Prove Reciprocal Cubic Lattice of Cubic Lattice is Also Cubic

[Petar Mali]

Homework Statement

Show that the reciprocal cubic of cubic lattice is also cubic.

Homework Equations

$$cos\alpha*=\frac{cos\beta cos\gamma-cos\alpha}{sin\beta sin\gamma}$$

$$cos\beta*=\frac{cos\alpha cos\gamma-cos\beta}{sin\alpha sin\gamma}$$

$$cos\gamma*=\frac{cos\alpha cos\beta-cos\gamma}{sin\alpha sin\beta}$$

$$\vec{a*}=\frac{\vec{b}\times\vec{c}}{V}$$

$$\vec{b*}=\frac{\vec{c}\times\vec{a}}{V}$$

$$\vec{c*}=\frac{\vec{a}\times\vec{b}}{V}$$

The Attempt at a Solution

If I use this formula I will show that $$\alpha*=\beta*=\gamma*=90^{\circ}$$

and $$a*=b*=c*=\frac{1}{a}$$

and so reciprocal lattice of cubic lattice is cubic. Q.E.D.

But I don't know from where I get this angle relations

$$cos\alpha*=\frac{cos\beta cos\gamma-cos\alpha}{sin\beta sin\gamma}$$

$$cos\beta*=\frac{cos\alpha cos\gamma-cos\beta}{sin\alpha sin\gamma}$$

$$cos\gamma*=\frac{cos\alpha cos\beta-cos\gamma}{sin\alpha sin\beta}$$

 

[nickjer]

Just take the dot products between the reciprocal lattice vectors. If they are 0, then you know the angle between them must be 90 degrees. And they will come out to be 0 if you do the math right.

 

[Petar Mali]

Just take the dot products between the reciprocal lattice vectors. If they are 0, then you know the angle between them must be 90 degrees. And they will come out to be 0 if you do the math right.

Did you read my question?

Do you know from where I can get this expression for angles?

I'm not sure how do you mean to get zero if this is for example some loxogonal system in reciprocal lattice?

This way is I think unique. Because of that I'm interesting in this angle relations.

 

[nickjer]

If you want to solve for those expressions, then you would do it the same way as I just described. Take the dot products of the reciprocal lattice vectors, then wade through all the trigonometry until you get the final solution. It will get messy, but you can get the same expressions.

 

[nickjer]

Your first equation is wrong. Use the identity:

$$(\vec{a}\times \vec{b})\cdot(\vec{c} \times \vec{d}) = (\vec{a}\cdot \vec{c})(\vec{b}\cdot \vec{d}) - (\vec{a}\cdot \vec{d})(\vec{b}\cdot \vec{c})$$

Also, set the angles between your a and b to alpha, b and c to beta, and a & c to gamma. That way things won't get too confusing.

 

[nickjer]

You will get the exact same identity I gave you using the vector identity you just listed, but you will have to go through more work since yours isn't fully expanded.

 

[Petar Mali]

You will get the exact same identity I gave you using the vector identity you just listed, but you will have to go through more work since yours isn't fully expanded.

$$\vec{a^*}\cdot{\vec{b^*}=\frac{1}{V^2}[(\vec{b}\times\vec{c})\cdot(\vec{c}\times\vec{a})]=\frac{1}{V^2}[(\vec{b}\cdot \vec{c})(\vec{c}\cdot\vec{a})-(\vec{b}\cdot \vec{a})(\vec{c}\cdot\vec{c})]=\frac{1}{V^2}[bccos\alpha accos\beta-abcos\gamma c^2]$$

$$a^*b^*cos\gamma^*=\frac{abc^2}{V^2}(cos\alpha cos\beta-cos\gamma)$$

$$cos\gamma^*=\frac{abc^2}{a^*b^*V^2}(cos\alpha cos\beta-cos\gamma)$$

I use different because I want to get solutions in same form. So angle between the a and b is gamma.

I'm pretty tired. I don't see how to get $$\frac{abc^2}{a^*b^*V^2}=\frac{1}{sin\beta sin\alpha}$$

 

[nickjer]

|a*| = |a x b|/V = a b sin(alpha)/V

Was too lazy to put it in tex, but you get the point.

 

[Petar Mali]

Of course! :)

$$\vec{a*}=\frac{\vec{b}\times\vec{c}}{V}$$

$$\vec{b*}=\frac{\vec{c}\times\vec{a}}{V}$$

From that

$$a^*=\frac{|\vec{b}\times\vec{c}|}{V}=\frac{bcsin\alpha}{V}$$

$$b*=\frac{|\vec{c}\times\vec{a}|}{V}=\frac{acsin\beta}{V}$$

$$cos\gamma^*=\frac{abc^2}{a^*b^*V^2}(cos\alpha cos\beta-cos\gamma)$$

$$cos\gamma^*=\frac{abc^2}{\frac{bcsin\alpha}{V}\frac{acsin\beta}{V} V^2}(cos\alpha cos\beta-cos\gamma)$$

$$cos\gamma*=\frac{cos\alpha cos\beta-cos\gamma}{sin\alpha sin\beta}$$

Thanks!