10^9 GeV dispersed collision energy

In summary: GeV of potential energy, it could lift 5 million metric tons! potential energy = m∙g∙h; where g = 9.8 m/s/s on Earth.If you drop a 1.02 kg brick on your foot, from a height of 1 metre, it delivers energy of;1.02 kg * 9.8 m/s/s * 1 m = 10 joule.That is 10 / 2.08e-6 = 5 million times as much as an electron... so theoretically, if an object had 10^19 GeV of potential energy, it could lift 5 million metric tons!
  • #1
jtlz
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When cars or trains collide.. does it exceed 10^19 for the total dispersed energy? I know that in particle accelerator you need to focus this much energy in the Planck scale to probe it. So I just want to know what is the equivalent everyday object that has this 10^19 GeV only its not focused at small area but dispersed (so useless as particle accelerator).

It is often asked what's it's like to have 13 TeV collision at the LHC.. like what bigger object has similar dispersed collision energy. So I want to know what bigger object has at least 10^19 GeV dispersed collision energy.
 
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  • #2
jtlz said:
When cars or trains collide.. does it exceed 10^19 for the total dispersed energy? I know that in particle accelerator you need to focus this much energy in the Planck scale to probe it. So I just want to know what is the equivalent everyday object that has this 10^19 GeV only its not focused at small area but dispersed (so useless as particle accelerator).

It is often asked what's it's like to have 13 TeV collision at the LHC.. like what bigger object has similar dispersed collision energy. So I want to know what bigger object has at least 10^19 GeV dispersed collision energy.
Do you know how to convert units of energy from eV to Joules? Using the SI system of units, you can calculate the KE of a train in Joules, which will go to zero in a collision. Use Google to find the mass of a typical heavy train. Or even of a fully-loaded container ship. Please show us your calculations so we can check them. Thanks! :smile:
 
  • #3
berkeman said:
Do you know how to convert units of energy from eV to Joules? Using the SI system of units, you can calculate the KE of a train in Joules, which will go to zero in a collision. Use Google to find the mass of a typical heavy train. Or even of a fully-loaded container ship. Please show us your calculations so we can check them. Thanks! :smile:

I can't find any converter of GeV to Joules... most of them have only eV to Joules.. and I don't want to keep typing too many zeros... so I was asking if anyone knows roughly from reading books of what's Planck scale energy is like compared to everyday objects...
 
  • #4
jtlz said:
I can't find any converter of GeV to Joules... most of them have only eV to Joules.. and I don't want to keep typing too many zeros... so I was asking if anyone knows roughly from reading books of what's Planck scale energy is like compared to everyday objects...
What does the letter "G" stand for as a prefix for a number?
 
  • #5
berkeman said:
What does the letter "G" stand for as a prefix for a number?

Ok i'll attempt manual convertion since I can't find a converter for this.

1 GeV = 1 billion eV or 1 x 10^9 eV

1J = 6.241509⋅10^18 eV

so 1J = 6.241509⋅10^18 eV * (1 GeV /10^9 eV) = 6.241509 x 10^9 GeV

so the Planck energy 10^19 GeV x (1J / 6.24 x 10^9 GeV) = 1.6 x 10^10 Joule

how big is 1.6 X 10^10 Joule? equivalent to a tornado or hurricane plucking a tree or lifting cars? what?
 
  • #6
jtlz said:
Ok i'll attempt manual convertion since I can't find a converter for this.

1 GeV = 1 billion eV or 1 x 10^9 eV

1J = 6.241509⋅10^18 eV

so 1J = 6.241509⋅10^18 eV * (1 GeV /10^9 eV) = 6.241509 x 10^9 GeV

so the Planck energy 10^19 GeV x (1J / 6.24 x 10^9 GeV) = 1.6 x 10^10 Joule

how big is 1.6 X 10^10 Joule? equivalent to a tornado or hurricane plucking a tree or lifting cars? what?

I'm trying out some online calculator site about mass, velocity and kinetic energy in joules... https://www.calculatorsoup.com/calculators/physics/kinetic.php

I entered the mass and energy to come up with the kinetic energy near 1.6 x 10^10 Joules...

I entered mass of 30,000 kilograms and velocity of 1000 meters /second..

really? you need that much weight and speed to reach Planck energy? Or is there some mistakes somewhere?
 
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  • #7
When an electron moves through 1 volt in an electric field, it changes energy by 1 eV.
Energy = Q * V. The charge on the electron; Qe = 1.60217653e-19 coulomb.
Therefore; 1 eV = 1.60217653e-19 coulomb * 1 volt = 1.60217653e-19 joule. So 1 joule = 6.241e+18 eV

The 13 TeV at the LHC = 13.0e12 * 1.602e-19 = 2.08 uJ

Potential Energy = m∙g∙h; where g = 9.8 m/s/s on Earth.
If you drop a 1.02 kg brick on your foot, from a height of 1 metre, it delivers energy of;
1.02 kg * 9.8 m/s/s * 1 m = 10 joule.

That is 10 / 2.08e-6 = 5 million times as much as an electron from the LHC.

Your 10^19 GeV = 1e28 eV = 1e28 * 1.602e-19 joule = 1.602 GJ
If a satellite weighing 1020 kg fell on you from 100 km above it would deliver a pain equivalent of;
1020 * 9.8 * 100e3 = 1 GJ
 
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  • #9
Guys.. what is the formula relating energy to distant scale.. for example. 1 GeV.. 1 TeV, 5 TeV.. etc. with the relevant scales.. for example 10^19 GeV related to Planck scale.. how about say 10^12 GeV. How do you compute the distant scale for this?
 
  • #10
There is the inverse square law that gives the energy per square metre of a radiating wavefront.
For a single photon, 1/r2 gives the probability of being hit, since photons maintain their energy until they hit something.
 
  • #11
Baluncore said:
There is the inverse square law that gives the energy per square metre of a radiating wavefront.
For a single photon, 1/r2 gives the probability of being hit, since photons maintain their energy until they hit something.

What is the formula that involves the energy...

In particle accelerator such as the LHC... what is the smallest distance the 1.5 TeV particle energy can probe? (noting that the 13 TeV is for the collision energy of the entire accelerating protons.. so each individual particle would have less energy than that.. so it's about 1.5 TeV maximum now)..

So given particle energy = 1.5 TeV..
what is the distance scale (in meters it can probe).. what's the exact formula?
 
  • #13
Baluncore said:
When an electron moves through 1 volt in an electric field, it changes energy by 1 eV.
Energy = Q * V. The charge on the electron; Qe = 1.60217653e-19 coulomb.
Therefore; 1 eV = 1.60217653e-19 coulomb * 1 volt = 1.60217653e-19 joule. So 1 joule = 6.241e+18 eV

The 13 TeV at the LHC = 13.0e12 * 1.602e-19 = 2.08 uJ

Potential Energy = m∙g∙h; where g = 9.8 m/s/s on Earth.
If you drop a 1.02 kg brick on your foot, from a height of 1 metre, it delivers energy of;
1.02 kg * 9.8 m/s/s * 1 m = 10 joule.

That is 10 / 2.08e-6 = 5 million times as much as an electron from the LHC.

Your 10^19 GeV = 1e28 eV = 1e28 * 1.602e-19 joule = 1.602 GJ
If a satellite weighing 1020 kg fell on you from 100 km above it would deliver a pain equivalent of;
1020 * 9.8 * 100e3 = 1 GJ

Can you think of a more everyday example of what 1.602 GJ can do? What examples do particle physicists use to illustrate what it's like to probe Planck scale with particle accelerator... like imagine 2 trains colliding and all the energy were focused on the Planck scale? but trains come in different masses.. perhaps a bus colliding.. or better yet. what examples were exactly used by physicists to illustrate the large Planck scale energy?
 
  • #14
jtlz said:
Can you think of a more everyday example of what 1.602 GJ can do?
If you turned on a 2.645 kW electric heater, then kept it going for one whole week, it would transfer 1.6 GJ.

Energy flow; 1 joule per second = 1 watt.
Electrical energy is sold by the; unit = 1 kW∙Hour = 1k * 60 min * 60 sec = 3.6 MJ
If one unit costs you $0.20, then $1.00 will get you 18 MJ.
So 1.6 GW would add $88.89 to your electricity bill.
 
  • #15
Let's compute some figures...

formula of Kinetic Energy = 1/2 m * v^2 (this is also used for momentum of a particle?)
given velocity = 250 meters / second (average speed of airliner)
and Kinetic Energy of =1.6 GJ
the mass needs to be 51,200 kilograms (approx. weight of a Boeing 737 I checked in the weights of airliner lists)

therefore to get Planck energy is like having a Boeing 737 flying at 250 meters/second...

but then would it still be right to say that to get probe Planck scale is like having Boeing 737 flying at 250 meters/second crashing into a building and the energy focusing into the Planck scale?
 

What is 10^9 GeV dispersed collision energy?

10^9 GeV dispersed collision energy refers to the amount of energy released during a high-energy particle collision. It is an extremely high energy level, equivalent to 1 billion gigaelectronvolts.

How is 10^9 GeV dispersed collision energy measured?

10^9 GeV dispersed collision energy is typically measured using particle accelerators, which can accelerate particles to these extremely high energies before colliding them with a target.

What is the significance of 10^9 GeV dispersed collision energy in particle physics?

10^9 GeV dispersed collision energy is significant in particle physics because it allows scientists to study the fundamental building blocks of the universe at a very small scale. At this energy level, particles can be broken down into their constituent parts, revealing new insights into the nature of matter and the universe.

What are the potential dangers associated with 10^9 GeV dispersed collision energy?

There are currently no known dangers associated with 10^9 GeV dispersed collision energy. Particle accelerators have numerous safety measures in place to ensure the protection of both scientists and the general public.

What are the practical applications of 10^9 GeV dispersed collision energy?

10^9 GeV dispersed collision energy has many practical applications in addition to its use in particle physics research. These include medical treatments, materials science, and industrial processes such as nuclear energy production.

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