Taking the derivative of a dot product

In summary: Therefore, the force, F, can be expressed as:F = -dV/dr = -4ε[(144σ^11(x2-x1)^13 - 36σ^5(x2-x1)^7(y2-y1)^8 - 36σ^5(x2-x1)^7(z2-z1)^8]In summary, to take the derivative of the Lennard-Jones potential, we can use the chain rule and the product rule to break it down into simpler terms. By rewriting the potential in terms of the distance between
  • #1
pholvey
6
0

Homework Statement



Hi, I need to take the derivative of the following energy potential (attached) in order to get an expression of the force so I can do a conjugate gradient minimization.

An explanation of the potential is found in the attached jpeg below the functional form.

Homework Equations



d/dx(a.b)=a′b+ab′

The Attempt at a Solution



I know that I should be able to take the derivative of a dot product as shown above, but in this case, the vectors being constructed of differences in xyz coordinates is confusing me. If anyone can give any help on how to approach this problem, I would be grateful.

Many thanks,

Pholvey
 

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  • #2


Dear Pholvey,

Thank you for reaching out for help with your problem. The potential you have attached is commonly known as the Lennard-Jones potential, which is used to model the interaction between two particles in a system. To take the derivative of this potential, we can use the chain rule and the product rule to break it down into simpler terms.

First, let's rewrite the potential in terms of the distance between the two particles, r, instead of the differences in xyz coordinates:

V(r) = 4ε[(σ/r)^12 - (σ/r)^6]

Using the chain rule, we can write the derivative as:

dV/dr = 4ε[-12(σ/r)^11 * (-1/r^2) - 6(σ/r)^5 * (-1/r^2)]

= -4ε[-12(σ/r)^11 * (-1/r^2) - 6(σ/r)^5 * (-1/r^2)]

= 4ε[(12σ^12/r^13) - (6σ^6/r^7)]

= 4ε[12σ^12/r^13 - 6σ^6/r^7]

Now, we can use the product rule to expand the terms:

dV/dr = 4ε[12σ^12/r^13 - 6σ^6/r^7]

= 4ε[12(12σ^11 * (1/r^14)) - 6(6σ^5 * (1/r^8))]

= 4ε[144σ^11/r^14 - 36σ^5/r^8]

Finally, we can rewrite this expression in terms of the differences in xyz coordinates:

dV/dr = 4ε[144σ^11/(x2-x1)^14 - 36σ^5/(x2-x1)^8]

= 4ε[144σ^11/(x2-x1)^14 - 36σ^5/(y2-y1)^8 - 36σ^5/(z2-z1)^8]

= 4ε[(144σ^11(x2-x1)^13 - 36σ^5(x2-x1)^7(y2-y1)^8 - 36σ^5(x2-x1)^7(z2-z1)^8]

= 4ε[(144σ^11(x2-x1)^13 -
 

1. What is the formula for taking the derivative of a dot product?

The formula for taking the derivative of a dot product is d/dx (u · v) = u' · v + u · v', where u and v are two vectors and the prime notation represents the derivative with respect to x.

2. How do you interpret the result of taking the derivative of a dot product?

The result of taking the derivative of a dot product is a vector that represents the rate of change of the dot product with respect to x. This vector is perpendicular to the original vectors u and v, and its magnitude is equal to the product of the magnitudes of u and v multiplied by the cosine of the angle between them.

3. Can you provide an example of taking the derivative of a dot product?

Sure, let's say we have two vectors u = (3x + 2, x^2) and v = (2x, 5). The dot product of these two vectors is u · v = (6x^2 + 10x, 5x^2 + 5). To find the derivative of this dot product, we use the formula d/dx (u · v) = u' · v + u · v', which gives us d/dx (u · v) = (12x + 10, 10x) + (6x^2 + 10x, 5x^2). Simplifying this, we get the derivative of the dot product as 12x + 10, 10x + 5x^2.

4. Are there any special rules or properties to keep in mind when taking the derivative of a dot product?

Yes, there are a few rules and properties to keep in mind when taking the derivative of a dot product. First, the derivative of a dot product is a vector, not a scalar. Second, the order of the vectors in the dot product does not matter, as the derivative of u · v is the same as the derivative of v · u. Finally, the derivative of a dot product is distributive, so the derivative of u · (v + w) is equal to u · v + u · w.

5. How is taking the derivative of a dot product useful in scientific applications?

Taking the derivative of a dot product is useful in many scientific applications, particularly in physics and engineering. For example, in mechanics, the derivative of the dot product of a force vector and a displacement vector can give us the work done by that force. In electromagnetics, the derivative of the dot product of an electric field and a displacement vector can give us the induced voltage. In general, the derivative of a dot product can help us analyze the relationship between two quantities and their rates of change.

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