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

## 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

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Delta2, sheepy_pasta, weirdoguy and 1 other person

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Ibix
Because observation tells us that ##mv## is a useful quantity, and ##m+v## makes no sense. You can't add quantities that have different dimensionalities.

PhDeezNutz, KDPhysics, sheepy_pasta and 4 others
Because observation tells us that ##mv## is a useful quantity, and ##m+v## makes no sense. You can't add quantities that have different dimensionalities.
Ok so as we know multiplication is repeated addition..
How would explain momentum according to multiplication definition ?
Does this mean we are scaling velocity by multiplying mass.

Delta2
Or more simply i can put the question like
What does multiplication mean? If we can't add two different types of quantities, then how can we multiply two different types of quantities, like p=mv ?

Ibix
You can't add quantities with different units because it makes no sense. ##m+v## is the mathematical expression of a question like "what's 1kg more than 10m/s", which makes no sense at all.

You can multiply quantities with different units because the units also multiply. That's possible because quantities with units are already a multiplication - for example 3m is just saying "three times the length of a metre rule" in a compact way. And constructing compound units such as m/s is also clearly meaningful - it's the mathematical expression of a question like "how many times the length of a metre rule did an object travel in the time it took my clock to tick once".

(I'm aware that modern SI defines units in a slightly more sophisticated way than "here's a lump of metal of a defined length". It doesn't make a difference to the point at hand.)

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PhDeezNutz, Infrared, KDPhysics and 5 others
You can't add quantities with different units because it makes no sense. ##m+v## is the mathematical expression of a question like "what's 1kg more than 10m/s", which makes no sense at all.

You can multiply quantities with different units because the units also multiply. That's possible because quantities with units are already a multiplication - for example 3m is just saying "three times the length of a metre rule" in a compact way. And constructing compound units such as m/s is also clearly meaningful - it's the mathematical expression of a question like "how many times the length of a metre rule did an object travel in the time it took my clock to tick once".

(I'm aware that modern SI defines units in a slightly more sophisticated way than "here's a lump of metal of a defined length". It doesn't make a difference to the point at hand.)
But definition of multiplication doesn't go hand in hand with this momentum thing.. that's all i am saying.

You can't add quantities with different units because it makes no sense. ##m+v## is the mathematical expression of a question like "what's 1kg more than 10m/s", which makes no sense at all.

You can multiply quantities with different units because the units also multiply. That's possible because quantities with units are already a multiplication - for example 3m is just saying "three times the length of a metre rule" in a compact way. And constructing compound units such as m/s is also clearly meaningful - it's the mathematical expression of a question like "how many times the length of a metre rule did an object travel in the time it took my clock to tick once".

(I'm aware that modern SI defines units in a slightly more sophisticated way than "here's a lump of metal of a defined length". It doesn't make a difference to the point at hand.)
m/s is how man times the length of a meter rule object travelled in one tick of my clock.

So how would mathematically define this "kg.m/s" ?

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 m/s we will get 2 kg+m/s = 501.5 g+m/s, which can't be derived independently from the original 2 kg+m/s without knowing the separate values that went into it. We'll have to keep track of this values all the time (kind of like that: 1kg+1m/s=1000g+1m/s) so there was no point in adding the numbers in the first place.
The multiplied value holds its own as an independent value under the change of units.

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Delta2
Stephen Tashi
m/s is how man times the length of a meter rule object travelled in one tick of my clock.
Physical units can have more than one interpretation. One example of the interpretation of m/s is as a velocity, but there can be different examples.

So how would mathematically define this "kg.m/s"
That question is not addressed by mathematics. For example, trigonometry does not tell us that the hypoteneuse of a triangle must be a ladder leaning against a wall. Properties of a ladder leaning against a wall might be analyzed by using trigonometry, but the mathematics of trigonometry does not tell us that only one particular application of trigonometry exists.

sysprog
A.T.
Ok so as we know multiplication is repeated addition..
How do you write √2 * π as repeated addition?

PeroK
Homework Helper
Gold Member
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
If you think of a mass ##M## as being composed of a lot of smaller masses, ##m##, all moving together at some velocity ##v##. Let's say 100 small masses make up the larger mass.

If you add mass and velocity, the the momentum of each component would be ##p = m + v## and the momentum of the whole thing would be ##P = 100(m + v) = 100m + 100v##.

But, if you calculate the momentum of the large mass directly you get ##P = M + v = 100m + v##. And you get a different answer.

Adding two quantities in this way makes no sense.

Whereas, if you multiply quantities, then we have ##p = mv##, and ##P = 100mv = Mv##, and it all makes sense.

Merlin3189 and Ibix
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
I think when people started studying motion,they didn't knew that their is quantity momentum which remains conserve but when they started doing maths they found that this quantity mv remains same and they gave name momentum to it.
It would be like saying something like why cucumber is cucumber not orange.

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A.T.
Whereas, if you multiply quantities, then we have ##p = mv##, and ##P = 100mv = Mv##, and it all makes sense.
It makes sense formally, but it also has to be useful somehow, which it is because it's conserved.

sophiecentaur
Gold Member
But definition of multiplication doesn't go hand in hand with this momentum thing.. that's all i am saying.
Take a step backwards. When did you first learn the definition of Multiplication as repeated additions? I suspect it was when you were less than ten years old. Since then, the more advanced concepts of Maths crept into your awareness - even if you never did an advanced Maths Analysis course and you just do it mechanically. However, once you go beyond integers in the application of that definition then you have to introduce implied multiplication and insert that in your 'multiple additions' definition. It has to pull itself up by its own bootstraps - especially when the numbers are not even rational.

The 'rules' about what you are allowed to do about dimensions are pretty straightforward and are actually pretty intuitive. You can multiply different quantities and get a meaningful answer. Amp Hours as the capacity of a battery make sense because you can use the same capacity battery for twice as long by taking half the current etc. etc. etc.. But what would adding possibly 10A to 5H mean? 15 'Somethings'? Equivalent to 1A plus 14H?? Clearly not.

Once you have established the principle of what does and what dean;t make sense then you OP question is answered.

Leo Liu, Ibix and PeroK
Dale
Mentor
But definition of multiplication doesn't go hand in hand with this momentum thing.. that's all i am saying.
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.

PhDeezNutz and Hemant
Meir Achuz
Homework Helper
Gold Member
How could a joke get so many answers?

Hamiltonian299792458, Leo Liu and Chestermiller
PeroK
Homework Helper
Gold Member
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.
That's how mutlipication is defined for positive integers. Multiplication of the rationals and the reals is defined by a systematic generalisation.

etotheipi, Dale and jbriggs444
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.
My whole life was a lie.

PeroK
Dale
Mentor
My whole life was a lie.
Well, I may be wrong about that. I cannot find an independent definition of real multiplication. So don’t question your life quite yet!

sophiecentaur and Hemant
Paul Colby
Gold Member
Well, in the grand scheme of things using dimensional real numbers to quantify things in the physical sciences is really a rather remarkable process. One could argue all day how "intuitive" it is but in the end the real reason is it's observed to work really well.

PhDeezNutz, etotheipi and akashpandey
PeroK
Homework Helper
Gold Member
Well, I may be wrong about that. I cannot find an independent definition of real multiplication. So don’t question your life quite yet!
Addition and multiplication of the rationals is, of course, defined by:
$$\frac a b + \frac c d = \frac {ad + bc}{bd}, \ \ \ \frac a b \times \frac c d = \frac {ac}{bd}$$
Where ##a, b, c, d## are integers, with ##bd \ne 0##.

Multiplication of the Real numbers depends on how you construct them from the rationals. One approach is to define a real number as an equivalence class of sequences of rationals. Then you have to show that if you take two such real numbers, then the product can be well-defined by multiplying the sequences term by term etc.

As has been mentioned before constructing the real numbers from the rationals is a non-trivial exercise.

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Stephen Tashi and Dale
martinbn
Multiplication of the rationals is, of course, defined by:
Where a,b,c,d are integers, with bd≠0.

Multiplication of the Real numbers depends on how you construct them from the rationals. One approach is to define a real number as an equivalence class of sequences of rationals. Then you have to show that if you take two such real numbers, then the product can be well-defined by multiplying the sequences term by term etc.

As has been mentioned before constructing the real numbers from the rationals is a non-trivial exercise.
Well, that is addition. multiplactions is ##\frac{ac}{bd}##.

PeroK
PeroK
Homework Helper
Gold Member
Well, that is addition. multiplactions is ##\frac{ac}{bd}##.
So it is! I better correct it.

martinbn
anorlunda
Staff Emeritus
Even with repeated addition, you are still not adding things with different units.

1 apple x 3 means 1 apple + 1 apple + 1 apple. You are not adding the 3 to the 1.

50 mph *2 hours = (50 mph *1 hour) + (50 mph * 1 hour) = 50 miles + 50 miles = 100 miles

cnh1995 and russ_watters