Understanding Multiplication: Its Role in Interactions and Beyond

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Discussion Overview

The discussion revolves around the use of multiplication versus addition in physics equations, particularly in the context of momentum, velocity, and kinetic energy. Participants explore the mathematical and conceptual implications of these operations in various physical laws and equations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question why multiplication is preferred over addition in equations like p=mv, suggesting that multiplication is used for convenience and due to definitions established by physicists.
  • Others argue that adding a scalar (mass) and a vector (velocity) is not valid in vector algebra, reinforcing that p=mv is a scalar multiplication of a vector.
  • There is a challenge regarding the equation v=u + at, with some asserting that it adheres to vector addition rules, while questioning the validity of v=u x at.
  • Concerns are raised about the implications of kinetic energy being proportional to the square of velocity, with participants noting that this leads to questions about fuel consumption and energy conservation.
  • One participant provides a derivation linking kinetic energy to Newton's second law, suggesting that the mathematical structure of these equations is dictated by physical principles.
  • Some participants express uncertainty about the implications of these mathematical relationships, particularly regarding energy and fuel consumption in practical scenarios.

Areas of Agreement / Disagreement

Participants express differing views on the appropriateness of using multiplication versus addition in physical equations. There is no consensus on the validity of proposed alternative formulations or the implications of kinetic energy's dependence on the square of velocity.

Contextual Notes

Participants note limitations in understanding the relationships between quantities, particularly regarding the addition of dissimilar units and the implications of energy conservation in practical applications.

Who May Find This Useful

This discussion may be of interest to students and enthusiasts of physics, particularly those exploring the mathematical foundations of physical laws and their applications in real-world scenarios.

Deepak Kapur
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When two bodies interact 'multilication' sign is mostly (always) used. It is used in various other instances also. Why not 'addition' sign?

( I know that multiplication can be understood as continued addition).

Some technical answers please.

For example

p=mv why not p=m+v
 
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For example

p=mv why not p=m+v

That's a violation of the rule of vector algebra. velocity v is a vector quantity and mass m is a scalar. How do you add them?
But p= mv which is a scalar multiplication of vector v. That's allowed in vector algebra.
The use of multiplication is more convenient with the vectors.
And some terms include 'multiplication' because they were defined by the physicist in that way. Sometimes, empirical data from experiments suggests to use multiplication.
 
Last edited:
Also, when you add feet + miles per hour, what do you get?
 
tensorbundle said:
That's a violation of the rule of vector algebra. velocity v is a vector quantity and mass m is a scalar. How do you add them?
But p= mv which is a scalar multiplication of vector v. That's allowed in vector algebra.
The use of multiplication is more convenient with the vectors.
And some terms include 'multiplication' because they were defined by the physicist in that way. Sometimes, empirical data from experiments suggests to use multiplication.

What about v=u + at

why not v=u x at
 
Deepak Kapur said:
What about v=u + at

why not v=u x at

Right, so I'm assuming you're not a troll now.

Check the units on both side. They differ, therefore it's nonsense.

When it comes to the use of the multiplication sign in physics, for example when computing forces of interaction, it is the only way to assign properties to objects which are independent of other objects' properties.
 
What about v=u + at
Well, by this example you are trying to say that on the right hand side we have added two non-similar quantities like you did in your previous example p=m+v.
But the quantity u and at are similar in nature. at is nothing but a velocity same as u. so the expression v= u+ at does hold the rule of addition in vector algebra.
why not v=u x at
And that does not convey any sensible argument. If you are genius enough to derive a new kind of equation of motion like that, derive it using rigorous mathematics. We shall accept it if you can. Otherwise, you have to accept that v=u x at can't be possible because it can't be derived. That's a pretty simple argument.
 
My friend you know Momentum is a vector so its mathematical qualities will satisfy laws of vector algebra.Now think that momentums direction is direction of velocity so according to the law of scalar multiplication of vectors "m" will be multiplied.
 
tensorbundle said:
Well, by this example you are trying to say that on the right hand side we have added two non-similar quantities like you did in your previous example p=m+v.
But the quantity u and at are similar in nature. at is nothing but a velocity same as u. so the expression v= u+ at does hold the rule of addition in vector algebra.

And that does not convey any sensible argument. If you are genius enough to derive a new kind of equation of motion like that, derive it using rigorous mathematics. We shall accept it if you can. Otherwise, you have to accept that v=u x at can't be possible because it can't be derived. That's a pretty simple argument.

Actually, I was just "tongue in cheek" but still had some kind of uneducated/trivial/unqualified doubt in my mind.

Now, consider

KE=1/2mv2

If m=1, v=2 , KE=2

Now just double the speed

m=1, v=4 , KE=8 that is 4 times increase.

According to the law of conservation of energy, this energy did not come from nowhere but was supplied by the fuel. So, fuel used should be in proportion to the gain in energy.

But as we all know we don't have to increase the amount of fuel intake to 4 times if we want to double the speed of our car, bike etc. (Fuel is actually only slighty increased by the accelerator for such low speeds.)

So, this 'square' of the velocity troubles me. It shouldn't be there.

Seriously, I am only very-2 less tongue in cheek this time.
 
For the equation v = u + at just think about it physically.

u = your initial velocity,
a= your constant acceleration,
t = the amount of time you've been moving,

every second you accelerate you get a faster velocity and because this equation only works for constant acceleration you are only adding more velocity to your beginning velocity u.

If you start accelerating from standing still then u = 0 so the equation is

v = 0 + at so we can just write it as v = at.

For the kinetic energy formula, it is really just a mathematical knock off of Newtons second law given a different name.

Start with F = ma and use one of the constant acceleration equations that can be easily derived;

v² = u² + 2a(x - x_0)

then use F= ma but solve for a, a = F/m and put it into the formula;


v² = u² + 2(F/m)(x - x_0) now solve for F(x - x_0);

F(x - x_0) = ½m(v² - u²)

we can also say that a force acting over a distance (x - x_0) (call this d!) is

Fd = ½mv² and we'll call Fd work and call ½mv² kinetic energy.

That's kind of why you are squaring it.

Personally I remember walking down the street trying to come up with explanations as to why kinetic energy and e = mc² and p = mv need to do what they do, there are three answers I've gotten so far, 1: The math dictates it to be so, 2: The units make it so, 3: It's something to do with the fact that the universe is in 3 dimensions and just like the area of a square is a² when you move into two dimensions it's somehow related.

That's about it so far :-p
 
  • #10
Deepak Kapur said:
But as we all know we don't have to increase the amount of fuel intake to 4 times if we want to double the speed of our car, bike etc. (Fuel is actually only slighty increased by the accelerator for such low speeds.)

What makes you say this though?
 

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