Exploring the Correct Calculation of Gradient on Dot Product

  • Thread starter Addez123
  • Start date
  • Tags
    Dot
In summary, using the convective operator gets you a different answer than using the material derivative.
  • #1
Addez123
199
21
Homework Statement



Calculate
Relevant Equations
Calculating dot product then doing gradient on it gets you:

which is the correct answer.

Using the formula, which you're required to do, gets a whole different answer.
Lets do each term individually.






For the cross product terms, and both gets you the zero vector, which cross with anything still just gives zero.

So you're left with


Which is nothing like the answer. I've recalculated every single piece of this equation 10 times and I can testify that the equation given in Relevant Equations is false.
 
Physics news on Phys.org
  • #2
What makes you think that ? It should be a differential operator.

Edit: Note that it is not , which is equal to one.
 
  • Like
Likes topsquark
  • #3
Addez123 said:


means
 
  • Like
Likes topsquark, PhDeezNutz, Addez123 and 1 other person
  • #4
pasmith said:
means
well that's a confusing use of parentheses.
 
  • #5
Addez123 said:
well that's a confusing use of parentheses.

We’ll get used to it. If you are going into physics you’re going to need it.For instance the force on an electric dipole is where is the external field that the dipole is immersed in.

But yeah it is perplexing when you first come across it. You’re unlikely to come across it in a first multivariable calculus class. I’d bet money you’re doing Griffiths E&M now.
 
  • Like
Likes topsquark
  • #6
You apply the expression to each component of

I believe this expression in general is called the “Material Derivative”
 
  • #7
pasmith said:
means
are you sure its not

?
Because the answer is suppose to be a vector.

@PhDeezNutz not doing Griffiths rn, but good guess! :p
Funny fact I've completed electro magnetic course with almost an A, but it was years ago and I never actually passed the vector.
 
  • #8
Addez123 said:
Because the answer is suppose to be a vector.
If is a vector then so is etc.

Addez123 said:
well that's a confusing use of parentheses.
Not as confusing as having differential operators acting to the left …
 
  • #9
PhDeezNutz said:
I believe this expression in general is called the “Material Derivative”
It is not. The material derivative is the derivative of some quantity along the flow of some velocity field and is given by
 
  • Like
Likes PhDeezNutz
  • #10
@Addez123

The full expression for is the following



Distributing that entire operation over each of the components

 
  • Like
Likes Addez123
  • #11
Thanks a ton, just figured that out and now I get the correct answer.
Thanks! :)
 
  • Like
Likes PhDeezNutz
  • #13
o,i calculate this way:
p =A \cdot B =x^2z+y^3+z^2x,
the answer should be Jacobian
(\frac{\partial p}{\partial x},\frac{\partial p}{\partial y},\frac{\partial p}{\partial z})
=(2xz+z^2,3y^2,x^2+2xz)
 
  • #14
alex_Hou said:
o,i calculate this way:
p =A \cdot B =x^2z+y^3+z^2x,
This was done in the OP already.
 

FAQ: Exploring the Correct Calculation of Gradient on Dot Product

What is the formula for calculating grad(A dot B)?

The formula for calculating grad(A dot B) is grad(A dot B) = grad(A) dot B + A dot grad(B). This formula is also known as the product rule for gradients.

What do the symbols in the formula for grad(A dot B) represent?

The symbol grad represents the gradient operator, which is a vector operator that calculates the rate of change of a scalar field. The symbol dot represents the dot product, which is a mathematical operation that calculates the scalar product of two vectors.

How do you calculate grad(A) and grad(B)?

The gradient of a scalar field A is calculated by taking the partial derivatives of A with respect to each variable and arranging them in a vector. Similarly, the gradient of a scalar field B is calculated by taking the partial derivatives of B with respect to each variable and arranging them in a vector. These gradients are then used in the formula for grad(A dot B).

What is the significance of calculating grad(A dot B)?

Calculating grad(A dot B) is useful in many areas of science and engineering. It allows us to determine the direction and magnitude of the greatest rate of change of a scalar field, which can provide insights into the behavior of physical systems and help in solving optimization problems.

Can grad(A dot B) be calculated for any type of vector fields A and B?

Yes, grad(A dot B) can be calculated for any type of vector fields A and B as long as they are both differentiable scalar fields. This means that they must have continuous partial derivatives with respect to all variables. If this condition is met, then the formula for grad(A dot B) can be applied.

Back
Top