The programming labs for Physics 1251 have slightly different units than the SI unit system that we typically use in the course. There are three issues that deserve mention: the length units (pretty simple), force units, and the electric field units (which is related to the force units).
In this course we typically think in meters. In the programming labs, the x and y variables are actually in pixel units and the width of the screen is 750 pixels. For example, if the position of the object (x) is less than zero or larger than 750, the object will move off of the screen. Typically the height of the screen is 500. If the vertical position of the object (y) is less than zero or larger than 500 the object will move off the screen.
How do we relate pixels to the typical meter units that we use in class? We have some options!
We could figure out how large each pixel actually is on your screen in meters and then multiply the x and y positions by this size. This involves a lot of multiplying and a lot of confusion about whether we are talking about meters or pixels. For example, we would have to multiply all the velocities by the pixel size to change the units from pixels per second to meters per second, and the same thing for the acceleration (to change it from pixels per second2 to meters per second2). Everyone's screen will have a slightly different pixel size. What a mess!
The simpler thing is to just think of pixels as meters and talk about velocities as meters per second and accelerations as meters per second squared. There is no harm in doing this! It just means that the screen (with the particle being accelerated between two plates) is showing something that is much larger than what actually appears on the screen. Since the screen is 750 pixels across, this means that you are looking down on something that is really 7.5 soccer fields across. This is a little bit weird to imagine but it is a lot simpler than the alternative.
In the accelerator exercise, the mass of the particle is equal to 1.0 and user is told that the particle is a proton and that the mass is in terms of atomic mass units (instead of kilograms). This means that the force is *not* in the usual Newton units! For example, if you want to accelerate a proton at 1 meter per second2, it only takes a force of 1.0 to do this as far as the computer program is concerned. Whereas in SI units, the proton mass is $1.67\cdot 10^{-27} \, {\rm kg}$. The force required to make the proton accelerate at 1 meter per second2 is $$F = ma = (1.67\cdot 10^{-27} \, {\rm kg})\cdot (1.0 \, {\rm m/s^2} ) = 1.67\cdot 10^{-27} \, {\rm kg \cdot m / s^2} = 1.67\cdot 10^{-27} \, {\rm N} $$ If this seems like a very small number of newtons, remember that protons have a very small mass, so it doesn't take much to get them to accelerate at 1.0 meters per second2.
Rather than deal with all these factors of $10^{-27}$ in the computer program, it would be more convenient in some ways to just use different force units. Instead of using Newtons, which is ${\rm kg \cdot m/s^2}$, we can use ${\rm amu \cdot m/s^2}$ (you can call this Orbans if you want or "natural" units). When people write computer programs to do simulations, they often change the units to do the calculations. You don't always change the units when you write a computer program, but sometimes people do. If you were only ever going to run simulations of protons, and other nuclei perhaps it would make good sense.
Normally we use electric field units of Newtons per Coulomb (N/C). But since we are using force units of $\rm amu \cdot m/s^2$ instead of Newtons we will have to use something different for our electric field units.
But as long as we are changing the units around, what if instead of using some force unit per Coulomb, we use some force unit per unit of elementary charge?
If we did this, defining the electric field to be $(\rm amu \cdot m/s^2) / e$ where $e$ is the elementary charge unit then an electric field of 1.0 would accelerate a proton (which has a charge of +1 elementary charge unit) at exactly $1.0 \, \rm m/s^2$. For comparison, if we did this in SI units and wanted to know how strong the electric field would need to be to accelerate a proton at $1.0 \, \rm m/s^2$ then we would have $$E = F / q = ma / q = (1.67 \cdot 10^{-27} \, \rm kg)\cdot(1.0 \, m/s^2) / (1.6 \cdot 10^{-19} \, {\rm C}) = 1.04 \cdot 10^{-8} \, N/C$$ Again, this is a pretty small number, but it doesn't take much of an electric field to accelerate a proton at $1.0 \, \rm m/s^2$
The accelerator exercise uses these $(\rm amu \cdot m/s^2)/e$ electric field units. The default electric field strength is 5. This means that a proton will be accelerated at $5 \, \rm m/s^2$ if it is placed in the field.
| SI units | "Natural" Units | |
| proton mass | $1.67 \cdot 10^{-27}$ | 1.0 |
| Force | $1.67 \cdot 10^{-27}$ | 1.0 |
| Electric Field | $1.04 \cdot 10^{-8}$ N/C | 1.0 |