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

In summary: What kind of unit would this hypothetical physical quantity have and what can it be used for? So we take 1 kilogram and 1 meter per second and add them and get 2 kg+m/s? What if we choose different units? Momentum will scale in the obvious manner, 1 kg·m/s = 1000 g·m/s, whether we multiplied 1kg by 1m or 0.5 kg by 2m/s, but your added quantity will depend on the values we adding up, so if we add 1kg and 1m/s it will be 2 kg+m/s = 1001 g+m/s, but if we add 0.5 kg and 1.5
  • #36
akashpandey said:
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?
 
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  • #37
PeroK said:
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.
 
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  • #38
PeroK said:
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
 
  • #39
akashpandey said:
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.
 
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  • #40
Nugatory said:
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.
 
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  • #41
Hemant said:
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.

:smile: :smile: :smile:
 
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  • #42
akashpandey said:
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.
 
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  • #43
akashpandey said:
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
 
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  • #44
Nugatory said:
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.

:smile: :smile: :smile:
Are you mocking me ?
 
  • #45
akashpandey said:
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!
 
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  • #46
etotheipi said:
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.
 
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  • #47
akashpandey said:
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.
 
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  • #48
Paul Colby said:
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 :smile:
 
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  • #49
akashpandey said:
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.
 
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  • #50
akashpandey said:
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.
Paul Colby said:
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.
 
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  • #51
etotheipi said:
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.
 
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  • #52
PeroK said:
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 $30? If not, why not?
akashpandey said:
No i know i will add 20$, 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

$$ \frac{$20}{h} \times 10h = \frac {$20 \times 10h}{h} = $20 \times 10 = $200 $$
 
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  • #53
akashpandey said:
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.
 
  • #54
Nugatory said:
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?
 
  • #55
sophiecentaur said:
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.
 
  • #56
Paul Colby said:
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.
 
  • #57
akashpandey said:
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.
 
  • #58
shivaprasadvh said:
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.

shivaprasadvh said:
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.
 
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  • #59
A.T. said:
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.
 
  • #60
shivaprasadvh said:
KE = (momentum)square / 2
You forgot to multiply it by 1/m😇.
 
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  • #61
Hemant said:
You forgot to multiply it by 1/m
Oh you it to be multiplied by 1/m. Thanks for that...
 
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  • #62
shivaprasadvh said:
Kinetic Energy and momentum are related.
Everything is "related" somehow. The question was about specifically about momentum, and why it has the formula mv.
 
  • #63
shivaprasadvh said:
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.
For a fixed non-zero momentum (mv) I can have any speed I like (greater than zero), any velocity I like (other than zero) and any energy I like (greater than zero).

So does it really make sense to say that any of those quantities are dependent on mv?

[Thanks to @Hemant for the correction on allowed speeds]
 
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  • #64
shivaprasadvh said:
Hence the final velocity with which it hits the ground is more
Actually, since you dropped them from the same height the final velocity is the same.
 
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  • #65
Dale said:
Actually, since you dropped them from the same height the final velocity is the same.
Got it final velocity is same but as the mass is different so the depth with which each penetrates is less and more.
 
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  • #66
A.T. said:
Everything is "related" somehow. The question was about specifically about momentum, and why it has the formula mv.
Through observations, there are several effects which vary with mass and vary with velocity but are the same if the product mv is same. This is observed by physicist. Then they termed this multiplication mv as momentum. It did not happen that momentum = mass * velocity was derived first and then they observed.
 
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  • #67
jbriggs444 said:
For a fixed non-zero momentum (mv) I can have any speed I like (other than zero), any velocity I like (other than zero) and any energy I like (other than zero).

So does it really make sense to say that any of those quantities are dependent on mv?
For a fixed non zero momentum = Does this mean the mass and velocity of the object remains same throughout or mass and velocity vary such a way that their momentum remains constant?
 
  • #68
shivaprasadvh said:
For a fixed non zero momentum = Does this mean the mass and velocity of the object remains same throughout or mass and velocity vary such a way that their momentum remains constant?
You pick a momentum. I am free to pick a mass and a velocity whose product is equal to that momentum.
 
  • #69
Dale said:
The repeated addition thing is not the definition of multiplication. It is the easiest way to teach multiplication to little children who only know addition. The repeated addition thing only even works for multiplication by integers.
And that only if they're non-zero integers.
 
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  • #70
shivaprasadvh said:
Through observations, there are several effects which vary with mass and vary with velocity but are the same if the product mv is same.
Which effects specifically?
 

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