Meaning of Multiplication and Division in Physics

In summary: So, for example, if you're studying the behavior of a mass in a gravitational field, then you would use the units of kilograms (or meters if you're measuring in space) to do your calculations.
  • #36
sophiecentaur said:
Isn't that more 'nuts and bolts' than the OP question needs, though? Whatever Maths you use is just a model of a particular accuracy and the general comments will apply.

Well yes. It's a personal thing. When considering general questions, I tend to fall back on the most fundamental way I think of something. I guess I am a bit too influenced by Landau - Mechanics I always recommend.

The interesting part is the second link, where mass depends on the position, which used Lagrangian formalism.

Thanks
Bill
 
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  • #37
One case where multiplication comes up is attenuation. If you have a filter that passes half the light I that enters it, then the light that emerges is 0.5 * I. If you added a second filter, then the light coming out would be 0.5 * 0.5 * I. Attenuation is one example where multiplication models the effect. Units don't really come up because the attenuation coefficient is unitless.
 
  • #38
mremadahmed said:
Mathematics is all around sciences.It is also considered a language.In language everything should have meaning.

I know addition and subtraction meaning.Same things[quantities ,etc.] can add or subtract . It cannot be done on different things.

What is the meaning of multiplication and division in sciences[Physics,.. etc].I know what it means in real life.Multiplication is higher form of addition.Division is higher form of subtraction.

What does F=ma means. m is in kg. a in ms-2.How can both multiply.They cannot add or subtract.

Why any quantities can multiply or divide,contrary to adding and subtracting.

Can anyone answer this question,or provide a book to read about this subject.
F= ma is Newton's most famous equation but he never used it! Instead he used m=F/a to define mass. F/a is a ratio (fraction) so you have to use multiplication or division. Suppose you buy a pie and cut it into 8 slices. Being a pig you ate 3 of the pieces then decided to be a hog and have another piece. Since the pie is in units of 8, you can't simply add or subtract. Suppose you decide to be a hog and have another piece. Now you ate 4/8th of the pie which you can simplify by dividing by 2. I hope this was your question and that I answered it.
 
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  • #39
I don't know whether this has been brought up already.
In physics many quantities are extensive, which means they are proportional to some amount.
For example mass and energy. If you add two amounts of matter with a certain mass each, the masses add (ignoring relativistic effects).
Adding and subtracting makes sense for extensive quantities.

Then there are intensive quantities such as temperature. They are not proportional to the amount of substance. You don't usually add these quantities unless you calculate an average.
Subtracting them only makes sense for example as a temperature difference. You're not actually removing one temperature from another.

I'm speculating quite a bit here but anyway:
Many multiplications in physics involve an extensive and an intensive quantity so that the result is extensive.
Divisions of extensive by extensive quantities give intensive quantities, such as some kind of density.
Intensive quantities can be freely divided and multiplied in many ways and the results are always intensive.

A multiplication of two extensive quantities doesn't make sense to me. The result would be proportional to the square of an amount.

I think these considerations limit which algebraic operations make sense when applied to physics.

Again: This is a lot of speculation.
 
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  • #40
The intellectual abstraction vexed me until the light bulb came on in my head: IT IS ALL ABOUT ENERGY AND REDISTRIBUTION THEREOF and nothing more. Mass is energy. Acceleration is energy... EVERYTHING is about energy and states thereof, which necessitates vector analysis.

The confusion comes from the way that physics is taught, (mostly because schools are largely interested in churning out "working scientists and engineers" rather than in producing great thinkers) in that we are introduced to "practical" problems with Newtonian "solutions" (yes, I used quotes, and real physicists know why) first. If we were to teach (1) ENERGY first, (2) then explain that every quantity we deal with is a form thereof, (3) constantly reiterate how every problem is about energy (4) then teach how SCALE dictates the level of necessary precision for practical results (i.e. you don't really need general relativity to figure out how much energy it takes to kick a football 30 meters to clear a goal post that is X meters high... but you could, if you were effing masochistic, and that you can't "do lasers" using Newton), the cognitive dissonance would be greatly diminished.

UNIT analysis is what confused our original inquirer mremadahmed . All units distill to energy, because that is all that there is, and we are merely adding and subtracting.

All of this is best contemplated over a pint of Guinness or a nice scotch, but not a pint of scotch, which is counter productive in all but the rarest of circumstances.
 
  • #41
Robert Stenton said:
F= ma is Newton's most famous equation but he never used it! Instead he used m=F/a to define mass.

It's a definition and has no physical content. It assumes something called mass that also appears in Newton's law of gravitation. People can use gravitation to give mass a value via, say, a mass balance. The status of F=ma as a law is as a paradigm that says - get thee to the forces in analysing mechanics. That makes it a bit subtle, which is why I prefer the Lagrangian formulation. However, you would have to have rocks in your head to start with that in studying mechanics. Physics sometimes is like that. You start with a basic concept that later has to be scrapped and replaced with something more advanced. Even math is sometimes like that. In calculus, you begin with an intuitive idea of the limit, then replace it with the standard epsilon definition later.

Thanks
Bill
 
  • #42
Physhead said:
The intellectual abstraction vexed me until the light bulb came on in my head: IT IS ALL ABOUT ENERGY AND REDISTRIBUTION THEREOF and nothing more. Mass is energy. Acceleration is energy... EVERYTHING is about energy and states thereof, which necessitates vector analysis.

There is a lot of confusion here. My suggestion is to study Noether's Theorem:
https://math.ucr.edu/home/baez/noether.html

Thanks
Bill
 
  • #43
bhobba said:
There is a lot of confusion here. My suggestion is to study Noether's Theorem:
https://math.ucr.edu/home/baez/noether.html

Thanks
Bill
I am terribly familiar with Em, and YES! Everyone needs to know and love her work!
I am confusioned about what you think there is a lot of confusion about. Please help deconfuse me as to my confusion over your allegation of confusion...or was this a reference to my statement about the original poster's confusion. I am so sorry. English is way difficulter than it looks.
 
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  • #44
I will address some considerations to the OP original request, because I’ve the sensation that no one has ever clarified him that Mathematics is NOT Physics. Math is only an instrument , a way to put physical ideas and experimental findings in formulae, in order to give us the faculty to manage them, do calculations and obtain results, that then are to be compared with the experimental results.

The second law of Newtonian dynamics : F=kma ( where k is a constant, that can be assumed equal to 1 by convention, with an opportune choice of units of measure: see later) is nothing else that the way to put mathematically the following experimental findings :

1) I suppose the OP knows the first law of Mechanics : a body (imagine a small rigid body, don’t want to be extremely detailed here...) remains "at rest” or in rectilinear uniform motion, if is not acted upon by any external action that changes this state ( a lot of contours should be better précised here, but I’ll leave them).

2) Let me also suppose that the OP knows what the acceleration is : change of speed with time. But to define it quantitatively I need a unit of measure; I decide that meters and seconds are good for measuring space and time, and assume that speed is measured in m/s : every quantity in physics has a physical significance only when measured , with conventional units. If you don’t measure, no one will appreciate you. Any physical quantity is made of a number and a unit of measure , the number alone doesn’t mean anything.
Therefore, acceleration being defined as the change of speed wrt time , I’ll find that, for dimensional reasons of coherence, the unit of measure of acceleration must be : m/s2.

2) So far so good. Now take a ball , kick it. Repeat this several times, with different “intensity” of the kick. You will notice that, the greater the intensity, the greater the acceleration taken by the ball. Suppose to repeat this very many times, and to also have invented a system for determining with precision the “intensity” of you kick: the experimental result is that , when you double the intensity, the acceleration doubles...and so on. There is a direct proportionality between the intensity of the kick and the acceleration assumed by the ball.
Now , it’s a matter of definitions and conventions.
The previous ideas suggest that there is, for the body ( better: material point; for systems, a lot of more precise concepts should be given) , a defined quantity, given by the intensity of the kick and the acceleration taken by the body. This quantity is called “inertial mass” of the body; the intensity of your kick is called “force” . So, To express the proportionality between force and acceleration, found experimentally, we simply write :

$$F = ma$$

mass “m” is given a proper unit, for example “kg” . Therefore “ma” is expressed in kg*m/s2. The LHS is given the name of force, and the unit takes the name of Newton N.

That is, it is needed the force of a N to give the mass of 1 kg the acceleration of 1 m/s2 .

This is the end of story , and the beginning of mechanics. As you can see, all quantities have units, conventionally chosen of course. All quantities can be someway measured, all observers must agree on the results.

I Hope I have been able to give you at least an idea on how things work, when speaking of physical relations. F=ma doesn’t mean that you take m factors equal to a ! . They are not simple numbers, are physical entities.
 
  • #45
Physhead said:
IPlease help deconfuse me as to my confusion over your allegation of confusion...or was this a reference to my statement about the original poster's confusion. I am so sorry. English is way difficulter than it looks.

Ah - English is not your primary language. That likely explains my 'confusion' over what you wrote. Acceleration, for example, is not energy. That was the sort of thing that was leading me to think you needed to investigate what things like energy are further. Since you know them already I will be more careful in reading your responses in future.

BTW welcome to Physics Forums. I see you just joined yesterday. It is always a joy to have new members.

Thanks
Bill
 
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  • #46
italicus said:
Suppose to repeat this very many times, and to also have invented a system for determining with precision the “intensity” of you kick

That's the bit I do not quite understand. What would be an example of a system of determining the intensity of the kick, other than of course defining it as ma? Would you use for example stretched springs? But that relies on Hooks law which requires for its statement the concept of force. You could take it as a given. But that would mean Hooks law is true by fiat - and we know it is true for only small displacements.

My personal view is that Feynman had the right approach:
https://www.feynmanlectures.caltech.edu/I_12.html

It is more than just a definition because it allows a precise statement of the third law and other laws like gravitation. Basically, that is the true content of F=ma. It is, as I said, the correct paradigm to analyse mechanical phenomena from experience. It is a law, but not of the usual type such as the third law which is a testable statement about nature. Rather it is saying when analysing problems in classical mechanics look at the forces. That is a statement about nature.

The interesting thing of course, and Feynman initially resisted this, is that the Lagrangian formulation is more like the usual testable statements about nature we call laws and often makes more difficult problems easier to solve. Evidently, Feynman would insist on using forces when the Lagrangian approach would be easier. The real twist of course is when he developed his path integral approach to QM the Lagrangian formulation popped out immediately. Just one of the many curious things in the life of that curious character. :DD:DD:DD:DD:DD

Thanks
Bill
 
  • #47
@bhobba

Feynman is a master scientist, but in the linked chapter he repeats several times that if one defines force as the mass times the acceleration, he has found out nothing. This is an abstract:

If we have discovered a fundamental law, which asserts that the force is equal to the mass times the acceleration, and then define the force to be the mass times the acceleration, we have found out nothing. We could also define force to mean that a moving object with no force acting on it continues to move with constant velocity in a straight line. If we then observe an object not moving in a straight line with a constant velocity, we might say that there is a force on it. Now such things certainly cannot be the content of physics, because they are definitions going in a circle. The Newtonian statement above, however, seems to be a most precise definition of force, and one that appeals to the mathematician; nevertheless, it is completely useless, because no prediction whatsoever can be made from a definition.

Furthermore, in one of the initial lines of his lesson he says that one has, more or less, an intuitive idea of what is “mass” of a body. Then I'd ask, if possible : what is mass, beyond intuition ? When it comes to definitions, some problems arise.

On another side, let’s take into consideration what Benjamin Crowell, estimated member of PF , says in his book on Mechanics :

0.6 The Newton, the metric unit of force
A force is a push or a pull, or more generally anything that can change an object’s speed or direction of motion. A force is required to start a car moving, to slow down a baseball player sliding into home base, or to make an airplane turn. (Forces may fail to change an object’s motion if they are canceled by other forces, e.g., the force of gravity pulling you down right now is being canceled by the force of the chair pushing up on you.) The metric unit of force is the Newton, defined as the force which, if applied for one second, will cause a 1-kilogram object starting from rest to reach a speed of 1 m/s. Later chapters will discuss the force concept in more detail. In fact, this entire book is about the relationship between force and motion.


May be this is more intuitive than the concept of mass, and IMO it gives a hint towards an operational definition of force.

As far as the Lagrange equations are concerned, one has first to define Energy, another not-easy task (look for Feynman chapter on energy) , and then apply other not so immediate concepts of the calculus of variations. I find it difficult to introduce this to students of a first course of Physics.

Thanks.
 
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  • #48
italicus said:
As far as the Lagrange equations are concerned, one has first to define Energy, another not-easy task (look for Feynman chapter on energy) , and then apply other not so immediate concepts of the calculus of variations. I find it difficult to introduce this to students of a first course of Physics.

I agree it is not for the first physics course. But once you move onto Lagrangians, Noether comes into play and what energy is and why it is conserved is a snap.

Thanks
Bill
 
  • #49
ergospherical said:
To the contrary, variable mass systems are way more subtle than you think. The expression ##\boldsymbol{F} = \dot{\boldsymbol{p}}## in fact only holds for systems of constant mass.

I won't go into details, but Spivak's Physics for Mathematicians: Mechanics I has some nice discussion on this issue, for example.

And a footnote, completely inappropriate for this thread so start a new thread for any followup:
In general treatments, the derivative of the momentum ##\boldsymbol{p} = \int_{\mathcal{D}(t)} \rho \boldsymbol{v} dV## of the medium within some domain ##\mathcal{D}(t) = g^t \mathcal{D}_0## must be evaluated using the Reynold's transport theorem of vector calculus, and the result involves a term ##-\int_{\partial \mathcal{D}(t)} \rho\boldsymbol{v}(\boldsymbol{v}-\boldsymbol{w}) \cdot d\boldsymbol{S}## in addition to the total force on the body. [Here ##\boldsymbol{w}## is the velocity of the boundary and ##\boldsymbol{v}## the velocity of the medium.]
Well, to be fair, I explicitly said "Newtonian mechanics" which allows us to figure out how to send rockets to the moon and such.
 
  • #50
valenumr said:
Well, to be fair, I explicitly said "Newtonian mechanics

What @ergospherical discussed is also Newtonian mechanics, just a bit advanced :wink:
 
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  • #51
weirdoguy said:
What @ergospherical discussed is also Newtonian mechanics, just a bit advanced :wink:
Fluid mechanics is a bit beyond what Newton had in mind. 😀
 
  • #52
I agree with the explanations by David Lewis and gleem. But I think they missed a point that acceleration is not only dimensionally different, it's also a vector, and so is F. So F=(m1+m2)a makes sense but F=m(a1+a2) doesn't.

I also noticed that no one suggested a book on this subject. There is an excellent one by Hart "Multidimensional Analysis" which will probably show you that it's more complicated than you thought and is commonly abused (especially by engineers).
 
  • #53
meekerdb said:
But I think they missed a point that acceleration is not only dimensionally different, it's also a vector, and so is F. So F=(m1+m2)a makes sense but F=m(a1+a2) doesn't.
I don't see why the latter form doesn't make sense. For example, you could have two separate forces being applied to an object of mass m -- ##F_1## and ##F_2##. We could have ##F = F_1 + F_2 = ma_1 + ma_2 = m(a_1 + a_2)##.
 
  • #54
meekerdb said:
F=m(a1+a2) doesn't.
Can you help me with why it doesn't when it appears to me to be fine? Or is there an 'always' that needs to be added somewhere.
 
  • #55
First thing to notice is that force and acceleration are vectors, mass is a scalar. So the multiplication is in vector space. Multiplying a vector by a scalar is quite different from multiplying two vectors which may be a dot product or a cross product. But you can't add things with different units and sometimes you can't add things with the same units, e.g. torque+energy.

Second it's a way of defining force. There's an excellent book by Hart "Multidimensional Analysis" which shows how dimensions are used (and abused) in physics and engineering.
 
  • #56
mremadahmed said:
Multiplication is higher form of addition.Division is higher form of subtraction.

What does F=ma means. m is in kg. a in ms-2.How can both multiply.They cannot add or subtract.

Why any quantities can multiply or divide,contrary to adding and subtracting.
It is not generally correct that multiplication is simply a higher form of addition. When you were in elementary school this is indeed how the topic is introduced, but that is simply because elementary school students only know addition and are not mentally prepared for an axiomatic approach to math. In the axiomatic approach, multiplication is not constructed from addition, but a set is proposed and multiplication is defined abstractly as an operation on elements of that set with certain axiomatically imposed properties. It is those axioms that define multiplication, not the elementary-school construction.

The reason for that is to allow the extension of the concept of multiplication to other sets besides the real numbers. For example, a mathematical field (not to be confused with a physical field) is a set that has addition, subtraction, multiplication, and division that all work just like real numbers. So, of course, the real numbers are a field, but so are the rational numbers and the complex numbers. It doesn't make sense to think of adding ##5+2i## to itself ##3-i## times, but for complex numbers ##(5+2i)(3-i)## is a perfectly valid operation defined axiomatically. Other more exotic sets include algebraic functions, where it likewise makes sense to consider multiplication between two functions, but it does not make sense to consider it as adding one function to itself a function number of times.

In the case of dimensional analysis the appropriate abstract mathematical formalism is that of vectors, which is described in section 2 here: https://terrytao.wordpress.com/2012/12/29/a-mathematical-formalisation-of-dimensional-analysis/. Vectors have their own scalar multiplication and vector addition axioms.

So, specifically for your question we have a one-dimensional real vector space ##V^M## of all possible masses, so ##5 \text{ kg}## is the vector in the one-dimensional mass space which is the scalar product of ##5## times the named vector ##\text{kg}##. Similarly we have one-dimensional real vector spaces ##V^L## and ##V^T## for all possible lengths and times. Then, we can use a standard tensor product to construct a new vector space ##V^{LT^{-2}}=V^L \otimes V^{T^{-1}}\otimes V^{T^{-1}} ## which is the space of all possible (1D) accelerations, and ##V^{MLT^{-2}}=V^M \otimes V^{LT^{-2}}## which is the space of all possible (1D) forces. That is what is implied by something like ##F=ma##.

So then the issue with addition of dimensionally inconsistent quantities is that you are trying to add vectors of different vector spaces, so that doesn't make any sense because no such operation is defined. Whereas multiplication of dimensionally inconsistent quantities makes sense because you use it to create a new vector space using the standard tensor product.
 
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  • #57
meekerdb said:
First thing to notice is that force and acceleration are vectors, mass is a scalar. So the multiplication is in vector space.
Yes, and so what? One of the axioms of any vector space has to do with the multiplication of a vector by a scalar.

meekerdb said:
Multiplying a vector by a scalar is quite different from multiplying two vectors which may be a dot product or a cross product.
The vector space axioms don't even define the multiplication of two vectors. The only operations defined in the vector space axioms are addition of vectors and multiplication by a scalar.

meekerdb said:
But you can't add things with different units and sometimes you can't add things with the same units, e.g. torque+energy.
This has nothing to do with your earlier statement in post #53 that ##F = m(a_1 + a_2)## doesn't make sense. I gave you a scenario in which it does make sense.
 
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  • #58
meekerdb said:
But you can't add things with different units and sometimes you can't add things with the same units, e.g. torque+energy.
Torque and energy have different units. And if you meant force and torque, they have different units, as well.
 
  • #59
Mark44 said:
torque and energy don't have the same units
They do
 
  • #60
ergospherical said:
They do
They have the same dimensionality. Not the same units.
 
  • #61
jbriggs444 said:
They have the same dimensionality. Not the same units.
Having equal dimensionality implies they can be expressed in the same units…

It is usually the Nm (Newton metre)
 
  • #62
ergospherical said:
Having equal dimensionality implies they can be expressed in the same units…
I distinguish between units and dimensions.
 
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  • #63
jbriggs444 said:
I distinguish between units and dimensions.
Indeed, because they are different things… (loosely, dimensionality is associated with an [infinite] set of possible units…)

Nonetheless torque and energy share the same set of units

dW = F . dr
t = r x F
 
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  • #64
ergospherical said:
Indeed, because they are different things… (loosely, dimensionality is associated with an [infinite] set of possible units…)

Nonetheless torque and energy share the same set of units

dW = F . dr
t = r x F
With radians (arguably a dimensionless unit) as a conversion factor between torque applied and energy expended through a unit rotation angle).
 
  • #65
This discussion has given me a headache. But it did remind me of a favorite quote (seems apropos):

"Under capitalism, man exploits man. Under communism, it's just the opposite."

John Kenneth Galbraith

.
 
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  • #66
AlexS said:
Interesting conversation, I think that the criterion with unit is very important.I actually think that after you have this fundamental defining formulas such as average speed=total distance/total time, you can obtain other formulas such as Galileo's formula(v square=v0 square+2ad) just doing the math,without any trouble. But,in my view, the problem is still with this fundamental equations. For instance, why the speed is distance per time and not distance multiplied with time? I see the proportionality between time and distance (if speed is constant), but I still cannot perfectly understand this problem.
The ideal model of the discussed experiment assumes that the first falling snowflakes (constant vertical precipitation velocity, constant density) touch the ground at the moment of starting the measurement of the dozer movement (horizontal, uniform rectilinear motion).
Which dozer, under the same conditions, will pick up more snow; the one who traveled 40 m in 30 s, or the one who traveled 50 m in 20 s?
The entire experience (with two dozers) repeated at a other snowfall speed will not change which of them will sweep more snow.
Imagine a right triangle against the falling snow. The triangle moves vertically down with the snow. One cathetus is the distance s, the other cathetus - height is proportional to the time t. The movement of the point of intersection of the hypotenuse with the ground line is the movement of the dozer. The precipitation pushing capacity x=st/2 could be a physical quantity, the measurement unit of which would be a meter-second.
 
  • #67
jbriggs444 said:
I distinguish between units and dimensions.
Different animals entirely; I agree
'They' had to invent dimensional analysis to take care of this and other problems. As far as I know, the US has to accept the same convention for dimensions as we use in UK, despite their continuing love affair with feet.
 
  • #68
sophiecentaur said:
Different animals entirely; I agree
'They' had to invent dimensional analysis to take care of this and other problems. As far as I know, the US has to accept the same convention for dimensions as we use in UK, despite their continuing love affair with feet.
Feet go up to twelve. That's two more.

With apologies to Spinal Tap.
 
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  • #69
jbriggs444 said:
Feet go up to twelve. That's two more.

With apologies to Spinal Tap.
I wonder where that movie would be now without the volume control gag.

One afternoon, I watched it with my teenage son in a cinema in Brighton (UK). There were just three of us in the audience; Tom and I and a little old man on the other side of the gangway. Without the '11' gag, he would probably not have been there.
 
  • #70
mremadahmed said:
Why any quantities can multiply or divide,contrary to adding and subtracting.
Let's take a simple equation from rectilinear motion: ##\Delta x=vt##. In other words, displacement equals the product of the velocity and the time. You know that multiplication is a series of additions. So if you have a velocity of ##2 \ \mathrm{m/s}## for 3 seconds you travel 2 meters in the first second, 2 meters in the next second, and finally 2 meters in the third second. ##2 \ \mathrm{m}##+##2 \ \mathrm{m}##+##2 \ \mathrm{m}##=##6 \ \mathrm{m}##.

Note that you never add two quantities with different units.
 
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