Why do most formulas in physics have integer exponents?

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Discussion Overview

The discussion revolves around the question of why most physical formulas, such as F=ma and the area of geometric shapes, utilize integer exponents. Participants explore the implications of proportional relationships in physics and mathematics, questioning the nature of these relationships and the reasoning behind the use of whole number powers.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions why formulas like F=ma do not include non-integer exponents, suggesting a broader inquiry into the nature of mathematical relationships in physics.
  • Another participant argues that integer powers arise from the relationships expressed in the formulas, emphasizing the importance of understanding how these formulas are derived.
  • Some participants discuss the specific case of the area of a square being proportional to the length squared, questioning why the exponent is 2 and not another value.
  • There is mention of the gravitational force being proportional to m/r², highlighting the need for constants to express quantitative relationships.
  • One participant asserts that Newton's second law is an axiom, suggesting that the integer powers are a result of how Newton formulated the law.
  • Several participants engage in a back-and-forth about the implications of doubling mass and its direct relationship to force, questioning whether non-integer relationships could apply.

Areas of Agreement / Disagreement

Participants express differing views on the nature of mathematical relationships in physics, with no clear consensus reached. Some agree on the importance of integer powers in certain contexts, while others challenge the reasoning behind these choices.

Contextual Notes

Participants note that the discussion involves assumptions about proportionality and the definitions of physical quantities, which may not be universally accepted or understood. There are also unresolved questions regarding the application of these principles to different formulas.

dushyanth
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I mean why is f=ma?
why not m0.123a1.43 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.
 
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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.
 
Tyrannosaurus_ said:
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
 
Tyrannosaurus_ said:
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]
 
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Tyrannosaurus_ said:
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
 
dushyanth said:
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.
 
Thanks DaveC! This must be too "deep" for me.
 
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DaveC426913 said:
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
 
Tyrannosaurus_ said:
Thanks DaveC! This must be too "deep" for me.
That was a good one.I can't stop laughing but please stop mocking me
 
  • #10
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.
 
  • #11
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!
 
  • #12
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.
 
  • #13
DaveC426913 said:
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
 
  • #14
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.
 
  • #15
DaveC426913 said:
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?
 
  • #16
dushyanth said:
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).
 
  • #17
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.
 
  • #18
DaveC426913 said:
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
 
  • #19
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.
 
  • #20
dushyanth said:
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?
 
  • #21
DaveC426913 said:
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
 
  • #22
brainpushups said:
That would not change the fact that the exponents are 1! that changes the proportionality constant.
no.
 
  • #23
dushyanth said:
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?
 
  • #24
brainpushups said:
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 m1.5, which is an exponential change, not a proportional change. You're describing F=1.5ma
 
  • #25
Right. I misread his post.
 
  • #26
DaveC426913 said:
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?
 
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  • #27
dushyanth said:
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.
dushyanth said:
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.
 
  • #28
DaveC426913 said:
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):oldgrumpy::oldfrown::rolleyes:
 
  • #29
DaveC426913 said:
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.
 
  • #30
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?
 

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