Kinetic Enrgey according to another viewer

  • Thread starter Thread starter ori
  • Start date Start date
  • Tags Tags
    Kinetic
ori
Messages
28
Reaction score
0
2 particles with mass M and Kinetic Engergy T moving toward each other.
what is the kinetic energy T' as it measured by the other particle
given data: T=...
m=...

thanks
 
Physics news on Phys.org
Why you proposed this question?
 
blue_sky said:
Why you proposed this question?
training for test

that's what i did, could anyone tell what's my prob here:
for first particle
T=mc^2(r-1)
since we're working with Mev , c=1
T=m(r-1)
r=T/M+1
therefore
(1) V^2=T/(T+M)
(V is speed of particle one according to "the lab")

therefore the speed of particle2 is -V (simetry)
now we use additional formula
u1=(u1'+u3)/(1+u1'u3)
whille u1 is velocity of particle1 according to the lab (u1=V)
u1' is velocity of particle 1 according to particle 2
u3 is the velocity between lab to particle2 (u3=-V)
therefore
u1'=2V/(V^2+1)

now i look on kinetic energy of particle 1 according to particle 2:
T'=m(r-1)
while r=1/sqrt(1-u1'^2)
i assigned u1' & V and got
T'=4/3M
this isn't correct..
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
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...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. The Relativator was sold by (as printed) Atomic Laboratories, Inc. 3086 Claremont Ave, Berkeley 5, California , which seems to be a division of Cenco Instruments (Central Scientific Company)... Source: https://www.physicsforums.com/insights/relativator-circular-slide-rule-simulated-with-desmos/ by @robphy

Similar threads

Replies
5
Views
2K
Replies
2
Views
1K
Replies
3
Views
3K
Replies
24
Views
4K
Back
Top