Relativistic mechanics

The relativistic four-velocity, that is the four-vector representing velocity in relativity, is defined as follows:

${\displaystyle {\boldsymbol {\mathbf {U} }}={\frac {d{\boldsymbol {\mathbf {X} }}}{d\tau }}=\left({\frac {cdt}{d\tau }},{\frac {d\mathbf {x} }{d\tau }}\right)}$

In the above, ${\displaystyle {\tau }}$ is the proper time of the path through spacetime, called the world-line, followed by the object velocity the above represents, and

${\displaystyle {\boldsymbol {\mathbf {X} }}=(ct,\mathbf {x} )}$

is the four-position; the coordinates of an event. Due to time dilation, the proper time is the time between two events in a frame of reference where they take place at the same location. The proper time is related to coordinate time t by:

${\displaystyle {\frac {d\tau }{dt}}={\frac {1}{\gamma (\mathbf {v} )}}}$

where ${\displaystyle {\gamma }(\mathbf {v} )}$ is the Lorentz factor:

${\displaystyle \gamma (\mathbf {v} )={\frac {1}{\sqrt {1-\mathbf {v} \cdot \mathbf {v} /c^{2}}}}\,\rightleftharpoons \,\gamma (v)={\frac {1}{\sqrt {1-(v/c)^{2}}}}.}$

(either version may be quoted) so it follows:

${\displaystyle {\boldsymbol {\mathbf {U} }}=\gamma (\mathbf {v} )(c,\mathbf {v} )}$

The first three terms, excepting the factor of ${\displaystyle {\gamma (\mathbf {v} )}}$, is the velocity as seen by the observer in their own reference frame. The ${\displaystyle {\gamma (\mathbf {v} )}}$ is determined by the velocity ${\displaystyle \mathbf {v} }$ between the observer's reference frame and the object's frame, which is the frame in which its proper time is measured. This quantity is invariant under Lorentz transformation, so to check to see what an observer in a different reference frame sees, one simply multiplies the velocity four-vector by the Lorentz transformation matrix between the two reference frames.

The mass of an object as measured in its own frame of reference is called its rest mass or invariant mass and is sometimes written ${\displaystyle m_{0}}$. If an object moves with velocity ${\displaystyle \mathbf {v} }$ in some other reference frame, the quantity ${\displaystyle m=\gamma (\mathbf {v} )m_{0}}$ is often called the object's "relativistic mass" in that frame.^[1] Some authors use ${\displaystyle m}$ to denote rest mass, but for the sake of clarity this article will follow the convention of using ${\displaystyle m}$ for relativistic mass and ${\displaystyle m_{0}}$ for rest mass.^[2]

Lev Okun has suggested that the concept of relativistic mass "has no rational justification today" and should no longer be taught.^[3] Other physicists, including Wolfgang Rindler and T. R. Sandin, contend that the concept is useful.^[4] See mass in special relativity for more information on this debate.

A particle whose rest mass is zero is called massless. Photons and gravitons are thought to be massless, and neutrinos are nearly so.

There are a couple of (equivalent) ways to define momentum and energy in SR. One method uses conservation laws. If these laws are to remain valid in SR they must be true in every possible reference frame. However, if one does some simple thought experiments using the Newtonian definitions of momentum and energy, one sees that these quantities are not conserved in SR. One can rescue the idea of conservation by making some small modifications to the definitions to account for relativistic velocities. It is these new definitions which are taken as the correct ones for momentum and energy in SR.

The four-momentum of an object is straightforward, identical in form to the classical momentum, but replacing 3-vectors with 4-vectors:

${\displaystyle {\boldsymbol {\mathbf {P} }}=m_{0}{\boldsymbol {\mathbf {U} }}=(E/c,\mathbf {p} )}$

The energy and momentum of an object with invariant mass ${\displaystyle m_{0}}$, moving with velocity ${\displaystyle \mathbf {v} }$ with respect to a given frame of reference, are respectively given by

${\displaystyle {\begin{aligned}E&=\gamma (\mathbf {v} )m_{0}c^{2}\\\mathbf {p} &=\gamma (\mathbf {v} )m_{0}\mathbf {v} \end{aligned}}}$

The factor ${\displaystyle \gamma }$ comes from the definition of the four-velocity described above. The appearance of ${\displaystyle \gamma }$ may be stated in an alternative way, which will be explained in the next section.

The kinetic energy, ${\displaystyle K}$, is defined as (see #Kinetic energy)

${\displaystyle K=(\gamma -1)m_{0}c^{2}=E-m_{0}c^{2}\,,}$

and the speed as a function of kinetic energy is given by

${\displaystyle v=c{\sqrt {1-\left({\frac {m_{0}c^{2}}{K+m_{0}c^{2}}}\right)^{2}}}={\frac {c{\sqrt {K(K+2m_{0}c^{2})}}}{K+m_{0}c^{2}}}={\frac {c{\sqrt {(E-m_{0}c^{2})(E+m_{0}c^{2})}}}{E}}={\frac {pc^{2}}{E}}\,.}$

The spatial momentum may be written as ${\displaystyle \mathbf {p} =m\mathbf {v} }$, preserving the form from Newtonian mechanics with relativistic mass substituted for Newtonian mass. However, this substitution fails for some quantities, including force and kinetic energy. Moreover, the relativistic mass is not invariant under Lorentz transformations, while the rest mass is. For this reason, many people prefer to use the rest mass and account for ${\displaystyle \gamma }$ explicitly through the 4-velocity or coordinate time.

A simple relation between energy, momentum, and velocity may be obtained from the definitions of energy and momentum by multiplying the energy by ${\displaystyle \mathbf {v} }$, multiplying the momentum by ${\displaystyle c^{2}}$, and noting that the two expressions are equal. This yields

${\displaystyle \mathbf {p} c^{2}=E\mathbf {v} }$

${\displaystyle \mathbf {v} }$ may then be eliminated by dividing this equation by ${\displaystyle c}$ and squaring,

${\displaystyle (pc)^{2}=E^{2}(v/c)^{2}}$

dividing the definition of energy by ${\displaystyle \gamma }$ and squaring,

${\displaystyle E^{2}\left(1-(v/c)^{2}\right)=\left(m_{0}c^{2}\right)^{2}}$

and substituting:

${\displaystyle E^{2}-(pc)^{2}=\left(m_{0}c^{2}\right)^{2}}$

This is the relativistic energy–momentum relation.

While the energy ${\displaystyle E}$ and the momentum ${\displaystyle \mathbf {p} }$ depend on the frame of reference in which they are measured, the quantity ${\displaystyle E^{2}-(pc)^{2}}$ is invariant. Its value is ${\displaystyle -c^{2}}$ times the squared magnitude of the 4-momentum vector.

The invariant mass of a system may be written as

${\displaystyle {m_{0}}_{\text{tot}}={\frac {\sqrt {E_{\text{tot}}^{2}-(p_{\text{tot}}c)^{2}}}{c^{2}}}}$

Due to kinetic energy and binding energy, this quantity is different from the sum of the rest masses of the particles of which the system is composed. Rest mass is not a conserved quantity in special relativity, unlike the situation in Newtonian physics. However, even if an object is changing internally, so long as it does not exchange energy or momentum with its surroundings, its rest mass will not change and can be calculated with the same result in any reference frame.

The relativistic energy–momentum equation holds for all particles, even for massless particles for which m₀ = 0. In this case:

${\displaystyle E=pc}$

When substituted into Ev = c²p, this gives v = c: massless particles (such as photons) always travel at the speed of light.

Notice that the rest mass of a composite system will generally be slightly different from the sum of the rest masses of its parts since, in its rest frame, their kinetic energy will increase its mass and their (negative) binding energy will decrease its mass. In particular, a hypothetical "box of light" would have rest mass even though made of particles which do not since their momenta would cancel.

Looking at the above formula for invariant mass of a system, one sees that, when a single massive object is at rest (v = 0, p = 0), there is a non-zero mass remaining: m₀ = E/c². The corresponding energy, which is also the total energy when a single particle is at rest, is referred to as "rest energy". In systems of particles which are seen from a moving inertial frame, total energy increases and so does momentum. However, for single particles the rest mass remains constant, and for systems of particles the invariant mass remain constant, because in both cases, the energy and momentum increases subtract from each other, and cancel. Thus, the invariant mass of systems of particles is a calculated constant for all observers, as is the rest mass of single particles.

For systems of particles, the energy–momentum equation requires summing the momentum vectors of the particles:

${\displaystyle E^{2}-\mathbf {p} \cdot \mathbf {p} c^{2}=m_{0}^{2}c^{4}}$

The inertial frame in which the momenta of all particles sums to zero is called the center of momentum frame. In this special frame, the relativistic energy–momentum equation has p = 0, and thus gives the invariant mass of the system as merely the total energy of all parts of the system, divided by c²

${\displaystyle m_{0,\,{\rm {system}}}=\sum _{n}E_{n}/c^{2}}$

This is the invariant mass of any system which is measured in a frame where it has zero total momentum, such as a bottle of hot gas on a scale. In such a system, the mass which the scale weighs is the invariant mass, and it depends on the total energy of the system. It is thus more than the sum of the rest masses of the molecules, but also includes all the totaled energies in the system as well. Like energy and momentum, the invariant mass of isolated systems cannot be changed so long as the system remains totally closed (no mass or energy allowed in or out), because the total relativistic energy of the system remains constant so long as nothing can enter or leave it.

An increase in the energy of such a system which is caused by translating the system to an inertial frame which is not the center of momentum frame, causes an increase in energy and momentum without an increase in invariant mass. E = m₀c², however, applies only to isolated systems in their center-of-momentum frame where momentum sums to zero.

Taking this formula at face value, we see that in relativity, mass is simply energy by another name (and measured in different units). In 1927 Einstein remarked about special relativity, "Under this theory mass is not an unalterable magnitude, but a magnitude dependent on (and, indeed, identical with) the amount of energy."^[5]

In a "totally-closed" system (i.e., isolated system) the total energy, the total momentum, and hence the total invariant mass are conserved. Einstein's formula for change in mass translates to its simplest ΔE = Δmc² form, however, only in non-closed systems in which energy is allowed to escape (for example, as heat and light), and thus invariant mass is reduced. Einstein's equation shows that such systems must lose mass, in accordance with the above formula, in proportion to the energy they lose to the surroundings. Conversely, if one can measure the differences in mass between a system before it undergoes a reaction which releases heat and light, and the system after the reaction when heat and light have escaped, one can estimate the amount of energy which escapes the system.

In both nuclear and chemical reactions, such energy represents the difference in binding energies of electrons in atoms (for chemistry) or between nucleons in nuclei (in atomic reactions). In both cases, the mass difference between reactants and (cooled) products measures the mass of heat and light which will escape the reaction, and thus (using the equation) give the equivalent energy of heat and light which may be emitted if the reaction proceeds.

In chemistry, the mass differences associated with the emitted energy are around 10⁻⁹ of the molecular mass.^[6] However, in nuclear reactions the energies are so large that they are associated with mass differences, which can be estimated in advance, if the products and reactants have been weighed (atoms can be weighed indirectly by using atomic masses, which are always the same for each nuclide). Thus, Einstein's formula becomes important when one has measured the masses of different atomic nuclei. By looking at the difference in masses, one can predict which nuclei have stored energy that can be released by certain nuclear reactions, providing important information which was useful in the development of nuclear energy and, consequently, the nuclear bomb. Historically, for example, Lise Meitner was able to use the mass differences in nuclei to estimate that there was enough energy available to make nuclear fission a favorable process. The implications of this special form of Einstein's formula have thus made it one of the most famous equations in all of science.

The equation E = m₀c² applies only to isolated systems in their center of momentum frame. It has been popularly misunderstood to mean that mass may be converted to energy, after which the mass disappears. However, popular explanations of the equation as applied to systems include open (non-isolated) systems for which heat and light are allowed to escape, when they otherwise would have contributed to the mass (invariant mass) of the system.

Historically, confusion about mass being "converted" to energy has been aided by confusion between mass and "matter", where matter is defined as fermion particles. In such a definition, electromagnetic radiation and kinetic energy (or heat) are not considered "matter". In some situations, matter may indeed be converted to non-matter forms of energy (see above), but in all these situations, the matter and non-matter forms of energy still retain their original mass.

For isolated systems (closed to all mass and energy exchange), mass never disappears in the center of momentum frame, because energy cannot disappear. Instead, this equation, in context, means only that when any energy is added to, or escapes from, a system in the center-of-momentum frame, the system will be measured as having gained or lost mass, in proportion to energy added or removed. Thus, in theory, if an atomic bomb were placed in a box strong enough to hold its blast, and detonated upon a scale, the mass of this closed system would not change, and the scale would not move. Only when a transparent "window" was opened in the super-strong plasma-filled box, and light and heat were allowed to escape in a beam, and the bomb components to cool, would the system lose the mass associated with the energy of the blast. In a 21 kiloton bomb, for example, about a gram of light and heat is created. If this heat and light were allowed to escape, the remains of the bomb would lose a gram of mass, as it cooled. In this thought-experiment, the light and heat carry away the gram of mass, and would therefore deposit this gram of mass in the objects that absorb them.^[7]

In relativistic mechanics, the time-varying mass moment

${\displaystyle \mathbf {N} =m\left(\mathbf {x} -t\mathbf {v} \right)}$

and orbital 3-angular momentum

${\displaystyle \mathbf {L} =\mathbf {x} \times \mathbf {p} }$

of a point-like particle are combined into a four-dimensional bivector in terms of the 4-position X and the 4-momentum P of the particle:^[8][9]

${\displaystyle \mathbf {M} =\mathbf {X} \wedge \mathbf {P} }$

where ∧ denotes the exterior product. This tensor is additive: the total angular momentum of a system is the sum of the angular momentum tensors for each constituent of the system. So, for an assembly of discrete particles one sums the angular momentum tensors over the particles, or integrates the density of angular momentum over the extent of a continuous mass distribution.

Each of the six components forms a conserved quantity when aggregated with the corresponding components for other objects and fields.

In special relativity, Newton's second law does not hold in the form F = ma, but it does if it is expressed as

${\displaystyle \mathbf {F} ={\frac {d\mathbf {p} }{dt}}}$

where p = γ(v)m₀v is the momentum as defined above and m₀ is the invariant mass. Thus, the force is given by

${\displaystyle \mathbf {F} =\gamma ^{3}m_{0}\,\mathbf {a} _{\parallel }+\gamma m_{0}\,\mathbf {a} _{\perp }\ \mathrm {where} \ \gamma =\gamma (\mathbf {v} )}$

Derivation

Starting from ${\displaystyle \mathbf {F} =m_{0}{\frac {d(\gamma (\mathbf {v} )\,\mathbf {v} )}{dt}}=m_{0}\left({\frac {d\gamma (\mathbf {v} )}{dt}}\,\mathbf {v} +\gamma (\mathbf {v} ){\frac {d\mathbf {v} }{dt}}\right).}$ Carrying out the derivatives gives ${\displaystyle \mathbf {F} ={\frac {\gamma (\mathbf {v} )^{3}m_{0}}{c^{2}}}\left(\mathbf {v} \cdot \mathbf {a} \right)\,\mathbf {v} +\gamma (\mathbf {v} )m_{0}\,\mathbf {a} .}$ If the acceleration is separated into the part parallel to the velocity (a∥) and the part perpendicular to it (a⊥), so that: ${\displaystyle \mathbf {a} =\mathbf {a} _{\parallel }+\mathbf {a} _{\perp }\,,\quad \mathbf {v} \cdot \mathbf {a} _{\perp }=0\,,\quad \mathbf {v} \cdot \mathbf {a} =\mathbf {v} \cdot \mathbf {a} _{\parallel }\,,}$ one gets ${\displaystyle \mathbf {F} ={\frac {\gamma (\mathbf {v} )^{3}m_{0}}{c^{2}}}\left(\mathbf {v} \cdot \mathbf {a} _{\parallel }\right)\,\mathbf {v} +\gamma (\mathbf {v} )m_{0}\,(\mathbf {a} _{\perp }+\mathbf {a} _{\parallel })\,.}$ By construction a∥ and v are parallel, so (v·a∥)v is a vector with magnitude v2a∥ in the direction of v (and hence a∥) which allows the replacement: ${\displaystyle (\mathbf {v} \cdot \mathbf {a} _{\parallel })\mathbf {v} =v^{2}\mathbf {a} _{\parallel }}$ then ${\displaystyle {\begin{aligned}\mathbf {F} &={\frac {\gamma (\mathbf {v} )^{3}m_{0}v^{2}}{c^{2}}}\,\mathbf {a} _{\parallel }+\gamma (\mathbf {v} )m_{0}\,(\mathbf {a} _{\parallel }+\mathbf {a} _{\perp })\\&=\gamma (\mathbf {v} )^{3}m_{0}\left({\frac {v^{2}}{c^{2}}}+{\frac {1}{\gamma (\mathbf {v} )^{2}}}\right)\mathbf {a} _{\parallel }+\gamma (\mathbf {v} )m_{0}\,\mathbf {a} _{\perp }\\&=\gamma (\mathbf {v} )^{3}m_{0}\left({\frac {v^{2}}{c^{2}}}+1-{\frac {v^{2}}{c^{2}}}\right)\mathbf {a} _{\parallel }+\gamma (\mathbf {v} )m_{0}\,\mathbf {a} _{\perp }\\&=\gamma (\mathbf {v} )^{3}m_{0}\,\mathbf {a} _{\parallel }+\gamma (\mathbf {v} )m_{0}\,\mathbf {a} _{\perp }\end{aligned}}\,}$

Consequently, in some old texts, γ(v)³m₀ is referred to as the longitudinal mass, and γ(v)m₀ is referred to as the transverse mass, which is numerically the same as the relativistic mass. See mass in special relativity.

If one inverts this to calculate acceleration from force, one gets

${\displaystyle \mathbf {a} ={\frac {1}{m_{0}\gamma (\mathbf {v} )}}\left(\mathbf {F} -{\frac {(\mathbf {v} \cdot \mathbf {F} )\mathbf {v} }{c^{2}}}\right)\,.}$

The force described in this section is the classical 3-D force which is not a four-vector. This 3-D force is the appropriate concept of force since it is the force which obeys Newton's third law of motion. It should not be confused with the so-called four-force which is merely the 3-D force in the comoving frame of the object transformed as if it were a four-vector. However, the density of 3-D force (linear momentum transferred per unit four-volume) is a four-vector (density of weight +1) when combined with the negative of the density of power transferred.

The torque acting on a point-like particle is defined as the derivative of the angular momentum tensor given above with respect to proper time:^[10][11]

${\displaystyle {\boldsymbol {\Gamma }}={\frac {d\mathbf {M} }{d\tau }}=\mathbf {X} \wedge \mathbf {F} }$

or in tensor components:

${\displaystyle \Gamma _{\alpha \beta }=X_{\alpha }F_{\beta }-X_{\beta }F_{\alpha }}$

where F is the 4d force acting on the particle at the event X. As with angular momentum, torque is additive, so for an extended object one sums or integrates over the distribution of mass.

The work-energy theorem says^[12] the change in kinetic energy is equal to the work done on the body. In special relativity:

${\displaystyle {\begin{aligned}\Delta K=W=[\gamma _{1}-\gamma _{0}]m_{0}c^{2}.\end{aligned}}}$

Derivation

${\displaystyle {\begin{aligned}\Delta K=W&=\int _{\mathbf {x} _{0}}^{\mathbf {x} _{1}}\mathbf {F} \cdot d\mathbf {x} \\&=\int _{t_{0}}^{t_{1}}{\frac {d}{dt}}(\gamma m_{0}\mathbf {v} )\cdot \mathbf {v} dt\\&=\left.\gamma m_{0}\mathbf {v} \cdot \mathbf {v} \right|_{t_{0}}^{t_{1}}-\int _{t_{0}}^{t_{1}}\gamma m_{0}\mathbf {v} \cdot {\frac {d\mathbf {v} }{dt}}dt\\&=\left.\gamma m_{0}v^{2}\right|_{t_{0}}^{t_{1}}-m_{0}\int _{v_{0}}^{v_{1}}\gamma v\,dv\\&=m_{0}\left(\left.\gamma v^{2}\right|_{t_{0}}^{t_{1}}-c^{2}\int _{v_{0}}^{v_{1}}{\frac {2v/c^{2}}{2{\sqrt {1-v^{2}/c^{2}}}}}\,dv\right)\\&=\left.m_{0}\left({\frac {v^{2}}{\sqrt {1-v^{2}/c^{2}}}}+c^{2}{\sqrt {1-v^{2}/c^{2}}}\right)\right|_{t_{0}}^{t_{1}}\\&=\left.{\frac {m_{0}c^{2}}{\sqrt {1-v^{2}/c^{2}}}}\right|_{t_{0}}^{t_{1}}\\&=\left.{\gamma m_{0}c^{2}}\right|_{t_{0}}^{t_{1}}\\&=\gamma _{1}m_{0}c^{2}-\gamma _{0}m_{0}c^{2}.\end{aligned}}}$

If in the initial state the body was at rest, so v₀ = 0 and γ₀(v₀) = 1, and in the final state it has speed v₁ = v, setting γ₁(v₁) = γ(v), the kinetic energy is then;

${\displaystyle K=[\gamma (v)-1]m_{0}c^{2}\,,}$

a result that can be directly obtained by subtracting the rest energy m₀c² from the total relativistic energy γ(v)m₀c².

The Lorentz factor γ(v) can be expanded into a Taylor series or binomial series for (v/c)² < 1, obtaining:

${\displaystyle \gamma ={\dfrac {1}{\sqrt {1-(v/c)^{2}}}}=\sum _{n=0}^{\infty }\left({\dfrac {v}{c}}\right)^{2n}\prod _{k=1}^{n}\left({\dfrac {2k-1}{2k}}\right)=1+{\dfrac {1}{2}}\left({\dfrac {v}{c}}\right)^{2}+{\dfrac {3}{8}}\left({\dfrac {v}{c}}\right)^{4}+{\dfrac {5}{16}}\left({\dfrac {v}{c}}\right)^{6}+\cdots }$

and consequently

${\displaystyle E-m_{0}c^{2}={\frac {1}{2}}m_{0}v^{2}+{\frac {3}{8}}{\frac {m_{0}v^{4}}{c^{2}}}+{\frac {5}{16}}{\frac {m_{0}v^{6}}{c^{4}}}+\cdots ;}$

${\displaystyle \mathbf {p} =m_{0}\mathbf {v} +{\frac {1}{2}}{\frac {m_{0}v^{2}\mathbf {v} }{c^{2}}}+{\frac {3}{8}}{\frac {m_{0}v^{4}\mathbf {v} }{c^{4}}}+{\frac {5}{16}}{\frac {m_{0}v^{6}\mathbf {v} }{c^{6}}}+\cdots .}$

For velocities much smaller than that of light, one can neglect the terms with c² and higher in the denominator. These formulas then reduce to the standard definitions of Newtonian kinetic energy and momentum. This is as it should be, for special relativity must agree with Newtonian mechanics at low velocities.