Fact-checked by Grok 2 weeks ago

Center-of-momentum frame

The center-of-momentum frame (COM frame), also referred to as the center-of-mass frame in relativistic mechanics, is a reference frame in which the total three-momentum of a system of particles is zero, ensuring that the system's overall motion is at rest while internal dynamics can be analyzed independently.^[1] This frame is defined such that the spatial components of the total four-momentum vector vanish, leaving only a timelike component equal to the total energy E divided by the speed of light c, which corresponds to the invariant mass M of the system via E = Mc^2.^[1] In special relativity, the COM frame simplifies calculations for processes like particle collisions and decays by maximizing the available energy for outcomes such as particle creation, as all kinetic energy associated with the system's bulk motion is eliminated.^[2] In the COM frame, conservation laws for energy and momentum become particularly tractable, especially for two-body interactions where incoming particles have equal and opposite momenta.^[3] For instance, in elastic collisions of identical particles, velocities transform under Lorentz boosts to yield symmetric configurations before and after the interaction, highlighting the frame's utility in verifying relativistic invariance.^[3] This frame contrasts with laboratory frames, where total momentum is nonzero, often complicating analyses; transforming to the COM frame aligns the total energy-momentum along the time axis, enabling the use of invariants like M^2 = (E_1 + E_2)^2/c^4 - (\mathbf{p}_1 + \mathbf{p}_2)^2/c^2 to predict possible reaction products.^[1] The COM frame is foundational in particle physics experiments, such as those at accelerators like the LHC, where it facilitates the interpretation of scattering cross-sections and decay kinematics by isolating intrinsic properties from boost effects.^[2] Its relativistic formulation extends classical center-of-mass concepts but accounts for the interdependence of energy and momentum, ensuring that the frame's velocity relative to the lab is given by the total momentum divided by the total energy over c^2.^[1]

Definition

In Classical Mechanics

In classical mechanics, the center-of-momentum frame is defined as the inertial reference frame in which the vector sum of the momenta of all particles in the system is zero.^[4] This frame moves with a constant velocity relative to any laboratory frame, ensuring that the total linear momentum vanishes, which simplifies the analysis of the system's internal dynamics.^[5] The velocity of the center-of-momentum frame, \vec{V}_c, is calculated as the total momentum divided by the total mass:

\vec{V}_c = \frac{\sum_i m_i \vec{v}_i}{\sum_i m_i},

where m_i is the mass of the i-th particle and \vec{v}_i is its velocity in the laboratory frame.^[4]^[5] To transform the velocities of the particles to this frame, apply the Galilean velocity shift:

\vec{v}'_i = \vec{v}_i - \vec{V}_c

for each particle, resulting in \sum_i m_i \vec{v}'_i = 0.^[4]^[5] Consider a simple two-particle system with masses m_1 and m_2. In the center-of-momentum frame, the velocities are \vec{v}'_1 = \frac{m_2}{m_1 + m_2} (\vec{v}_1 - \vec{v}_2) and \vec{v}'_2 = -\frac{m_1}{m_1 + m_2} (\vec{v}_1 - \vec{v}_2), ensuring the momenta are equal in magnitude but opposite in direction, such that m_1 \vec{v}'_1 + m_2 \vec{v}'_2 = 0.^[5] This configuration highlights how the frame isolates the relative motion of the particles./15:_Collision_Theory/15.07:_Two-Dimensional_Collisions_in_Center-of-Mass_Reference_Frame) In non-relativistic physics, the center-of-momentum frame is identical to the center-of-mass frame.^[4] This foundational concept provides the basis for its extension to relativistic regimes without altering the core idea of zero total momentum.^[5]

In Relativistic Mechanics

In special relativity, the center-of-momentum (COM) frame for an isolated system of particles is defined as the inertial reference frame in which the total three-momentum \vec{P} of the system vanishes.^[6] This condition is equivalently expressed using the total four-momentum P^\mu = \sum_i p_i^\mu, where p_i^\mu = (E_i/c, \vec{p}_i) is the four-momentum of the i-th particle, with E_i its total energy and \vec{p}_i its three-momentum; in the COM frame, the spatial components of P^\mu sum to zero, so P^\mu = (E_\text{total}/c, \vec{0}).^[6] The temporal component then corresponds to the total energy in this frame, related to the invariant mass of the system by M = E_\text{total}/c^2, where M is the total rest mass.^[7] The existence of a unique COM frame requires the total four-momentum to be time-like, meaning the system possesses a non-zero invariant rest mass Mc^2 = \sqrt{E_\text{total}^2 - (c|\vec{P}|)^2}.^[8] For systems composed entirely of massless particles, such as two photons propagating in the same direction, the total four-momentum is null (light-like) with invariant mass zero, precluding a frame where \vec{P} = \vec{0} because no Lorentz boost can reduce the magnitude of the total momentum below E_\text{total}/c.^[1] However, for counter-propagating photons of equal energy, the invariant mass is non-zero in their symmetric frame, allowing a well-defined COM frame.^[1] To reach the COM frame from a laboratory frame where the total three-momentum is \vec{P} and total energy is E_\text{total}, apply a Lorentz boost with velocity \vec{v} = \frac{\vec{P} c^2}{E_\text{total}} directed along \vec{P}.^[6] This velocity follows from the Lorentz transformation of the total four-momentum: under a boost \vec{v}, the transformed spatial momentum is \vec{P}' = \gamma (\vec{P} - \frac{\vec{v}}{c^2} E_\text{total}) + cross terms that vanish for collinear boost, setting \vec{P}' = \vec{0} yields the required \vec{v}.^[6] The Lorentz factor is then \gamma = E_\text{total} / (Mc^2), ensuring the boost aligns with the system's overall motion.^[7] In general relativity, the COM frame concept extends to locally inertial frames for isolated systems in curved spacetime, where the total energy-momentum tensor defines an analogous zero-momentum condition, though global frames may not exist due to spacetime curvature. For example, consider a high-energy proton-proton collision in a laboratory frame where one proton is at rest and the other approaches with significant momentum, resulting in net total momentum \vec{P} \neq \vec{0}; boosting to the COM frame symmetrizes the incoming particles with equal and opposite momenta, simplifying the analysis of decay products like antiproton production by conserving four-momentum without directional bias.^[7] In the non-relativistic limit, where velocities are much less than c, the relativistic COM frame reduces to the classical center-of-mass frame.^[6]

Key Properties

Total Momentum and Energy

In the center-of-momentum (COM) frame, the total three-momentum of the system is defined to be zero, \sum_i \vec{p}_i = 0, where \vec{p}_i is the three-momentum of the i-th particle.^[9] This condition implies that the momenta of the particles must cancel vectorially; for a balanced two-body system, such as colliding particles of equal mass, the individual momenta are equal in magnitude but opposite in direction, \vec{p}_1 = -\vec{p}_2.^[9] In multi-particle systems, the momenta arrange such that their vector sum vanishes, simplifying the analysis of internal dynamics without net translational motion.^[9] In relativistic mechanics, the total energy in the COM frame is given by E_\text{total} = \sum_i \sqrt{m_i^2 c^4 + p_i^2 c^2}, where m_i is the rest mass of the i-th particle, p_i = |\vec{p}_i|, and c is the speed of light.^[9] For systems with non-zero total rest mass, this total energy represents the minimum possible value observable from any inertial frame, as boosts to frames with net momentum add kinetic energy associated with the system's overall motion.^[9] Specifically, in a laboratory frame where the total momentum is \vec{P} \neq 0, the observed total energy is E_\text{lab} = \gamma E_\text{COM}, where \gamma = 1/\sqrt{1 - v^2/c^2} and \vec{v} is the velocity of the COM frame relative to the laboratory frame, given by \vec{v} = \vec{P} c^2 / E_\text{lab}.^[9] Conservation laws hold straightforwardly in the COM frame: both the total three-momentum remains zero before and after interactions, and the total energy is conserved, as required by the invariance of the energy-momentum four-vector under Lorentz transformations.^[9] This frame thus simplifies the description of dynamics, avoiding complications from net momentum that arise in other frames, such as the laboratory frame, where energy and momentum conservation must account for the system's bulk velocity.^[9] The COM frame differs from the center-of-mass (CM) frame in relativistic contexts. While both aim to eliminate net motion, the COM frame is strictly the inertial frame where total three-momentum vanishes, focusing on momentum balance.^[10] The CM frame, by contrast, centers on the rest position of the system's center of mass and may accelerate due to internal relativistic effects, even without external forces; the two coincide only if all particles share the same velocity relative to the CM, which is not generally true.^[10] To illustrate, consider a system of three particles with masses m_1, m_2, m_3 and momenta \vec{p}_1, \vec{p}_2, \vec{p}_3 in the COM frame such that \vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0. For example, if \vec{p}_1 points along the positive x-axis, \vec{p}_2 could align oppositely but with adjusted magnitude, and \vec{p}_3 in the y-z plane to ensure vectorial cancellation, maintaining zero total momentum while allowing independent internal interactions.^[9]

Invariant Quantities

In special relativity, the center-of-momentum (COM) frame is characterized by several Lorentz-invariant quantities that remain unchanged under boosts between inertial frames, providing a frame-independent description of the system's dynamics. The most fundamental of these is the invariant mass M of the system, defined as

M = \frac{1}{c^2} \sqrt{E_\text{total}^2 - ( \mathbf{p}_\text{total} c )^2 },

where E_\text{total} is the total energy and \mathbf{p}_\text{total} is the total three-momentum of all particles in the system, with c denoting the speed of light.^[11] This quantity represents the effective rest mass of the entire system as "seen" from the COM frame, where \mathbf{p}_\text{total} = 0, simplifying the expression to M = E_\text{total} / c^2./09:_Relativistic_Kinematics/9.02:_Invariant_Mass) The invariance of M arises from its connection to the total four-momentum P^\mu = (E_\text{total}/c, \mathbf{p}_\text{total}), the Minkowski-space sum of the individual four-momenta p_i^\mu. The squared magnitude of the four-momentum, P^\mu P_\mu = (E_\text{total}/c)^2 - \mathbf{p}_\text{total}^2 = M^2 c^2, is a Lorentz scalar, unchanged by frame transformations because the metric tensor \eta_{\mu\nu} = \operatorname{diag}(1, -1, -1, -1) preserves the interval under Lorentz boosts.^[12] To derive this, consider the four-momentum conservation: in any frame, P^\mu = \sum_i p_i^\mu, and since each p_i^\mu p_{i\mu} = m_i^2 c^2 (with m_i the rest mass of particle i) is invariant, the total P^\mu P_\mu inherits this property, yielding M as frame-independent despite varying E_\text{total} and \mathbf{p}_\text{total} across frames.^[13] In particle physics, this invariant mass plays a crucial role in analyzing virtual particles and resonances, where the distribution of the invariant mass of decay products peaks at the resonance mass when reconstructed in the COM frame. For instance, in electron-positron (e^+ e^-) annihilation, the invariant mass \sqrt{s} = \sqrt{(P^\mu P_\mu) c^2} (with s the Mandelstam variable) sets the total center-of-mass energy available for producing particle-antiparticle pairs, such as muon pairs, with each particle receiving energy \sqrt{s}/2 in the COM frame due to momentum balance.^[14] This framework ensures that threshold energies and kinematic constraints are universally applicable, independent of the laboratory frame.

Frame Transformations

From Laboratory Frame to COM Frame

In typical experimental setups in particle and nuclear physics, the laboratory frame is defined such that one particle (the projectile) is incident on a stationary target, leading to a net total momentum for the system due to the target's rest condition.^[15] This asymmetry complicates kinematic analysis, as the initial velocities and subsequent scattering angles are biased toward the direction of the incoming particle.^[16] The center-of-momentum (COM) frame, characterized by zero total momentum for the system, provides a more balanced perspective where interactions appear isotropic and symmetric.^[17] To shift from the laboratory frame to the COM frame, the total momentum \vec{p}_\text{lab} and total energy E_\text{lab} of all particles are first calculated in the lab frame. The required boost velocity parameter \beta (where \beta = v/c and c is the speed of light) is then determined as

\beta = \frac{|\vec{p}_\text{lab}| c}{E_\text{lab}}.

This boost aligns the frame such that the total momentum vanishes, as referenced in the key properties of the COM frame.^[18] In the resulting COM frame, processes like collisions exhibit enhanced symmetry, with outgoing particles distributed more uniformly in angle compared to the forward-peaked distributions often seen in the lab frame.^[19] The adoption of the COM frame gained prominence in 20th-century nuclear physics, particularly for simplifying the interpretation of scattering experiments by reducing kinematic asymmetries and enabling clearer comparisons with theoretical predictions.^[15] A representative example is a proton-proton collision in the lab frame, where one proton is at rest: the incident proton carries all the initial momentum, leading to asymmetric scattering. Transforming to the COM frame symmetrizes the event, with both protons approaching head-on at equal speeds and opposite directions, which facilitates analysis of isotropic angular distributions in the final state.^[19]

Galilean and Lorentz Boosts

In classical mechanics, the Galilean boost describes the transformation of coordinates between two inertial frames moving at constant relative velocity \mathbf{V}. For a boost along the x-direction, the position and time coordinates transform as x' = x - V t, y' = y, z' = z, and t' = t.^[20] The velocity components of a particle then transform as v_x' = v_x - V, v_y' = v_y, v_z' = v_z.^[21] For momentum, assuming non-relativistic particles, the transformation for a single particle of mass m is \mathbf{p}' = \mathbf{p} - m \mathbf{V}, where \mathbf{p} = m \mathbf{v}.^[21] To reach the center-of-momentum (COM) frame for a system of particles, the boost velocity is chosen as \mathbf{V} = \mathbf{P}/M, where \mathbf{P} = \sum \mathbf{p}_i is the total momentum and M = \sum m_i is the total mass; this ensures the total momentum in the new frame is \mathbf{P}' = \sum (\mathbf{p}_i - m_i \mathbf{V}) = \mathbf{P} - M \mathbf{V} = 0.^[21] In relativistic mechanics, the Lorentz boost replaces the Galilean transformation to preserve the spacetime interval and account for the speed of light c. The four-momentum of a particle, p^\mu = (E/c, \mathbf{p}) where E = \gamma m c^2 and \gamma = 1/\sqrt{1 - v^2/c^2}, transforms under a boost along the z-direction with velocity V (so \beta = V/c, \gamma = 1/\sqrt{1 - \beta^2}) as:

\begin{align*} p'^0 &= \gamma (p^0 - \beta p^z), \\ p'^x &= p^x, \\ p'^y &= p^y, \\ p'^z &= \gamma (p^z - \beta p^0), \end{align*}

or in three-momentum form, E' = \gamma (E - V p_z) and p_z' = \gamma (p_z - V E / c^2), with transverse components unchanged.^[22] This can be written compactly using the Lorentz transformation matrix \Lambda^\mu{}_\nu acting on the four-momentum.^[22] For a multi-particle system, the COM frame is obtained by boosting with velocity \mathbf{V} such that the total three-momentum vanishes: \mathbf{P}' = \sum \mathbf{p}_i' = 0. The required \beta satisfies \beta = |\mathbf{P}| c / E_\text{total}, where E_\text{total} = \sum E_i, derived from the longitudinal boost equation applied to the total four-momentum; \gamma follows from \beta.^[17] For ultra-relativistic cases where \beta \approx 1, direct computation of \gamma becomes numerically unstable due to large values (\gamma \gg 1); instead, the boost is parameterized by rapidity \eta = \artanh(\beta), where \beta = \tanh \eta, \gamma = \cosh \eta, and \beta \gamma = \sinh \eta.^[23] Rapidities add under successive collinear boosts, simplifying compositions: \eta_{12} = \eta_1 + \eta_2, which is advantageous for high-energy systems like particle collisions.^[23] As a numerical example, consider a two-body decay K^{*-} \to K^- + \pi^0 analyzed in the reverse direction to illustrate boosting from lab to COM frame. In the lab frame, the parent K^{*-} has momentum p = 5.5 GeV/c along z, so total lab momentum P_z = 5.5 GeV/c and E_\text{total} \approx m_{K^*} \gamma c^2 with \gamma = 6.2485 (since rest mass m_{K^*} c^2 = 892 MeV, E_{K^*} = 5.572 GeV). The boost to COM (rest frame of K^{*-}) uses V = -p c^2 / E_{K^*} \approx -0.9871 c, \beta = -0.9871, yielding total P_z' = 0. In the COM frame, the daughters have equal and opposite momenta: p_{K^-} = (0.2371, 0, 0.1661) GeV/c, p_{\pi^0} = (-0.2371, 0, -0.1661) GeV/c, with energies E_{K^-} = 0.5723 GeV and E_{\pi^0} = 0.3194 GeV.^[24]

Applications

Two-Body Collisions

In the center-of-momentum (COM) frame for a two-body collision, the total momentum vanishes, so the momenta of the two particles are equal in magnitude and opposite in direction: \vec{p}_1 = -\vec{p}_2 = \vec{p}.^[5] In the non-relativistic case, the magnitude |\vec{p}| is determined by the reduced mass \mu = \frac{m_1 m_2}{m_1 + m_2} and the relative velocity \Delta \vec{u} = \vec{u}_1 - \vec{u}_2, yielding \vec{p}_1 = -\mu \Delta \vec{u} and \vec{p}_2 = \mu \Delta \vec{u}.^[25] This formulation separates the center-of-mass motion from the relative motion, simplifying the analysis to an effective one-body problem with mass \mu.^[5] In the relativistic extension, the magnitudes of the three-momenta remain equal before and after the collision in the COM frame: |\vec{p}_1| = |\vec{p}_2|, with the total energy given by E_1 + E_2 = \sqrt{s}, where s = (p_1 + p_2)^2 is the Mandelstam variable representing the square of the total four-momentum.^[26] This invariant s quantifies the available energy in the COM frame and sets the threshold for the reaction, such as s \geq (m_1 + m_2)^2 c^4 (with c=1 in natural units).^[27] The COM frame simplifies conservation laws for both elastic and inelastic scattering by reducing the problem to one dimension along the line of centers.^[5] Momentum and energy conservation then dictate that the relative speed remains constant in elastic collisions (preserving kinetic energy in the non-relativistic limit) or changes according to the coefficient of restitution e in inelastic cases, where e = \sqrt{1 - \frac{\Delta E}{T_{\rm CM}}} and T_{\rm CM} = \frac{1}{2} \mu |\Delta u|^2 is the kinetic energy in the COM frame.^[25] In relativistic elastic scattering, the magnitudes |\vec{p}| are unchanged, with only the scattering angle \Theta^* determining the post-collision directions.^[27] A classic example is the elastic collision of two billiard balls of equal mass m, where the laboratory frame often has one ball at rest.^[5] Transforming to the COM frame (via a Galilean boost), the balls approach with equal and opposite velocities \vec{v}' = -\vec{v}', and after collision, they recede at equal speeds along directions separated by the scattering angle \Theta^*, simplifying the prediction of trajectories using reduced mass \mu = m/2 and relative velocity.^[25] In neutron-proton scattering, a foundational case in nuclear physics, the COM frame treats the interaction as rotationally symmetric with equal opposite momenta.^[28] For a head-on elastic collision, the neutron transfers all its energy to the proton in the non-relativistic approximation (using reduced mass \mu \approx m_n/2), but in general, the post-collision energies are E' = E \frac{1 + \cos \phi}{2} (with \phi the COM scattering angle), highlighting the frame's utility in analyzing energy sharing.^[28]

Multi-Particle Systems

In a system consisting of N particles, the center-of-momentum (COM) frame is defined such that the vector sum of the individual momenta vanishes: \sum_{i=1}^N \mathbf{p}_i = 0. This condition directly enforces the conservation of total momentum, making the COM frame particularly useful for analyzing multi-particle decays and reactions where initial and final states must balance vectorially without net motion. The total energy in the COM frame is given by E_\mathrm{COM} = \sum_{i=1}^N E_i, representing the invariant mass of the system, but unlike two-body collisions, there exists no straightforward reduced mass for partitioning kinetic energy among multiple particles. To approximate internal dynamics, techniques such as Jacobi coordinates are employed, which recursively define relative position vectors between particle clusters after isolating the overall center-of-mass motion, facilitating the separation of translational and rotational degrees of freedom. Cluster decompositions further simplify calculations by grouping particles into subsystems treated as effective two-body interactions. In relativistic multi-particle systems, significant challenges arise because highly energetic particles, for which |\mathbf{p}_i| \approx E_i/c, dominate the momentum balance, leading to asymmetric distributions even when the total momentum is zero. For instance, in three-body decays, the COM frame elucidates angular distributions of the decay products, which encode information about particle polarizations, spin alignments, and potential CP-violating effects beyond simple kinematic reconstruction.^[29]^[30] In quantum field theories featuring non-abelian gauge symmetries, the COM frame applies to multi-particle configurations, but the non-commutative nature of the interactions complicates the isolation of color-neutral states. A illustrative case is the three-body beta decay of the neutron, n \to p + e^- + \bar{\nu}_e, where analysis in the neutron's COM frame (rest frame) reveals the angular correlations between the proton, electron, and antineutrino momenta, aiding in the extraction of branching ratios and form factors that test the weak interaction's V-A structure.^[30]

In Particle Physics

In high-energy particle physics experiments at colliders such as the Large Hadron Collider (LHC), events are routinely reconstructed in the center-of-momentum (COM) frame to facilitate the computation of invariant masses of decay products and the application of jet clustering algorithms. At the LHC, the laboratory frame coincides with the COM frame for proton-proton collisions due to the equal and opposite momenta of the beams, allowing direct kinematic reconstruction of particle momenta and energies. Invariant masses, defined as the rest mass of a system in its COM frame, are calculated from the four-momenta of reconstructed objects like leptons or jets to identify resonances, while infrared- and collinear-safe jet algorithms (e.g., anti-kT) cluster hadronic activity in this frame to isolate high-momentum jets from quantum chromodynamics backgrounds.^[31]^[32]^[33] The use of the COM frame offers key advantages in experimental analysis, particularly for studying decay patterns and modeling interaction rates. In the COM frame of a decaying particle or resonance, angular distributions of decay products are often isotropic for unpolarized initial states, simplifying the extraction of spin-parity quantum numbers and reducing biases from boost effects that complicate laboratory-frame analyses. Cross-sections, being Lorentz-invariant quantities, are theoretically frame-independent, but perturbative quantum chromodynamics calculations and Monte Carlo simulations are more straightforward in the COM frame, where azimuthal symmetry aids in validating models against data.^[34]^[35]^[36] However, practical limitations arise in determining the precise COM frame for individual events. Beam energy spreads, typically on the order of 0.1% at the LHC, and detector resolution effects—such as momentum smearing from tracking inefficiencies—can introduce uncertainties in the reconstructed total event energy and momentum, blurring the effective COM boost and degrading invariant mass resolutions. To mitigate these, event generators like PYTHIA simulate full collision events including parton showers and hadronization in the COM frame, enabling detailed comparisons with experimental data after accounting for detector responses.^[37]^[38]^[39] The adoption of the COM frame in particle physics traces back to cosmic ray studies in the 1950s, where it was essential for analyzing high-energy interactions lacking controlled beams, and became a standard tool in accelerator physics following the rise of colliding-beam experiments in the 1970s, which naturally operate in the COM frame to maximize available energy. A prominent example is the discovery of the Higgs boson, where the invariant mass of its decay products into four leptons (H → ZZ → 4ℓ) exhibits a clear peak at 125 GeV in the COM frame reconstruction, confirming the boson's mass and properties amid continuum backgrounds.^[40]^[41]^[32]

References

[1]
https://phys.libretexts.org/Bookshelves/Relativity/Special_Relativity_(Crowell)/04:_Dynamics/4.03:_Relativistic_Momentum
[2]
[PDF] SPECIAL RELATIVITY - UCSB Physics
○ Relativistic system – CM yields non-zero momentum. – More useful: “center of momentum” frame. ○ Example: – Newtonian CM frame moves to right at speed c/4.
[3]
[PDF] Lecture 06: Kinematics in Spacetime - MIT OpenCourseWare
p center of momentum frame: v = -uxex , and so γ = 1/ 1 - (ux)2/c2 . Notice that the velocity components in the vertical direction are no longer equal and.
[4]
[PDF] Classical Mechanics LECTURE 6: THE CENTRE OF MASS
The centre of mass (CM) is the point where the mass-weighted. position vectors (moments) relative to the point sum to zero ; the CM is the mean location of a ...
[5]
[PDF] 8.01SC S22 Chapter 15: Collision Theory - MIT OpenCourseWare
Jun 15, 2022 · In the CM frame the momentum of the system is zero before the collision and hence the momentum of the system is zero after the collision. For an ...
[6]
[PDF] Relativity II: Dynamics - Kevin Zhou
(e) A system's center of mass frame is the one where its momentum is zero. For a system with total energy E and momentum p, show that the center of mass has ...
[7]
[PDF] 8.033 (F24): Lecture 08: Using 4-Momentum - MIT OpenCourseWare
8.3.5 The center of momentum (COM) frame. Some problems can be greatly simplified by changing the reference frame in which we do the calculation. A frame ...
[8]
[PDF] Relativistic Kinematics - Caltech
Apr 29, 2015 · reference frame is called the system's center of momentum frame or rest frame; in this frame ... massless particles such as photons (for which 1).
[9]
[PDF] RELATIVISTIC ENERGY AND MOMENTUM - UT Physics
or anything else besides the center of mass motion — is the Ecm = √s. Let's calculate the center-of-mass ...
[10]
None
Summary of each segment:
[11]
Four-vectors in Relativity - HyperPhysics
The invariance of the energy-momentum four-vector is associated with the fact that the rest mass of a particle is invariant under coordinate transformations.
[12]
Electrodynamics in Relativistic Notation - Feynman Lectures - Caltech
The four-velocity is a useful quantity; we can, for instance, write pμ=m0uμ. This is the typical sort of form an equation which is relativistically correct ...
[13]
https://www.phys.ufl.edu/~korytov/phz4390/note_02_kinematics.pdf
[14]
[PDF] Handout 4 : Electron-Positron Annihilation - Particle Physics
q Calculate decay rate/cross section using formulae from handout 1. •e.g. for a decay. •For scattering in the centre-of-mass frame ... Invariant (scalar product ...
[15]
[PDF] SubatomicPhysicsHenleyGarcia.pdf - faculty.washington.edu
Jun 7, 2007 · ... center-of-momentum frame, both particles approach each other with equal but opposite momenta. The two frames are defined by lab frame : plab.
[16]
Rel Dynamics - Galileo and Einstein
Go to the center of mass frame, where initially two protons are moving towards each other with equal and opposite velocities, there being no total momentum.<|control11|><|separator|>
[17]
[PDF] Special Relativity 3
We will call this the center of momentum (cm) frame. It is useful to measure the total energy in the cm frame which is easy to do using dot product invariants.
[18]
[PDF] PHYS 4D Solution to HW 9 - Physics Courses
Mar 6, 2011 · Find the velocity of the center of momentum frame, in which the total momentum is zero. Solution: ET. = E1 + E2 + E3 + ... ;. pT. = p1 + p2 + ...
[19]
[PDF] 1 Laboratory frame and center of mass frame - bingweb
Feb 1, 2021 · We note that the center of mass frame is favorable compared to the laboratory frame from a theoretical view-point. The separation of the system ...
[20]
[PDF] Galilean Transformations
Sep 24, 2012 · But Galilean transformations, boosts, involve both time and space. Under a time translation,. ¯t= t − δt, ¯r = r,. (17.59) while under a boost,.
[21]
[PDF] 1 Introduction 2 Boost operator - University of Oregon
Apr 4, 2011 · Then we switch to a. “moving frame” using coordinates (t0,x0,y0,z0). The transformation is t0 = t, x 0 = x + v t. Note that if we were doing ...
[22]
[PDF] 7.1 Transforming energy and momentum between reference frames
T = pz L . In other words, the relativistic formulations of energy and momentum form a set of quantities that transform under a Lorentz transformation.
[23]
[PDF] Lorentz Symmetry of Particles and Fields - UT Physics
For successive boosts in the same direction, the rapidities add up, r1+2 = r1 + r2. Consequently, a finite Lorentz boost of rapidity r in the direction n is ...
[24]
[PDF] Relativistic Kinematics II - UF Physics
Aug 26, 2015 · Relativistic kinematics uses 4-vectors to simplify problems, including particle decay and scattering, and accounts for conservation of energy ...
[25]
[PDF] Classical Mechanics LECTURE 9: INELASTIC COLLISIONS
These expressions give the CM kinetic energy in terms of the relative velocities in the CM & Lab and the reduced mass µ.<|control11|><|separator|>
[26]
[PDF] 46. KINEMATICS
Aug 21, 2014 · The velocity of the center-of-mass in the lab frame is ... energy is 1.699 GeV and the center of mass momentum of either particle is 0.442 GeV/c.
[27]
[PDF] Chapter 2 Relativistic kinematics
Note also that s + t + u = . 4 i=1 m2 i . The center of mass frame is defined by. #»p1 + #»p2 =0= #»p3 + #»p4. (2.8). The corresponding variables are ...
[28]
[PDF] Part Fourteen Kinematics of Elastic Neutron Scattering - DSpace@MIT
Neutron scattering is most easily analyzed in a center-of-mass frame of reference. (COM). Collisions in a COM frame appear as rotations with the speeds the same.Missing: reduced | Show results with:reduced
[29]
Systematics of Angular and Polarization Distributions in Three-Body ...
A general method is described for the measurement of the polarization and alignment of a particle of arbitrary spin from the analysis of its three-body decays.Missing: frame | Show results with:frame
[30]
Angular distribution as an effective probe of new physics in ...
Jul 9, 2019 · Our methodology can be used for detection of new physics in experimental study of various three-body pseudoscalar meson decays at various ...<|separator|>
[31]
Jet substructure at the Large Hadron Collider | Rev. Mod. Phys.
Dec 12, 2019 · This review focuses on the development and use of state-of-the-art jet substructure techniques by the ATLAS and CMS experiments.<|control11|><|separator|>
[32]
Observation of a new boson at a mass of 125 GeV with the CMS ...
Sep 17, 2012 · In the H → Z Z → 4 ℓ decay mode a search is made for a narrow four-lepton mass peak in the presence of a small continuum background. Early ...
[33]
[PDF] Fundamentals of LHC Experiments - arXiv
Jun 13, 2011 · With this trick, it is possible to reconstruct the full tau momentum and calculate the invariant mass of the Higgs candidate (see Exercise 8).
[34]
[PDF] Phenomenology of Particle Physics
Jun 9, 2020 · ... decay, so the final state particles must be distributed isotropically in the center- of-momentum frame. One then obtains the total decay rate ...
[35]
[PDF] 49. Cross-Section Formulae for Specific Processes
Dec 6, 2019 · We provide the production cross sections calculated within the Standard Model for several such processes. 49.1 Resonance Formation. Resonant ...
[36]
[PDF] The Lorentz transformation - Physics Department, Oxford University
The Lorentz transformation is a coordinate transformation replacing the Galilean transformation, relating coordinates between reference frames.
[37]
Study the effect of beam energy spread and detector resolution on ...
Feb 18, 2017 · The impact of the energy spread of the FCC-ee beam and of the resolution in the reconstruction of the leptons is discussed. The minimum ...
[38]
[PDF] A comprehensive guide to the physics and usage of PYTHIA 8.3 ...
Sep 7, 2022 · PYTHIA 8.3 is an event generator used to simulate high-energy particle collisions, aiming to accurately reproduce experimental collision ...
[39]
[PDF] Further investigation of dilepton-based center-of-mass energy ...
Aug 22, 2023 · collision events are further developed with an emphasis on accelerator, detector, and physics limitations. We discuss two main estimators, ...
[40]
[PDF] Evolution of Particle Accelerators & Colliders
After 1970 colliders—machines using two accelerator beams in collision—entered the picture. Since then most, but certainly not all, new revelations in particle ...
[41]
[PDF] Pions to Quarks: Particle Physics in the 1950s
During the same period, cosmic rays gave way to accelerators as the major source of high-energy particles. Accelerator energies increased by an order of ...
[42]
[PDF] Lecture Notes on High Energy Cosmic Rays - arXiv
Jan 29, 2008 · Consider for instance the squared center-of-mass energy s = (pa + pb)2 = (pc + pd)2 with four-momenta pi = (Ei,pi) in the 2 → 2 scattering ...