# for a polydisperse polymer, why is its weight average molecular weight always bigger than its number average molecular weight?

The statement comes from my lecture slides on polymers. I hope someone can show me the maths proof for the inequality. I will try come up with the proof, too. It is actually a maths question:D Here is the basic definition for the two average weights: https://en.wikipedia.org/wiki/Molar_mass_distribution

• Surely it's axiomatic that for a positive number (in this case the molecular mass), that the squared value is larger? By definition if a polymer is described as polydisperse, the PDI is >1, so Mw must be greater than Mn Nov 12 '16 at 10:51

Mathematics has many examples of identities and concepts which are extremely powerful due to their simplicity, such as the pigeonhole principle and the triangle inequality. One such identity will guide us through the problem.

We start with the very innocent identity, that for a real number $z$, the following is true:

$$z^2 \geq 0$$

Now let $z = x-y$. We have: $$(x-y)^2 \geq 0$$ $$x^2 -2xy + y^2 \geq 0$$ $$x^2 + y^2 \geq 2xy$$ $$\frac{x^2 + y^2}{2} \geq xy$$

Now this last inequality has a very powerful interpretation. To show it more clearly, we can replace $x^2$ with $a$ and $y^2$ with $b$, yielding:

$$\frac{a + b}{2} \geq \sqrt{ab}$$

In words, the arithmetic mean of two positive real numbers will always be greater than or equal to their geometric mean. This is called the AM-GM inequality, and it often shows up in the most surprising of places.

Let's get started. We define the number average molecular weight ($M_{n}$) and the weight average molecular weight ($M_{w}$) of a polydisperse polymer mixture as the following, respectively:

$$M_{n}=\frac {\sum\limits_{i} N_{i}M_{i}}{\sum\limits_{i} N_{i}} \quad , \quad M_{w}=\frac {\sum\limits_{i} N_{i} M_{i}^2 }{\sum\limits_{i} N_{i}M_{i}}$$

We then define the dispersity index $Đ$ (or $PDI$) as:

$$Đ = \frac{M_w}{M_n} = \frac{\sum\limits_{i} N_{i} M_{i}^2 \sum\limits_{i} N_{i}}{(\sum\limits_{i} N_{i}M_{i})^2}$$

To make things easier, let us start by analyzing the first non-trivial case ($i=2$):

$$Đ = \frac{M_w}{M_n} = \frac{(N_{1} M_{1}^2 + N_{2} M_{2}^2)(N_{1} + N_{2}) }{(N_{1} M_{1} + N_{2} M_{2})^2}$$

We multiply the bracketed terms and expand the binomial to obtain:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{1}N_{2}M_{1}^2 + N_{1}N_{2}M_{2}^2}{N_{1}^2 M_{1}^2 + 2N_{1}N_{2}M_{1}M_{2} + N_{2}^2 M_{2}^2}$$

Rearranging the terms slightly:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}}$$

Notice that the terms in the numerator and denominator match except for the coloured parts.

Here is where the magic happens. From the AM-GM inequality, we must have:

$$x^2 + y^2 \geq 2xy \quad \Rightarrow \quad M_{1}^2 + M_{2}^2 \geq 2M_{1}M_{2}$$

We then replace the term $(M_{1}^2 + M_{2}^2)$ in the numerator with the term $2M_{1}M_{2}$. By doing so, the new numerator is smaller than or equal to the old numerator. Thus:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}} \geq \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}}=1$$

$$Đ = \frac{M_w}{M_n} \geq 1$$

What about the next non-trivial case ($i=3$)? We have:

$$Đ = \frac{M_w}{M_n} = \frac{(N_{1} M_{1}^2 + N_{2} M_{2}^2 + N_{3} M_{3}^2)(N_{1} + N_{2} + N_{3}) }{(N_{1} M_{1} + N_{2} M_{2} + N_{3} M_{3})^2}$$

Multiplying everything out and performing the appropriate arrangements yields:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3} M_{3}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2) + N_{1}N_{3}(M_{1}^2 + M_{3}^2) + N_{2}N_{3}(M_{2}^2 + M_{3}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3}^2M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}}$$

This time, we can use the AM-GM inequality thrice:

$$M_{1}^2 + M_{2}^2 \geq 2M_{1}M_{2} \quad , \quad M_{1}^2 + M_{3}^2 \geq 2M_{1}M_{3} \quad , \quad M_{2}^2 + M_{3}^2 \geq 2M_{2}M_{3}$$

The end result is the same:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3} M_{3}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2) + N_{1}N_{3}(M_{1}^2 + M_{3}^2) + N_{2}N_{3}(M_{2}^2 + M_{3}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3}^2M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}}$$ $$\geq \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3} M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3}^2M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}} = 1$$

$$Đ = \frac{M_w}{M_n} \geq 1$$

Now stating the general case:

$$Đ = \frac{M_w}{M_n} = \frac{\sum\limits_{i} N_{i} M_{i}^2 \sum\limits_{i} N_{i}}{(\sum\limits_{i} N_{i}M_{i})^2} = \frac{\sum\limits_{i} N_{i}^2 M_{i}^2 + \color{#ff0000}{\sum\limits_{i,j\ (j>i)} N_{i}N_{j}(M_{i}^2 + M_{j}^2)}}{\sum\limits_{i} N_{i}^2 M_{i}^2 + \color{#0000ff}{2\sum\limits_{i,j\ (j>i)} N_{i}N_{j}M_{i}M_{j}}} \geq \frac{\sum\limits_{i} N_{i}^2 M_{i}^2 + \color{#0000ff}{2\sum\limits_{i,j\ (j>i)} N_{i}N_{j}M_{i}M_{j}}}{\sum\limits_{i} N_{i}^2 M_{i}^2 + \color{#0000ff}{2\sum\limits_{i,j\ (j>i)} N_{i}N_{j}M_{i}M_{j}}}=1$$

$$Đ = \frac{M_w}{M_n} \geq 1$$

• I was thinking about explaining it in terms of the Cauchy-Schwarz inequality, but your way is more convincing. Nov 15 '16 at 6:18