# Why is momentum considered a vector and kinetic energy a scalar?

• I
• regine22

#### regine22

I'm not interested in the mathematical derivation, the mathematical derivation already is based on the assumption that momentum is a vector and kinetic energy is a scalar, thus it proves nothing.

Specifically, what happens if we discuss scalarized momentum? What happens if we discuss vectorized energy? Why don't we?

For momentum, vectorized total momentum is retained in a system, ie. sum of all momentum in one direction is the same before and after an event. Now, just like scalarized total energy is retained, there's no immediately apparent reason scalarized total momentum wouldn't be retained. Where you observe an apparent loss of larger-scale momentum, it can have been converted to momentum elsewhere, eg. momentum of atomic particles, in the form of heat. Is it proven that scalarized total momentum is not retained?

For energy, scalarized total energy is retained in a system, ie. sum of all direction-independent energy is the same before and after an event. Now, there's no immediately apparent reason vectorized total energy wouldn't be retained, just like vectorized total momentum is retained. Momentum and energy are both simply descriptions of movement, they just describe different aspects of the movement, it'd require justification to say one is inherently directional, the other inherently isn't directional. Is it proven that vectorized total energy is not retained?

Sorry if this question is blatantly wrong for obvious reasons, there's no good way to look up answers for this (including asking at university, at which I am studying physics).

I'm not interested in the mathematical derivation, the mathematical derivation already is based on the assumption that momentum is a vector and kinetic energy is a scalar, thus it proves nothing.
A vector is defined in terms of Mathematics (likewise KE and Momentum) so why would you want to have (or even conceive) a non-mathematical based argument about it? Imo, you can't pick and choose when to use Maths as it's just a very convenient language with which to discuss Science.

I think you need to go over your argument, take each arm waving statement and replace it with Maths (that would be only a form of translation into a more concise language which would involve no loss of meaning).

PS JWST was navigated into the right orbit using Maths - not with a qualitative discussion.

• ### Why is momentum considered a vector and kinetic energy a scalar?​

by definition

what happens if we discuss scalarized momentum? What happens if we discuss vectorized energy?
provide a definition of these objects first

• robphy, Vanadium 50, nasu and 2 others
Specifically, what happens if we discuss scalarized momentum? What happens if we discuss vectorized energy? Why don't we?
Possibly the best answer to this comes from Special Relativity. There we have a single four-dimensional energy-momentum four-vector, with energy corresponding to the "time" component and three momentum components, one for each spatial dimension. Vectorising energy, as you put it, would require vectorising time.

For momentum, vectorized total momentum is retained in a system, ie. sum of all momentum in one direction is the same before and after an event. Now, just like scalarized total energy is retained, there's no immediately apparent reason scalarized total momentum wouldn't be retained.
If momentum in each direction is conserved, then the magnitude of the total momentum is conserved. You get that for free. But, conservation of the magnitude of momentum itself is not enough to have the specific law of each component of momentum being conserved. The latter is a much stronger statement.

• Vanadium 50, sophiecentaur, Delta2 and 1 other person
A vector is defined in terms of Mathematics (likewise KE and Momentum) so why would you want to have (or even conceive) a non-mathematical based argument about it?
A vector, defined not using mathematics, is a direction (observational concept, such as rightwards), a base measurement (observational concept, such as ye long), and a quantity (more than one of the base measurement, such as a ye long and then another ye long).

The fact that it's also defined as [a, b, c] doesn't remove from the fact that it has a non physical (edit: non mathematical) definition.

The non physical definition is necessary to understand it. You can have the skill to be able to add two vectors together numerically and not know what the implication of the mathematical definition is for reality.

You can also add two vectors together non mathematically; you just take a ye long in that direction and then a ye long and a ye long in that other direction, and then find out how to get from the starting point to the finishing point, and you have a new vector.

Vectors, as with all quantities in physics I'm aware of, originate in observations. For instance, kinetic energy and momentum both originate in observing movement.

provide a definition of these objects first

In both of these cases, heat is viewed as mechanical movements of atomic-level particles.

Scalarized momentum would be absolute values of momentum. For a mechanical system, scalarized momentum would be sum of absolute values of momentum of particles in the system.

Vectorized energy would be energy in a direction (eg. kinetic energy that is in x direction in a (x,y,z) coordinate system). For a mechanical system, vectorized energy would be sum of energy adjusted for direction. For examples, two objects moving in opposite directions with equal energy would be summed to have zero energy. An object moving upwards in a gravitational field and slowing down would be losing kinetic energy in direction upwards, and gaining potential kinetic energy in direction downwards. Simultaneously, the counterpart (Earth) would likely be gaining equal and opposite energies in the other direction.

• weirdoguy
Vectorized energy would be energy in a direction (eg. kinetic energy that is in x direction in a (x,y,z) coordinate system). For a mechanical system, vectorized energy would be sum of energy adjusted for direction.
Vectorised energy is not conserved. Consider a two-dimensional collision between mass ##m_1## with velocity ##u_1## in the x-direction and a stationary mass ##m_2##. Before the collision, all the energy is in the x-direction. After the collision, there is non-zero energy in the y-direction.

You can see this from conservation of y-momentum, where ##v_1, v_2## are the y-components of velocity after the collision:
$$m_1v_1 = -m_2v_2$$And, unless ##m_1 = m_2## we have ##v_1 \ne v_2## and:
$$\frac 1 2 m_1 v_1^2 \ne \frac 1 2 m_2 v_2^2$$
And we see that energy is not conserved in each direction.

• regine22
In both of these cases, heat is viewed as mechanical movements of atomic-level particles.

Scalarized momentum would be absolute values of momentum. For a mechanical system, scalarized momentum would be sum of absolute values of momentum of particles in the system.
ok There is a very simple way to test whether your theories have any value or not: solve a known problem that has not been solved by classical methods yet. In physics and math only results matter eventually

• robphy
Scalarized momentum would be absolute values of momentum. For a mechanical system, scalarized momentum would be sum of absolute values of momentum of particles in the system.
It's obvious from looking at 2D collisions that scalarized momentum is not conserved. For example, for a collision between equal masses, ##m##, with one at rest (as above). Conservation of x-momentum gives:
$$mu = mv_{1x} + mv_{2x}$$And, conservation of y-momentum gives:
$$mv_{1y} = -mv_{2y}$$And, conservation of energy gives:
$$\frac 1 2 mu^2 = \frac 1 2 m v_1^2 + \frac 1 2 mv_2^2$$And, unless there is no momentum in the y-direction after the collision, we see that your scalarized momentum is not conserved. I.e.
$$mu \ne mv_1 + mv_2$$

• And we see that energy is not conserved in each direction.

And, unless there is no momentum in the y-direction after the collision, we see that your scalarized momentum is not conserved.
Maths checks out on both of these. That rejects all my current ideas for uses for these values.

ok There is a very simple way to test whether your theories have any value or not: solve a known problem that has not been solved by classical methods yet. In physics and math only results matter eventually
The question's not pertaining to the value of the quantities. They could both for example in combination be solving the exact same problems normal, vector, kinetic energy, and normal, scalar, momentum, solve, but be more time-consuming and add no extra functionality, and thusly have no value to learn in physics, but, regardless, could have been interesting to note to be valid.

what is "valid"? Formula ##T=\frac{1}{2}m\boldsymbol v^2## can not be valid or invalid. It is an equation on ##T## can be valid or invalid form the experimental point of view for example

Maths checks out on both of these. That rejects all my current ideas for uses for these values.
Exactly. So conservation of vector momentum (in all cases) and conservation of magnitude of momentum (in all cases) are incompatible. It can only be, at most, one or the other.

As @wrobel notes, you need experimental evidence to decide the issue.

• vanhees71
A vector, defined not using mathematics, is a direction (observational concept, such as rightwards), a base measurement (observational concept, such as ye long), and a quantity (more than one of the base measurement, such as a ye long and then another ye long).
Inside that statement is an implied relationship. Whether you admit to using mathematical operators to describe it, your 'quantities' have no meaning, taken together, unless you include the relationship.
For instance, I could take my mass, The direction you are looking at this moment and the speed that the dog outside is running. There is nothing significant that can be said about those quantities unless you can link them. The Maths is the relationship.

Why do you not want to include Maths - except that, now you have established your position in this discussion, you are not keen on letting go? Could the fact that PF is not agreeing with you be a reason for thinking again? Or can you summon some support to your argument from somewhere?

• weirdoguy and vanhees71
Momentum and energy are both simply descriptions of movement, they just describe different aspects of the movement,
Only momentum is a description of movement. Energy is the quantitative property that must be transferred to a body or physical system to perform work on the body, or to heat it.

How do we measure the quantity of movement, i.e. momentum? From simple observation, we know that the faster something moves, the more movement it has. If there is more mass, then there is also more movement. So the simple definition arises:
$$p = mv$$
The question is now this one: Do we use speed (a scalar) or velocity (a vector)?

We know that when one mass with a certain momentum going against another mass of equal and opposite momentum (but not necessarily the same speed) results in a bigger mass with zero momentum because the speed is now zero. We know that two masses traveling at perpendicular paths and colliding together will result in a bigger mass with the same momentum in their respective direction. So, from simple observations, we know that direction of motion does matter, therefore we must use velocity, not speed, such that:
$$\vec{p} = m\vec{v}$$
And a force, defined as the variation of momentum with respect to time, must therefore also be a vector:
$$\vec{F} = \frac{d\vec{p}}{dt}$$
Work, by definition, is proportional to a force (a vector) and the displacement (also a vector) it traveled, which is represented by a product. But which product? Vectors can be multiplied by a dot product (giving a scalar) or a cross product (resulting in another vector).

Let's look at both cases:
• Cross product: ##\vec{F} \times d\vec{s}##. If the force and the displacement have the same direction, the work done is zero. If they are perpendicular, the work done is proportional to both.
• Dot product: ##\vec{F} \cdot d\vec{s}##. If the force and the displacement have the same direction, the work done is proportional to both. If they are perpendicular, the work done is zero.
Which of these corresponds better to what we can observe? Plus, what does a direction perpendicular to both vectors add as information? If I use heat as my energy source, how the heck do I determine the direction of an equivalent work vector?

Momentum and energy are not assumed to be vector or scalar. They are chosen to be, based on observations.

Bonus: If you want to have more fun, try defining rotational kinetic energy as a vector and show us what the direction of the resulting work represents.

The question's not pertaining to the value of the quantities.
I would say the question is only about the value. If an idea or concept is useful, we keep it around. If it's not, we forget about it. In this particular case, as you and others have noted, momentum as a scalar and kinetic energy as a vector just isn't very useful. They don't match up well with experiments, and you lose the conservation laws on top of that.

• • • 