Momentum paradox: Why can't we write it as p=m+v ?

[Frigus]

I think this will solve your problem,
Suppose their is a quantity m/v and you named it jake,suresh or any other name and one day someone comes to you and ask why jake or suresh is m/v not mv.
Then how will you answer that?

 

[Paul Colby]

So how could you explain him that p=mv and p is not m+v.

Has anyone pointed out that the net $mv$ in a collision is conserved while the net $m+v$ isn't?

 

[akashpandey]

I think this will solve your problem,
Suppose their is a quantity m/v and you named it jake,suresh or any other name and one day someone comes to you ask why jake or suresh is m/v not mv.
Then how will you answer that?

I just don't want to accept the that this is definition that's how it is.
I want to know the real meaning or experiment why it was define that way and not the other.
So just saying this is the definition and accepting it what school does.
Sorry i mean no disrespect to any member of community.
If you guys this question dumb or something else just ignore it.

 

[jbriggs444]

So tell me how else you define multiplication ?

Try this.

One standard approach has already been described in rough outline.

You start with multiplication of the ordinary positive integers defined in terms of repeated addition. Optionally you add a rule for multiplication by zero.

You extend this to signed integers. There are several ways to do this, but they all yield equivalent results. Perhaps the simplest is to simply say by fiat that positive times positive = positive, positive times negative = negative and negative times negative = positive.

You extend this to rational numbers as in #22 (equivalence classes of ordered pairs of signed integers).

You extend this to real numbers using equivalence classes of Cauchy sequences of rationals. Or Dedekind cuts.

 

[Nugatory]

So tell me how else you define multiplication ?

Without mathematical rigor, but at a level that is useful to a child who ready to move past integers and the "repeated addition" definition:

$a\times b$ is the area of a rectangle with sides of length $a$ and $b$.

This works as a lie to children because every naif has an intuitive notion of length and area so there's no need to explore the difficulty of formally defining either. Likewise we gloss over the problem of extending the rational numbers to the reals by blithely stating that $\sqrt{2}\times\pi$ is a rectangle with sides of those lengths.

 

[PeroK]

I just don't want to accept the that this is definition that's how it is.
I want to know the real meaning or experiment why it was define that way and not the other.
So just saying this is the definition and accepting it what school does.
Sorry i mean no disrespect to any member of community.
If you guys this question dumb or something else just ignore it.

If you took a job at $20 an hour and worked 10 hours and the manager said "that's $20 + 10 = $30". Would you go away happy with your $30? If not, why not?

 

[jbriggs444]

If you took a job at $20 an hour and worked 10 hours and the manager said "that's $20 + 10 = $30". Would you go away happy with your $30? If not, why not?

I want to work for 36 billion microseconds.

 

[akashpandey]

If you took a job at $20 an hour and worked 10 hours and the manager said "that's $20 + 10 = $30". Would you go away happy with your $30? If not, why not?

No i know i will add 20$, 10 times that's the basic definition of multiplication

Wel i guess i understood that multiplying with a number is repeated addition cause they are of same type.

But when it comes to two different type of quantity we cannot say that one quantity is added to other multiple times

 

[russ_watters]

Yeah got your point..
But definition of multiplication doesn't go hand in hand with this momentum thing.. that's all i am saying.

Sure it does. You're just having trouble with momentum because it is more abstract.

For example, there's multiplication-by-addition problems you probably do every day: drive 60 mph For 1 hour each way too and from work. How long did you drive and how far did you go?

One soda costs $3 and you are about to buy your third. How much have you spent and will you spend?

Wel i guess i understood that multiplying with a number is repeated addition cause they are of same type.

No they aren't. One has units of dollars, another hours, and the relation combines them.

 

[Frigus]

This works as a lie to children

So until now I was living in a lie.
From now onwards I am even doubting even if I have studied anything right.

 

[Nugatory]

So until now I was living in a lie.

That's OK, you're in good company - essentially every adult who has managed to avoid college-level abstract number theory and therefore accepts "two plus two equals four" as a truism.

 

[russ_watters]

I just don't want to accept the that this is definition that's how it is.
I want to know the real meaning or experiment why it was define that way and not the other.
So just saying this is the definition and accepting it what school does.
Sorry i mean no disrespect to any member of community.
If you guys this question dumb or something else just ignore it.

It's not dumb, but it it is a fundamental misunderstanding of what science and math are. Science is a way to understand the world around us and math is a tool/language to describe what we see.

mv is a quantity that scientists hundreds of years ago discovered was often conserved, and that's useful. There isn't, nor does there need to be any deeper meaning to it than that.

You may also want to look into the history of the development of momentum vs conservation of energy. We've had several discussions of it. There was a lot of feeling in the dark for relationships between velocity and mass, and debate over the value of the two relations. At the time they were competing as much as complimentary.

 

[Dale]

I just don't want to accept the that this is definition that's how it is.

That is silly. There is no other reason for definitions. You need to accept definitions just because “that’s how it is” or you will be unable to communicate.

Once you have agreed to use the definition then you can ask things like why is this concept useful? But a definition is just an arbitrary convention with no further justification needed

 

[akashpandey]

That's OK, you're in good company - essentially every adult who has managed to avoid college-level abstract number theory and therefore accepts "two plus two equals four" as a truism.

Are you mocking me ?

 

[etotheipi]

But why we can't write it as p=m+v ?

It is not dimensionally homogenous but maybe more alarmingly you are adding a scalar to a vector... $m + \vec{v}$... that can't be good!

Maybe we might ask why velocity is to the first power? $mv^1$ and $mv^2$ are meaningful, and we associate the first one with the idea of momentum and the second with the idea of kinetic energy, but I have no idea if $mv^3$, $mv^4$, etc. are of any use. Perhaps it's a complete fluke, and those two combinations just happened to play nicely. In any case, those two useful ones are themselves just approximations in th low speed regime of the relativistic formulae which are a little more ugly, so $mv$ is not even the full story for momentum!

 

[Paul Colby]

It is not dimensionally homogenous but maybe more alarmingly you are adding a scalar to a vector... m+v→... that can't be good!

Actually, in Clifford Algebras this kind of thing is permitted and meaningful for some value of meaning. And of course not meaningful here. The real answer to the OPs question is, as others have pointed out, in the application of algebra to physical theories. Why is the quantity $mv$ an algebraically useful concept in physics.

 

[A.T.]

I want to know the real meaning or experiment why it was define that way and not the other.

There are plenty of experiments that show the conservation of "it" and not "the other". You can look up simple collisions for a start.

 

[etotheipi]

Actually, in Clifford Algebras this kind of thing is permitted and meaningful for some value of meaning.

That is very cool, I have not heard of this! I have found an introduction to the topic here so that should make for interesting reading... thanks

 

[jbriggs444]

Are you mocking me ?

No. The response about 2+2=4 was a good humored jab at the fact that something seemingly obvious and well accepted by the general public can nonetheless become dark, mysterious and counter-intuitive once mathematicians have worked to explore the full generality of things.

It was clearly not intended to mock yourself, @Hemant, the general public or even mathematicians. The parable of the blind men and the elephant comes to mind: the same thing can appear differently depending on which aspect one considers.

 

[sophiecentaur]

I just don't want to accept the that this is definition that's how it is.

There are some aspects of Maths that you either just have to accept or go to the trouble of learning Maths to a much greater depth. You can't just expect a quick hand waving explanation of 'why' we can use Maths the way we do.
Your comment is a bit like saying that you don't want to watch TV until you are capable of designing and building the whole chain from camera to display. You just have to treat things as black boxes if you haven't the time or inclination to get in deep.

Why is the quantity mv an algebraically useful concept in physics.

Extremely useful and, in fact the conservation of momentum is one of those principles that 'works, whatever', which is pretty rare.
It has to be remembered that "mv" is not a full description of momentum. For a start, that formula doesn't work for massless particles like photons. So it's not at all obvious how a mass times a velocity suddenly becomes Planck's constant divided by a wavelength when a photon is involved in a collision. Getting too precious about mv won't get one very far.

 

[Paul Colby]

I have not heard of this!

It is neat. A book you might want on your shelf is

"Clifford (Geometric) Algebras with Applications in Physics, Mathematics, and Engineering" William E. Baylis, Editor. 1996 Birkhauser Boston.

The book is packed with enjoyable articles on all sorts of Clifford Algebra applications.

 

[hmmm27]

If you took a job at $20 an hour and worked 10 hours and the manager said "that's $20 + 10 = $30". Would you go away happy with your $30? If not, why not?

No i know i will add 20$, 10 times that's the basic definition of multiplication

Well, no it isn't : look at the statement ; it does not say "If you took a job at $20 and worked 10...".

It says

$$ \frac{$20}{h} \times 10h = \frac {$20 \times 10h}{h} = $20 \times 10 = $200 $$

 

[Nugatory]

Are you mocking me ?

No, that post was sympathizing with @Hemant... it is somewhat remarkable that the mathematical underpinnings of basic arithmetic are so obscure when the applications are so readily accessible.

 

[sophiecentaur]

it is somewhat remarkable that the mathematical underpinnings of basic arithmetic are so obscure when the applications are so readily accessible.

It's more than "remarkable" I find it quite unsettling if I try to think too deeply about it. Where is the real link between the Physical world and the symbolic processes of our Mathematics? How can we rely on the two never parting company?

 

[Paul Colby]

How can we rely on the two never parting company?

What's left after the two part? I think they are joined at the hip.

 

[sophiecentaur]

What's left after the two part? I think they are joined at the hip.

Experience tells me not to worry but I know of some fractal functions that look well behaved until some point and then there’s chaos. But I imagine anything like that would happen before we would have time to worry about it.

 

[shivaprasadvh]

Summary:: Why we multiply mass and velocity ?

So as we know momentum has a formula p=mv right ?
But why we can't write it as p=m+v ?
The real question is why we multiply both mass and velocity quantity
And not add them ?

Lot of entities / situations around us are dependent on mass and velocity.
Eg:
Consider a ball of mass 0.5 kg. Throw it from a height of say 2 m on a sand pit. The ball enters into the sand to some depth. Now from the same height throw a ball with weight 1 kg on a sand pit.The ball enters the sand to a greater depth. We know this happens through our experience from childhood. Why this happens?This is because PE of the ball is more of second ball than the first ball. Hence the final velocity with which it hits the ground is more and hence more depth.
Now by varying m, v varied. Lot of things happen because of the variable of this product mv.
By controlling this product mv we can bring lot of things under control or out of control. Hence mv is significant variable.

Hope I have cleared your doubt.

 

[A.T.]

Lot of entities / situations around us are dependent on mass and velocity.
Eg:
Consider a ball of mass 0.5 kg. Throw it from a height of say 2 m on a sand pit. The ball enters into the sand to some depth. Now from the same height throw a ball with weight 1 kg on a sand pit.The ball enters the sand to a greater depth. We know this happens through our experience from childhood. Why this happens?This is because PE of the ball is more of second ball than the first ball. Hence the final velocity with which it hits the ground is more and hence more depth.

Sounds more like an explanation of kinetic energy, not momentum.

Now by varying m, v varied. Lot of things happen because of the variable of this product mv.
By controlling this product mv we can bring lot of things under control or out of control. Hence mv is significant variable.

There are many other possible combinations m and v, so this doesn't really explain why specifically mv is significant. The significance of mv comes from the fact that its total is conserved for isolated systems.

 

[shivaprasadvh]

Sounds more like an explanation of kinetic energy, not momentum.There are many other possible combinations m and v, so this doesn't really explain why specifically mv is significant. The significance of mv comes from the fact that its total is conserved for isolated systems.

Kinetic Energy and momentum are related. KE = (momentum)square / 2. This is a formula. So like kinetic energy is dependent on equation mv which we call momentum, say speed and velocity are also dependent on equation of mv. So KE was just one example.

 

[Frigus]

KE = (momentum)square / 2

You forgot to multiply it by 1/m.