Why do most formulas in physics have integer exponents?

[dushyanth]

I mean why is f=ma?
why not m^0.123a^1.43 or some random non integers?
I hope you understand that my doubt doesn't limit just to force or energy or velocity e.t.c.
it also extends to area of a square,circle e.t.c and all other formulae
i think whole thing starts with direct proportionality.

 

[Tyrannosaurus_]

For the same reason that the area of a square is L². You're taking the lengths of the sides of a square and multiplying them together, thus the same value multiplied by the same value.

The reason for whole number powers, & whole number coefficients is due to the relationship that is expressed. You need to understand how the formula is created, not simply what the formula looks like.

Furthermore, C = πd, where π is 3.14159... an irrational number! Furthermore, look at the formula for the volume of a cylinder. A fraction is involved there. But not for mystical reasons, but because it is a way to express the relationships between parts that can be measured.

I think your question lacks depth. Learn why the formulas are what they are. Continue your studies to find situations without whole number coefficients and powers. Check out the universal gravitation equation. The gravitation constant is not a "nice" number.

 

[dushyanth]

C = πd, where π is 3.14159... an irrational number!

I know why it has pi.it's because we chose the constant for area square to be one.If we chose it to be 1 for circle then area of square will have a constant=1/pi

 

[dushyanth]

To be correct, the area of a square is directly proportional to the length of the side, but rather, the area is directly proportional to the length of a side squared. That is, to the power of 2.

Based on your comments, I do not believe you understand the relationship between the length of a square's sides and the area of that square.

i know that A=L² ,but why 2?
and why is it 3 for volume of a cube.
i understand

[Mentor's note: Edited to remove references to some deleted posts]

 

[dushyanth]

To be correct, the area of a square is directly proportional to the length of the side, but rather, the area is directly proportional to the length of a side squared. That is, to the power of 2.

Based on your comments, I do not believe you understand the relationship between the length of a square's sides and the area of that square.

i know that A=L² ,but why 2?
don't say that it is because L*L=L²

 

[DaveC426913]

don't say that it is because L*L=L²

Can you explain what it is about this answer that is not satisfactory?

I too am having difficulty with the source of your confusion.

 

[Tyrannosaurus_]

Thanks DaveC! This must be too "deep" for me.

 

[dushyanth]

Can you explain what it is about this answer that is not satisfactory?

i mean how did we chose a physical quantity such that it's square is equal to area of square

 

[dushyanth]

Thanks DaveC! This must be too "deep" for me.

That was a good one.I can't stop laughing but please stop mocking me

 

[DaveC426913]

The definition of a square is that both sides are the same. Instead of saying l*l, we can say l². We cannot say the same for any other rectangular shape - it would be l*w.

 

[Tyrannosaurus_]

Your question is like asking, why is the colour green the colour green and not the colour purple. Simply!, by definition that is what it is.

The relationship between area of a square, is found by squaring the length of a side. If you don't believe that's true, measure it! Prove it to yourself! There is nothing mystical about this relationship!

 

[DaveC426913]

You mentioned proportionality.

Gravitational force is proportional to m/r². This is only a qualitative relationship.
To show a quantitative relationship, we need to add a constant: 6.677 x10^-11.
So, G = g*m/r².

F=ma is actually an oversimplification. At leas the way I was taught.
Is is more correctly F~m*a. These reads as F is proportional to m*a.
F does not equal m*a unless you use the correct units.

 

[dushyanth]

You mentioned proportionality.

Gravitational force is proportional to m/r². This is only a qualitative relationship.
To show a quantitative relationship, we need to add a constant: 6.677 x10^-11.
So, G = g*m/r².

F=ma is actually an oversimplification. At leas the way I was taught.
Is is more correctly F~m*a. These reads as F is proportional to m*a.
F does not equal m*a unless you use the correct units.

i completely understand that a constant is necessary for that.i have previously mentioned a similar thing on the area of circle above

 

[DaveC426913]

OK. Can you reformulate your question a little more descriptively? Pick a particular formula that concerns you and describe why it does not behave the way you expect.

 

[dushyanth]

OK. Can you reformulate your question a little more descriptively? Pick a particular formula that concerns you and describe why it does not behave the way you expect.

thank you very much for asking.Lets say f=ma
why does mass and aceeleration have integer powers to them?

 

[DaveC426913]

thank you very much for asking.Lets say f=ma
why does mass and aceeleration have integer powers to them?

Put in the simplest way possible: every doubling kilogram of moving mass will double the force it applies when it hits something.

That word "every" means that F always directly and exactly varies as mass varies. And that means the exponent is 1.
If you double the mass, you don't get four times the force applied (m²) or a quarter of the force applied (m^.5).

 

[brainpushups]

Newton's second law is an axiom. This is one of the starting points necessary to deduce the consequences of classical physics. So, in a sense, the answer to 'why' are the powers in this equation integer powers is that because that is the way Newton formulated it (actually it was Euler who wrote the law algebraically - Newton only used geometric and paragraph proofs).

Your question is similar to asking of Euclid why a straight line segment can be drawn between two points.

 

[dushyanth]

every doubling kilogram of moving mass will double the force it applies when it hits something

every doubling kilogram of moving mass will double the force it applies when it hits something why not 1.5 times or some other non integer

 

[brainpushups]

That would not change the fact that the exponents are 1! that changes the proportionality constant. Pick a different system of units (or base your units on this) and that can be true.

 

[DaveC426913]

every doubling kilogram of moving mass will double the force it applies when it hits something why not 1.5 times or some other non integer

Does this make any sense though?

Say you had a one kg cannonball in a bag, and swinging created a force of f.
Then you added a second one kg cannonball to the bag. Would you exepct that the second ball (which is, in every way identical to the first) would only add .5f?

 

[dushyanth]

Does this make any sense though?

Say you had a one kg cannonball in a bag, and swinging created a force of f.
Then you added a second one kg cannonball to the bag. Would you exepct that the second ball (which is, in every way identical to the first) would only add .5f?

can i apply the same to velocity in k.e=mv²/2

 

[dushyanth]

That would not change the fact that the exponents are 1! that changes the proportionality constant.

no.

 

[DaveC426913]

can i apply the same to velocity in k.e=mv²/2

Hey. Don't change the subject. ;) Does 1.5 make sense or no?

 

[DaveC426913]

That would not change the fact that the exponents are 1! that changes the proportionality constant. Pick a different system of units (or base your units on this) and that can be true.

bpu: dushyanth is correct on this one. He is asking about m^1.5, which is an exponential change, not a proportional change. You're describing F=1.5ma

 

[brainpushups]

Right. I misread his post.

 

[dushyanth]

Hey. Don't change the subject. ;) Does 1.5 make sense or no?

No sir I don't want to change the subject but you are saying that adding one more kg should only increase the force by 1f is because you already know that f is directly proportional to mass.why don't we apply the same for v in the other formula?

 

[DaveC426913]

No sir I don't want to change the subject but you are asking why one more kg should only increase the force by 1f is because you already know that f is directly proportional to mass.

The question is: do you know that f is directly proportional to m? That seems to be the crux of your question.

why don't we apply the same for v in the other formula?

That is a good question. Why is kinetic energy proportional to the square of velocity? I don't have a concise answer yet.

And apparently, no one else does either.

 

[dushyanth]

The question is: do you know that f is directly proportional to m? That seems to be the crux of your question.
That is a good question. Why is kinetic energy proportional to the square of velocity? I don't have a concise answer yet.

And apparently, no one else does either.

at least you understood the "depth" of my questio(no pun inteded)

 

[russ_watters]

That is a good question. Why is kinetic energy proportional to the square of velocity? I don't have a concise answer yet.

And apparently, no one else does either.

No, that's a fairly straightforward issue of math to apply the definition of work to a moving object accelerated under a constant force.

To me it seems like there are two different issues here:

1. Not understanding basic mathematical relations and definitions (like what a square is).
2. Looking for meaning where none need be.

 

[russ_watters]

Consider f=ma a little more closely with this example:

Say, you have two 1kg weights and you apply a 1N force to each of them. They each accelerate at 1m/s^2.

Now you tie them together with a string and repeat. Why should the acceleration be different if they are connected than if they weren't?

 

[GiuseppeR7]

i know that A=L² ,but why 2?
and why is it 3 for volume of a cube.
i understand

[Mentor's note: Edited to remove references to some deleted posts]

The area is DEFINED to be that way.

 

[Stephen Tashi]

i think whole thing starts with direct proportionality.

That's a good insight. I agree that much of it does. If y = kt , k = constant, t = time (or some other important physical quantity) then the antiderivatives of y have integer powers. If one applies calculus to physics, the integer powers arise naturally. Linear relationships like y = kt + b give rise to integer powers in the same way.

You could go further and say the whole thing starts with constants (i.e. constant functions).

Dimensional analysis might offer another explanation for integer exponents. If we have the equation y = f(x,y,..) describing a physical situation then the units on the left hand side and right hand side must be the same. If y is some quantity involving units with non-integer, but rational exponents like Newtons^(2/3) meters^(1/2) / sec^(2/10) then you can raise y to a power that makes all the exponents integers. If raise both sides of the equation to that power, you have a physical law. The right hand side of the new equation may be a complicated function but dimensional analysis says the units on the right hand side must work out to match the units on the left hand side.

 

[sophiecentaur]

I mean why is f=ma?
why not m^0.123a^1.43 or some random non integers?
I hope you understand that my doubt doesn't limit just to force or energy or velocity e.t.c.
it also extends to area of a square,circle e.t.c and all other formulae
i think whole thing starts with direct proportionality.

There are a great many formulae that do not use integer exponents. The best known examples would probably be those describing Exponential Decay of Temperature, Concentration, Radioactivity and Charge on a discharging Capacitor. To answer your question, you could say that simple formulae have integer exponents.

 

[mic*]

I know why it has pi.it's because we chose the constant for area square to be one.If we chose it to be 1 for circle then area of square will have a constant=1/pi

This is the only thing in this thread that doesn't seem to be able to be resolved by simply deriving formulas and their units.

Why do we use units that are based on square 1 (no pun intended)? I suspect it may have been more prosperous than other proposed systems. Surely it must have been historically better at getting us from point A to point B than any circular alternative? ...Producing 0 results won't get a concept adopted (there may have been a pun in there that time).

BTW; if a circle with radius 'a' were defined as the unit for 2 dimensional analysis, how does a square of side length '2*a' have area 1/pi?

 

[anorlunda]

I mean why is f=ma?
why not m^0.123a^1.43 or some random non integers?
I hope you understand that my doubt doesn't limit just to force or energy or velocity e.t.c.
it also extends to area of a square,circle e.t.c and all other formulae
i think whole thing starts with direct proportionality.

Have you thought about units? What would the units of m^0.123a^1.43 be?

 

[jbriggs444]

Have you thought about units? What would the units of m^0.123a^1.43 be?

kg^0.123meters^1.43/sec^2.86

Try it. If you convert a measured numeric value of mass expressed in kg to a corresponding numeric value expressed in pounds (multiply it by factor of 2.2), by what factor will the computed value of m^0.123 change?

 

[Quantum Defect]

I think the original poster is asking the question why so many of the laws of physics are "simple." I believe that s/he gets that this is what is true, based upon experiment i.e. doubling the mass for a constant force will halve the acceleration, but s/he wonders why everything works out so cleanly for us.

i.e. why is Coulomb's Law: F = constant * q_1*q_2/R^2 -- why not more gnarly exponents? Is there some deeper reason for why it all works out so prettily for us?

In the old days, an acceptable answer for most people (including Newton) would have said that the simple exponents reveal to us the hand of god.

 

[TeethWhitener]

Wikipedia has a pretty good illustration of the connection between the inverse square law found in Newton's law and Coulomb's law and the geometry of 3 dimensions:
http://en.wikipedia.org/wiki/Inverse-square_law
In fact, it doesn't have to be this way. Coulomb's law, e.g., could have the form F \propto \frac{1}{r^{2+\epsilon}} where \epsilon is some tiny number. In fact, that will be the case if the photon has a mass. The fact that we observe that \epsilon \approx 0 is one verification that, if the photon does have a mass, it is very very small. See the first chapter of Jackson's electrodynamics book for a better explanation than I could possibly hope to give.

 

[David Nassau]

...
I think your question lacks depth. Learn why the formulas are what they are. Continue your studies to find situations without whole number coefficients and powers. Check out the universal gravitation equation. The gravitation constant is not a "nice" number.

Your answer is condescending. It's actually a very deep question. Why are the laws of nature such that they can be described with simple math? There's not an obvious reason why this should be the case.

 

[zahbaz]

I think a good read, if you haven't done so yet, would be Feynman's lectures. He often addresses these issues.
Lectures 9 and 10 on dynamics come to mind.

http://www.feynmanlectures.caltech.edu/I_toc.html

 

[zahbaz]

Also, a lot of formulas do not have integer exponents. Radicals are fairly common. But what would you rather read:

$$K = \frac{1}{2}mv^2$$ or $$v = \sqrt{\frac{2K}{m}}$$ ?

Perhaps the refined question would be, "why not irrational exponents?"

 

[Ken G]

I mean why is f=ma?
why not m^0.123a^1.43 or some random non integers?

Here's my take on that-- we look to define concepts that work simply, and when we succeed, it ends up meaning that the formulae have simple exponents.

Look at the F = ma example. Notice that we could just as easily define "Lorce" as e raised to what we now call force. To fix the units, we would have some reference value of "Lorce = e", and that would correspond to some reference force, where Lorce = e^{(force / reference force)} . Now we get the exact same physics, all the same predictions to every observation, using the law:
(ma / reference m * reference a) = ln(Lorce)
That doesn't look anything like an integer power, but it's the same law. An important thing to note is how "Lorces" would work. When you have two lorces on the same body, you don't add them, you multiply them. All the physics is the same, every prediction just as accurate.

But if we really did have that law, what would happen? It would not take long for someone to notice that instead of defining this "Lorce" concept, it would make more sense to define the "force" concept by taking the natural log of the Lorce, because that new concept would be additive-- the force on a body is then the sum of all the forces on its parts. We look for concepts that have nice properties, and the formulas they engender will reflect those nice properties. Why there is anything with "nice properties" is a very deep and difficult question-- why does the universe make sense at all to intelligent apes? Either it really does have some kind of mathematical design that our minds are programmed to understand somehow, or perhaps we merely focus our attention on the things that we have noticed we are able to understand, choose idealizations and simplifications that connect with what we can understand, and lo and behold, we find that the formulas we get are understandable.

 

[Dadface]

I think in many cases the integer exponents occur in basic simplified forms of the equation.

Take for example Ohms law: V=IR. This simple equation applies only if R is constant, but R depends on factors such as temperature.

We could write a more detailed equation which takes into account factors such a linear dimensions,resistivity and temperature coefficient of resistance but that equation would itself be simplified because, for example, the temperature of the sample can vary with the temperature of the surroundings. Amongst other things a temperature change in the surroundings will introduce a time varying exponent into the equation.

 

[sophiecentaur]

People seem to be missing part of the point here. By choosing a small enough range of your variables, you can make a formula with integer exponents - or even straight line relationships, to describe any relationship. But that formula will apply only over that range. No simple formula works over an unlimited range of variable values. There is no magic involved - it's just pragmatism. Imo, the OP is putting the (integer) cart before the (real) horse.

 

[Demystifier]

I mean why is f=ma?
why not m^0.123a^1.43 or some random non integers?
I hope you understand that my doubt doesn't limit just to force or energy or velocity e.t.c.
it also extends to area of a square,circle e.t.c and all other formulae
i think whole thing starts with direct proportionality.

This is actually a good question, and I think I have a good answer.
In fact, most mathematical relations between physical quantities can not be written by simple equations involving only integer powers. Most realistic systems in nature are non-linear or complicated by other means, so that we cannot write down a complete mathematical expression that describes them. It's only a relatively small subset of all phenomena that can actually be described by simple equations, and only in that cases we actually write down the equations. The result is that most written equations are simple, despite the fact that most equations (written or not) are not simple. In other words, we tend to explicitly write down equations only when they are simple, which, among other things, means - only when their exponents are simple. And of course, integer exponents are simpler that non-integer ones, so we tend to write down equations only when the exponents are integer.

 

[[Quadratic]]

F, m, a, L, etc are just symbols we use to represent concepts. If m=2 then m^0.123=2^0.123.

If you are asking why the equations that describe the universe are not strictly linear then it's an interesting question but not one I'm certain can be answered.

 

[russ_watters]

Let me repeat: most equations have to be integers due to the requirements of mathematical logic and geometry. F=m^1.2 * a (for example) isn't logically possible:

Consider f=ma a little more closely with this example:

Say, you have two 1kg weights and you apply a 1N force to each of them. They each accelerate at 1m/s^2.

Now you tie them together with a string and repeat. Why should the acceleration be different if they are connected than if they weren't?

 

[sophiecentaur]

Let me repeat: most equations have to be integers due to the requirements of mathematical logic and geometry. F=m^1.2 * a (for example) isn't logically possible:

You can't be sure that statement would apply near a black hole, for instance and it definitely wouldn't apply 'near' the quantum level. All formulae are simplifications and the extreme simplification is a linear relationship. Next simplest involves integer exponents. etc. etc.
You don't have to go far to find non integer exponents. Take the voltage decay in an RC discharge.

 

[Ken G]

Let me repeat: most equations have to be integers due to the requirements of mathematical logic and geometry. F=m^1.2 * a (for example) isn't logically possible:

Logic does not rule out an expression like that, without implicit assumptions. Let's go back to your example of two masses falling together, with a string between them. I can easily define the "mash" of those objects to be what you would call mass^1/1.2, and if I use the letter "m" to denote "mash", then I do get F = m^1.2 a. So there is nothing logically impossible about that formula. What you are saying is that the "mash" won't have the property that the mash of a system is the sum of the mash of all its parts, but so what? Who said that must be true-- as sophiecentaur points out, that's actually not true for mass in high-gravity systems, and it's not true for the quarks inside nucleons. Still, we can agree that we are never going to define "mash" this way, if we have a perfectly working concept of "mass" that does, at weak gravity levels and outside nucleons, work like the mass of the whole is the sum of the masses of the parts. So that's what I mean that we are the source of our own simple equations-- we build the equations around concepts that obey simple rules, like the whole is the sum of the parts, and if we make our building blocks have simple properties, we end up with simple equations that invoke those building blocks.

 

[russ_watters]

Logic does not rule out an expression like that, without implicit assumptions. Let's go back to your example of two masses falling together, with a string between them. I can easily define the "mash" of those objects to be what you would call mass^1/1.2, and if I use the letter "m" to denote "mash", then I do get F = m^1.2 a. So there is nothing logically impossible about that formula.

When you try to apply that to reality, it creates a logical contradiction: stringing a slack rope between the two objects would cause their acceleration to increase by 44% without any identifiable cause.