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?