Dimensional analysis of an equation of motion

[JD_PM]

Homework Statement

The evolution of the density in a system of attractive spheres can be described by the following dynamic equation.

$$\frac{\partial}{\partial t} \rho (r,t) = D_o [\nabla^2 \rho (r,t) + \beta \nabla \rho (r,t) \int dr' [\nabla V (|r-r'|)] \rho (r',t) g(r,r',t)]$$

a) Identify each term in this equation.

b) Show this equation holds using dimensional analysis.

Homework Equations

$$\frac{\partial}{\partial t} \rho (r,t) = D_o [\nabla^2 \rho (r,t) + \beta \nabla \rho (r,t) \int dr' [\nabla V (|r-r'|)] \rho (r',t) g(r,r',t)]$$

The Attempt at a Solution

Before answering the explicit questions I made some research.

This is the Smoluchowski Equation, which is the equation of motion for the probability density function (pdf) of the position coordinates of the Brownian particles. Besides, it applies on the Brownian (or diffusive) time scale.

a)
- On the left hand side of the equation there is the derivative of the pdf with respect to time.
- On the right hand side of the equation we can distinguish two main parts:
1) $D_o \nabla^2 \rho (r,t)$ is related to the Brownian motion
2) $D_o\beta \nabla \rho (r,t) \int dr' [\nabla V (|r-r'|)] \rho (r',t) g(r,r',t)$ is related to the effect of the direct interactions. g(r,r',t) is the pair correlation function.

b)
$$[D_o] = \frac{L^2}{T}$$

$$[\beta] = \frac{ML^2}{T^2}$$

$$[\rho] = LT$$

$$[g] = LT$$

So:

$$LT = L^3 + \frac{M^2 L^7}{T}$$

It is clear something is wrong. I think it has to be related to the dimensions of the pdf and the pair correlation function, which would not be LT.

 

[nrqed]

Homework Statement

The evolution of the density in a system of attractive spheres can be described by the following dynamic equation.

$$\frac{\partial}{\partial t} \rho (r,t) = D_o [\nabla^2 \rho (r,t) + \beta \nabla \rho (r,t) \int dr' [\nabla V (|r-r'|)] \rho (r',t) g(r,r',t)]$$

a) Identify each term in this equation.

b) Show this equation holds using dimensional analysis.

Homework Equations

$$\frac{\partial}{\partial t} \rho (r,t) = D_o [\nabla^2 \rho (r,t) + \beta \nabla \rho (r,t) \int dr' [\nabla V (|r-r'|)] \rho (r',t) g(r,r',t)]$$

The Attempt at a Solution

Before answering the explicit questions I made some research.

This is the Smoluchowski Equation, which is the equation of motion for the probability density function (pdf) of the position coordinates of the Brownian particles. Besides, it applies on the Brownian (or diffusive) time scale.

a)
- On the left hand side of the equation there is the derivative of the pdf with respect to time.
- On the right hand side of the equation we can distinguish two main parts:
1) $D_o \nabla^2 \rho (r,t)$ is related to the Brownian motion
2) $D_o\beta \nabla \rho (r,t) \int dr' [\nabla V (|r-r'|)] \rho (r',t) g(r,r',t)$ is related to the effect of the direct interactions. g(r,r',t) is the pair correlation function.

b)
$$[D_o] = \frac{L^2}{T}$$

$$[\beta] = \frac{ML^2}{T^2}$$

$$[\rho] = LT$$

$$[g] = LT$$

So:

$$LT = L^3 + \frac{M^2 L^7}{T}$$

It is clear something is wrong. I think it has to be related to the dimensions of the pdf and the pair correlation function, which would not be LT.

It looks like you did not take into account the dimensions of the derivatives.

 

[JD_PM]

It looks like you did not take into account the dimensions of the derivatives.

Absolutely, my bad. Doing so we have on the right hand side dimensions of L. On the first term of the right hand side of the equation I got L as well but in the second one I got:

$$\frac{M^2 L^5}{T}$$

So I have to be missing something here...

 

[RPinPA]

I had to think a little about the dimensions of a pdf. At first I wanted to say it's unitless, but that's not correct. If you integrate a pdf over its entire support, you get a unitless 1. So with a density that depends on $r$ and $t$, you'd be integrating $dr$ and $dt$, which means your integral would have units of $\rho * L * T$ and since that's unitless, then $[\rho] = L^{-1} T^{-1}$.

So your left hand side has units of $L^{-1} T^{-2}$.

Keep in mind how to handle the units of the spatial derivatives and the integral on the right hand side.

 

[JD_PM]

Thank you for pointing that out. Besides, I fixed some mistakes:

$$[\beta] = \frac{T^2}{ML^2}$$

$$[g] = \frac{1}{L^2T}$$

So now I get:

$$\frac{1}{LT^2} = \frac{L^2}{L^3T^2} + \frac{T^2ML^2}{ML^8T^5}$$

I simplified it and got:

$$\frac{1}{LT^2} = \frac{1}{LT^2} + \frac{1}{L^6T^3}$$

As you can see, the second term on the right hand side is still wrong.

Note I regarded $\nabla$ as:

$$\nabla = \frac{\partial}{\partial x}$$

 

[RPinPA]

That's the right way to handle $\nabla$ dimensionally. So perhaps as you guessed in your initial post, there's something wrong with the units of the other functions in that second term. You need to check where they come from and how they're used to make sure you've got the units right. Is V a potential for instance? Since I see M in your analysis, I guess it's a potential energy per unit mass?

 

[JD_PM]

That's the right way to handle $\nabla$ dimensionally. So perhaps as you guessed in your initial post, there's something wrong with the units of the other functions in that second term. You need to check where they come from and how they're used to make sure you've got the units right. Is V a potential for instance? Since I see M in your analysis, I guess it's a potential energy per unit mass?

Yes V is potential due to interaction, so its dimensions are energy ones. $\beta$ is the thermodynamic constant: https://en.wikipedia.org/wiki/Thermodynamic_beta. So by definition has dimensions of inverse energy.

g is the pair correlation function, so its dimensions should be:

$$[g] = \frac{1}{L^2T}$$