Vector Differentiation: Solving for Partial Derivatives

In summary, the vector S is defined as (S_x,S_y,S_z) and it has a function that calculates E_h=-\frac{\partial E_h}{\partial S_x},-\frac{\partial E_h}{\partial S_y},-\frac{\partial E_h}{\partial S_z}
  • #1
ilvreth
33
0
Hi there to all. I am stuck and i want some help to clear my view

When you have defined the vector S=(S_x ,S_y ,S_z) and you have a function like
E_h = k1 (S_x^2 S_y^2 + S_x^2 S_z^2 + S_y^2 S_z^2) + k2 S_z - k3 S_x S_y

and you try to calculate -\frac{\partial E_h}{\partial S} what the result is?

(the S is vector)

My mind has been blocked... help me please.
 
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  • #2
You can't differentiate with respect to a vector. You need to differentiate with respect to the coordinates.
 
  • #3
I don't understand your notation. For example, what does S_x2 mean?

Are you trying to describe a directional derivative?
 
  • #4
LCKurtz said:
I don't understand your notation. For example, what does S_x2 mean?

Are you trying to describe a directional derivative?
If you have S=(Sx,Sy,Sz) then S_x is the x component of S so S_x^2 is (S_x)^2 correctly .
 
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  • #5
Pengwuino said:
You can't differentiate with respect to a vector. You need to differentiate with respect to the coordinates.

The landau-lifgarbagez equation is

[tex] \frac{\partial S}{\partial t}= -\gamma S \times H_{eff} [/tex] where [tex] S [/tex] is the magnetization vector and [tex] H_{eff} [/tex] is the effective magnetic field. More often you meet the [tex] H_{eff} [/tex] to be

[tex] H_{eff}= - \frac{\partial f }{\partial S} [/tex] where [tex] f [/tex] is a function like [tex] E_h [/tex]

[tex] E_h = k_1 (S_x^2 S_y^2 + S_x^2 S_z^2 + S_y^2 S_z^2) + k_2 S_z - k_3 S_x S_y [/tex]

and S is the magnetization vector.

How do you write down the result of [tex] -\frac{\partial E_h}{\partial S} [/tex]?
 
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  • #6
Pengwuino said:
You can't differentiate with respect to a vector. You need to differentiate with respect to the coordinates.
Sure you can. Welcome to the wonderful world of tensors.

ilvreth said:
Hi there to all. I am stuck and i want some help to clear my view

When you have defined the vector S=(S_x ,S_y ,S_z) and you have a function like
E_h = k1 (S_x^2 S_y^2 + S_x^2 S_z^2 + S_y^2 S_z^2) + k2 S_z - k3 S_x S_y

and you try to calculate -\frac{\partial E_h}{\partial S} what the result is?

(the S is vector)

My mind has been blocked... help me please.

Try to learn the LaTeX system here; it looks like you already know it. Something like this:

When you have defined the vector [tex]S=(S_x ,S_y ,S_z)[/tex] and you have a function like
[tex]E_h = k_1 (S_x^2 S_y^2 + S_x^2 S_z^2 + S_y^2 S_z^2) + k_2 S_z - k_3 S_x S_y[/tex]

and you try to calculate [tex]-\frac{\partial E_h}{\partial S}[/tex] what the result is?
What you want to calculate are the components of the vector (actually, a Cartesian tensor) [tex]-\frac{\partial E_h}{\partial S_{\mu}}[/tex] where [tex]\mu[/tex] indices the vector elements.
 
  • #7
D H said:
What you want to calculate are the components of the vector (actually, a Cartesian tensor) [tex]-\frac{\partial E_h}{\partial S_{\mu}}[/tex] where [tex]\mu[/tex] indices the vector elements.
So is this the result correct [tex]-\frac{\partial E_h}{\partial S}= ( -\frac{\partial E_h}{\partial S_x}, -\frac{\partial E_h}{\partial S_y}, -\frac{\partial E_h}{\partial S_z} ) [/tex] ??
 
  • #8
Correct. So what is the result?

BTW, this looks like homework. If it is, you should have posted this thread in the appropriate homework section. If not, ignore my remark.
 
  • #9
D H said:
Correct. So what is the result?

BTW, this looks like homework. If it is, you should have posted this thread in the appropriate homework section. If not, ignore my remark.

It is not homework. This is an example to clear some things in my mind. Thank you.
 

1. What is differentiation with vectors?

Differentiation with vectors is a mathematical operation that involves finding the rate of change of a vector function with respect to its independent variable. It is a way to measure how much a vector changes over a small interval of time or space.

2. How is differentiation with vectors different from regular differentiation?

The main difference is that while regular differentiation involves finding the derivative of a scalar function, differentiation with vectors involves finding the derivative of a vector function. This means that instead of a single value, the result is a vector of derivatives for each component of the vector function.

3. What are the applications of differentiation with vectors?

Differentiation with vectors is used in many fields such as physics, engineering, and economics. It is used to analyze the motion of objects, optimize systems, and solve problems involving vectors such as velocity, acceleration, and force.

4. What are some common techniques for differentiating vectors?

The most common technique is to use the chain rule, which involves taking the derivative of each individual component of the vector function and combining them using the chain rule. Other techniques include the product rule, quotient rule, and vector identities such as the dot and cross product.

5. How can I practice and improve my skills in differentiation with vectors?

You can practice by solving various problems and exercises involving vector functions and their derivatives. You can also use online resources such as tutorials, videos, and interactive quizzes to improve your understanding and skills in differentiation with vectors.

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