Is the Quotient Theorem Applicable to 4th Rank Tensors?

In summary, the author is asking how to transform partial derivatives, and is frustrated with the lack of response to his question.
  • #1
latentcorpse
1,444
0
Prove that [itex]b_{ijkl}=\int_{r<a} dV x_i x_j \frac{\partial^2}{\partial_k \partial_l} (\frac{1}{r})[/itex] where [itex]r=|x|[/itex] is a 4th rank tensor.

i've had a couple of bashes and got nowhere other than to establish that its quotient theorem.

can i just pick a tensor of rank 3 to multiply it with or something?
 
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  • #2
hmm … do you mean …
Prove that bijkl = ∫r<a dV xi xj2(1/r)/∂kl, where r=|x|, is a 4th rank tensor.
 
  • #3
yep.
 
  • #4
I have no idea how to solve this too, can you give me some idea please?
 
  • #5
welcome to pf!

hi nhanle! welcome to pf! :wink:

ok, what is the test for something being a tensor? :smile:
 
  • #6
hi tiny-tim,
thank you for your reply. This is how vague the definition of tensor I am holding at the moment.
I am also confused about the Affine connection. Can you help me clarify this?

Thank you
 
  • #7
?? :confused:

i'm not going to type out a lecture on tensors and connections :redface:

please go back to your book or your lecture notes, and read up about tensors
 
  • #8
those appears on my lecture notes and also my book (general relativity - M.P.Hobson, G. Efstathiou, A.N. Lasenby) with very vague definitions.

From my understanding, if one is to be a rank N-tensor, it should expect to have N derivative summations under coordinate transformation. Is that right?
 
  • #9
does your book show why the Christoffel symbols aren't tensors?

if so, that should show you how to do it :smile:
 
  • #10
it does but only with a few special case. So, I stumpled on this question "Prove that bijkl = ∫r<a dV xi xj ∂2(1/r)/∂k∂l, where r=|x|, is a 4th rank tensor."

How to transform the partial derivatives? Thank you for being so patient with me
I also have question about the affine connection https://www.physicsforums.com/showthread.php?t=189456 which was raised long ago but no one seems to be interested in answering :(
 

Related to Is the Quotient Theorem Applicable to 4th Rank Tensors?

1. What is the Quotient Theorem (Tensors)?

The Quotient Theorem (Tensors) is a mathematical theorem that states that the quotient of two tensors is equal to the quotient of their components if and only if the two tensors have the same basis and the denominator tensor is invertible.

2. How is the Quotient Theorem (Tensors) used in physics?

In physics, the Quotient Theorem (Tensors) is used to simplify calculations involving tensors. It allows for the manipulation of tensors by converting them into a simpler form, making it easier to solve complex equations and understand physical phenomena.

3. Can the Quotient Theorem (Tensors) be applied to tensors of any dimension?

Yes, the Quotient Theorem (Tensors) can be applied to tensors of any dimension as long as the tensors have the same basis and the denominator tensor is invertible.

4. Is the Quotient Theorem (Tensors) related to the concept of covariant and contravariant tensors?

Yes, the Quotient Theorem (Tensors) is closely related to the concept of covariant and contravariant tensors. It is often used to convert between these two types of tensors and simplify equations involving them.

5. How does the Quotient Theorem (Tensors) differ from the Tensor Product Theorem?

The Quotient Theorem (Tensors) and the Tensor Product Theorem are two different theorems that involve tensor operations. While the Quotient Theorem deals with dividing tensors, the Tensor Product Theorem deals with multiplying tensors. However, both theorems are important in tensor calculations and are often used together.

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