Re: How can I tell if F is a string or if it is a number?
tc...@lsa.umich.edu
Pioneer1 <1pio...@gmail.com> wrote:
>On Apr 9, 10:55 am, tc...@lsa.umich.edu wrote:
>>If "F" were just an abbreviation for
>> "m*a" then that would be just a purely logical fact and would not tell us
>> anything about what physical laws happen to hold in our universe.
>
>Why logical? I would say that an abbreviation is notation, it is a
>time saving device.
"Notational" is fine; maybe even better than my word "logical." I meant to
contrast it with "empirical."
>If F is an abbreviation it would tell a lot about what physical laws happen
>to hold in our universe. It would tell us that those laws do not involve force.
No. If F were an abbreviation for "m*a" then that's all it would be: an
abbreviation. I could just as well use "G" or "H" as an abbreviation for
"m*a". All I would be doing is playing with notation. I would be saying
nothing about force, because force does not show up anywhere. I have m,
I have a, and I have an abbreviation for "m*a". Who said anything about
force? Force does not enter the picture at all in this setup.
Only if you use "F" to represent force can you claim that assertions about F
are assertions about force. If "F" is merely a notational abbreviation for
concepts not involving force, then talking about "F" tells you nothing about
force.
Anyway, this is all moot, because "F" is not, in fact, merely an abbreviation
for "m*a." But maybe this discussion will help you see *why* F is not merely
a notational abbreviation. F = ma is supposed to be a law of nature, and for
it to be a law of nature, it must assert something more than "F is an
abbreviation for m*a" since the latter tells us nothing about nature.
>But, I now think I understand the problem. F=ma is actually two
>expressions:
>
>F_occult = m_occult a
>F_contact = m_weight a
>
>F_contact is the force used in the science of mechanics and it is the
>force of weights and pulleys. Newton overlayed his occult force
>F_occult on this contact force and utterly confused the issue. I am
>talking about F_occult. This is the force used in Newtonian
>theoretical astronomical derivations but F_occult always cancels out
>of the formulas used to calculate orbits. So F_contact may be an
>empirical force but F_occult is not. It has never be observed. It
>always cancels out of the orbital formulas. That's why I am
>questioning the assumption that force is a "physical quantity."
Your metaphysical assertions here are a little odd, but I will not criticize
them here; instead I will try to rephrase your argument more coherently.
What I believe you are trying to say is this. Newton's second law F = ma
and Newton's law of gravitation F = Gmm'/r^2 are commonly taken to be two
statements. But in fact, there is a third, hidden statement here that is
rarely made explicit: namely, that the "F" in Newton's second law is the
same as the "F" in Newton's law of gravitation. So we should really write
three equations:
F_contact = ma
F_occult = Gmm'/r^2
F_contact = F_occult
Again, I'm not trying to defend your argument, but trying to help you
articulate it more clearly.
>If so, do you think it is relevant that if F can be replaced for ma
>it's an abbreviation? Thanks for the help.
No, F is not an abbreviation for "m*a". However, according to Newton's
second law, the value of F is the same as the value of m*a. So if you
are doing calculations where all that matters are the *values* of the
quantities you are manipulating (and not what they "mean"), then you can
substitute one for the other, even though one is not merely an abbreviation
for the other.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
John Forkosh
> John Forkosh <j...@please.see.sig.for.email.com> wrote:
>> Sure, there's ambiguity in number vs. string, but I think that's
>> because these kinds of things are just representations. ?And physics
>> is usually formulated as representation-independent (i.e., as
>> generally) as possible.
>
> If I understand you correctly, if physics is "representation-
> independent" in this sense, then in physics symbols are loaded and
> they can represent either a string or a number and a physicist will
> know how to interpret the symbol correctly depending on the context.
> Is this correct?
Well, yes, that's correct and I think it answers the question
you were originally asking. However, it's not quite what I
was talking about. So, I'll ignore my point and try to clarify
the point you just made above.
Context is important, but it's a little more complicated
than you're imagining. Consider your F=ma and the electrostatic
force F=qE on a charge q in an electric field E. That's two
different contexts for F, in the following way.
F=ma is kind of "definitional". If you see a mass m
with an acceleration a, then you infer that there must be
some force F=ma somewhere, somehow, of some sort, acting on m.
But you have no clue what F is or where it comes from.
In this sense F might be called a "label" (like you said
in a previous post) in the F=ma context.
F=qE is a quantity. If you put (as opposed to see)
a charge q in an electric field E, then that charge q will
experience an electrostatic force F=qE. Now, if that charge
q also has mass m, then you'll then see an acceleration given
by F=ma=qE so a=(q/m)E.
--
John Forkosh ( mailto: j...@f.com where j=john and f=forkosh )
Pioneer1
> Force does not enter the picture at all in this setup.
I agree. If F is an abbreviation it would have no meaning other than
being an abbreviation. Since in physics meaning comes from observation
and F does not enter computations of orbits, it is a meaningless
abbreviation, as I see it. For instance, if I start by writing mR/T2 =
ma, are there two different physical quantities here? One mR/T2 and
another ma? To me, No. ma is an abbreviation for mR/T2. The same way F
is an abbreviation for ma. mR/T2 --> ma --> F.
>. . . it must assert something more than "F is an
> abbreviation for m*a" since the latter tells us nothing about nature.
But force does not enter any formulas used to describe orbits. It's
written down as an abbreviation and then cancelled. It does not have
an observational presence, it does not assert something more than "F
is an abbreviation for m*a." I don't know if you agree with this.
> Again, I'm not trying to defend your argument, but trying to help you
> articulate it more clearly.
Yes, thanks, that's helpful. But I think about it this way. There is a
true proportionality discovered by Kepler from observations. This
proportionality is R3=T2. If we break this proportionality and write
it as 1/R2 no matter what we label 1/R2 this 1/R2 will not work. This
is not a true proportionality. Same is true for R/T2. In order to do
orbit computations we must combine these two by eliminating the
placeholders or abbreviations we named these non-working statements.
This is how I see the problem. Correct me if I am wrong. That's why
I've been looking for an analogous situation from computer science
because in programming types are well-defined unlike in physics.
Also I don't believe that F_contact = F_occult is possible. You cannot
equate an occult quantity with its non-occult version. If we could do
that we can reinstate astrology to be physics. Astrology too is based
on an occult force very similar to Newton's occult force. The occult
astrological influx is assumed to exist between humans and celestial
objects. Newton generalized this occult influx between humans and
celestial objects to be an influx between the Earth and celestial
objects and then between all objects.
> then you can
> substitute one for the other, even though one is not merely an abbreviation
> for the other.
If I can substitute F for ma then it appears that F = ma is an
identity: ma = ma? Is this correct?
Thanks again.
tc...@lsa.umich.edu
Pioneer1 <1pio...@gmail.com> wrote:
>I agree. If F is an abbreviation it would have no meaning other than
>being an abbreviation. Since in physics meaning comes from observation
>and F does not enter computations of orbits, it is a meaningless
>abbreviation, as I see it.
No. F is not an abbreviation, not even a meaningless one.
>For instance, if I start by writing mR/T2 =
>ma, are there two different physical quantities here? One mR/T2 and
>another ma? To me, No. ma is an abbreviation for mR/T2. The same way F
>is an abbreviation for ma. mR/T2 --> ma --> F.
No. ma is not an abbreviation for mR/T^2. F is not an abbreviation
for ma.
>But force does not enter any formulas used to describe orbits. It's
>written down as an abbreviation and then cancelled.
No. Force is not an abbreviation.
>It does not have
>an observational presence, it does not assert something more than "F
>is an abbreviation for m*a." I don't know if you agree with this.
I do not agree. F is not an abbreviation.
You might disagree with Newtonian mechanics; that is your choice. However,
you should at least form a correct understanding of what it is that you are
disagreeing with. By asserting that F is an abbreviation, you are simply
demonstrating your lack of understanding of what Newtonian mechanics
asserts, and therefore you cannot coherently formulate any objection to it.
>If I can substitute F for ma then it appears that F = ma is an
>identity: ma = ma? Is this correct?
No. The assertion that F = ma is not the same as the assertion that ma = ma.
For example, consider "George Bush is the president of the United States."
The name "George Bush" is *not* just an abbreviation for "the president
of the United States." The statement that "George Bush is the president
of the United States" conveys information that the statement "the president
of the United States is the president of the United States" does not. Just
because you can take the sentence "The president of the United States will
greet the Pope at the airport" and replace "the president of the United
States" with "George Bush" to obtain "George Bush will greet the Pope at
the airport" does *not* mean that "George Bush" is an *abbreviation* for
"the president of the United States." The ability to substitute does not
imply that one is an *abbreviation* of the other.
In contrast, "GWB" *is* an abbreviation of "George W. Bush."
Barb Knox
> Also I don't believe that F contact = F occult is possible. You cannot
> equate an occult quantity with its non-occult version. If we could do
> that we can reinstate astrology to be physics. Astrology too is based
> on an occult force very similar to Newton's occult force. The occult
> astrological influx is assumed to exist between humans and celestial
> objects. Newton generalized this occult influx between humans and
> celestial objects to be an influx between the Earth and celestial
> objects and then between all objects.
Note that "contact" forces are actually "occult" electrostatic ones. If
you consider gravitational attraction to be suspect, what about
electrostatic repulsion (which is responsible for "contact"
interactions).
--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum viditur.
| BBB aa a r bbb |
-----------------------------
Pioneer1
wrote:
> If you see a mass m
> with an acceleration a, then you infer that there must be
> some force F=ma somewhere, somehow, of some sort, acting on m.
> But you have no clue what F is or where it comes from.
I don't see it this way. Formulas describing orbits are independent of
the term F. Therefore, to me orbits are independent of F. The orbit is
independent of m too. Not that we don't know what force is or where it
comes from but we know that, by looking at formulas, F is not a factor
in orbital motion. Please let me know what's wrong with this
reasoning.
> Now, if that charge
> q also has mass m, then you'll then see an acceleration given
> by F=ma=qE so a=(q/m)E.
The way I read it, this proves my point. You used F as a placeholder
to obtain an equation that does not involve F. Looking at your final
equation a=(q/m)E, there is no F in it. The same result can be
obtained without using force. The fundamental quantity here is R/T2.
This quantity is found to be proportional to E, therefore, R/T2 :: E.
Constant terms are added as units to scale for measurements.
You were right about units and dimensions. I didn't see it at first,
but this problem is related to the use of units in physics. For
instance, in R/T2 :: E both terms have dimensions R/T2. To me, this
means that this is an identity, not a proportionality. If this were a
computer program there would be an error because the wrong assignment
operator is used.
As I understand it E is proportional to 1/R2 (the intensity around
charge falls off as 1/R2). So here too E is a placeholder for 1/R2.
Thanks again.
John Forkosh
Maybe I'm wrong.
> As I understand it E is proportional to 1/R2 (the intensity around
> charge falls off as 1/R2).
For point charges. How does E behave between capacitor plates?
> So here too E is a placeholder for 1/R2.
E and 1/R^2 have different units.
Are apples placeholders for oranges?
Look, if you think you know what you're doing,
and have found something worth doing, then do it.
I see nothing, but you don't have to convince me.
Pioneer1
> No. F is not an abbreviation, not even a meaningless one.
Okay. What is it? Does it have a meaning? If it does, its meaning is
not coming from observations. Then, where does it come from?
> No. ma is not an abbreviation for mR/T^2. F is not an abbreviation
> for ma.
So why does it cancel? It's not a placeholder. It's not a temporary
variable. It's not a label. It's supposed to be a "physical" quantity
but it cancels out of formulas.
> You might disagree with Newtonian mechanics;
No. I neither disagree or agree with Newtonian mechanics. To me,
Newtonian mechanics is a meaningless generalization. I'm questioning
one expression: F=ma. I cannot understand the types of the symbols in
this expression.
Another example: Physicists call this expression a=v^2/R an
"equation." To me this is not an equation, it is an identity: a= aR/R.
> you should at least form a correct understanding of what it is that you are
> disagreeing with.
I think this is what I've been trying to do but regarding F=ma, not
Newtonian mechanics.
> In contrast, "GWB" *is* an abbreviation of "George W. Bush."
Okay. F is an abbreviation for Force. Force is an abbreviation for
that occult quality that emanates from the sun and travels the earth-
sun distance at zero second by breaking all of the conservation laws
of physics and then sets the earth in motion by instantaneously
communicating with the earth's center about how much of itself to
apply and then keeps the earth in orbit. . . It does all those
fantastic things at no time at all but this force cannot be found in
the formulas we use to compute the orbit of the earth. To me if a
quantity does not exist in formulas then it does not govern orbits.
M is an abbreviation for mass of the earth. Whatever the mass of the
earth is it has no relevance in orbit calculations because mass too
cancels out. Two out of three terms of F=ma cancel out of the
formulas. We are left with acceleration a. No. That's a goner too.
Because acceleration a stands for R/T2. R and T are the only two
quantities necessary and sufficient to compute orbits.
Each and every term of the fundamental equation of physics cancel out
of the formulas to compute orbits. Why did physicists write three
terms that cancel and made the cancelling terms the cornerstone of
physics? What am I missing here?
Pioneer1
> Note that "contact" forces are actually "occult" electrostatic ones.
I don't understand why electrostatic force is considered occult. Can
you explain a bit more?
Barb Knox
<209da200-4767-4b5b...@m1g2000pre.googlegroups.com>,
Pioneer1 <1pio...@gmail.com> wrote:
You consider gravitational attraction to be an "occult" force. What in
your view makes it essentially different from electrostatic forces?
tc...@lsa.umich.edu
Pioneer1 <1pio...@gmail.com> wrote:
>On Apr 15, 12:08 pm, tc...@lsa.umich.edu wrote:
>
>> No. F is not an abbreviation, not even a meaningless one.
>
>Okay. What is it? Does it have a meaning? If it does, its meaning is
>not coming from observations. Then, where does it come from?
F stands for the numerical value of the force, where "force" is a concept in
our physical theory of the world.
>> No. ma is not an abbreviation for mR/T^2. F is not an abbreviation
>> for ma.
>
>So why does it cancel? It's not a placeholder. It's not a temporary
>variable. It's not a label. It's supposed to be a "physical" quantity
>but it cancels out of formulas.
Canceling in formulas happens whenever you have more than one equation for an
unknown. Suppose you have two coins of unknown weight. By doing some
experiments you find that the difference in weight is 4 and the sum of the
weights is 10. So I can let X be the weight of one coin and let Y be the
weight of the other coin. I get the equations
X = Y + 4
X = 10 - Y
I can cancel out X to obtain the equation Y + 4 = 10 - Y, whence Y = 3
and X = 7. Does the fact that I have canceled out X mean that X is an
"abbreviation" for "Y + 4"? No. X is the weight of the coin. X = Y + 4
is not a definition, but a fact that you obtained from experiment.
Elimination of variables is just a mathematical technique for solving
equations. The fact that you can eliminate a variable doesn't mean that
the variable doesn't mean anything or is just an abbreviation; it just
means that you can make progress in *solving* the equations.
Perhaps you understand the coin-weighing example but still have trouble
making the leap to F = ma. So let's take another coin-weighing example.
Suppose I have a balance beam that has unequal length arms. One arm has
length A and the other arm has length B. If I put a weight X on one side
and a weight Y on the other side and they balance, then I have the "physical
law" AX = BY. Here we have an equation involving multiple variables. Does
this mean that "A" is an abbreviation for "BY/X"? No. A is the length of
the first arm. AX = BY is a generalization from many empirical observations.
It is not a mathematical identity, but a physical law.
What if I find some other physical laws that also involve A, X, B, and Y,
and that allow me to solve for these quantities by eliminating variables?
Would that mean that A suddenly becomes an abbreviation for "BY/X"? No.
A is the length of the first arm. Even if you manage to find some other
equation for A that allows you to eliminate A and solve for some other
unknowns, A continues to denote the length of the first arm.
Patricia Shanahan
...
> No. I neither disagree or agree with Newtonian mechanics. To me,
> Newtonian mechanics is a meaningless generalization. I'm questioning
> one expression: F=ma. I cannot understand the types of the symbols in
> this expression.
I think you are making a mistake in even trying to understand one
equation in isolation from the theory of which it is part. The "F" in
"F=ma" carries meaning from static mechanics.
Patricia
Pioneer1
> You consider gravitational attraction to be an "occult" force. What in
> your view makes it essentially different from electrostatic forces?
I think they are not different, they are the same phenomenon. This is
how I understand it:
http://www.densytics.com/wiki/index.php?title=Force#Force_and_electrostatics
Comments are welcome.
tc...@lsa.umich.edu
Pioneer1 <1pio...@gmail.com> wrote:
>On Apr 22, 12:44 am, Barb Knox <s...@sig.below> wrote:
>
>> You consider gravitational attraction to be an "occult" force. What in
>> your view makes it essentially different from electrostatic forces?
>
>I think they are not different, they are the same phenomenon.
I think you're losing the thread of the discussion. If you agree that
electrostatic forces are the same sort of thing as gravity, then "contact"
forces in mechanics are no less "occult" than Newtonian action-at-a-distance.
Go back and read the thread again if you've lost the chain of thought.
Pioneer1
> Pioneer1 wrote:
>
> I think you are making a mistake in even trying to understand one
> equation in isolation from the theory of which it is part. The "F" in
> "F=ma" carries meaning from static mechanics.
Okay, I understand but the meaning from statics does not carry to
orbital motion. In statics force is communicated by pulleys and
strings. In orbital motion there are no strings, force acts
instantaneously from a distance. And the F term cancels out of the
formulas used to compute orbits. I am just looking at formulas for
orbits and those are not the formulas of static mechanics. Is this
observation wrong?
tc...@lsa.umich.edu
Pioneer1 <1pio...@gmail.com> wrote:
>On Apr 24, 9:46 am, Patricia Shanahan <p...@acm.org> wrote:
>> Pioneer1 wrote:
>>
>> I think you are making a mistake in even trying to understand one
>> equation in isolation from the theory of which it is part. The "F" in
>> "F=ma" carries meaning from static mechanics.
>
>Okay, I understand but the meaning from statics does not carry to
>orbital motion. In statics force is communicated by pulleys and
>strings. In orbital motion there are no strings, force acts
>instantaneously from a distance.
The force that is "communicated by pulleys and strings" still acts at a
distance---it's just that the distances are at the molecular level. You
seem to have missed the point of the discussion of electrostatic forces.
>And the F term cancels out of the formulas used to compute orbits.
I explained elsewhere what cancellation means. Did you see my explanation?
>I am just looking at formulas for orbits and those are not the formulas
>of static mechanics. Is this observation wrong?
If you're asking whether F = ma and F = Gmm'/r^2 are distinct formulas,
then the answer is yes. They represent different physical laws. But
in both cases, F represents the numerical value of the force. By
generalizing from observation, we are able to make two distinct statements
about F, namely F = ma and F = Gmm'/r^2.
The fact that we have two different equations for the same quantity is a
good thing, because it means that we can use the mathematical technique
of eliminating variables (what you called "cancellation") to solve for
some of the quantities we're interested in.
Patricia Shanahan
Consider the transition between a static system and a dynamic one.
Suppose, for example, I have a weight suspended, in a vacuum, near the
surface of the Earth, by some system of pulleys and strings. I think you
would agree that there is a downwards force on the weight, and that in a
static system it is exactly balanced by the upwards force the suspension
system is applying to the weight. That force can be measured.
[It is also an action-at-a-very-small-distance force, but Tim Chow has
already responded on that issue.]
Now I cut the last piece of string, putting the weight into free fall. I
would like to predict how long it will take to fall a certain distance.
I could do that if I knew its acceleration. Fortunately, I could measure
the force in the static system. Assuming I know the mass of my weight, I
can get the acceleration from a=m/F, a variant of F=ma for non-zero force.
What's the difference between the force making the weight travel towards
the Earth and the force making the moon travel towards the Earth? Or
did the nature of the downwards force on the weight change when I cut
the string attaching it to the pulleys?
Of course, you can remove a lot of the meaning from any formula by
looking at it in isolation, but that does not seem to me to be a
particularly useful or interesting exercise.
Patricia
Patricia Shanahan
That should have been a=F/m, for non-zero mass.
Pioneer1
>You seem to have missed the point of the discussion of electrostatic forces.
I think this is true. I don't understand how electrostatic forces are
related to planetary orbits. Are there any observed orbits that are
held together by electrostatic forces? But I will go back and read the
whole thread, as suggested in your earlier post, and I might have a
more informed reply to this. I would also like to thank you and
everyone else who commented, this has been very helpful.
> in both cases, F represents the numerical value of the force. By
> generalizing from observation, we are able to make two distinct statements
> about F, namely F = ma and F = Gmm'/r^2.
I think this is at the heart of my confusion. You are saying that "by
generalizing from observation" we are able to write the two force
definitions. But I believe that observations of orbits do not support
this statement. Observed orbits are not dependent on F, m, m' or G
terms. Observations of orbits lead to the proportionality discovered
by Kepler that the period is proportional to the 1.5 power of the
radius of the orbit. "Generalizing from observations" leads to the
result that force does not enter the description of orbital motion.
Force in orbital motion exists only as a definition; observed orbits
are independent of force. Physicists start from Newton's definitions
of force (F=ma and F=GMm/R2) and fit observations to those
definitions. I start from observations and don't see force in the
observations. T = R^1.5 is the observation and it doesn't have a force
term in it.
> The fact that we have two different equations for the same quantity is a
> good thing,
I am not sure that the assumption that F_ma and F_GM are the same
quantity is justified. F_am is proportional to R, while F_GM is
proportional to 1/R2. I don't believe that the same quantity can
increase and decrease with R at the same time. If we were to equate a
force proportional to R and another force proportional to 1/R2, this
should lead to an absurd result. But it doesn't. Why? As far as I
understand, there is one possibility for this elimination to work: R/
T2 and 1/R2 must form a proportionality. In other words, physicists
put together the original proportionality that Newton split in two
parts by eliminating the placeholder F and recover the original rule,
R3/T2 to make computations. But this proves that force is not used in
orbit calculations, it's eliminated.
> . . . because it means that we can use the mathematical technique
> of eliminating variables (what you called "cancellation") to solve for
> some of the quantities we're interested in.
Okay, thanks for clarifying this. I looked up the way we eliminate
variables in simultaneous linear equations. I assume that this is what
you were referring to. I would appreciate your comments on the
following.
For instance, let's say in a given set of simultaneous equations of x
and y, we eliminated y and obtained this expression for x: x = a + b.
Can you tell what y was? I don't think this is possible. The variable
y was eliminated, it doesn't exist in the formula anymore. In our
case, similarly, we eliminated F and we obtained R3/T2. When we use R3/
T2 the force term we eliminateed is no longer a factor in our
calculations of orbits.
But I don't think force is like the variable y in the above example.
For instance, if the given simultaneous equations are 2x + y = 8 and x
+ y = 6, let's write them as Equation1 = 2x + y – 8 and Equation2 = x
+ y – 6. To me, intuitively, the terms Equation1 and Equation2 are not
like the other variables x and y. Equation1 and Equation2 are names
that we assigned to the equations. We need to eliminate them to solve
the equations. Do you mind commenting on this? Is this intutition
correct? Force terms appear to me to be like Equation1 and Equation2.
Can we write R3/T2 as simultaneous equations? For instance, a/R2 = y
and bR/T2 = x. Are these now simultaneous equations? I think, in the
case of force, probably, we should write, a/x2 = A and bx/y2 = B. And
we must eliminate A and B by saying that they are equal. And indeed,
they must be equal since the ratios in an equality of ratios are equal
by definition. Thanks for helping to formulate the question this way.
Pioneer1
> Suppose, for example, I have a weight suspended, in a vacuum, near the
> surface of the Earth, by some system of pulleys and strings. I think you
> would agree that there is a downwards force . . .
If you assume that this force is the Newtonian occult attraction
emanating from the center of the earth and attracting the weight with
its occult qualities, then, I don't agree. To me, the Newtonian force
is action-at-a-distance and it's supernatural and does not exist. So,
I don't make that assumption.
> . . . on the weight . . .
This is really Newtonian language. This occult attraction "on the
weight" is the exact thing I am questioning and I will not assume it.
> . . . and that in a
> static system it is exactly balanced by the upwards force . . .
Are you saying that there is an upward force of attraction attracting
the weight upwards the way the assumed occult force attracts the
weight downwards? No, of course not. That upward force is just a
physics fiction invented by physicists because force by definition is
the universal cause of everything. The assumption of balancing a non-
existing occult force with another non-existing force only tells me
that the paradigm of force is broken.
> the suspension
> system is applying to the weight. That force can be measured. . .
I believe that what is measured is tension. The chain of reasoning
based on the assumption of various types of forces to explain a
natural phenomenon that does not involve an occult force is not
natural. You are assuming that the tension on the string is a
manifestation of the occult force defined in F=ma because you assume
that F is proportional to acceleration a. So measuring the
acceleration gives you the force. But in the equation
acceleration = a = R/T2 = weight/force = g
the only measurable quantities are the radius R and period T of the
orbit. We measure R and T and call it acceleration, (which is the
label for the ratio radius/period squared) and then call the
acceleration force. Force is not necessary to explain what is observed
but it must be included because it is the universal cause of physics.
> What's the difference between the force making the weight travel towards
> the Earth and the force making the moon travel towards the Earth?
You are assuming the force. According to observations, neither the
moon nor the weight are set in motion by the assumed occult force.
> Or
> did the nature of the downwards force on the weight change when I cut
> the string attaching it to the pulleys?
No. There is no such force acting between the weight and the Earth.
This "force" experienced by a weight is analogous to the "force"
experienced by the passengers in an accelerating airplane. Passengers
observe that they are pulled towards the back of the plane. Some of
them conclude that there is a Newtonian mass at the back of the plane
that attracts them with the Newtonian force. Others think that they
are pressed to the back of their seat because of the motion of the
airplane. They don't believe they are attracted by a mass in the tail
section of the plane. Who's right? To me, passengers who reject the
Newtonian occult attraction are right. The same with free fall. I
reject on principle the occult attraction postulated by Newton. There
is no occult attraction between the falling weight and the Earth.
From the point of view of the weight nothing changes when the string
is cut. It's motion is always described by the same two terms: Radius
R and period T of the orbit. Newton hid these two quantities in his
definitions of force in order to design a Newtonian dynamical orbit.
The motion of the Moon is described by the same two terms R and T
related by R3/T2. This rule does not have a term called force,
therefore, the assumption of force moving the weight and the Moon does
not come from observations, it comes from Newton's authority, i.e.,
his definitions stated as laws of motion.
> Of course, you can remove a lot of the meaning from any formula by
> looking at it in isolation . . .
I am not removing meaning, I am removing superflous authority terms to
recover the meaning. The meaning is there hidden under these authority
terms which physicists discard anyway. Newtonian symbolic definition
of force F=ma is discarded in its entirety to recover the rule R3/T2.
tc...@lsa.umich.edu
Pioneer1 <1pio...@gmail.com> wrote:
>On Apr 28, 1:24 pm, tc...@lsa.umich.edu wrote:
>I think this is true. I don't understand how electrostatic forces are
>related to planetary orbits. Are there any observed orbits that are
>held together by electrostatic forces?
The point is that there is a very close analogy between electrostatic forces
and gravitational forces. In (Newtonian) gravitation we have the equation
F = Gmm'/r^2 and in electrostatics we have the equation F = qq'/r^2. Since
"contact" forces are really electrostatic forces, they actually act at a
distance (but the distances are microscopic). If "action at a distance"
is occult, then contact forces are occult too.
But this is something of a tangent. I wouldn't worry about this point too
much if you have trouble understanding it.
>I think this is at the heart of my confusion. You are saying that "by
>generalizing from observation" we are able to write the two force
>definitions. But I believe that observations of orbits do not support
>this statement. Observed orbits are not dependent on F, m, m' or G
>terms.
Although orbital motion is perhaps the most important example of an
observation supporting Newton's law of gravitation, it is not the only
example. There is, for example, Cavendish's experiment.
>I start from observations and don't see force in the observations.
This is perhaps the crucial point. You could, perhaps, develop a theory
of gravitation and mechanics without any reference to forces. In a sense,
that is what general relativity theory does. So if you ever pursue your
study of physics that far, you may experience a sense of satisfaction
and vindication when you get to general relativity---"Aha! I *knew*
there was something fishy about those Newtonian forces!"
Forces are, as you say, not directly observed. However, the way Newtonian
theory works is that physicists *postulate* that there is something called a
"force" that obeys certain laws. By calculating with the equations one can
derive experimental predictions that can be tested against experiment. In
this way one develops confidence that one is on the right track with the
theory.
This is a general fact about how physics works. Physicists are constantly
devising theories containing entities that are not directly observed. We
can't "directly" observe subatomic particles; what we see are tracks in a
cloud chamber, which we postulate are caused by particles zooming through the
chamber.
I think that once you accept the idea that one can hypothesize the existence
of things that are not directly observed, in order to help develop a
conceptually satisfactory model of how the world works, you will be more
comfortable with "force."
>I am not sure that the assumption that F_ma and F_GM are the same
>quantity is justified. F_am is proportional to R, while F_GM is
>proportional to 1/R2.
I don't understand why you think that "F_ma is proportional to R."
F_ma is proportional to *acceleration*, which is a very different thing
from the *distance between the objects*.
>For instance, let's say in a given set of simultaneous equations of x
>and y, we eliminated y and obtained this expression for x: x = a + b.
>Can you tell what y was? I don't think this is possible. The variable
>y was eliminated, it doesn't exist in the formula anymore.
You can, in fact, usually tell what y was, by substituting x = a+b back
into one of your original equations.
>But I don't think force is like the variable y in the above example.
>For instance, if the given simultaneous equations are 2x + y = 8 and x
>+ y = 6, let's write them as Equation1 = 2x + y – 8 and Equation2 = x
>+ y – 6. To me, intuitively, the terms Equation1 and Equation2 are not
>like the other variables x and y. Equation1 and Equation2 are names
>that we assigned to the equations. We need to eliminate them to solve
>the equations. Do you mind commenting on this? Is this intutition
>correct? Force terms appear to me to be like Equation1 and Equation2.
No, force is like x or y.
Part of your problem, it seems to me, is that you don't see how to calculate
the value of the force. After it's eliminated, and we solve for the motion,
we never seem to care about it any more. So wasn't it redundant in the first
place?
For the particular problems you're currently trying to solve, it may appear
that the force term is redundant. However, Newton's laws apply much more
generally, to other physical situations, and in other situations, you will
be able to calculate the force. For example, you can calculate the force
if you have some way of determining the mass of the object in question
as well as its acceleration. Presumably you don't have any problems with
measuring acceleration, which boils down to measuring distance and time?
Do you have any problem with measuring mass, by using a balance pan and a
"standard mass" like the one they keep locked up in Paris? If you can
measure mass and measure acceleration then you can calculate the value
of the force by multiplying the mass by the acceleration.
Pioneer1
> This is perhaps the crucial point. You could, perhaps, develop a theory
> of gravitation and mechanics without any reference to forces. In a sense,
> that is what general relativity theory does.
I understand that general relativity excludes forces in its realm. But
that doesn't help in this case. As you mentioned above Newton's force
is an experimental quantity routinely measured in Cavendish
experiments even by students. Physics admits the action-at-a-distance
Newtonian force as an experimental quantity and also general
relativity's no force explanation.
General relativity specifies a different mechanism for gravity but
that does not remove Newtonian action-at-a-distance force from
physics. But my point is that, the force term is eliminated and
therefore the orbit is independent of force. Since force is
eliminated, the fact that it is action-at-a-distance becomes
irrelevant.
> Forces are, as you say, not directly observed. However, the way Newtonian
> theory works is that physicists *postulate* that there is something called a
> "force" that obeys certain laws.
Okay, I understand that Newton postulated something called "force"
that obeys Newton's laws and physicists start their derivation from
the assumption that something called force is needed to describe
orbits. But it is not true that orbital predictions are made by using
equations that contain force. The force terms are eliminated during
the derivation process:
1. Force is assumed
2. Force terms are eliminated
3. Orbits are computed with formulas that do not contain force terms.
From the above list I conclude that orbits are independent of force.
How do physicists conclude that orbits are described by the assumed
force after they eliminate the force from equations?
From the above I also conclude that force is not a quantity that is
observed indirectly, but it is a quantity that is eliminated, in other
words, it does not exist in orbital motion. To assume that force
remains in the equations after it is eliminated appears to me to be an
unjustified assumption that does not have a physical basis.
For instance, in the Newtonian description of orbits we assume that
force deflects rectilinear motion into curved orbits. If force is not
represented in the equations how does it do that? This seems like
magic to me. But first thing we learn in physics is that magic and
similar non-physical things are not part of physics.
> I think that once you accept the idea that one can hypothesize the existence
> of things that are not directly observed, in order to help develop a
> conceptually satisfactory model of how the world works, you will be more
> comfortable with "force."
In the case of orbits, there is a conceptually satisfactory model
describing orbits well without an assumption of force. Why is there a
need to assume a hidden cause not visible in the formulas? To me the
assumption of force adds nothing to the model and it is eliminated and
therefore is not needed.
> I don't understand why you think that "F_ma is proportional to R."
> F_ma is proportional to *acceleration*, which is a very different thing
> from the *distance between the objects*.
I expand acceleration a by writing it as a = R/T2 or F = m R/T2. Then
for a given period, force F is proportional to R. And this is true by
observation because in a sling, as Newton mentioned in Definition 5
while defining this type of force, for a given angle the force on the
string increases with R. This follows from radian motion theta = s/r
or for a given angle, s is proportional to r.
Newton then generalized this sling-type rotation to all orbits by
saying that "the same applies to all bodies that are made to move in
orbits." But planetary orbits do not obey this rule, planets do not
rotate by radian rule as a sling does, they revolve by Kepler's rule.
In other words, in planetary motion, for a given angle radius is not
respected as in radian motion, and radius and arc are not directly
proportional. Increasing the radius r decreases the arc s according to
Kepler's rule.
When physicists equate F_ma and F_GM they are saying that orbits
rotate the way slings rotate. This is not true. The sling rotation and
planetary revolutions are different. I realize that this was very
ingenious on Newton's part. He created this confusion to let his force
and his dynamical acceleration concept to be the rule also for
Keplerian orbits. But orbits are still Keplerian and not Newtonian and
dynamical. We must eliminate Newton's force to describe orbits with
Kepler's rule. Planetary orbits and a stone rotating on a sling are
not the same type of motion.
> You can, in fact, usually tell what y was, by substituting x = a+b back
> into one of your original equations.
Yes, that was my point. If you know the original equation you can
substitute y. In this case you do not know the original equations.
Similarly, the orbit does not know that physicists write F terms in
intermediary equations and then cancel them. For the orbit, there
exists no force because it was eliminated.
> Part of your problem, it seems to me, is that you don't see how to calculate
> the value of the force.
This is true. I don't see how to compute the value of force by using
the same formula we used to compute the orbit because the formula we
used to compute the orbit does not include a term for force.
> After it's eliminated, and we solve for the motion,
> we never seem to care about it any more. So wasn't it redundant in the first
> place?
Yes. This is how I see the problem. But not only we don't care about
force after we eliminate it, the orbit does not care about it either.
The orbit does not know that physicists wrote equations with force
terms in them.
> For the particular problems you're currently trying to solve, it may appear
> that the force term is redundant.
But do you think it is redundant in the orbital motion? If it is
redundant, would it be correct to say that "orbits are independent of
force" because we eliminated the force term?
> However, Newton's laws apply much more
> generally, to other physical situations . . .
I understand this. But instead of considering Newton's laws as a
general framework, I'm just looking at the specific assumption of
force in the computation of orbital motion. The assumption is not
justified, as far as I understand it, because we assume it but then we
eliminate it, and compute orbits without it. If it's not used why
assume it?
> . . . and in other situations, you will
> be able to calculate the force.
I would like to make sure that this is the same force that enters
orbital calculations but no force term enter orbital calculations.
> For example, you can calculate the force
> if you have some way of determining the mass of the object in question
> as well as its acceleration.
But there is no way to compute masses of celestial objects because
they don't appear in formulas alone. As far as I understand, I could
only compute force, a quantity that does not enter orbital
computations, from another quantity, the mass m, that does not enter
orbital computations. Given that neither is orbital, this computation
will not be relevant to orbit computations.
> Presumably you don't have any problems with
> measuring acceleration, which boils down to measuring distance and time?
I believe that in the case of orbits, the time in the a = R/T2 refers
to the period of the orbit. And I'm having, I think, the same problem
as force regarding acceleration as well. As far as I understand, in
physics, it is considered that a = R/T2 refers to two different
quantities, acceleration and R/T2. I believe that since only R and T
are measured, the live physical quantity is R/T2, not a different
quantity called acceleration. I would say that I am measuring the
ratio R/T2 and calling it acceleration.
> Do you have any problem with measuring mass, by using a balance pan and a
> "standard mass" like the one they keep locked up in Paris?
No problem. But this mass is not an orbital quantity because it is
eliminated. And in this case physicists say that the orbit is
independent of m because it is eliminated.
> If you can
> measure mass and measure acceleration then you can calculate the value
> of the force by multiplying the mass by the acceleration.
Okay. Say F_ma = X Newtons. This quantity is not a quantity that
enters in orbital computations because F_ma is eliminated.
There is a reason why F is eliminated from orbital computations:
Orbits are not rotational motion like a sling where R/T2 alone defines
the orbit. In definition 5 Newton called sling motion an "orbit" and
then generalized to all orbits. This way Newton confused the issue by
making acceleration a different property than the orbital quantity R/
T2. The fact that Newton calls R/T2 "acceleration" does not change the
fact that the orbital arc traveled in unit time is proportional to the
radius. Newton wants to project this sling "acceleration" to orbital
motion in order to say that orbits are formed by this force which
bends rectilinear inertial motion into curved orbits. But this does
not work.
The sling-type rotation is described by radian motion, theta =s/r or s
= r theta. Dividing by time gives velocity: s 1/t = r theta/t.
Dividing again by time: s 1/tt = r theta/tt is called acceleration. In
other words, however many times you divide by time will never change
the fact that the arc s is proportional to r for a given angle theta.
And the rotational motion will never become orbital motion. The
acceleration is true for sling-type rotational motion but not true for
revolutionary motion that obeys Kepler's rule.
The R/T2 hidden in acceleration a in F = ma is really half of Kepler's
rule, R/T2=1/R2. R/T2 alone works only for rotational motion. To get
orbital Keplerian motion we must eliminate Newton's force and obtain
Kepler's rule R/T2=1/R2. And this is no longer rotational radian
motion, now, R^1.5 is proportional to the period T. Or rotational
motion is R proportional to s and orbital motion is R^1.5 is
proportional to 1/s. Or in the planetary motion, the arc traveled per
unit time is no longer directly proportional to R. In orbital motion
radius is not respected. If you increase R, the arc s does not
increase directly as R, it decreases according to Kepler's rule. This
is the point of using Kepler's rule to describe orbits, instead of
radian motion.
But Newton confused the two type of motion, rotational and orbital,
and physics tradition still follows Newton and writes Newton's force
terms and then cancels them. Newton's projection from radian rotation
to orbital revolution is wrong. We don't need to write and cancel
Newton's force just because he did so. This is how I understand it.
Thanks again for all your help.
Patricia Shanahan
...
> I understand this. But instead of considering Newton's laws as a
> general framework, I'm just looking at the specific assumption of
> force in the computation of orbital motion. The assumption is not
> justified, as far as I understand it, because we assume it but then we
> eliminate it, and compute orbits without it. If it's not used why
> assume it?
...
That misses the key point of Newtonian physics, and indeed of much of
science. The main objective of the game is to find simple theories that
explain and predict a lot of observations. By that standard, Newtonian
physics is one of the most spectacularly successful theories ever developed.
You might be able to build a reasonably simple forceless theory for
orbital motion, but it would probably be at least as complex as
Newtonian physics and be much less useful for even closely related
questions, such as deciding how heavy components of a satellite need to
be to survive accelerations during launch.
Patricia
tc...@lsa.umich.edu
Pioneer1 <1pio...@gmail.com> wrote:
>But it is not true that orbital predictions are made by using
>equations that contain force. The force terms are eliminated during
>the derivation process:
You're contradicting yourself. Something cannot be "eliminated" unless it is
there in the first place. The calculation *does* use equations that contain
force, and then eliminates them.
In a sense, you are right that if all you care about are orbits, then you get
by with Kepler's laws. But as Patricia Shanahan pointed out, physicists seek
to find laws that govern *all* physical phenomena. It's true that you can
view the world as a disjointed set of isolated, unrelated problems. Today
I'm interested in orbits. Tomorrow I may be interested in airplane design.
The next day I may be interested in semiconductors. I could treat each of
these problems separately, seeking to find the minimum number of assumptions
required to explain each situation, and ignoring the fact that each time I am
inventing a new set of assumptions that has nothing to do with the assumptions
I used yesterday for a different problem.
Alternatively, I could seek a few very general principles that apply in *all*
situations. Most scientists find this approach much more satisfying than the
piecemeal, ad hoc approach above.
Newton's law of gravitation is called the *universal* law of gravitation for
this reason. In a particular problem, say the calculation of a planetary
orbit, you might be able to get away without assuming forces. But to
conclude from that one example that forces are *in general* an unnecessary
theoretical device is to take a very narrow view of science. Before you
discard the notion of force, ask yourself if you could calculate the dynamics
of two colliding galaxies without assuming a theory of gravitation. Kepler's
laws won't be good enough any more. If a simple theory of gravitation
explains both galaxy collision and orbital motion, then it makes sense to
claim that gravity is at work in orbits, even if for some particularly simple
problems you don't need the general theory.
Pioneer1
> By that standard, Newtonian
> physics is one of the most spectacularly successful theories ever developed.
Yes, I agree with this and I thank you for helping me clarify some of
the confusion I'm having. Thinking about it a bit more, and based on
your comment, I believe that I have a better picture. I agree with the
success of the Newtonian mechanics and I'm not questioning it. I see
Newtonian mechanics as a consistent system of units made of physical
quantities. It is a computational device, or an algorithm, that takes
given values of physical quantities as its arguments and transforms
them according to some rule. Much the same way that Ptolemaic model
predicts planetary orbits.
What I see now is that physicists are making the assumption that this
computational algorithm is coupled to the axioms they call Newton's
laws. This is the same claim Ptolemaic astronomers made about their
model. They believed that their trigonometric computational device was
coupled to the axiom of the stationary Earth and that was why it
worked and that in turn proved that the Earth must be stationary. Only
centuries later Copernicus realized that algorithm was independent of
the axiom and that he could use the same geometric model to design a
system with a moving earth. This is my point. Newtonian physics is
independent of Newton's laws. Newtonian physics is a computational
device just like the Ptolemaic model. It works because it is
independent of Newton's metaphysical laws.
In computer languages the separation of commentary and code is
absolute. The computer never sees the comments, they are clearly
marked as comments. In physics the metaphysical assumption that force
is the universal cause is freely mixed with the computational
algorithm in order to make the algorithm look Newtonian. This is the
question I'm investigating. Does the computational part of Newtonian
mechanics use force? If we do not write "force" and do not start from
F=ma, can we still predict astronomical motions as well as Newtonian
mechanics? If we start directly from Kepler's rule R3/T2 can we obtain
the same results? My answer is related to the second part of your
comment:
> You might be able to build a reasonably simple forceless theory for
> orbital motion, but it would probably be at least as complex as
> Newtonian physics . . .
You are right. Actually, my impression is that it would be not only
more complex but impossible to use in engineering applications because
it would be based on using proportions and not units. Thanks for
mentioning this, because it clarifies to me that what makes Newtonian
physics a useful paradigm in engineering is not force but the unit
system. Newtonian mechanics works because it is a consistent system of
units. Trying to use Kepler's rule without units is cumbersome and
difficult. So physicists write Kepler's rule R3/T2 with a Newtonian
unit as a = GM/R2 and believe that they are using Newtonian mechanics
and Newtonian force. This is due to historical reasons and to Newton's
immense authority. I think when physicists will realize that Newtonian
physics is independent of Newtonian doctrine of force, a new
scientific revolution will happen. It happened in astronomy and
mathematics. I think it will happen in physics too.
Pioneer1
> The calculation *does* use equations that contain
> force, and then eliminates them.
Yes. You are right. I realize that physicists sometimes call the
derivation process "calculations." Maybe I should have asked if the
orbit knows about preparatory calculations and temporary terms we
might write and then eliminate. My personal opinion is that such
transitionary terms are not visible to orbits once they are
eliminated.
> . . . physicists seek
> to find laws that govern *all* physical phenomena.
I understand that it makes sense to have a unifying principle, if
possible. But are you saying that the force that Newton assumed
described orbits is the unifying principle for aerodynamics, materials
science and the rest of physics? Maybe I'm misunderstanding you but
that force assumed by Newton does not even enter the orbital
formulas.
I am not advocating inventing a new set of assumptions to explain each
situation. I agree with you that the business of science should be to
look for general principles. But I disagree that instead of looking
for such principles we should fit everything into the Newtonian
doctrine and always assume that the doctrine is right. In the case of
orbital equations mathematical description of orbits says that orbits
are independent of force. Newtonian doctrine says that orbits are
caused by force. I choose to respect the authority of mathematics
instead of the authority of Newton. This may not be how physicists
reason. But I was always taught that science means to respect the
authority of mathematics and reject any other kind. If my
understanding of mathematics is wrong, that's another thing and I
would appreciate corrections.
> Alternatively, I could seek a few very general principles that apply in *all*
> situations.
I agree with this view. But to me there is a general principle that
comes before Newton's doctrine of force, and that is Kepler's rule
because it came from observations. I believe this because Newton
defined force on Kepler's rule in such a way that the force term must
be cancelled in order to make orbital computations. This fact tells me
that unless Newton discovered a new proportionality replacing Kepler's
rule, astronomical computations must be done with Kepler's rule. As
far as I know, Newton did not discover a new rule, so physicists
compute with Kepler's rule and call it Newton's laws.
But as I wrote to Patricia Shanahan, finally I realized that the
difference is with the way physicists use Kepler's rule. They do not
use the proportion, which would be very cumbersome, but they convert
the proportion to an equation by writing it with a unit and because
they wrote with a unit named after Newton and because of historical
reasons they believe that they are applying Newton's laws to solve
physical problems.
Newton's axiom of the doctrine of force as the universal cause is
believed by physicists to be the reason why the computational device
called Newtonian mechanics works. I am saying that this is not true
because algorithms are independent of axioms. In the case of orbits I
see that indeed computations are independent of Newtonian force.
> But to
> conclude from that one example that forces are *in general* an unnecessary
> theoretical device is to take a very narrow view of science. Before you
> discard the notion of force, ask yourself if you could calculate the dynamics
> of two colliding galaxies without assuming a theory of gravitation. Kepler's
> laws won't be good enough any more.
I don't know how galaxy collisions work, so you may be right. But I am
not reasoning from an example. I am looking at Newton's definition of
force. I am looking at the origin of force. I see that Newton's
definition of force adds nothing to Kepler's rule because what Newton
adds, the force term, must always be eliminated to recover Kepler's
rule. And there is no other rule that Newton discovered. What drives
"Newtonian mechanics" is not force, it is Kepler's rule written with a
unit so that it fits into the unit system of physics. This is what
makes Kepler's rule written with GM to appear as if it were an
integral part of physics. Is there another origin for force?
> If a simple theory of gravitation
> explains both galaxy collision and orbital motion . . .
The theory of gravitation that you mention is the consistent system of
units that was developed over many centuries by Laplace, Euler, Gauss,
Lagrange, Hamilton and many others. This system is made of many
algorithms developed to use the Keplerian proportion with units within
physics. These algorithms are independent of the doctrine of causes
and forces invented by Newton. I believe that the general principle
that unifies physics is not the Newtonian force but the unit system.
Do you have any comments about this?
> then it makes sense to
> claim that gravity is at work in orbits, even if for some particularly simple
> problems you don't need the general theory.
The gravity term that you say explains the orbit is eliminated. How do
you justify the claim that gravity explains orbits? The mathematical
formulas say that orbit is independent of Newtonian gravity.
Patricia Shanahan
> In computer languages the separation of commentary and code is
> absolute. The computer never sees the comments, they are clearly
I very strongly disagree with this, though I am glad to see something
computer-related in this thread, even if it is more a matter of practice
than theory.
Every scrap of code should be written with two audiences in mind, the
compiler and all programmers who will ever read it. The whole program is
commentary on what the program does and how it does it. That does not
just require meaningful identifiers. The structure of the code should
show as clearly as possible what is really going on. Recently designed
programming languages tend to give a lot of weight to aiding human to
human communication in the code.
Explicit comments should only be needed in a few situations, such as
stating preconditions and postconditions for functions. Too many
comments suggest that either the code is not well designed, or that the
comments are junk comments, merely repeating what the code already says.
There are also programs, such as Javadoc, that read and use special
comments.
[http://java.sun.com/j2se/1.5.0/docs/tooldocs/windows/javadoc.html]
Most of the commentary on my Java programs is in the code, and a high
proportion of the comments in them are intended for Javadoc, not direct
human reading.
Patricia
tc...@lsa.umich.edu
Pioneer1 <1pio...@gmail.com> wrote:
>Maybe I'm misunderstanding you but that force assumed by Newton does not
>even enter the orbital formulas.
It *does* enter the formulas, in Newton's formulation of the situation.
>I don't know how galaxy collisions work, so you may be right. But I am
>not reasoning from an example.
Yes, you *are* reasoning from just an example, namely the simple case of a
two-body problem. Kepler's rules are good enough for such a simple case.
But what happens when you have more than two bodies? What is Kepler's rule
in that case? There is no Kepler law for three bodies. How do you propose
to account for three-body motion with appealing to the concept of force?
>I see that Newton's
>definition of force adds nothing to Kepler's rule because what Newton
>adds, the force term, must always be eliminated to recover Kepler's
>rule. And there is no other rule that Newton discovered.
Not true. I think it would help your understanding greatly if you made a
serious attempt to analyze the motion of three bodies of roughly equal size.
You will then see that Kepler's laws don't work. They work in the solar
system only because the sun is so much more massive than any other body, so
you can approximate the situation by assuming that the sun is fixed.
Newton's laws, however, work just fine.
>The theory of gravitation that you mention is the consistent system of
>units that was developed over many centuries by Laplace, Euler, Gauss,
>Lagrange, Hamilton and many others. This system is made of many
>algorithms developed to use the Keplerian proportion with units within
>physics. These algorithms are independent of the doctrine of causes
>and forces invented by Newton. I believe that the general principle
>that unifies physics is not the Newtonian force but the unit system.
>Do you have any comments about this?
Your comments about "Keplerian proportion with units" are simply false.
Again, try to analyze three bodies.
However, suppose I correct that error for you, and rephrase what you are
saying in a way that is not obviously false. There is a distinction in
the philosophy of science between "realism" and "anti-realism." Roughly
speaking, realism is the doctrine that the objects in our physical theories
correspond to real things in the physical world, and anti-realism is the
doctrine that our physical theories are merely computational algorithms
for predicting observations.
One can, of course, be a realist about some things and an anti-realist about
others. Presumably, you are a realist about concepts like "distance." That
is, the earth really is some distance from the sun in the real world, and
"distance" is not merely a convenient computational concept that we use in
our physical theories, that has no correlate in the real world. On the other
hand, you seem to be arguing for an anti-realist attitude towards Newtonian
forces. That is, according to you, Newtonian forces do not exist in the real
world; they are merely a convenient computational device that we use to
predict our observations.
It can be fruitful, as a thought experiment, to adopt realist and
anti-realist attitudes towards various concepts in physics, to see where that
leads you. Adopting an anti-realist attitude towards Newtonian force might
lead you to think of the principle of equivalence between gravitation and
acceleration and therefore to general relativity. Adopting a realist
attitude towards the wavefunction in quantum mechanics might lead you to the
Everett or "many-worlds" interpretation of quantum mechanics. Adopting an
anti-realist attitude towards motion might lead you to Zeno's paradoxes. So,
I am not discouraging you from exploring these ideas.
However, regardless of your stance on realism versus anti-realism, you still
have to get your calculations right. And to repeat, Kepler won't be good
enough when you go beyond two bodies.
Patricia Shanahan
...
> algorithm in order to make the algorithm look Newtonian. This is the
> question I'm investigating. Does the computational part of Newtonian
> mechanics use force? If we do not write "force" and do not start from
> F=ma, can we still predict astronomical motions as well as Newtonian
> mechanics? If we start directly from Kepler's rule R3/T2 can we obtain
The answer is a resounding "No!".
I've just finished reading an article in the May 2008 issue of
Scientific American on how systems of planets develop. It discusses, for
example, effects of existing planets on clouds of gas and dust,
affecting when and where additional planets develop. I don't think
Kepler's laws would get very far with that problem.
Just looking at our own solar system, Kepler's laws work reasonably well
in the very short term, but repeated or sustained small effects can
accumulate over time, and become significant.
Long term behavior of a complex system is a much harder problem than the
three body problem. However, I agree with Tim Chow's suggestion that you
should begin by looking into the three body problem for similar masses.
Patricia
rpost
>[...] I see
>Newtonian mechanics as a consistent system of units made of physical
>quantities. It is a computational device, or an algorithm, that takes
>given values of physical quantities as its arguments and transforms
>them according to some rule.
This is a big mistake. An algorithm is directed: it transforms input
into output. Newton's laws are undirected: they relate quantities
without saying that one of them results from the other. Certain
specification methods for computer systems work with the same kind
of undirected correspondence, but the approach isn't very common.
--
Reinier Post
Paul E. Black
> Pioneer1 wrote:
>>[...] I see
>>Newtonian mechanics as a consistent system of units made of physical
>>quantities. It is a computational device, or an algorithm, that takes
>>given values of physical quantities as its arguments and transforms
>>them according to some rule.
>
> This is a big mistake. An algorithm is directed: it transforms input
> into output. Newton's laws are undirected: they relate quantities
Expressed another way, Newtonian mechanics is a method for building
mathematical models for physical events. (In fact, much of basic
college Physics is not calculating - the equations are pretty simple -
but learning how to use which equation.)
Suppose I want to know how high my model rocket will go if I shoot it
straight up. I could try launching it and measure the height somehow.
Or I could do some physics ...
The rocket weighs 23 grams.
http://www.estesrockets.com/products.php?number=1225
I'll use a C6-5 engine, which produces an average force of 6 newtons.
I'll simplify by leaving out air friction and using a gravitational
force of 9.8 m/s^2.
The (average) initial acceleration of the rocket by the engine is
6 N / 23 g = 260 m/s^2
subtracting gravity, the net upward acceleration is
260 - 9.8 = 250 m/s^2
The engine fires for 1 2/3 seconds. At burn-out the rocket is going
250 m/s^2 * 1.666 s = 420 m/s at a height of
1/2 * 250 m/s^2 * (1.666... s)^2 = 350 m.
Ignoring air friction, which is actually substantial in this case,
the rocket coasts up for
420 m/s / 9.8 m/s^2 = 43 s
At 43 seconds, the rocket reaches a height of
350 m + 43 s * 420 m/s - 1/2 * (43 s)^2 * 9.8 m/s^2 = 9350 m
(The force of air friction is so great that the engine is only has a 5
second delay. The simplifying assumption of no air friction is far
from realistic. http://my.execpc.com/~culp/rockets/rckt_sim.html has
equations that account for drag.)
Classical mechanics is what leads me to multiple, divide, or add this
and that.
-paul-
Pioneer1
First I want to thank you for the positive encouragement at the end of
your post and for not dismissing this research without seeing
computations with Kepler's rule. I agree that computations must work.
But even though Kepler's rule is a simple proportion (R^3/T^2)
relating radius and period of the orbit, "Newton's equations of
motion" is not that simple. R^3/T^2 includes only two terms and it is
itself a rule. Newton's equations of motion is said to be tied to
"Newton's laws." I need to understand how Newton's laws exist in a
formula used for measurement of orbits.
Below are my replies:
> It *does* enter the formulas, in Newton's formulation of the situation.
Maybe I am not using the word formula in its proper sense. Please
correct if there is another word for what I am trying to say. I
checked "formula" in Wikipedia and an example of a formula would be
computation of volume: V = 4/3 pi R^3. This is the type of formula I
have in mind. If someone decides to derive volume from Newton's laws
and derives it from force, I would still say that volume is
proportional to R^3 and volume is independent of force. In the orbital
case, the final formula is R^3/T^2 and this does not contain a term
for force because we eliminated it to obtain this formula. The orbit
defined by R^3/T^2 does not know that we wrote force terms. So I say
that R^3/T^2 is independent of force. I believe that you would agree
that volume is independent of force. Why is it that R^3/T^2 is
different than V = 4/3 pi R^3? They are both independent of force.
Should I be using another word than "formula?"
> Yes, you *are* reasoning from just an example, namely the simple case of a
> two-body problem.
But perturbations are additive. As Newton did, I can compute
perturbations between three bodies by computing perturbations between
each pair and add. If force does not enter two-body problem what is
the justification that it will enter three-body problem?
As I see it, the observational rule is for orbits, not for bodies. The
orbit is described by R and T, both are properties of the orbit, not
of a body. This did not fit Newton's beliefs so he defined orbits in
terms of mass (R^3/T^2) and forces (R/T^2) and (1/R^2) but because the
rule is for the orbit and not for force and mass both Newtonian terms
cancel.
> Kepler's rules are good enough for such a simple case.
But this is about the absence of force in Newton's equations of
motion. Not about the precision of Kepler's rule. Without force and
mass Newton's equations of motion reduce to Kepler's rule.
> There is no Kepler law for three bodies.
I believe that the rule R^3/T^2 is valid for the entire solar system,
not only for pairs of bodies. Adding computational sugar to R^3/T^2
does not make R^3/T^2 Newtonian. It just makes it easier to compute
with. The rule is still the same. For instance, to apply Kepler's rule
to a three-body problem I look at the perturbations in the orbits of
Saturn and Jupiter due to each other and the Sun. This is Newton's
proposition 13 in Book III. Newton computes R/T^2 = 1/R^2 for Sun-
Saturn, Jupiter-Saturn and Jupiter-Sun and gets 16, 81 and 156,609. I
am sure that you could compute the same with equational form of R^3 =
T^2 by using textbook methods. You will get the same results, except
that Newton's Sun-Saturn distance is too low (9) compared to modern
value (10.4). Newton uses Earth-Venus distance as his unit and that
complicates everything. We use AU and kilometers and seconds. Modern
methods too use Kepler's rule, there is no other rule. I would
appreciate if you can tell me what rule textbook version is using
different than R^3/T^2.
> How do you propose
> to account for three-body motion with appealing to the concept of force?
By applying Kepler's rule the way Newton has done in Principia. Force
terms cancel because we must eliminate them to obtain R^3/T^2. We
don't need the force terms if we are not using them in calculations. I
assume that your "with" was a typo and you meant "without" but to me
it makes perfect sense: "How do you propose to account for orbital
motion by using concept of force?" You cannot because you must
eliminate force terms.
> >And there is no other rule that Newton discovered.
>
> Not true.
If it is not true that Newton did not discover a new rule, what is the
new rule that Newton discovered? Force is not a rule, it is a cause,
and because it is a cause, it does not enter formulas used in
measurements.
> I think it would help your understanding greatly if you made a
> serious attempt to analyze the motion of three bodies of roughly equal size.
> You will then see that Kepler's laws don't work.
I am assuming simple circular orbits, so, all three of "Kepler's laws"
do not apply in this case. When I say "Kepler's rule" I mean the
proportionality R^3 = T^2, or Kepler's third law. This is the rule
Newton used by writing it as R/T^2 = 1/R^2.
> They work in the solar
> system only because the sun is so much more massive than any other body, so
> you can approximate the situation by assuming that the sun is fixed.
The way I understand it, this is not about approximation. When we
eliminate force, force is gone, it is not approximated. Perturbations
in Saturn's orbit were observed in Newton's time and as I showed above
Newton computed the perturbations from Kepler's rule. So it is not
true that Kepler's rule works in the solar system "only because the
sun is so much more massive than any other body." The effect is
observable and it is computed by using Kepler's rule. This makes sense
to me, because to compute orbits we only need to know R and T. All
other names Newton invented for combinations of R and T can be
ignored. Mass is R^3/T^2. You may include it or not.
> Newton's laws, however, work just fine.
"Newton's equations of motion" do not include "Newton's laws."
Calculations involve equations of motions, not laws. And it has been
proven that Newton did not solve the three-body problem. He faked it.
"Newton never managed to get the numbers to come out right, and
finally had to resort to the ancient Greek device of the epicycle to
complete his lunar theory."(From Dana Densmore, Newton's Principia:
The central argument. p. 509) This is my point. To this day three-body
problem is solved by geometric or trigonometric methods and the result
is attributed to Newton's laws. The three-body problem itself is a
problem created by the assumption of force. Without force there is no
three-body problem.
> Your comments about "Keplerian proportion with units" are simply false.
Can you please give more detail on this? Why do you think a = GM/R^2
is a different rule than R/T^2 = 1/R^2? In other words why do you
think if I compute the period T of an orbit from its given radius R, I
would get two different results if I use the equational version and
proportional version? I believe that we would get the same results.
> However, suppose I correct that error for you, and rephrase what you are
> saying in a way that is not obviously false. There is a distinction in
> the philosophy of science between "realism" and "anti-realism." . . .
Thanks for this example, it was thought provoking but I do not agree
that Newtonian forces
> are merely a convenient computational device that we use to
> predict our observations.
I object that forces "are merely a convenient computational device." I
believe that force terms are not used to compute orbits, they are
eliminated. Force is not computational, it is doctrinal. Observing
this fact I say that orbits are independent of force. The orbit sees
only R and T. We cannot have a formula which contains F, T and R and
use it to compute orbits. We can write F and then eliminate it but the
orbit does not know that.
> It can be fruitful, as a thought experiment, to adopt realist and
> anti-realist attitudes towards various concepts in physics, to see where that
> leads you. Adopting an anti-realist attitude towards Newtonian force might
> lead you to think of the principle of equivalence between gravitation and
> acceleration and therefore to general relativity.
As I mentioned above, thanks for this positive encouragement and I
agree that such questioning of physics concepts at least will help
understanding them better. But I am not sure that my attitude is
reality or anti-reality. I don't know what it is called, maybe it is
rationality and an absolute belief in the authority of measurement and
observation and total denial of doctrine. If the formula we use to
measure a quantity does not include a term, I believe that the problem
is independent of that term. I believe this to be true absolutely, not
case by case. So if I use the formula Volume :: R^3 then I believe
that volume is independent of force, no matter what Newton says. If
Density :: 1/T^2 then I believe that density is independent of force,
no matter what Newtonian doctrine says. If R^3 and 1/T^2 are
independent of force then R^3 :: T^2 is independent of force too. Do
you believe in this principle? If you do, what makes you think that
there is force in R^3 :: T^2?
I also like to mention that the lack of force in orbital formulas is
not a thought experiment, it is an observation. That orbital formulas
must have force is doctrine. I go with observations.
> I am not discouraging you from exploring these ideas.
>
> However, regardless of your stance on realism versus anti-realism, you still
> have to get your calculations right.
Yes, thanks again. I agree but to get the calculations right and
compare Kepler's rule with Newton's equations of motion I will have to
state clear definitions of both. I don't believe that
(R/T^2 = 1/R^2) + (units and constants) + (coordinates) + (calculus
notation)
turns (R/T^2 = 1/R^2) magically into a Newtonian rule.
Pioneer1
> The answer is a resounding "No!".
If I compute the period T of a planet from a given R by using Kepler's
rule in its proportional form R_0^3/T_0^2 = R^3/T^2 and by using its
Newtonian form R/T^2 = GM/R^2, can you explain why these would give
different results? The rule is the same rule. Newtonian form has a
standard unit in it, Kepler's rule has an arbitrary unit R_0^3/T_0^2.
I don't see any other difference.
> It discusses, for
> example, effects of existing planets on clouds of gas and dust,
> affecting when and where additional planets develop. I don't think
> Kepler's laws would get very far with that problem.
I haven't read the article and I don't know what kind of model they
base their speculations on. I have a simple claim: Newton's equations
of motion is free of force terms. Then orbits computed with them must
be free of force. Why is this wrong?
> Just looking at our own solar system, Kepler's laws work reasonably well
> in the very short term, but repeated or sustained small effects can
> accumulate over time, and become significant.
As far as I can understand you are saying that Kepler's rule fails
because it does not have a force term, while Newton's equations of
motion explain "repeated or sustained small effects" because they have
a force term in them. But we eliminate force F from formulas. Neither
has force in it. And they don't have to. (R0/T0^2)(T1^2/R1) = (R1/R0)
compares two systems with different densities. Their interaction
results in observed effects. This is not a forceful interaction
because there are no force terms in the formula. It is difficult to
compute with (R0/T0^2)(T1^2/R1) = (R1/R0). So It has been incorporated
into physics by writing it with standard units in a coordinate system.
But adding units and coordinate system does not add a force term. And
I believe that if force is not included then it doesn't have any
effect on orbits.
> you
> should begin by looking into the three body problem for similar masses.
Can you help by writing the three-body problem for similar masses that
has a term F in it and corresponding terms for masses that do not
cancel? As far as I know F is written and then eliminated. How does it
govern orbits after it is eliminated?
Pioneer1
> This is a big mistake. An algorithm is directed:
Thanks for correcting this. Can you suggest an alternative wording?
What about using "computational device?"
> it transforms input
> into output. Newton's laws are undirected: they relate quantities
I don't believe laws have direction or relate quantities. What we use
in computations do not have laws in them.
> without saying that one of them results from the other.
Is the rule R^3 = T^2 directional? Since this is a rule, maybe we
should say that physics is a "measurement system" instead of an
"algorithm" or "computational system." We define a unit to be used
with R^3 = T^2 and use this rule to measure orbits.
But the question under debate is about the input parameters. I claim
that force F is not an input parameter. Only R and T are input and
they are transformed according to the rule R^3 = T^2. So orbits are
independent of F. Do you agree with this?
tc...@lsa.umich.edu
Pioneer1 <1pio...@gmail.com> wrote:
>But perturbations are additive. As Newton did, I can compute
>perturbations between three bodies by computing perturbations between
>each pair and add. If force does not enter two-body problem what is
>the justification that it will enter three-body problem?
The justification for introducing forces into the three-body problem is that
there isn't any other good way to solve the three-body problem.
Three-body motion isn't necessarily periodic. So there aren't any "orbits."
If there are no orbits, how are you going to apply Kepler's laws?
>But this is about the absence of force in Newton's equations of motion.
Newton's equation of motion has force in it. F = ma. F is force.
When you solve certain simple problems, you can eliminate F. That does not
mean that F is not present. This is a simple matter of clear thinking and
clear speaking, not physics. The meaning of the word "eliminate" is to get
rid of something that is there. If something is not there, then you cannot
"eliminate" it, or "cancel" it. It has to be there in the first place
before it makes sense to talk about getting rid of it. Force is there in
the equation. To say otherwise is doubletalk. Don't use doubletalk. Speak
precisely, or you will confuse yourself.
>Not about the precision of Kepler's rule. Without force and
>mass Newton's equations of motion reduce to Kepler's rule.
Not true, except in the extremely special case of two bodies.
Again, you're avoiding doing the hard work of actually solving the
three-body problem, in favor of handwaving. Write down the positions,
velocities, and masses of three approximately equal bodies in space. Now
tell me what the positions and velocities of these bodies are, say, one
year later. Don't tell me in vague terms how you would go about doing this,
applying Kepler's laws---actually do it. Pick specific numbers for the
positions and velocities and masses, and then tell me what the positions
and velocities are one year later. When you have finished your calculation,
post the numbers here. Then we can talk.
Pioneer1
> Expressed another way, Newtonian mechanics is a method for building
> mathematical models for physical events. . . .
> Suppose I want to know how high my model rocket will go if I shoot it
> straight up.
Thanks for this example. I have no problem with using methods that
have been developed over centuries to solve quickly problems such as
this one. We now call this system of methods "Newtonian mechanics." I
am only objecting to the assumption that because this system is called
"Newtonian" it must prove Newton's religious doctrines and
metaphysical laws expressed in his book. The name of the method does
not solve the problem, the method does. And there is nothing in the
method invented or discovered by Newton without which this problem
could not be solved. Galileo could have solved this problem without
Newtonian mechanics.
> I could try launching it and measure the height somehow.
> Or I could do some physics ...
You are saying that instead of trial and error you can use ready-made
methods and units of Newtonian mechanics. I agree with this. But I
don't agree with the hidden assumption that I believe is hidden in the
word "physics." When you say "you could do some physics," you mean
"Newtonian physics," implying that the method works because it uses
Newton's laws and Newton's force. The implication is that this problem
cannot be solved without assuming Newton's occult force. Is this
correct? You are not simply using a mathematical formula, but you are
implying to use a method that works because it is based on the
Newtonian doctrine. Without the doctrine this method will not work and
the problem could not be solved. This is the assumption I am objecting
to.
> The rocket weighs 23 grams.
Rocket's weight is irrelevant to this problem. What we are weighing,
or comparing, is g_earth and g_rocket. Newtonian mechanics mentions
weight because it is necessary to convert units.
> I'll use a C6-5 engine, which produces an average force of 6 newtons.
Here all you are saying is that the engine produces some multiple of
g_earth. There is nothing discovered by Newton here. Newton did not
know about the unit Newton. It is false to imply that Newton's laws
govern projectile motion just because Newtonian physicists labeled R_e/
T_e^2 force and named its unit Newton.
The relevant quantity is g_rocket/g_earth and that is not force.
> I'll simplify by leaving out air friction and using a gravitational
> force of 9.8 m/s^2.
Calling a specific value of R/T^2, namely, g_e = R_e/T_e^2,
"gravitational force" only serves to assert Newton's authority.
Everything is called force so that unit system will work. Otherwise,
where is force in g_e = R_e/T_e^2? And you will eliminate the label
force and its unit Newton in the next step to get the relevant
quantity which is independent of force and weight.
> The (average) initial acceleration of the rocket by the engine is
> 6 N / 23 g = 260 m/s^2
> subtracting gravity, the net upward acceleration is
> 260 - 9.8 = 250 m/s^2
Force is now eliminated because it is not needed to solve this
problem. What is at work here is the consistent system of units. Not
force. Not Newton's laws.
In this case we need to know g_r/g_e and this is what you have found
expressed with units as 250 m/s^2. But you had to write Newton's name
several times before this quantity appeared.
> The engine fires for 1 2/3 seconds. At burn-out the rocket is going
> 250 m/s^2 * 1.666 s = 420 m/s at a height of
> 1/2 * 250 m/s^2 * (1.666... s)^2 = 350 m.
This is Galileo's formula. Why is this method called "Newtonian
mechanics?" There is nothing Newtonian here. Force doesn't exist in
formulas used to compute the height. Force enters only as a unit
conversion factor. This is the way physicists relate g_e and g_r
because they don't work with ratios but with units.
> Classical mechanics is what leads me to multiple, divide, or add this
> and that.
I agree. And as this example shows, the operations are made possible
by the unit system, not by the names of the units. Force is stated in
the beginning of the problem as if it were a given but it was later
eliminated. So it was not a given, it was a unit conversion factor
reified and turned into doctrine.
Maybe your initial definition of Newtonian mechanics needs to be
amended: Newtonian mechanics does not let us build models, because we
did not build a model in this case, we applied the Newtonian model
defined with Newtonian units and physical quantities to solve a
problem involving projectiles. Then we concluded that because we used
units named after Newton the projectile motion must be Newtonian and
the entire Newtonian doctrine must be true. Newtonian mechanics lets
us use its standard units and methods as long as we agree to uphold
Newton's authority and believe in Newtonian doctrines.
Just to repeat, I am not questioning the convenience of the method you
are using. I am just questioning the implication that the problem is
solved by Newton because force is called Newton. Force is assumed but
does not enter this problem. Force assures that physical quantities
have consistent units so that they can be manipulated. Thanks again
for this example.
Pioneer1
I've been trying to understand data types in physics.
http://www.densytics.com/wiki/index.php?title=Data_types_in_Physics
I see that there are similarities with data types used in computer
science. I am confused about what is number and what is string in
physics. When I look at F=ma I see F as a string. F is a label for ma.
But physicists, like this one, http://globalpioneering.com/wp02/science-is-legal-physics/#comment-16779
see it differently and say it is a number or a quantity.
And indeed F stands for a number, ma, but to me F is just a
placeholder, not a quantity.
I know I'm not using the correct terminology. Can anyone help me
understand this? At least, help me state the problem clearly?
Many thanks.
Richard Harter
Oh dear, oh dear. You appear to be confusing the idiosyncratic
notation of imperative programming languages without ever having
passing through mathematics and physics.
The = sign in many languages, e.g. C and Fortran, means
assignment. In physics and mathematics generally it means
identity. Secondly, F=ma is not a statement in the sense of a
statement in a programming language; rather it is a formula,
i.e., a governing relationship between the three quantities, F,
m, and a. Thirdly m and a are separate quantities. Fourthly, it
is normal in maths and physics to use single letter variables
(using a variety of alphabets, subscripting, and superscripting
as needed) and to indicate multiplication by concatenation.
Fifthly, more subtly, variables stand for something specific; the
association is intrinsic (albeit by definition) rather than
happenstance (as is so often the case in computer programs).
To recapitulate: F = ma is a formula that expresses the
relationship between force, F, mass, m, and acceleration, a. It
says force is equal to the product of mass times acceleration.
I hope this helps.
Richard Harter, c...@tiac.net
http://home.tiac.net/~cri, http://www.varinoma.com
Save the Earth now!!
It's the only planet with chocolate.
Pioneer1
Thanks for the help.
>
> The = sign in many languages, e.g. C and Fortran, means
> assignment. In physics and mathematics generally it means
> identity.
I think in physics the = sign is context sensitive and has many
meanings. I think this is part of my problem.
http://www.densytics.com/wiki/index.php?title=Equality_sign
>Secondly, F=ma is not a statement in the sense of a
> statement in a programming language; rather it is a formula,
Ok. I understand this. I was hoping to use concepts of computer
language and math to decipher physicists' F=ma
> i.e., a governing relationship between the three quantities, F,
> m, and a.
I think here the assumption is that F is a quantity like m and a. But
I am questioning this assumption. How do I test mathematically that F
is not just a label or a placeholder or a pointer to ma?
Let's say I want to write a program to compute F, it should be
something like this:
(def force (m a) (* m a))
I assume in this case force is merely the name of the function (* m
a). There is not something distinct from (* m a) called force. Is this
correct?
> Fifthly, more subtly, variables stand for something specific; the
> association is intrinsic (albeit by definition) rather than
> happenstance (as is so often the case in computer programs).
So in this case, does F stand for anything other than ma?
Thanks again.
Chris Smith
>> i.e., a governing relationship between the three quantities, F, m, and
>> a.
>
> I think here the assumption is that F is a quantity like m and a. But I
> am questioning this assumption. How do I test mathematically that F is
> not just a label or a placeholder or a pointer to ma?
This isn't making any sense. Can you express more clearly what you are
looking for? I am not sure of this, as I don't understand what you're
saying yet; but I strongly suspect you are asking meaningless questions.
In general, when you've got several related concepts, it's a rather
arbitrary choice what you choose to call a definition, and what you
choose to call a fact.
> (def force (m a) (* m a))
>
> I assume in this case force is merely the name of the function (* m a).
> There is not something distinct from (* m a) called force. Is this
> correct?
What do you mean? As a matter of programming language semantics, you
want to know whether 'force' (by the quotes I mean I'm talking about your
function, not the concept from physics) has some identity that would let
you distinguish it from a different function that returns the same value
for the same inputs? That depends on your programming language: usually
no, but occasionally yes. If you intend to infer something about
physical quantities from such accidents of language semantics, though,
I'll join in warning you against it.
--
Chris Smith
Barb Knox
<871757b5-ab88-4134...@1g2000prg.googlegroups.com>,
Pioneer1 <1pio...@gmail.com> wrote:
> On Apr 5, 11:00 am, c...@tiac.net (Richard Harter) wrote:
>
> Thanks for the help.
> >
> > The = sign in many languages, e.g. C and Fortran, means
> > assignment. In physics and mathematics generally it means
> > identity.
>
> I think in physics the = sign is context sensitive and has many
> meanings. I think this is part of my problem.
>
> <http://www.densytics.com/wiki/index.php?title=Equality sign>
Your problem here is treating that page is authoritative. Pretend you
never saw it.
For your purposes, equality is unambiguous: if alpha and beta are
mathematical expressions, then alpha=beta means that any alpha(s) may be
replaced by beta in any expression, and vice versa, without changing the
value of the expression. Basically, alpha and beta are expressions that
represent the same value. E.g.:
2 + 2 = 3 + 1
F = m*a
Note that unlike programming, alpha=beta means *exactly* the same thing
as beta=alpha.
3 + 1 = 2 + 2
m*a = F
> >Secondly, F=ma is not a statement in the sense of a
> > statement in a programming language; rather it is a formula,
>
> Ok. I understand this. I was hoping to use concepts of computer
> language and math to decipher physicists' F=ma
Do NOT attempt to use programming conventions to understand physics. As
Richard wrote:
[unsnip]
> > Oh dear, oh dear. You appear to be confusing the idiosyncratic
> > notation of imperative programming languages without ever having
> > passing through mathematics and physics.
[/unsnip]
> > i.e., a governing relationship between the three quantities, F,
> > m, and a.
>
> I think here the assumption is that F is a quantity like m and a. But
> I am questioning this assumption. How do I test mathematically that F
> is not just a label or a placeholder or a pointer to ma?
The letter 'F' is a variable that denotes some value. (This is a
*mathematical* variable, NOT a programming one. For example, you can
not change its value.) Similarly, 'm' and 'a' denote values. F=m*a
constrains those 3 values to have the relationship that F's value is the
same as m's value times a's value.
NOTHING in physics is analogous to a C pointer.
> Let's say I want to write a program to compute F, it should be
> something like this:
>
> (def force (m a) (* m a))
>
> I assume in this case force is merely the name of the function (* m
> a). There is not something distinct from (* m a) called force. Is this
> correct?
>
> > Fifthly, more subtly, variables stand for something specific; the
> > association is intrinsic (albeit by definition) rather than
> > happenstance (as is so often the case in computer programs).
>
> So in this case, does F stand for anything other than ma?
None of 'F', 'm', or 'a' *stand for* any combination of the others.
They do however *relate* in a particular way in F=m*a. Note that when m
is non-zero, F=m*a is EXACTLY equivalent to
F/m = a
Say to yourself several times: "Programming is worse than useless for
understanding first-year physics". Repeat this until you believe it.
> Thanks again.
--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum viditur.
| BBB aa a r bbb |
-----------------------------
Patricia Shanahan
...
> Say to yourself several times: "Programming is worse than useless for
> understanding first-year physics". Repeat this until you believe it.
When I started learning to program, in 1967, my teacher had to explain
to me that my prior understanding of equality, from mathematics and
scientific formulas, was worse than useless for understanding Fortran
assignment statements.
Patricia
Pioneer1
> This isn't making any sense. Can you express more clearly what you are
> looking for?
Yes, I don't understand the concepts to ask it clearly. Sorry about
that. Maybe I can explain geometrically?
If AB is a line, I think that AB = F = ma. In this case, there is just
one quantity, ma, and AB and F are two names for it.
But according to physics convention, there are two lines, say, CD = F
and GH = ma. So, F is not merely a label but a line itself.
Also, if I want to create this same formula in a spreadsheet I would
have three columns with names, mass, acceleration and force. The
contents of the columns will be the value of mass, m, the value of
acceleration, a, and the third column will be m*a.
So the content of the cells, where you can only put quantities, do not
contain a term for force. Then force is not a quantity but just a
label for m*a. If I put "force" in the third column I would get an
error message where a number would be expected.
Does this makes sense? Reading other responses I see that both
mathematics and programming may be useless in trying to decide such a
question. Thanks again for the help.
Barb Knox
<6684be35-f6c3-4601...@s39g2000prd.googlegroups.com>,
Pioneer1 <1pio...@gmail.com> wrote:
> On Apr 5, 5:52 pm, Chris Smith <cdsm...@twu.net> wrote:
>
> > This isn't making any sense. Can you express more clearly what you are
> > looking for?
>
> Yes, I don't understand the concepts to ask it clearly. Sorry about
> that. Maybe I can explain geometrically?
>
> If AB is a line, I think that AB = F = ma. In this case, there is just
> one quantity, ma, and AB and F are two names for it.
>
> But according to physics convention, there are two lines, say, CD = F
> and GH = ma. So, F is not merely a label but a line itself.
>
> Also, if I want to create this same formula in a spreadsheet I would
> have three columns with names, mass, acceleration and force. The
> contents of the columns will be the value of mass, m, the value of
> acceleration, a, and the third column will be m*a.
>
> So the content of the cells, where you can only put quantities, do not
> contain a term for force. Then force is not a quantity but just a
> label for m*a. If I put "force" in the third column I would get an
> error message where a number would be expected.
>
> Does this makes sense?
Not really. Given *any* 2 of 'F', 'm', and 'a', you can calculate the
third one. So you could have a spreadsheet with 3 columns, with 'F' and
'm' entered as values, and 'a' calculated as F/m.
And F=m*a is still meaningful even if you don't know any of the specific
values. It describes a *relationship* between 3 items, which holds even
when you don't know what their specific values are.
To avoid your apparent fixation that some variable "stands for" or
"points to" the rest of the formula, you would do better by pondering
formulae which have composite terms on both sides of the '='. E.g.:
m * delta_v = F * delta_t (equivalent to m*a = F in the limit)
P*V = n*R*T (the ideal gas law)
You can of course use algebra to re-arrange these in many ways, all of
which are equivalent (since that's what algebra does).
> Reading other responses I see that both
> mathematics and programming may be useless in trying to decide such a
> question.
No. Programming is indeed worse than useless, since its notions of
assignment, pointers, and strings are irrelevant to physics. But
mathematics is exactly what you need. Why do you think that mathematics
in general (e.g. algebra) is useless for understanding physics formulae,
just because your one attempt to use geometry didn't make sense?
> Thanks again for the help.
I'd also be interested in your response to my previous post. If you
found my language too harsh then please don't take it personally -- I'm
just trying to penetrate what seem to be some strongly-held
misconceptions on your part.
Barb Knox
Patricia Shanahan <pa...@acm.org> wrote:
That was certainly the case in 1967 (which, coincidentally, is about
when I first learned programming, also in FORTRAN), but these days
things such as Hoare's axiomatisation make it feasible to relate the
semantics of programs to a student's existing background in mathematics.
> Patricia
John Forkosh
I tried answering your question posted in sci.physics.research
but the moderators haven't posted that yet.
I'm not sure the following will satisfy your spreadsheet
model (didn't see that in s.p.r), but I think you may want to
look at dimensional analysis. That is, what you're calling data
types is what physics calls units.
Consider your F=ma example, and interpret the rhs, ma,
as a lambda abstraction in a typed lambda calculus. Then
it's "input" would be a variable with type mass and a variable
with type acceleration (or would be lambda expressions that
reduce to these types). And the lambda abstraction itself
would have type force, or more accurately, it would have type
mass cross acceleration --> force.
--
John Forkosh ( mailto: j...@f.com where j=john and f=forkosh )
John Forkosh
> Patricia Shanahan <pa...@acm.org> wrote:
>> When I started learning to program, in 1967, my teacher had to explain
>> to me that my prior understanding of equality, from mathematics and
>> scientific formulas, was worse than useless for understanding Fortran
>> assignment statements.
>
> That was certainly the case in 1967 (which, coincidentally, is about
> when I first learned programming, also in FORTRAN),
Ditto. Same year, same language. Some coincidence.
NYU anybody???
> but these days
> things such as Hoare's axiomatisation make it feasible to relate the
> semantics of programs to a student's existing background in mathematics.
--
Pioneer1
> Not really. Given *any* 2 of 'F', 'm', and 'a', you can calculate the
> third one. So you could have a spreadsheet with 3 columns, with 'F' and
> 'm' entered as values, and 'a' calculated as F/m.
Yes, you are right. But this reminds me that F=ma is a circular double
definition. I know that Ernst Mach was concerned about this as
referenced here:
http://www.physicsforums.com/showthread.php?t=201903
But I think that this may be relevant because unless F is a label a
double definition would not be possible. (I don't know why, only
intituitively.) Both F and m cancel, again strongly suggesting that
they are labels.
> And F=m*a is still meaningful even if you don't know any of the specific
> values. It describes a *relationship* between 3 items, which holds even
> when you don't know what their specific values are.
As I see it, there are three mathematical objects, F, m and a. But
there is only one mathematical quantity, ma. F is the name of ma. The
relationship assigned by "=" is not a proportionality between two
quantities but simply a naming relationship. It means that anywhere I
see ma I can replace it with F. The fact that I can replace ma with F
tells me that this is naming and not proportionality. For instance in
the true proportionality RRR=TT, I cannot replace R with T and keep
the meaning of the proportionality intact. Does this make sense?
> To avoid your apparent fixation that some variable "stands for" or
> "points to" the rest of the formula, you would do better by pondering
> formulae which have composite terms on both sides of the '='. E.g.:
> m * delta_v = F * delta_t (equivalent to m*a = F in the limit)
To me F in m * delta_v = F * delta_t looks like a mismatch of types
like 2*2 = Four. Four is a placeholder for the quantity 4.
> P*V = n*R*T (the ideal gas law)
I don't seem to have a problem with the ideal gas law. As I understand
it, it says that for a given volume, pressure P is proportional to
temperature T. I believe that both pressure and temperature are
relevant and do not cancel the way F and m cancel.
But if you wrote the ideal gas law as PVK = nRT, where K is the
material the vessel is made of, then I would question the relevance of
K, because, like F, it is irrelevant to the proportionality and it
must be eliminated.
>Why do you think that mathematics
> in general (e.g. algebra) is useless for understanding physics formulae,
> just because your one attempt to use geometry didn't make sense?
I don't understand why algebra is relevant because, for instance, in
your ideal gas example, I can put the extra term K and algebra will
still work perfectly well. That's why I think mathematics is not
useful to decide if F is a label or a quantity. Mathematics does not
seem to care.
And also, the lack of a precisely defined equality sign in physics the
way assignment, pointers and strings are precisely defined in computer
science tells me a lot about physics. The science pretending to be the
most precise science is unable to distinguish between a label (a
string) and a quantity (a number). Or alternatively, that may be my
own inability to understand what's going on here. Thanks again this
has been very useful.
Pioneer1
wrote:
> Pioneer1 <1pione...@gmail.com> wrote:
>I think you may want to
> look at dimensional analysis. That is, what you're calling data
> types is what physics calls units.
I answered this in sci.physics.research as well but to repeat, in your
example Force is already defined as number. I am trying to understand
if it is a number or a string. I think there is no ambiguity in
programming but there is in physics.
Pioneer1
> I'd also be interested in your response to my previous post.
Okay, I read your earlier post again. My only question is about the
ambiguity of the equality sign in physics. One of the meaning of the
equality sign is that it says, "any time you see ma you can replace it
with F." But, as I understand it, it also defines a proportionality.
tc...@lsa.umich.edu
Pioneer1 <1pio...@gmail.com> wrote:
>Okay, I read your earlier post again. My only question is about the
>ambiguity of the equality sign in physics. One of the meaning of the
>equality sign is that it says, "any time you see ma you can replace it
>with F." But, as I understand it, it also defines a proportionality.
Let me try to restate your question using somewhat more standard terminology.
Your question is whether the symbol "F" is merely an *abbreviation* for the
longer expression "m*a".
The answer is no. There isn't any way to "test this mathematically"; you
just have to learn what notational convention people are using.
The fact that, during calculations, you can substitute "F" for "m*a" or
vice versa does not in itself imply that "F" is an abbreviation for "m*a".
It may help to think of the situation this way. Acceleration and force are
separate physical concepts. Acceleration is the rate of change of velocity.
Force is a more slippery concept but intuitively we all have a sense of what
the force of gravity or a magnetic force is. From the standpoint of pure
logic, there is no necessary connection between force and acceleration. The
fact that, in the actual universe that we live in, force is proportional to
acceleration, is somewhat surprising. Logically, things could have turned
out otherwise; it's only empirically, by performing experiments, that we
know that force is proportional to acceleration. This empirical observation
is known as Newton's second law of motion: F = m*a.
The assertion that F = m*a is therefore telling us something about the
universe that we happen to live in. If "F" were just an abbreviation for
"m*a" then that would be just a purely logical fact and would not tell us
anything about what physical laws happen to hold in our universe.
John Forkosh
> On Apr 8, 12:17?pm, John Forkosh> wrote:
>> Pioneer1 <1pione...@gmail.com> wrote:
>> I think you may want to
>> look at dimensional analysis. That is, what you're calling data
>> types is what physics calls units.
>
> I answered this in sci.physics.research as well but to repeat, in your
> example Force is already defined as number. I am trying to understand
> if it is a number or a string. I think there is no ambiguity in
> programming but there is in physics.
Sure, there's ambiguity in number vs. string, but I think that's
because these kinds of things are just representations. And physics
is usually formulated as representation-independent (i.e., as
generally) as possible. Now, exactly what numbers or strings might
be representations of depends on how you choose to interpret them.
Ultimately, in a computer's memory, everything's represented
as (finite-length) bit strings. So that's really the only data
type there is. You can then map bit strings into (finite) scalars,
integers or reals (or complex), or vectors/tensors/etc.
More foundationally, there are no directly measurable numbers,
per se, in nature (though there are dimensionless ratios).
Numbers, at least vis-a-vis physics, are a completely human
construction (despite Gauss' claim that "God invented the integers").
We find numbers very useful to describe measurements, but that's
because we first found numbers useful to calibrate our measuring
instruments in the first place. And then we find numbers useful
to formulate theories correlating our already-numerical measurements.
Now, I'm not sure how else all that might be done without numbers,
but maybe that's just how human brains work. In that sense, we're all
trapped in Whorf's "conceptual boat" (lots of google hits explaining
that).
By looking for numeric-like (including numbers, strings, addresses,
whatever) data types that are physically fundamental, I think you're
mistakenly elevating numeric-like entities to a theoretical status
they don't really occupy. For a recent (last month) discussion along
these lines, try http://arxiv.org/abs/0803.0417/
Jamie Andrews; real address @ bottom of message
> Numbers, at least vis-a-vis physics, are a completely human
> construction (despite Gauss' claim that "God invented the integers").
I think you're referring to Leopold Kronecker's quip
"God invented the natural numbers; all the rest is the work of man".
--Jamie. (efil4dreN)
andrews .uwo } Merge these two lines to obtain my e-mail address.
@csd .ca } (Unsolicited "bulk" e-mail costs everyone.)
John Forkosh
> John Forkosh <jo...@please.see.sig.for.email.com> wrote:
>> Numbers, at least vis-a-vis physics, are a completely human
>> construction (despite Gauss' claim that "God invented the integers").
>
> I think you're referring to
> Leopold Kronecker's quip "God invented the
> natural numbers; all the rest is the work of man".
You think, therefore you're right.
My bad. Thanks for the correction.
Pioneer1
wrote:
> Sure, there's ambiguity in number vs. string, but I think that's
> because these kinds of things are just representations. And physics
> is usually formulated as representation-independent (i.e., as
> generally) as possible.
If I understand you correctly, if physics is "representation-
independent" in this sense, then in physics symbols are loaded and
they can represent either a string or a number and a physicist will
know how to interpret the symbol correctly depending on the context.
Is this correct?
> Ultimately, in a computer's memory, everything's represented
> as (finite-length) bit strings. So that's really the only data
> type there is.
I find this relevant, thanks for the explanation. So, 1s and 0s are
strings not numbers? Or they are bits, neither number nor strings?
>
> More foundationally, there are no directly measurable numbers,
> per se, in nature (though there are dimensionless ratios).
I agree with this. There are only ratios, i.e. comparison of lengths.
For a recent (last month) discussion along
> these lines, tryhttp://arxiv.org/abs/0803.0417/
Thanks for this reference. Just looking at the introduction I see that
they ask the question What is a thing, then assume the question when
they assume "physical quantity." They also say classical physics uses
topos of sets. I think that classical physics uses tables. All
equations of physics can be reduced to tabular data or database.
Interesting paper, thanks again.
Pioneer1
> In article <c3d12d4f-028d-4568-87eb-c7254c6f2...@v26g2000prm.googlegroups.com>,
> Your question is whether the symbol "F" is merely an *abbreviation* for the
> longer expression "m*a".
Thank you! Abbreviation was the word that was eluding me!
>it's only empirically, by performing experiments, that we
> know that force is proportional to acceleration. This empirical observation
> is known as Newton's second law of motion: F = m*a.
>If "F" were just an abbreviation for
> "m*a" then that would be just a purely logical fact and would not tell us
> anything about what physical laws happen to hold in our universe.
Why logical? I would say that an abbreviation is notation, it is a
time saving device. If F is an abbreviation it would tell a lot about
what physical laws happen to hold in our universe. It would tell us
that those laws do not involve force.
But, I now think I understand the problem. F=ma is actually two
expressions:
F_occult = m_occult a
F_contact = m_weight a
F_contact is the force used in the science of mechanics and it is the
force of weights and pulleys. Newton overlayed his occult force
F_occult on this contact force and utterly confused the issue. I am
talking about F_occult. This is the force used in Newtonian
theoretical astronomical derivations but F_occult always cancels out
of the formulas used to calculate orbits. So F_contact may be an
empirical force but F_occult is not. It has never be observed. It
always cancels out of the orbital formulas. That's why I am
questioning the assumption that force is a "physical quantity."
If so, do you think it is relevant that if F can be replaced for ma
it's an abbreviation? Thanks for the help.