Английская Википедия:Four-momentum
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In special relativity, four-momentum (also called momentum–energy or momenergy[1]) is the generalization of the classical three-dimensional momentum to four-dimensional spacetime. Momentum is a vector in three dimensions; similarly four-momentum is a four-vector in spacetime. The contravariant four-momentum of a particle with relativistic energy Шаблон:Mvar and three-momentum Шаблон:Math, where Шаблон:Math is the particle's three-velocity and Шаблон:Mvar the Lorentz factor, is <math display="block">p = \left(p^0 , p^1 , p^2 , p^3\right) = \left(\frac E c , p_x , p_y , p_z\right).</math>
The quantity Шаблон:Math of above is ordinary non-relativistic momentum of the particle and Шаблон:Mvar its rest mass. The four-momentum is useful in relativistic calculations because it is a Lorentz covariant vector. This means that it is easy to keep track of how it transforms under Lorentz transformations.
Minkowski norm
Calculating the Minkowski norm squared of the four-momentum gives a Lorentz invariant quantity equal (up to factors of the speed of light Шаблон:Math) to the square of the particle's proper mass: <math display="block">p \cdot p = \eta_{\mu\nu} p^\mu p^\nu = p_\nu p^\nu = -{E^2 \over c^2} + |\mathbf p|^2 = -m^2 c^2</math> where <math display="block"> \eta_{\mu\nu} = \begin{pmatrix}
-1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1
\end{pmatrix} </math> is the metric tensor of special relativity with metric signature for definiteness chosen to be Шаблон:Math. The negativity of the norm reflects that the momentum is a timelike four-vector for massive particles. The other choice of signature would flip signs in certain formulas (like for the norm here). This choice is not important, but once made it must for consistency be kept throughout.
The Minkowski norm is Lorentz invariant, meaning its value is not changed by Lorentz transformations/boosting into different frames of reference. More generally, for any two four-momenta Шаблон:Mvar and Шаблон:Mvar, the quantity Шаблон:Math is invariant.
Relation to four-velocity
For a massive particle, the four-momentum is given by the particle's invariant mass Шаблон:Mvar multiplied by the particle's four-velocity, <math display="block">p^\mu = m u^\mu,</math> where the four-velocity Шаблон:Mvar is <math display="block"> u = \left(u^0 , u^1 , u^2 , u^3\right) = \gamma_v \left(c , v_x , v_y , v_z\right), </math> and <math display="block">\gamma_v := \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}</math> is the Lorentz factor (associated with the speed <math>v</math>), Шаблон:Math is the speed of light.
Derivation
There are several ways to arrive at the correct expression for four-momentum. One way is to first define the four-velocity Шаблон:Math and simply define Шаблон:Math, being content that it is a four-vector with the correct units and correct behavior. Another, more satisfactory, approach is to begin with the principle of least action and use the Lagrangian framework to derive the four-momentum, including the expression for the energy.[2] One may at once, using the observations detailed below, define four-momentum from the action Шаблон:Mvar. Given that in general for a closed system with generalized coordinates Шаблон:Math and canonical momenta Шаблон:Math,[3] <math display="block">p_i = \frac{\partial S}{\partial q_i} = \frac{\partial S}{\partial x_i}, \quad E = -\frac{\partial S}{\partial t} = - c \cdot \frac{\partial S}{\partial x_0},</math> it is immediate (recalling Шаблон:Math, Шаблон:Math, Шаблон:Math, Шаблон:Math and Шаблон:Math, Шаблон:Math, Шаблон:Math, Шаблон:Math in the present metric convention) that <math display="block">p_\mu = -\frac{\partial S}{\partial x^\mu} = \left({E \over c}, -\mathbf p\right)</math> is a covariant four-vector with the three-vector part being the (negative of) canonical momentum. Шаблон:Hidden begin Consider initially a system of one degree of freedom Шаблон:Mvar. In the derivation of the equations of motion from the action using Hamilton's principle, one finds (generally) in an intermediate stage for the variation of the action, <math display="block">\delta S = \left. \left[ \frac{\partial L}{\partial \dot q}\delta q\right]\right|_{t_1}^{t_2} + \int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot q}\right)\delta q dt.</math>
The assumption is then that the varied paths satisfy Шаблон:Math, from which Lagrange's equations follow at once. When the equations of motion are known (or simply assumed to be satisfied), one may let go of the requirement Шаблон:Math. In this case the path is assumed to satisfy the equations of motion, and the action is a function of the upper integration limit Шаблон:Math, but Шаблон:Math is still fixed. The above equation becomes with Шаблон:Math, and defining Шаблон:Math, and letting in more degrees of freedom, <math display="block">\delta S = \sum_i \frac{\partial L}{\partial \dot{q}_i}\delta q_i = \sum_i p_i \delta q_i.</math>
Observing that <math display="block">\delta S = \sum_i \frac{\partial S}{\partial {q}_i}\delta q_i,</math> one concludes <math display="block">p_i = \frac{\partial S}{\partial q_i}.</math>
In a similar fashion, keep endpoints fixed, but let Шаблон:Math vary. This time, the system is allowed to move through configuration space at "arbitrary speed" or with "more or less energy", the field equations still assumed to hold and variation can be carried out on the integral, but instead observe <math display="block">\frac{dS}{dt} = L</math> by the fundamental theorem of calculus. Compute using the above expression for canonical momenta, <math display="block">
\frac{dS}{dt} = \frac{\partial S}{\partial t} + \sum_i \frac{\partial S}{\partial q_i}\dot{q}_i =
\frac{\partial S}{\partial t} + \sum_i p_i\dot{q}_i = L.
</math>
Now using <math display="block">H = \sum_i p_i \dot{q}_i - L,</math> where Шаблон:Mvar is the Hamiltonian, leads to, since Шаблон:Math in the present case, <math display="block">E = H = -\frac{\partial S}{\partial t}.</math>
Incidentally, using Шаблон:Math with Шаблон:Math in the above equation yields the Hamilton–Jacobi equations. In this context, Шаблон:Mvar is called Hamilton's principal function.
The action Шаблон:Mvar is given by <math display="block">S = -mc\int ds = \int L dt, \quad L = -mc^2\sqrt{1 - \frac{v^2}{c^2}},</math> where Шаблон:Mvar is the relativistic Lagrangian for a free particle. From this, Шаблон:Hidden begin The variation of the action is <math display="block">\delta S = -mc\int \delta ds.</math>
To calculate Шаблон:Math, observe first that Шаблон:Math and that <math display="block">\delta ds^2
= \delta \eta_{\mu\nu}dx^\mu dx^\nu
= \eta_{\mu\nu} \left(\delta \left(dx^\mu\right) dx^\nu + dx^\mu \delta \left(dx^\nu\right)\right)
= 2\eta_{\mu\nu} \delta \left(dx^\mu\right) dx^\nu.
</math>
So <math display="block">\delta ds = \eta_{\mu\nu} \delta dx^\mu \frac{dx^\nu}{ds} = \eta_{\mu\nu} d\delta x^\mu \frac{dx^\nu}{ds},</math> or <math display="block">\delta ds = \eta_{\mu\nu} \frac{d\delta x^\mu}{d\tau} \frac{dx^\nu}{cd\tau}d\tau,</math> and thus <math display="block">\delta S =
-m\int \eta_{\mu\nu} \frac{d\delta x^\mu}{d\tau} \frac{dx^\nu}{d\tau}d\tau =
-m\int \eta_{\mu\nu} \frac{d\delta x^\mu}{d\tau} u^\nu d\tau =
-m\int \eta_{\mu\nu} \left[\frac{d}{d\tau} \left(\delta x^\mu u^\nu\right) - \delta x^\mu\frac{d}{d\tau}u^\nu\right] d\tau
</math> which is just <math display="block">\delta S = \left[-mu_\mu\delta x^\mu\right]_{t_1}^{t_2} + m \int_{t_1}^{t_2} \delta x^\mu\frac{du_\mu}{ds}ds</math>
Шаблон:Hidden end <math display="block">\delta S = \left[ -mu_\mu\delta x^\mu\right]_{t_1}^{t_2} + m\int_{t_1}^{t_2}\delta x^\mu\frac{du_\mu}{ds}ds = -mu_\mu\delta x^\mu = \frac{\partial S}{\partial x^\mu}\delta x^\mu = -p_\mu\delta x^\mu,</math>
where the second step employs the field equations Шаблон:Math, Шаблон:Math, and Шаблон:Math as in the observations above. Now compare the last three expressions to find <math display="block">p^\mu = -\partial^\mu[S] = -\frac{\partial S}{\partial x_\mu} = mu^\mu = m\left(\frac{c}{\sqrt{1 - \frac{v^2}{c^2}}}, \frac{v_x}{\sqrt{1 - \frac{v^2}{c^2}}}, \frac{v_y}{\sqrt{1 - \frac{v^2}{c^2}}}, \frac{v_z}{\sqrt{1 - \frac{v^2}{c^2}}}\right),</math> with norm Шаблон:Math, and the famed result for the relativistic energy,
Шаблон:Equation box 1} = m_{r}c^2,</math>
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where Шаблон:Math is the now unfashionable relativistic mass, follows. By comparing the expressions for momentum and energy directly, one has
that holds for massless particles as well. Squaring the expressions for energy and three-momentum and relating them gives the energy–momentum relation,
Substituting <math display="block">p_\mu \leftrightarrow -\frac{\partial S}{\partial x^\mu}</math> in the equation for the norm gives the relativistic Hamilton–Jacobi equation,[4]
It is also possible to derive the results from the Lagrangian directly. By definition,[5] <math display="block">\begin{align}
\mathbf p &= \frac{\partial L}{\partial \mathbf v}
= \left({\partial L\over \partial \dot x}, {\partial L\over\partial \dot y}, {\partial L\over\partial \dot z}\right)
= m(\gamma v_x, \gamma v_y, \gamma v_z) = m\gamma \mathbf v
= m \mathbf u , \\[3pt]
E &= \mathbf p \cdot \mathbf v - L = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}},
\end{align}</math> which constitute the standard formulae for canonical momentum and energy of a closed (time-independent Lagrangian) system. With this approach it is less clear that the energy and momentum are parts of a four-vector.
The energy and the three-momentum are separately conserved quantities for isolated systems in the Lagrangian framework. Hence four-momentum is conserved as well. More on this below.
More pedestrian approaches include expected behavior in electrodynamics.[6] In this approach, the starting point is application of Lorentz force law and Newton's second law in the rest frame of the particle. The transformation properties of the electromagnetic field tensor, including invariance of electric charge, are then used to transform to the lab frame, and the resulting expression (again Lorentz force law) is interpreted in the spirit of Newton's second law, leading to the correct expression for the relativistic three- momentum. The disadvantage, of course, is that it isn't immediately clear that the result applies to all particles, whether charged or not, and that it doesn't yield the complete four-vector.
It is also possible to avoid electromagnetism and use well tuned experiments of thought involving well-trained physicists throwing billiard balls, utilizing knowledge of the velocity addition formula and assuming conservation of momentum.[7][8] This too gives only the three-vector part.
Conservation of four-momentum
As shown above, there are three conservation laws (not independent, the last two imply the first and vice versa):
- The four-momentum Шаблон:Mvar (either covariant or contravariant) is conserved.
- The total energy Шаблон:Math is conserved.
- The 3-space momentum <math>\mathbf{p} = \left(p^1, p^2, p^3\right)</math> is conserved (not to be confused with the classic non-relativistic momentum <math>m\mathbf{v}</math>).
Note that the invariant mass of a system of particles may be more than the sum of the particles' rest masses, since kinetic energy in the system center-of-mass frame and potential energy from forces between the particles contribute to the invariant mass. As an example, two particles with four-momenta Шаблон:Nowrap and Шаблон:Nowrap each have (rest) mass 3Шаблон:NbspGeV/c2 separately, but their total mass (the system mass) is 10Шаблон:NbspGeV/c2. If these particles were to collide and stick, the mass of the composite object would be 10Шаблон:NbspGeV/c2.
One practical application from particle physics of the conservation of the invariant mass involves combining the four-momenta Шаблон:Math and Шаблон:Math of two daughter particles produced in the decay of a heavier particle with four-momentum Шаблон:Math to find the mass of the heavier particle. Conservation of four-momentum gives Шаблон:Math, while the mass Шаблон:Math of the heavier particle is given by Шаблон:Math. By measuring the energies and three-momenta of the daughter particles, one can reconstruct the invariant mass of the two-particle system, which must be equal to Шаблон:Mvar. This technique is used, e.g., in experimental searches for Z′ bosons at high-energy particle colliders, where the Z′ boson would show up as a bump in the invariant mass spectrum of electron–positron or muon–antimuon pairs.
If the mass of an object does not change, the Minkowski inner product of its four-momentum and corresponding four-acceleration Шаблон:Math is simply zero. The four-acceleration is proportional to the proper time derivative of the four-momentum divided by the particle's mass, so <math display="block">p^\mu A_\mu = \eta_{\mu\nu} p^\mu A^\nu = \eta_{\mu\nu} p^\mu \frac{d}{d\tau} \frac{p^{\nu}}{m} = \frac{1}{2m} \frac{d}{d\tau} p \cdot p = \frac{1}{2m} \frac{d}{d\tau} \left(-m^2c^2\right) = 0 .</math>
Canonical momentum in the presence of an electromagnetic potential
For a charged particle of charge Шаблон:Math, moving in an electromagnetic field given by the electromagnetic four-potential: <math display="block"> A = \left(A^0 , A^1 , A^2 , A^3\right) = \left({\phi \over c}, A_x , A_y , A_z\right) </math> where Шаблон:Mvar is the scalar potential and Шаблон:Math the vector potential, the components of the (not gauge-invariant) canonical momentum four-vector Шаблон:Mvar is <math display="block"> P^\mu = p^\mu + q A^\mu. </math>
This, in turn, allows the potential energy from the charged particle in an electrostatic potential and the Lorentz force on the charged particle moving in a magnetic field to be incorporated in a compact way, in relativistic quantum mechanics.
See also
References
- Шаблон:Cite book
- Шаблон:Cite book
- Шаблон:Cite book
- Шаблон:Cite book
- Шаблон:Cite book
- Шаблон:Cite journal Wikisource version
