In classical mechanics, momentum (pl. momenta; SI unit kg m/s, or, equivalently, N·s) is the product of the mass and velocity of an object. For more accurate measures of momentum, see the section "modern definitions of momentum" on this page. It is sometimes referred to as linear momentum to distinguish it from the related subject of angular momentum. Linear momentum is a vector quantity, as it takes into account the direction of the value.
Momentum is a conserved quantity, meaning that the total momentum of any closed system (one not affected by external forces) cannot change.
The concept of momentum in classical mechanics was originated by a number of great thinkers and experimentalists. The first of these was Ibn Sina (Avicenna) circa 1000, who referred to impetus as proportional to weight times velocity.[1] René Descartes later referred to mass times velocity as the fundamental force of motion. Galileo in his Two New Sciences used the term "impeto" (Italian), while Newton's Laws of Motion uses motus (Latin), which has been interpreted by subsequent scholars to mean momentum.[citation needed]
Contents
[hide]
* 1 Momentum in Newtonian mechanics
* 2 Momentum for a system
o 2.1 Relating to mass and velocity
o 2.2 Relating to force
* 3 Conservation of momentum
o 3.1 Elastic collisions
+ 3.1.1 Head-on collision (1 dimensional)
+ 3.1.2 Multi-dimensional collisions
o 3.2 Inelastic collisions
* 4 Modern definitions of momentum
o 4.1 Momentum in relativistic mechanics
o 4.2 Momentum in quantum mechanics
o 4.3 Momentum in electromagnetism
* 5 Figurative use
* 6 See also
* 7 Notes
* 8 References
* 9 External links
[edit] Momentum in Newtonian mechanics
If an object is moving in any reference frame, then it has momentum in that frame. It is important to note that momentum is frame dependent. That is, the same object may have a certain momentum in one frame of reference, but a different amount in another frame. For example, a moving object has momentum in a reference frame fixed to a spot on the ground, while at the same time having 0 momentum in a reference frame attached to the object's center of mass.
The amount of momentum that an object has depends on two physical quantities: the mass and the velocity of the moving object in the frame of reference. In physics, the symbol for momentum is usually denoted by a small bold p (bold because it is a vector); so this can be written:
\mathbf{p}= m \mathbf{v}
where:
p is the momentum
m is the mass
v the velocity
(using bold text for vectors).
The origin of the use of p for momentum is unclear. It has been suggested that, since m had already been used for "mass", the p may be derived from the Latin petere ("to go") or from "progress" (a term used by Leibniz).[citation needed]
The velocity of an object at a particular instant is given by its speed and the direction of its motion at that instant. Because momentum depends on and includes the physical quantity of velocity, it too has a magnitude and a direction and is a vector quantity. For example the momentum of a 5-kg bowling ball would have to be described by the statement that it was moving westward at 2 m/s. It is insufficient to say that the ball has 10 kg m/s of momentum because momentum is not fully described unless its direction is given.
[edit] Momentum for a system
[edit] Relating to mass and velocity
The momentum of a system of objects is the vector sum of the momenta of all the individual objects in the system.
\vec\mathbf{p}= \sum_{i = 1}^n m_i \vec\mathbf{v}_i = m_1 \vec\mathbf{v}_1 + m_2 \vec\mathbf{v}_2 + m_3 \vec\mathbf{v}_3 + ... + m_n \vec\mathbf{v}_n
where
\vec\mathbf{p} is the momentum
mi is the mass of object i
\vec\mathbf{v}_i the vector velocity of object i
n\ is the number of objects in the system
It can be shown that, in the center of mass frame the momentum of a system is zero. Additionally, the momentum in a frame of reference that is moving at a speed \vec\mathbf{v} with respect to that frame is simply:
\vec\mathbf{p}= M\vec\mathbf{v}
where:
M=\sum_{i = 1}^n m_i.
[edit] Relating to force
Force is equal to the rate of change of momentum:
\mathbf{F} = {\mathrm{d}\mathbf{p} \over \mathrm{d}t}.
In the case of constant mass, and velocities much less than the speed of light, this definition results in the equation \mathbf{F} = {\mathrm{d}\mathbf{p} \over \mathrm{d}t} = {\mathrm{d}m \over \mathrm{d}t}\mathbf{v}+ {\mathrm{d}\mathbf{v} \over \mathrm{d}t}m=0+ {\mathrm{d}\mathbf{v} \over \mathrm{d}t}m = m\mathbf{a} , commonly known as Newton's second law.
If a system is in equilibrium, then the change in momentum with respect to time is equal to 0:
\mathbf{F} = {\mathrm{d}\mathbf{p} \over \mathrm{d}t}=\ m\mathbf{a}= 0
[edit] Conservation of momentum
The law of conservation of momentum is a fundamental law of nature, and it states that the total momentum of a closed system of objects (which has no interactions with external agents) is constant. One of the consequences of this is that the centre of mass of any system of objects will always continue with the same velocity unless acted on by a force outside the system.
Conservation of momentum is a mathematical consequence of the homogeneity (shift symmetry) of space (position in space is canonical conjugate quantity to momentum). So, momentum conservation can be philosophically stated as "nothing depends on location per se".
In an isolated system (one where external forces are absent) the total momentum will be constant: this is implied by Newton's first law of motion. Newton's third law of motion, the law of reciprocal actions, which dictates that the forces acting between systems are equal in magnitude, but opposite in sign, is due to the conservation of momentum.
Since position in space is a vector quantity, momentum (being canonical conjugate of position) is a vector quantity as well - it has direction. Thus, when a gun is fired, the final total momentum of the system (the gun and the bullet) is the vector sum of the momentums of these two objects. Assuming that the gun and bullet were at rest prior to firing (meaning the initial momentum of the system was zero), the final total momentum must also equal 0.
In an isolated system with only two objects, the amount of momentum gained by one object must equal the loss of momentum by the other object. Mathematically,
\Delta \mathbf{p}_1 = -\Delta \mathbf{p}_2
Momentum has the special property that, in a closed system, it is always conserved, even in collisions and separations caused by explosive forces. Kinetic energy, on the other hand, is not conserved in collisions if they are inelastic. Since momentum is conserved it can be used to calculate an unknown velocity following a collision or a separation if all the other masses and velocities are known.
A common problem in physics that requires the use of this fact is the collision of two particles. Since momentum is always conserved, the sum of the momenta before the collision must equal the sum of the momenta after the collision:
m_1 \mathbf u_{1} + m_2 \mathbf u_{2} = m_1 \mathbf v_{1} + m_2 \mathbf v_{2} \,
where:
u signifies vector velocity before the collision
v signifies vector velocity after the collision.
Usually, we either only know the velocities before or after a collision and would like to also find out the opposite. Correctly solving this problem means you have to know what kind of collision took place. There are two basic kinds of collisions, both of which conserve momentum:
* Elastic collisions conserve kinetic energy as well as total momentum before and after collision.
* Inelastic collisions don't conserve kinetic energy, but total momentum before and after collision is conserved.
[edit] Elastic collisions
A collision between two pool or snooker balls is a good example of an almost totally elastic collision. In addition to momentum being conserved when the two balls collide, the sum of kinetic energy before a collision must equal the sum of kinetic energy after:
\begin{matrix}\frac{1}{2}\end{matrix} m_1 v_{1,i}^2 + \begin{matrix}\frac{1}{2}\end{matrix} m_2 v_{2,i}^2 = \begin{matrix}\frac{1}{2}\end{matrix} m_1 v_{1,f}^2 + \begin{matrix}\frac{1}{2}\end{matrix} m_2 v_{2,f}^2 \,
Since the 1/2 factor is common to all the terms, it can be taken out right away.
[edit] Head-on collision (1 dimensional)
In the case of two objects colliding head on we find that the final velocity
v_{1,f} = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) v_{1,i} + \left( \frac{2 m_2}{m_1 + m_2} \right) v_{2,i} \,
v_{2,f} = \left( \frac{2 m_1}{m_1 + m_2} \right) v_{1,i} + \left( \frac{m_2 - m_1}{m_1 + m_2} \right) v_{2,i} \,
which can then easily be rearranged to
m_{1,f} \cdot v_{1,f} + m_{2,f} \cdot v_{2,f} = m_{1,i} \cdot v_{1,i} + m_{2,i} \cdot v_{2,i}\,
Special Case: m1>>m2
Now consider if the mass of one body, say m1, is far greater than that of m2 (m1>>m2). In that case m1+m2 is approximately equal to m1. And m1-m2 is approximately equal to m1.
Put these values in the above equation to calculate the value of v2 after collision. The expression changes to v2 final is 2*v1-v2. Its physical interpretation is in case of collision between two body one of which is very heavy, the lighter body moves with twice the velocity of the heavier body less its actual velocity but in opposite direction.
Special Case: m1==m2
Another special case is when the collision is between two bodies of equal mass.
Say body m1 moving at velocity v1 strikes body m2 that is at rest (v2). Putting this case in the equation derived above we will see that after the collision, the body that was moving (m1) will start moving with velocity v2 and the mass m2 will start moving with velocity v1. So there will be an exchange of velocities.
Now suppose one of the masses, say m2, was at rest. In that case after the collision the moving body, m1, will come to rest and the body that was at rest, m2, will start moving with the velocity that m1 had before the collision.
Note that all of these observations are for an elastic collision.
This phenomenon is demonstrated by Newton's cradle, one of the best known examples of conservation of momentum, a real life example of this special case.
[edit] Multi-dimensional collisions
In the case of objects colliding in more than one dimension, as in oblique collisions, the velocity is resolved into orthogonal components with one component perpendicular to the plane of collision and the other component or components in the plane of collision. The velocity components in the plane of collision remain unchanged, while the velocity perpendicular to the plane of collision is calculated in the same way as the one-dimensional case.
For example, in a two-dimensional collision, the momenta can be resolved into x and y components. We can then calculate each component separately, and combine them to produce a vector result. The magnitude of this vector is the final momentum of the isolated system.
See the elastic collision page for more details.
[edit] Inelastic collisions
A common example of a perfectly inelastic collision is when two snowballs collide and then stick together afterwards. This equation describes the conservation of momentum:
m_1 \mathbf v_{1,i} + m_2 \mathbf v_{2,i} = \left( m_1 + m_2 \right) \mathbf v_f \,
It can be shown that a perfectly inelastic collision is one in which the maximum amount of kinetic energy is converted into other forms. For instance, if both objects stick together after the collision and move with a final common velocity, one can always find a reference frame in which the objects are brought to rest by the collision and 100% of the kinetic energy is converted. This is true even in the relativistic case and utilized in particle accelerators to efficiently convert kinetic energy into new forms of mass-energy (i.e. to create massive particles).
See the inelastic collision page for more details.
[edit] Modern definitions of momentum
[edit] Momentum in relativistic mechanics
In relativistic mechanics, in order to be conserved, momentum must be defined as:
\mathbf{p} = \gamma m_0\mathbf{v}
where
m_0\, is the invariant mass of the object moving,
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} is the Lorentz factor
v\, is the relative velocity between an object and an observer
c\, is the speed of light.
Relativistic momentum becomes Newtonian momentum: m\mathbf{v} at low speed \big(\mathbf{v}/c \rightarrow 0 \big).
The diagram can serve as a useful mnemonic for remembering the above relations involving relativistic energy , invariant mass , and relativistic momentum . Please note that in the notation used by the diagram's creator, the invariant mass is subscripted with a zero, .
The diagram can serve as a useful mnemonic for remembering the above relations involving relativistic energy E\,, invariant mass m_0\,, and relativistic momentum p\,. Please note that in the notation used by the diagram's creator, the invariant mass m\, is subscripted with a zero, m_0\,.
Relativistic four-momentum as proposed by Albert Einstein arises from the invariance of four-vectors under Lorentzian translation. The four-momentum is defined as:
\left( {E \over c} , p_x , p_y ,p_z \right)
where
p_x\, is the x\, component of the relativistic momentum,
E \, is the total energy of the system:
E = \gamma m_0c^2 \,
The "length" of the vector is the mass times the speed of light, which is invariant across all reference frames:
(E/c)^2 - p^2 = (mc)^2\,
Momentum of massless objects
Massless objects such as photons also carry momentum. The formula is:
p = \frac{h}{\lambda} = \frac{E}{c}
where
h\, is Planck's constant,
\lambda\, is the wavelength of the photon,
E\, is the energy the photon carries and
c\, is the speed of light.
Generalization of momentum
Momentum is the Noether charge of translational invariance. As such, even fields as well as other things can have momentum, not just particles. However, in curved space-time which is not asymptotically Minkowski, momentum isn't defined at all.
[edit] Momentum in quantum mechanics
In quantum mechanics, momentum is defined as an operator on the wave function. The Heisenberg uncertainty principle defines limits on how accurately the momentum and position of a single observable system can be known at once. In quantum mechanics, position and momentum are conjugate variables.
For a single particle with no electric charge and no spin, the momentum operator can be written in the position basis as
\mathbf{p}={\hbar\over i}\nabla=-i\hbar\nabla
where:
\nabla is the gradient operator
\hbar is the reduced Planck constant.
i = \sqrt{-1} is the imaginary unit.
This is a commonly encountered form of the momentum operator, though not the most general one.
[edit] Momentum in electromagnetism
When electric and/or magnetic fields move, they carry momenta. Light (visible, UV, radio) is an electromagnetic wave and also has momentum. Even though photons (the particle aspect of light) have no mass, they still carry momentum. This leads to applications such as the solar sail.
Momentum is conserved in an electrodynamic system (it may change from momentum in the fields to mechanical momentum of moving parts). The treatment of the momentum of a field is usually accomplished by considering the so-called energy-momentum tensor and the change in time of the Poynting vector integrated over some volume. This is a tensor field which has components related to the energy density and the momentum density.
The definition canonical momentum corresponding to the momentum operator of quantum mechanics when it interacts with the electromagnetic field is, using the principle of least coupling:
\mathbf P = m\mathbf v + q\mathbf A,
instead of the customary
\mathbf p = m\mathbf v,
where:
\mathbf A is the electromagnetic vector potential
m the charged particle's invariant mass
\mathbf v its velocity
q its charge.
[edit] Figurative use
A process may be said to gain momentum. The terminology implies that it requires effort to start such a process, but that it is relatively easy to keep it going. Alternatively, the expression can be seen to reflect that the process is adding adherents, or general acceptance, and thus has more mass at the same velocity; hence, it gained momentum.