I don't think it is the same thing. Particle accelerors are far more efficient with twin opposing colliding beams than with a single beam and a stationary target:
Particle accelerator - Colliding-beam storage rings | Britannica
Very good 11 minute explanation of velocity addition under relativity, and I got 12/13 with Discopete's hypothetical 2/3 lightspeed particles.
Relativistic Addition of Velocities - YouTube
The rocket velocities are relative to central observer on Earth.
Sorry to rant earlier. The constant changes of topic are frustrating me. I prefer to do one PROPERLY.
Apology accepted, but why didn't you answer Bonsai's question on high energy particles PROPERLY?Sorry to rant earlier. The constant changes of topic are frustrating me. I prefer to do one PROPERLY.
Ohh my Goodness! It took this guy over 11 minutes to make three simple statements! How many times did he repeat himself. Anybody take a count? Like talking to my wife...overandoverandoverandoverand... And he didn't even touch on travel in opposite directions!! Talk about puttin the kibosh on interestingI don't think it is the same thing. Particle accelerors are far more efficient with twin opposing colliding beams than with a single beam and a stationary target:
Particle accelerator - Colliding-beam storage rings | Britannica
Very good 11 minute explanation of velocity addition under relativity, and I got 12/13 with Discopete's hypothetical 2/3 lightspeed particles.
Relativistic Addition of Velocities - YouTube
The rocket velocities are relative to central observer on Earth.
Sorry to rant earlier. The constant changes of topic are frustrating me. I prefer to do one PROPERLY.
By aiming a particle beam to collide head on with another beam, the 'centre-of-mass' energy is increased compared to aiming the beam at a fixed target.
It's complicated and beyond my descriptive abilities!: http://www.personal.soton.ac.uk/ab1u06/teaching/phys3002/course/13_accelerators_a.pdf
I don't think it is the same thing. Particle accelerors are far more efficient with twin opposing colliding beams than with a single beam and a stationary target:
Particle accelerator - Colliding-beam storage rings | Britannica
Very good 11 minute explanation of velocity addition under relativity, and I got 12/13 with Discopete's hypothetical 2/3 lightspeed particles.
Relativistic Addition of Velocities - YouTube
The rocket velocities are relative to central observer on Earth.
Sorry to rant earlier. The constant changes of topic are frustrating me. I prefer to do one PROPERLY.
Particle beams approach the speed of light but don't reach C. Undetermined but less than 11 minutes.
But do the calculation, Discopete. It doesn't matter which direction the 2/3 lightspeed rockets are going. They approach other, cross, then move away. Nothing changes their relative speed Vab.
Relativistic Addition of Velocities - YouTube
Relativistic sum of velocities:
Vab = (Va - Vb) / (1 - (Va x Vb / c^2))
For Va = 2/3 and Vb = -2/3, Vab = 12/13.
I was surprised how big Vab was. But the maths does not lie.
Relativistic Addition of Velocities - YouTube
Relativistic sum of velocities:
Vab = (Va - Vb) / (1 - (Va x Vb / c^2))
For Va = 2/3 and Vb = -2/3, Vab = 12/13.
I was surprised how big Vab was. But the maths does not lie.
Last edited:
That makes no sense. This has to be a logical concept. Two cars passing in opposite directions are going to have a faster relative recession than if one just whizzed by a stalled one.
To avoid confusion, this should read 'For Va and Vb = 2c/3, Vab = 12c/13.'
12c/13 is equal to 0.92c.
E=(mA+mB)c^2 in this case so the total mass involved is 20 tons.
c remains 298 million meters/sec
There is no difference in the impact energy other than the increased mass from mA + mB
In the LHC, the reason why particles are fired in opposition and then collide together is that you can, by doing this, get even closer the the limit of c. Maybe Galu can do the calc but it’s like instead of 99.999% of C it’s 99.9999999%. Since the required energies for the experiments are huge, this is a smart way to get closer to c than just firing a particle at a stationary target
All back of a beermat stuff with me...
But interesting discoveries today.
I had a terrible time trying to solve the Galu Hypothesis on summing closing relativistic velocities of 2/3 lightspeed.
Tried E^2 = (p^2 x c^2) + m^2 x c^4, thought (p^2 x c^2) would reveal all.
and Kinetic Energy K = (Gamma - 1) m x c^2 where Gamma is the Lorentz square root factor.
Got about 4/5 Lightspeed each time. WRONG! 12/13 is right, as we know.
Did it with Galu's velocity transform and it came out better.
Centre of mass, the kinetic energy of the two particles is 0.6832 mc^2.
But from the different frame of one of the particles being stationary and the other one moving at 0.92 lightspeed the kinetic energy is 3.605 mc^2.
Bit weird you get different answers, but what reconciles it is the of the 3.605 mc^2, only a fraction is available to an impact, as we know from particle accelerators. The rest gets carried away in the debris.
But interesting discoveries today.
I had a terrible time trying to solve the Galu Hypothesis on summing closing relativistic velocities of 2/3 lightspeed.
Tried E^2 = (p^2 x c^2) + m^2 x c^4, thought (p^2 x c^2) would reveal all.
and Kinetic Energy K = (Gamma - 1) m x c^2 where Gamma is the Lorentz square root factor.
Got about 4/5 Lightspeed each time. WRONG! 12/13 is right, as we know.
Did it with Galu's velocity transform and it came out better.
Centre of mass, the kinetic energy of the two particles is 0.6832 mc^2.
But from the different frame of one of the particles being stationary and the other one moving at 0.92 lightspeed the kinetic energy is 3.605 mc^2.
Bit weird you get different answers, but what reconciles it is the of the 3.605 mc^2, only a fraction is available to an impact, as we know from particle accelerators. The rest gets carried away in the debris.
Why me? Steve's our resident mathematician!
I could do it, but would need to look it up. I suspect you or Steve have this stuff in your memory banks
Thank you, but my figure for the rate of arrival of high energy particles at the Earth's surface is probably way too high.
It depends on where on the Earth's surface we are, since the figure varies with latitude and altitude.
However, if our average man lives at a very high elevation and very near to the Earth's poles, the daily total I've calculated may not be too far out!
Whatever, it's safe to say that my figure of a man's daily dose of 1,641,600,000 particles is way too precise!
You give me too much credit. Steve's your man!I suspect you or Steve have this stuff in your memory banks
- Status
- Not open for further replies.