In special relativity, four-momentum 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 E and three-momentum p = (px, py, pz) = γmv, where v is the particle's three-velocity and γ the Lorentz factor, is
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{\displaystyle p=\left(p^{0},p^{1},p^{2},p^{3}\right)=\left({E \over c},p_{x},p_{y},p_{z}\right).}
The quantity mv of above is ordinary non-relativistic momentum of the particle and m 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.
The above definition applies under the coordinate convention that x0 = ct. Some authors use the convention x0 = t, which yields a modified definition with p0 = E/c2. It is also possible to define covariant four-momentum pμ where the sign of the energy (or the sign of the three-momentum, depending of the chosen metric signature) is reversed.
I was going through Spacetime Physics by Taylor and Wheeler and came to a point where they said, and I quote,
This part feels too abrupt for me and I am looking for a more elaborated explanation.
Here is a link to that chapter.
The information I have are the following:
##p^\mu=(E, p, 0, 0)##
##p'^\mu=(E', p'\cos\beta, -p'\sin\beta,0)##
##k^\mu=\tilde{E}(1, \cos\alpha, \sin\alpha, 0)##
Where:
##E=\sqrt{M^2+p^2}##
##E'=\sqrt{m^2+p'^2}##
Using the conservation of the four-momentum
##p^\mu=p'^\mu+k^\mu##...
So if i had this problem where i am squaring a four momentum vector with itself which gives
P2 = (##\gamma mc## )2 - ##\gamma##2## m ##2##\vec v## *##\vec v##
I have been told that the gamma factor is not considered at all. why would the gamma factor drop off? Does this rule apply to any...
So i have taken a beginner course on relativity, first year physics student. I am confused as to why four momentum squared simply gives
m2* c2*ϒ2 -(three vector multiplied and added with corresponding parts) *ϒ2
so as the three vector part which is being subtracted, is the same as - (P...
Homework Statement
A particle with mass M and speed v along the positive x-axis hits a stationary mass m. Two particles, each with mass µ, emerge from the collision, at angles with respect to the x-axis.
(a) Write the equation for conservation of the 4-momenta, for arbitrary angles θ_1, θ_2 of...
I have been reading about four momentum. There are four component vectors, three spatial, momentum, components and a time, energy, component. They each have a direction. I understand direction for the momentum components, being in the direction of the respective spatial components of the...
In particle phyisics four-momentum is used and De Broglie relation is used to understand what lenghts can be "seen" in an experiment.
Here (page 6) https://people.phys.ethz.ch/~pheno/PPP/PPP2.pdf it is claimed
Where ##Q^2## is not actually "momentum" but its the square of the four momentum...
This is not coursework, it is from a past paper (which we don’t have solutions for) and am doing preparation for this years exam.
1. Homework Statement
Particle ##A## with energy ##E_A## hits particle ##B## (at rest), producing particles ##C## and ##D## in the reaction
A+B \longrightarrow C...
Homework Statement
A particle of mass M and 4-moment P decays into two particles of masses m1 and m2
1) Find the total energy of each particle (lab frame).
2) Show that the kinetic energy T1 of the first particle in the same reference frame is given by
$$T_1= \Delta M (1 - \frac{m_1}{M} -...
Homework Statement
$$ E = -\vec{v_{obs}} \cdot \vec{p} $$
Where ## \vec{p} ## is the four momentum, and ## \vec{v_{obs}}## the velocity of the observer.
Homework EquationsThe Attempt at a Solution
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This was a stated result in a GR course. I look through my SR notes and find that I...
Homework Statement
In a fixed target experiment a particle of mass M and kinetic energy T strikes a stationary particle of mass M. By evaluating s, t and u in the laboratory frame and using the above relation, or otherwise, show that the kinetic energy T' of the particle scattered elastically...
Alright, I'm rather new to General Relativity, and I'm getting confused with four momentum. Back in SR, p^α=mU^α, but, this relationship doesn't hold in curved space, does it? Because, now I'm seeing that four momentum is somehow a covector in GR, and p_0=-E, so the time component of the...
1. A particle of mass M decays from rest into two particles. One particle has mass m and the other particle is massless. The momentum of the massless particle is...
2. Ei = Ef, Pi= Pf
3. This is a GRE practice problem. I can solve this problem using the old method as listed in the step 2, but I...
Homework Statement
a) Suppose that the total three-momentum of an isolated system is conserved in all inertial frames. Show that if this is true (which it is), then the fourth component of the total four-momentum has to be conserved as well. b) Using the zero-component theorem you can prove...
The question goes like this: Prove that conservation of four momentum forbids a reaction in which an electron and positron annihilate and produce a single photon(gamma ray). Prove that the production of two photons is not forbidden.
The solution is to work in the centre of momentum frame. I...
Homework Statement
two particles of masses m1 and m2 move at speeds u1, u2 respectively collide and fuse. If α is the angle between the two directions of motion before the collision, find an expression for the new mass m, in terms of m1 m2 u1 u2 and α
Homework Equations
Etot2 =...
Hi, I'm trying to reconcile some general relativity stuff and I'm getting confused on a few topics.
Would someone be able to explain the properties of the four momentum of a photon? The way I understand it is that you take the four wave-vector of the photon and multiply it by \hbar. The four...
Hi physics people,
This is a past (3rd year university level) exam question, so I hope it's ok that I didn't post this in the homework section even if it's set out like a homework question.
The Question:
Suppose we are observing the collision
Anti-electron-neutrino + electron ---> W-minus...
Homework Statement
To write the expression of force in STR
\ F=\frac{dp}{dt}=\ m\gamma\ a +\ m\gamma\frac{\ u .\ a}{\ c^2 -\ u^2}\ u
Here a is acceleration
Homework Equations
I used the equation \ p=\gamma\ m\ u
I interpreted F as four force,p as four momentum, a as four-acceleration, u...
Hi,
for a real scalar field one has the energy momentum tensor from Noethers theorem
T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial \partial_\mu \phi} \partial^\nu \phi - \eta^{\mu\nu} \mathcal{L}
and the conserved quantities
P^\nu = \int d^3 x \ T^{0\nu}
Now, how can one show that P...
Hi,
I have just been pondering the problem of electron - positron annihilation into a single photon in the CM frame.
I was stuck at a discrepancy - that in the center of mass frame, the total momentum of the particles was zero, but the energy is the sum of the energies of the original...
Suppose that some relativistic field has energy density and momentum density. Integrating these, we get the total energy E and momentum p of the field. But does it make sense to call (E,p) a four momentum of the field? At quick glance it looks like it would be impossible to derive any...
A particle of initial kinetic energy T0 and rest energy E0 strikes a like particle at rest. The initial particle is scattered at an agle theta to its original direction. Show that the final kinetic energy T is
T = T0cos2(theta)/(1+ (T0sin2(theta)/2E0))
what I have so far:
We know that...
I've been trying to see how relativity shows that the quantity gamma*m*c^2 (total energy) is conserved. I assumed that this would proceed from the conservation of momentum. So I researched momentum in relativity, and noticed that it has a time-component: gamma*m*c (which is E/c). So this...