# Why is a spatial grid necessary in plasma simulations?

I am new to the topic and I am reading Plasma Physics via Computer Simulations by Birdsall and Langdon. The authors claim that working with a spatial grid is more efficient opposed to considering the $$10^{25}$$ particle interactions. Additionally the problem of $$10^{25}$$ singularities ar $$r \to 0$$ also vanishes.

I understand how the use of a charge density across a grid is more efficient than considering all the 1-1 interactions, nevertheless I don't understand why do we need a spatial grid for this. Why can't the actual charge distribution be used? Why is it better to perform a weighting to the charges and approximate it to a distribution on the grid?

Additionally what is the problem with the singularities at $$r \to 0$$? Charges simply ignore their own electric fields.

Edit: I understand how dealing with $$10^{25}$$ items is not computational possible, nevertheless these simulations work with 1D-2D approximations that reduces the number of particles, say to $$10^6$$.

What is my problem? If we are working with a grid we still have to go through the $$10^6$$ particles and distribute their charge along the grid appropriately. We still have to go through all the particles, we just lose accuracy by dividing the charges along a grid.

Why don't we just use the actual particles positions? Does it make easier to numerically solve the fields for the grid opposed to non-uniform positions distribution?

• By "the actual charge distribution" do you mean $\rho=\sum_{n=0}^{10^{25}} q_n\delta(\vec{r}-\vec{r_n})$, where $\vec{r_n}$ is the position of the $n$th charge $q_n$? Do you see how dealing with a charge distribution that has $10^{25}$ distinct terms might not be amenable to computation? – probably_someone Sep 25 at 10:47
• Yes, that is what I mean by the actual charge distribution. Please look to my edit and let me know if it is more clear what my concerns is. – Daniel Duque Sep 25 at 10:55
• @alephzero please look at my edit; you still need to go through all particles with a grid – Daniel Duque Sep 25 at 11:20
• To see why your idea doesn't work, consider say 100 particles, and then try to simulate them by "grouping" them into say 10 groups of 10. The problem is that the 10 particles you put in each group don't stay close to each other as the simulation progresses. $10^6$ big particles, each with a fixed point charge, don't behave the same way as $10^{25}$ small ones. If you use a charge distribution, in effect you have an infinite number of small particles, independent of the number of grid points in your model. – alephzero Sep 25 at 11:26
• On a grid, you're not simulating $10^n$ particles but the bulk properties of the particles as a fluid. Two different things to consider there. – Kyle Kanos Sep 25 at 11:35

Ok, let us entertain the idea that you can take $$10^{25}$$ particles and evolve them numerically. Storing each position and velocity is 6 doubles, which gives a whopping 80 yottabytes or $$10^{12}$$ terrabytes for all the particles. Now you want to evolve these by one time-step. This requires computing the force on each individual particle by each individual particle three times (because of dimension). So that gives you $$3\cdot10^{50}$$ double operations or $$\sim 10^{41}$$ corehours or $$\sim 10^{28}$$ years ($$10^{11}$$ ages of the Universe) on the most powerful supercomputers to date. You could argue that you do not need to evaluate the force between every two particles, but even then it would require at least thousands of years to evaluate even a single time step on the most powerful supercomputers.
• @DanielDuque So it is possible to take the Boltzmann equation and discretize it in a way so that you are evolving, say, $10^6$ "superparticles" in phase space. But they do not really represent the real particles or clumps of particles, it is really a Monte Carlo method of evolving the Boltzmann equation (and it is used in certain contexts). But if you want to represent the actual particles for more than a few grams of plasma, you are still talking about $\gtrsim 10^{23}$ particles and nothing will save you from that. – Void Sep 25 at 12:05