference when writing
F = ma
a = F/m
m = F/a
assuming scalar quantities. (In vector quantities, the
third must be tossed, for obvious reasons; F and a are
inherently vectors.) Use whichever one is appropriate
for the problem solution.
Side issue: I'll admit it is a pity we're not willing to
spend money on the gigantic conversion efforts necessary
to bring our roadway and speedometer systems consistent
with Europe, though. :-/ But oh well. :-)
>
> Gene Nygaard
> http://ourworld.compuserve.com/homepages/Gene_Nygaard/
Edward Green
> Double-A wrote:
>
> > What about the very common case where a force pushes an object to
> > accelerate until friction matches the force, at which point the object
> > continues moving at a constant velocity. The force must still be
> > there since without it the object would stop moving. But there is no
> > further acceleration.
Good example.
> A body with no force acting on it will either be at rest or move
> uniformly in a straight line. Newton's first law. So if a body is moving
> uniformly there is no net force acting on it.
N.B. _net_
> That means either there is
> no force, periord, or all forces are cancelling out which is the same
> thing.
Well, not quite. It's the same thing for acceleration of the center
of mass, but it's not the same thing for stress and strain, and as
Double-A points out, it's not the same thing for systems where
terminal velocity depends on an applied force countered by friction,
even though the net force acting on the body is always zero at
terminal velocity.
Another example of a physical affect of applied forces independent of
the net force F is torque: it certainly makes a big difference if we
flip the action of two equal and opposite forces on the ends of a
lever, though the net force remains zero.
Donald Shead
> "Donald G. Shead" <u10...@snet.net> wrote in message news:<BGGib.10961$dS4....@newssvr33.news.prodigy.com>...
> > The total force [F] used to move an object against friction or any other
> > restraint, is the total force exerted on it minus the frictional - or other
> > restraining force; so that the _net force_ is what does the actual work, and
> > is [f = F-uw]; which leaves F = wa/g + uw; or f = wa/g!
> >
> > The coefficient of restraint [u] against moving something on a level surface
> > is usually a fraction of the object's weight [w], so that the product [uw]
> > is the measure of the restraining frictional force: On slopes the
> > coefficient of restraint is proportional to the sin of the angle of the
> > slope: For a 90° vertical slope the coefficient of restraint against driving
> > something pointed directly into it will depend on the hardness of the
> > material: As is the case for driving stakes and such into the ground.
>
>
> Yes, the total force needed to move an object would seem to include
> the force necessary to overcome friction, and any other impeding
> resistance, as well as the inertia of the object that results from
> mass. But in physics, is force really defined to include the total
> effort, or just that part that overcomes inertia to give acceleration?
>
If it isn't, it sure ought to be: The total force is the measure of
the total effort; including the force necessary to overcome friction,
and any other impeding resistance, as well as the inertia of the
object..
> When pushing a rock across the ground, frictional force usually seems
> to arise quickly
Frictional force doesn't _arise_, iI's already there, as 'static
friction'; preventing the rock from moving on its own. In order to
start the rock moving it's necessary to apply a force _greater_ than
the static force:
Immediately after the movement begins, the rocks begins to accelerate,
and will continue to do so as long as the force is maintained.
When the force is reduced the rock will slow its acceleration, and
when the force ceases, the rock will be stopped: Brought to rest by
'kinetic' friction.
to prevent the rock moving beyond a certain speed
> without applying additional force. When force is lessened, the rock
> slows immediately. It is easy to see how Aristotle came to think that
> force was needed to keep an object moving at a constant speed. But
> perhaps this is not what modern physics considers force.
>
> If you are pushing an object along the ground, there is also air
> resistance, which is another form of friction. If you pound a stake
> into the ground, you get ground resistance, which is really the forces
> of atomic electric fields that you are running up against.
>
Oh horsefeathers: Pounding stakes into the ground simply requires
pushing asside the particles of dirt that have settled due to gravity.
If you hit a rock with a steel point, then there might be some
electricity generated: Ever seen sparks and smelled brimstone, while
working a pick and shovel to dig a hole or trench?
> If I push against a brick wall, nothing will move. Am I really
> exerting a force, or is this only tension, or stress?
>
If you push with a force beyond the wall's ability to withstand lt;
you'll move that darn wall; in a quick fashon.
> But when we consider an object in a frictionless state, there is
> another question that arises. Newton said in his third law of motion:
>
> "To every action there is always opposed an equal reaction; or the
> mutual actions of two bodies upon each other are always equal, and
> directed to contrary parts."
>
> Now when a force is applied to a mass in a frictionless state, the
> inertial force opposes the motive force, which limits the acceleration
> according to a=F/m. But if the inertial force is an equal and
> opposite force to the motive force, as Newton's 3rd law seems to say,
> then why do we have acceleration at all?
We have acceleration because two or more particles of physicsal
matter, and/ or masses therof cannot simultaneously occupy or pass
through the exact same place; they exert a mutual, equal and
oppositely directed force on each other and cause each other to
accelerate; by spinning slowing down or changing direction; in
proportion to their elasticity and massiveness.
Don't forget Newton and Einstein were only human, and they put their
pants on one leg at a time, just like the rest of us.
Double-A
I agree with you that the frictional force must be there to begin with
as static friction, but the frictional force must increase as a
function of velocity, otherwise the rock wouldn't reach any terminal
velocity, which experience shows us it usually does.
> to prevent the rock moving beyond a certain speed
> > without applying additional force. When force is lessened, the rock
> > slows immediately. It is easy to see how Aristotle came to think that
> > force was needed to keep an object moving at a constant speed. But
> > perhaps this is not what modern physics considers force.
> >
> > If you are pushing an object along the ground, there is also air
> > resistance, which is another form of friction. If you pound a stake
> > into the ground, you get ground resistance, which is really the forces
> > of atomic electric fields that you are running up against.
> >
> Oh horsefeathers: Pounding stakes into the ground simply requires
> pushing asside the particles of dirt that have settled due to gravity.
But what do you suppose causes particles of dirt to stick together and
resist being pushed apart by the stake? It's the electromagnetic
forces of the atoms. If it were only gravity holding them together,
it would be like pushing a stake through a pile of flour.
> If you hit a rock with a steel point, then there might be some
> electricity generated: Ever seen sparks and smelled brimstone, while
> working a pick and shovel to dig a hole or trench?
Not my line of work.
>
> > If I push against a brick wall, nothing will move. Am I really
> > exerting a force, or is this only tension, or stress?
> >
> If you push with a force beyond the wall's ability to withstand lt;
> you'll move that darn wall; in a quick fashon.
>
> > But when we consider an object in a frictionless state, there is
> > another question that arises. Newton said in his third law of motion:
> >
> > "To every action there is always opposed an equal reaction; or the
> > mutual actions of two bodies upon each other are always equal, and
> > directed to contrary parts."
> >
> > Now when a force is applied to a mass in a frictionless state, the
> > inertial force opposes the motive force, which limits the acceleration
> > according to a=F/m. But if the inertial force is an equal and
> > opposite force to the motive force, as Newton's 3rd law seems to say,
> > then why do we have acceleration at all?
>
> We have acceleration because two or more particles of physicsal
> matter, and/ or masses therof cannot simultaneously occupy or pass
> through the exact same place; they exert a mutual, equal and
> oppositely directed force on each other and cause each other to
> accelerate; by spinning slowing down or changing direction; in
> proportion to their elasticity and massiveness.
Again, it is the electromagnetic fields of the atoms that cause this.
But I still can't quite see how Newton's third law applies.
>
> Don't forget Newton and Einstein were only human, and they put their
> pants on one leg at a time, just like the rest of us.
From the stories I've heard about Newton, he didn't always remember to
even put his pants on in the morning!
Double-A
tadchem
"Double-A" <doub...@hush.com> wrote in message
news:79094630.03101...@posting.google.com...
<snip>
> > Don't forget Newton and Einstein were only human, and they put their
> > pants on one leg at a time, just like the rest of us.
>
>
> From the stories I've heard about Newton, he didn't always remember to
> even put his pants on in the morning!
Nor did Einstein. As I heard it, he met Queen Elizabeth II wearing a suit
he had slept in, and once was escorted home by a Princeton policeman who
found him deep in thought on a park bench one Sunday morning, sans trousers.
I prefer to believe these incidents were not accidental, and that he did
these thing tongue in cheek because he knew that he, of all people in the
world, could get away with it. With his exquisitely subtle sense of humor,
he was having a private joke on all of US.
Tom Davidson
Richmond, VA
pete
>
> u10...@snet.net (Donald Shead) wrote in message news:<d38ce474.03101...@posting.google.com>...
> > doub...@hush.com (Double-A) wrote in message news:<79094630.03101...@posting.google.com>...
> > > When pushing a rock across the ground,
> > > frictional force usually seems to arise quickly
> >
> > Frictional force doesn't _arise_, iI's already there, as 'static
> > friction'; preventing the rock from moving on its own. In order to
> > start the rock moving it's necessary to apply a force _greater_ than
> > the static force:
> >
> > Immediately after the movement begins,
> > the rocks begins to accelerate,
> > and will continue to do so as long as the force is maintained.
> >
> > When the force is reduced the rock will slow its acceleration, and
> > when the force ceases, the rock will be stopped: Brought to rest by
> > 'kinetic' friction.
> >
>
> I agree with you
Don't ever do that again.
> that the frictional force must be there to begin with
> as static friction, but the frictional force must increase as a
> function of velocity, otherwise the rock wouldn't reach any terminal
> velocity, which experience shows us it usually does.
Friction only occurs as a reaction force.
If static friction was there to begin with,
with no other forces to oppose it,
then it would have to cause acceleration.
--
pete
Laurel Amberdine
> In article <pukanv09rgc0j0487...@4ax.com> HW@.. writes:
>>The only relationship that works is a=F/m.
>
> In general, for a system deriveable from a (non-singular) Lagrangian
> L = L(q,v), if describeable by a finite number of coordinates
> q=(q^1,q^2,..,q^N) you can write:
> dq^A/dt = v^A, dv^A/dt = sum (W^AB F_B).
>
> The W matrix is the inverse of the mass matrix W = m^{-1}, given by
> the second derivatives of L with respect to v:
> m_{AB} = @^2L/@v^A@v^B
> and the force by:
> F_B = @L/@q^B - (@/@t + sum v^C @/@q^C)@L/@v^B.
> In fact, what's not widely-known is that the mere requirement that
> a relation of this form exist in conjunction with a 2nd order system
> of equations of motion
> q''(t) = a(q(t),q'(t))
> FORCES there to be a Lagrangian (in the classical limit) whose
> dispersion matrix will be lim ([q,v]/(i h-bar)) as h-bar -> 0.
Wow. Do you give live performances too? :)
--
- Laurel * * * http://amberdine.com
Double-A
Then why does more than a certain amount of force have to be applied
to the rock before it begins to move at all?
Double-A
Randy Poe
> > If static friction was there to begin with,
> > with no other forces to oppose it,
> > then it would have to cause acceleration.
>
>
> Then why does more than a certain amount of force have to be applied
> to the rock before it begins to move at all?
For the same reason that more than a certain amount of
force has to be applied to a wall before you can break
it. The wall reacts back with only enough force to
counter your push. Exactly similar things involving
molecular bonding forces and barriers are happening
with static friction.
- Randy
Pmb
"Jeff Relf" <__.Jef...@NCPlus.NET> wrote in message
news:1syt3ujkrkk35.dlg@__.Jeff.Relf...
> Hi Pete ,
> You say :
> " 4-momentum , P , can never be zero . "
>
> Maybe so ,
> But in an ideal orbit , the centrifugal force is
> counterbalanced by the gravitational force .
>
> So the local forces do in fact add up to a zero 4-force ,
Don't confuse force with 4-force. They are different animals.
>
> From Wiki
> http://www.wikipedia.org/wiki/Four-force :
>
> " In the special theory of relativity ,
> four-force is a four-vector that
> replaces the classical force ;
> The four-force of the four-vector is defined as
> the change in four-momentum
> over the particle's own time . "
One has to be careful on this point. If by "change in 4-momentum over the
particle's own time" one means something like an absolute changed then that
is true. But if they mean F^u = dP^u/dT then that is not true.
Take a look at what it means for 4-force to be a time rate of change in
momentum.
http://www.geocities.com/physics_world/gr/grav_force.htm
The math is a bit tricky.
> I said that 4-forces are net forces ... and local ,
> quite unlike gravitational forces .
Then why do you say that then?
>
> You say :
> " Force = dp/dt and that is not zero
> when a particle is in free-fall
> in a gravitational field . "
>
>
> Again , In an ideal orbit , the centrifugal force is
> counterbalanced by the gravitational force .
That is not true either. That holds true *only* for the *radial* component
of the force. If the force were truly zero then the particle would move in a
straight line.
> The gravitational force still exists .
I agree.
> Would it them be proper to say that :
> In the coordinates of the satellite's free fall frame ,
> The satellite has no vertical movement ?
I'd say that the satellite is not accelerating at all in its own free-fall
frame. In fact in that frame the satellite is at rest.
Pete
Jeff Relf
You say :
" Don't confuse force with 4-force .
They are different animals . "
Right , and the sign of the 4-force's time component
is the opposite of the space components . Right ?
Pmb
"Jeff Relf" <__.Jef...@NCPlus.NET> wrote in message
No. If F^u is 4-force, P^u = 4-momentum, f is 3-force and p = 3-momentum, m
= (relativistic) mass, W = dE/dt, gamma = dt/dT, then, in Minkowski
coordinates (t, x, y, z)
P^u = (mc, p) = (E/c, p)
F^u = dP^u/dT = d/dT(mc, p) = (c*dm/dT, dp/dT) = [c*dm/dT, (dt/dT)dp/dt) =
(W/c, gamma*f)
F^u = (W/c, gamma*f)
Time component of 4-force ~ Power
Space component of 4-force = gamma*(3-force)
In curvilinear coordinates its not that simple.
Pmb
Jeff Relf
You say :
is not accelerating at all in its own free-fall frame .
In fact in that frame the satellite is at rest . "
That satellite , in that ideal orbit ,
would be at rest vertically ...
But it wouldn't be at rest horizontally . Right ?
Pmb
"Jeff Relf" <__.Jef...@NCPlus.NET> wrote in message
At rest refers to a particular frame of referance. Which frame of referance
are you refering to? And if you mean "at rest horizontally" you mean that an
Earth observer measures no radial velocity with respect to the center of
Earth then yes. But if you're an observer sitting on the Moon - What does
that observer mean by "vertical"?
Pmb
Jeff Relf
You say :
http://www.geocities.com/physics_world/gr/grav_force.htm "
At that site you say :
" Just as the magnetic force is velocity dependant ,
so too is the gravitational force . "
I agree with that ,
but the math doesn't catch my interest .
( Please don't be offended )
P.S. Your " Home " link is :
file:///C:/Documents%20and%20Settings/Peter%20Brown/Local%20Settings/Temporary%20Internet%20Files/Content.IE5/Physics%20World/gr/01_grav_force/Physics%20Worldgrgrav_forcegr.htm
That does nothing on my computer .
Jeff Relf
You say :
That holds true * only * for
the * radial * component of the force .
then the particle would move in a straight line . "
Huh ? That centrifugal force is a " Radial Component " .
The vector points
in the exact opposite direction of the gravitational force .
Jeff Relf
" Which frame of reference are you referring to ? :
The satellite's free fall reference frame .
I'm calling points along the radius of that ideal orbit
" The Vertical " .
And I'm calling the angles of the orbit " The Horizontal " .
So while it's vertical position is at rest ,
it's horizontal position is not ... It seems to me .
Jeff Relf
You say :
Space component of 4-force = gamma*(3-force) . "
The spatial sign is not the opposite of the temporal sign ?
Pmb
"Jeff Relf" <__.Jef...@NCPlus.NET> wrote in message
You're confusing the components of a 4-vector with the components of the
metric tensor.
Pmb
Jeff Relf
You say :
with the components of the metric tensor "
Ok , Thanks .
Why isn't the time component of a 4-force equal to energy ,
instead of " ~ Power " ?
Is there any difference between
the 4-force of a tiny particle
and the 4-force of a massive star ?
Pmb
"Jeff Relf" <__.Jef...@NCPlus.NET> wrote in message
> Hi Pete , Pmb ,
> You say :
> " You're confusing the components of a 4-vector
> with the components of the metric tensor "
>
> Ok , Thanks .
You're welcome.
> Why isn't the time component of a 4-force equal to energy ,
> instead of " ~ Power " ?
Because the time component of 4-momentum is energy and 4-force is the
derivative of 4-momentum. I.e.
Let (c=1)
p = momentum
E = free-particle energy = mc^2 = m
f = dp/dt = force
W = dE/dt = power
P = 4-momentum = (E, p)
F = dP/dT = (dE/dT, dp/dT) = gamma(dE/dt, dp/dt) = gamma(W,f)
>
> Is there any difference between
> the 4-force of a tiny particle
> and the 4-force of a massive star ?
Yes. Strictly speaking it's not meaningful to define a 4-momentum and
therefore it's not possible to define a 4-force. Especially since the star
is lossing rest mass. To define a 4-momentum the components must transform
as a 4-vector. That does not work for a non-conservative extended system
like a star.
"An Incorrect Application of Proper Mass" - I discuss the mass of a cooling
rod
http://www.geocities.com/physics_world/sr/invariant_mass.htm
This is an instance of the troubles that arise when we attempt to define
energy and momentum for a system in which the energy and momentum are not
conserved. The usual demonstration of the four-vector character of energy
and momentum fails when these quantities are not conserved. And then you
can't use the usual formulas for calculating the energy in one reference
frame from that in another. The general proposition that systems with
variable energy and momentum are trouble is known and has been known since
the days of Wolfgang Pauli's book.
Pete
Jeff Relf
You say :
" Strictly speaking it's not meaningful to define
So it's seldom or never done then ?
You say :
" the time component of 4-momentum is energy
and 4-force is the derivative of 4-momentum
[ with respect to time ] "
Very interesting , Thanks .
Pmb
"Jeff Relf" <__.Jef...@NCPlus.NET> wrote in message
> Hi Pmb ,
> You say :
> " Strictly speaking it's not meaningful to define
> [ a massive star's ] 4-momentum
> and therefore it's not possible to define a 4-force . "
>
> So it's seldom or never done then ?
That is like asking me h ow often a rock is considered big. I don't know and
could have no way of knowing that.
>
> You say :
> " the time component of 4-momentum is energy
> and 4-force is the derivative of 4-momentum
> [ with respect to time ] "
>
> Very interesting , Thanks .
You're welcome
Pete
Don...@mac.com
writes:
>Friction only occurs as a reaction force.
>If static friction was there to begin with,
>with no other forces to oppose it,
>then it would have to cause acceleration.
>
>--
>pete
Aww g'wan pete: The static friction is called static; because the acceleration
it would cause is resisted by the terra firma surface or other support or a
weight-scale on which it "rests".
You do know that rest and motion are relative don't you?
----- Posted via NewsOne.Net: Free (anonymous) Usenet News via the Web -----
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Double-A
John Christiansen
"Double-A" <doub...@hush.com> skrev i en meddelelse
news:79094630.03111...@posting.google.com...
> Don...@mac.com wrote in message news:<bp3prn$f6$1...@news.netmar.com>...
> > In article <3F8F4A...@mindspring.com>, pete
<pfi...@mindspring.com>
> > writes:
> >
> > >Friction only occurs as a reaction force.
> > >If static friction was there to begin with,
> > >with no other forces to oppose it,
> > >then it would have to cause acceleration.
> > >
> > >--
> > >pete
> >
> > Aww g'wan pete: The static friction is called static; because the
acceleration
> > it would cause is resisted by the terra firma surface or other support
or a
> > weight-scale on which it "rests".
> >
> > You do know that rest and motion are relative don't you?
> >
> > ----- Posted via NewsOne.Net: Free (anonymous) Usenet News via the
> > http://newsone.net/ -- Free reading and anonymous posting to 60,000+
groups
> > NewsOne.Net prohibits users from posting spam. If this or other
posts
> > made through NewsOne.Net violate posting guidelines, email
ab...@newsone.net
>
>
> That you Shead?
>
>
> Double-A
It must be Shead, who else would use a pleonasme like terra firma surface.
John Christiansen