Help velocity vectors, electrons, and KeV.

[m3the01]

Please correct anything i have wrong, I am not a physics guy :-). So my skills are lacking.

Say i have two points a and b. I wish to have a velocity vector in the direction from a to b with a 1-100 KeV range.

c = b - a;

0.5 mV^2 = eV

V = sqrt(2eV/m)

m = 9.1094e-31; //electron weight

mineV = sqrt(2000/m);

maxeV = sqrt(200000/m);


Range = maxeV - mineV;

value = mineV + (range * (float)rand()/((float)RAND_MAX+1))

So this gives me my random velocies in the range of 1-100 KeV.

However i have a velocity vector defined in cartesian space (xyz).

So how could i insure all particles have a velocity of 1-100KeV?

Also, if i have a velocity vector V how do i get the overall velocity of the particle?

I was thinking it should just be sqrt(V.x^2 + V.y^2 + V.z^2);

Thanks and sorry for the long post,

I greatly appreciate the help!

- m3the01

 

[LightHero]

To ensure that all particles have a velocity of 1-100 KeV, you should first calculate the magnitude of the velocity vector V using the formula sqrt(V.x^2 + V.y^2 + V.z^2). Then, use the equation value = mineV + (range * (float)rand()/((float)RAND_MAX+1)) to generate a random value in the range of 1-100 KeV. Finally, normalize the velocity vector V by dividing it by its magnitude to get a unit vector and then multiply the unit vector by the random value in the 1-100 KeV range to get the desired velocity vector.

 

[sir-pinski]

Hi m3the01,

First of all, great job on trying to understand and implement these concepts even though you may not have a strong background in physics. Let me try to clarify and correct some of the points you mentioned in your post.

1. Velocity vectors: A velocity vector is a mathematical representation of the speed and direction of an object's motion. It is typically denoted by an arrow pointing in the direction of motion with a length representing the speed. In your example, the velocity vector from point a to b can be denoted as v = b - a, where v is the velocity vector and a and b are the positions of the points.

2. Electrons: Electrons are subatomic particles with a negative charge and a mass of approximately 9.1094 x 10^-31 kilograms (kg). They are one of the fundamental building blocks of matter.

3. KeV: KeV stands for kiloelectron volts, which is a unit of energy commonly used in particle physics. It is equal to one thousand electron volts (eV).

4. The equation you mentioned, 0.5 mV^2 = eV, is known as the kinetic energy equation, where m is the mass of the object, V is its velocity, and eV is the energy in electron volts.

5. The equation V = sqrt(2eV/m) is the rearranged form of the kinetic energy equation, which gives the velocity (V) in terms of energy (eV) and mass (m).

6. In your example, you have correctly calculated the minimum and maximum velocities (mineV and maxeV) for a 1-100 KeV range. However, the units for these velocities should be meters per second (m/s) instead of KeV.

7. To ensure that all particles have a velocity within the range of 1-100 KeV, you can use the equation you mentioned, value = mineV + (range * (float)rand()/((float)RAND_MAX+1)), where value is the randomly generated velocity within the specified range.

8. To calculate the overall velocity of a particle with a velocity vector v = (Vx, Vy, Vz), you can use the Pythagorean theorem, which states that the magnitude of a vector is given by the square root of the sum of the squares of its components. In this case, the overall velocity would be sqrt(Vx^2