Questions of calculus of crystal structures

In summary, the rule of cross product states that |a||b|sinθ is the area of the base and also the magnitude of the cross product of two vectors, while |c|cosΦ gives the height of the parallelepiped. The product of the base area and height gives the volume of the cell. In this context, absinΘ = |axb| and ccosΦ = n·c, where both sides should be scalars.
  • #1
mysci
9
0
upload_2015-3-3_22-26-6.png


We know the rule of cross product
upload_2015-3-3_22-27-38.png
or
2d308f37dd82911690b919157eace04d.png


Why here |absinΘ| =
upload_2015-3-3_22-39-31.png
, and
upload_2015-3-3_23-13-6.png
= c cos Φ in the above picture?

Thanks for explanation.
 
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  • #2
mysci said:
View attachment 79881

We know the rule of cross productView attachment 79882 or
2d308f37dd82911690b919157eace04d.png


Why here |absinΘ| = View attachment 79883 , and View attachment 79884 = c cos Φ in the above picture?
|a||b|sinθ is the area of the base, and is also the magnitude of the cross product of ##\vec{a}## and ##\vec{b}## -- i.e., |##\vec{a}## X ##\vec{b}|##. |c|cosφ gives the height of the parallelipiped. The product of the area of the base and the height gives the volume of the cell.
 
  • #3
The base is formed by two vectors with lengths ||a|| and ||b|| and angle between them [itex]\theta[/itex]. Draw a line from the tip of line b to the base, the line forming the vector a. That gives you a right triangle with hypotenuse of length ||b||. So the "opposite side", the height of the parallelogram forming the base of the figure. The "opposite side over the hypotenuse" is sine of the angle so [itex]height/||b||= sin(\theta)[/itex] and [itex]height= ||b|| sin(\theta)[/itex]. The area of a parallelogram is "height times base" so [itex]||a||||b|| sin(\theta)[/itex].
 
  • #4
Mark44 said:
|a||b|sinθ is the area of the base, and is also the magnitude of the cross product of ##\vec{a}## and ##\vec{b}## -- i.e., |##\vec{a}## X ##\vec{b}|##. |c|cosφ gives the height of the parallelipiped. The product of the area of the base and the height gives the volume of the cell.

Thanks.

Yes, but why not |axb|, is |axb|(unit vector n) in third step?

absinΘ and |axbl are also magnitudes, but |axb|(unit vector n) is a vector. absinΘ = |axb| ≠ |axb|(unit vector n) = vector a x vector b.

However, here absinΘ = |axb|(unit vector n). I don't understand this.

On the other hand, ccosφ is the height of parallelogram, how to change it to vector c? I don't understand it as well.

Thanks.
 
  • #5
mysci said:
Thanks.

Yes, but why not |axb|, is |axb|(unit vector n) in third step?
|a x b| is a scalar, while |a x b|n is a vector that points straight up, and that whose magnitude is the area of the base. If you dot this vector (|a x b|n) with c, you get the volume. One definition for the dot product of a and b is ##a \cdot b = |a| |b| cos(\theta)##, where ##\theta## is the angle between the two vectors. In your problem, the angle is ##\phi##.
mysci said:
absinΘ and |axbl are also magnitudes, but |axb|(unit vector n) is a vector. absinΘ = |axb| ≠ |axb|(unit vector n) = vector a x vector b.

However, here absinΘ = |axb|(unit vector n). I don't understand this.

On the other hand, ccosφ is the height of parallelogram, how to change it to vector c? I don't understand it as well.

Thanks.
 
  • #6
I may get something.
In fact,
absinΘ = |axb|
ccosΦ = n·c
Is it right?

I thought absinΘ = |axb|n and ccosΦ = c before I get the above thinking.
 
  • #7
Mark44 said:
|a x b| is a scalar, while |a x b|n is a vector that points straight up, and that whose magnitude is the area of the base. If you dot this vector (|a x b|n) with c, you get the volume. One definition for the dot product of a and b is ##a \cdot b = |a| |b| cos(\theta)##, where ##\theta## is the angle between the two vectors. In your problem, the angle is ##\phi##.

By the way, how do you type the vector symbol in here? I can't find this symbol. Thanks.
 
  • #8
I use LaTeX. Put either two # symbols at the front and two more at the end (for inline) or two $ symbols front and back (for standalone).

Here I'm adding an extra space between each pair so you can see what it looks like without being rendered: # #\vec{a}# #
Removing the spaces gives ##\vec{a}##
 
  • #9
Mark44 said:
I use LaTeX. Put either two # symbols at the front and two more at the end (for inline) or two $ symbols front and back (for standalone).

Here I'm adding an extra space between each pair so you can see what it looks like without being rendered: # #\vec{a}# #
Removing the spaces gives ##\vec{a}##
Thanks.

Then
I got following,
absinΘ = |axb|
ccosΦ = n·c
Is it right?
 
  • #10
mysci said:
Thanks.

Then
I got following,
absinΘ = |axb|
Should be |a||b|sinθ = |a x b|. a and b are vectors, so ab is not defined. Both sides of the equation should be scalars, which is why you have the magnitudes (absolute values).
mysci said:
ccosΦ = n·c
The right side is a scalar because it's a dot product, so the left side needs to be a scalar as well.
The left side should be |c|cosΦ.
mysci said:
Is it right?
 
  • #11
Mark44 said:
Should be |a||b|sinθ = |a x b|. a and b are vectors, so ab is not defined. Both sides of the equation should be scalars, which is why you have the magnitudes (absolute values).
The right side is a scalar because it's a dot product, so the left side needs to be a scalar as well.
The left side should be |c|cosΦ.
Thank you.:wink:
 

FAQ: Questions of calculus of crystal structures

What is the purpose of studying calculus of crystal structures?

The purpose of studying calculus of crystal structures is to understand the mathematical principles and relationships that govern the atomic and molecular arrangement in a crystal. This knowledge is essential for predicting and manipulating the properties of materials, such as strength, conductivity, and optical properties.

What are the basic concepts of calculus in crystal structures?

The basic concepts of calculus in crystal structures include mathematical operations like differentiation and integration, vector calculus, and tensor calculus. These concepts are used to describe the geometry and symmetry of crystals, as well as the behavior of atoms and molecules within them.

How is calculus applied in crystallography?

Calculus is applied in crystallography to analyze diffraction patterns and determine the atomic structure of crystals. It is also used to calculate physical properties of crystals, such as their elastic constants and thermal conductivity.

What are the challenges of using calculus in crystal structures?

One of the main challenges of using calculus in crystal structures is the complex and irregular nature of crystal surfaces and defects. This can make it difficult to accurately model and predict the behavior of crystals using calculus-based equations.

How does calculus of crystal structures relate to other fields of science?

Calculus of crystal structures is closely related to other fields of science, such as materials science, physics, and chemistry. It provides a mathematical foundation for understanding the structure and properties of materials, which is essential for advancements in technology and scientific research.

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