How to Calculate the Length of Energy-Momentum Four-Vectors?

  • Thread starter Thread starter Jamesss
  • Start date Start date
  • Tags Tags
    Energy-momentum
Jamesss
Messages
13
Reaction score
0
Does anyone know how you find the length of the energy-momentum four-vector for a system of particles?
p_mu=(E/c,p)
where length is:
length(p_mu)=-(E/c)^2+(p)^2

Do you first add the corresponding vector elements then find the length
OR
find the length of each particle first then sum the individual lengths.

Cheers,
Jimmy
 
Physics news on Phys.org
Both values will give you invariants, although the energy-momentum four-vector (p^\mu) of the whole system is equal to the sum of all the individual p^\mu, and therefore the length of p^\mu for the system is the length of the sum of all the individual p^\mu.
 
It's exactly analogous to finding the magnitude of the total three-momentum of a system of particles. In that case, you find the total x-momentum, total y-momentum, and total z-momentum of momentum, then use them to find the magnitude of the total-momentum vector.
 
Jamesss said:
Does anyone know how you find the length of the energy-momentum four-vector for a system of particles?
p_mu=(E/c,p)
where length is:
length(p_mu)=-(E/c)^2+(p)^2
Do you first add the corresponding vector elements then find the length
OR
find the length of each particle first then sum the individual lengths.
Cheers,
Jimmy
If the particles are interacting through when they are separated (e.g. two charged particles) then the addition of the two 4-vectors is meaningless. Only systems of non-interacting particles anmd systems of particles which interact only through contact forces can be added in a meaningful way. To add the vectors you add components and then take find the magnitude.

This web page I created will get into great detail regarding this. See
http://www.geocities.com/physics_world/sr/invariant_mass.htm

Pete
 
Is this method ok?

Thanks for the clarification...

What I was trying to do was find the lengths of the four-vectors of this reaction before and after.
p + p ==> p + p + Z

Where a proton with 300GeV hits a stationary proton, then producing a particle Z.

I calculated the length of the Four-vector before the reaction in the stationary proton's frame.

I then equate this to the length of the four-vector after the collision in the
center-of-mass frame to extract the rest mass of the Z particle.

length(p_mu1+p_mu2)=length(p_mu3+p_mu4+p_muZ)

Question, is there anything wrong with my method?

I have assumed that after the collision the two protons and the Z particle are at rest, since I want the maximum possible rest mass of Z. Momentum in the COM from is zero so it should be ok?

Jimmy
 
Yes, that's a reasonable way to proceed. What you end up with is the largest mass the Z can have, and still be produced under these initial conditions.
 
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Thread 'Dirac's integral for the energy-momentum of the gravitational field'
See Dirac's brief treatment of the energy-momentum pseudo-tensor in the attached picture. Dirac is presumably integrating eq. (31.2) over the 4D "hypercylinder" defined by ##T_1 \le x^0 \le T_2## and ##\mathbf{|x|} \le R##, where ##R## is sufficiently large to include all the matter-energy fields in the system. Then \begin{align} 0 &= \int_V \left[ ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g}\, \right]_{,\nu} d^4 x = \int_{\partial V} ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g} \, dS_\nu \nonumber\\ &= \left(...
Abstract The gravitational-wave signal GW250114 was observed by the two LIGO detectors with a network matched-filter signal-to-noise ratio of 80. The signal was emitted by the coalescence of two black holes with near-equal masses ## m_1=33.6_{-0.8}^{+1.2} M_{⊙} ## and ## m_2=32.2_{-1. 3}^{+0.8} M_{⊙}##, and small spins ##\chi_{1,2}\leq 0.26 ## (90% credibility) and negligible eccentricity ##e⁢\leq 0.03.## Postmerger data excluding the peak region are consistent with the dominant quadrupolar...
Back
Top