All Questions
12
questions
1
vote
2
answers
262
views
Determinant of Rank-2 Tensor using Levi-Civita notation
In my Professor's notes on Special Relativity, the determinant of a rank-two tensor $[T]$ (a $4\times 4$ matrix, basically) is given using the Levi Civita Symbol as: $$T=-\epsilon_{\mu\nu\rho\lambda}T^...
- 442
6
votes
1
answer
300
views
Inverse of anti-symmetric rank 4 tensors?
I am trying to find an inverse of a tensor of the form
$$M_{\mu\nu\rho\sigma}$$ such that $M$ is anti-symmetric in the $(\mu, \nu)$ exchange and $(\rho, \sigma)$ exchange. The inverse should be such ...
- 1,200
0
votes
0
answers
432
views
How to know if a matrix is a (0,2) tensor, a (2,0) tensor or a (1,1) tensor?
For example:
$$
X = \begin{bmatrix}
1 & -1 & 0 & 0 \\
-1 & 0 & 5 & 3 \\
-2 & 1 & 0 & 0 \\
0 & 1 & 0 & 2
...
- 157
0
votes
2
answers
151
views
Dimension of a vector space of all tensors of rank $(k,l)$ in 4D
Dual vector space is the set of all linear functionals defined on a given vector space. The vector space and dual vector space is isomorphic and hence have the same dimension. A rank $(k,l)$ tensor is ...
- 310
3
votes
2
answers
193
views
Problem with proving the invariance of dot product of two four vectors
I am having a spot of trouble with index manipulation (its not that I am very unfamiliar with this, but I keep losing touch). This is from an electrodynamics course - we're just getting started with 4 ...
- 322
2
votes
2
answers
106
views
Tensorial direct product
The direct product of two tensors is also a tensor. I would like to know if we can write a tensor as a product of only two other tensors. For example, how to find $A^{\mu}$ and $ B^{\nu}$ so that $\...
- 31
0
votes
1
answer
494
views
General Relativity: change of coordinates in tangent space
For starters, in the context of the tangent space of a manifold in GR, we can derive that:
$$g'_{\mu \nu}=\frac{\partial x^\rho}{\partial x'^\mu}\frac{\partial x^\sigma}{\partial x'^\nu}g_{\rho \sigma}...
- 4,577
7
votes
4
answers
1k
views
Tensorial Proof that $\det(F^{\mu\nu})=(\vec{E}\cdot\vec{B})^2$?
I am trying to understand why
$$\det(F^{\mu\nu})=(\vec{E}\cdot\vec{B})^2\tag{1}.$$
Of course one can just calculate the determinant of $F^{\mu\nu}$ expressed as a matrix with components given in ...
- 3,407
0
votes
1
answer
165
views
A silly doubt on notation of Hartle's GR book
Quoting from the text
[...] Evaluating (20.46) in such coordinates is just like evaluating the derivative of a function (20.1):
$$\left( \mathbf{\nabla_t v}\right)^\alpha = t^\beta \frac{\...
- 3,085
0
votes
2
answers
822
views
Lorentz tensor - always outer product of two four-vectors?
A 2nd rank Lorentz tensor is defined as one that transforms as:
$$T'=\Lambda T\Lambda^T$$
It is clear that the quantity:
$$\tilde A \tilde B^T$$
where $\tilde A$ and $\tilde B$ are 4-vectors always ...
1
vote
2
answers
2k
views
Transformation of four-velocity in special relativity
I am revising special relativity introducing more matrix form in the equation. Currently I am reading book in which transformation matrix is defined as
$${\Lambda=
\begin{bmatrix}
\gamma & -v\...
- 197
8
votes
2
answers
6k
views
The definition of transpose of Lorentz transformation (as a mixed tensor)
In the appendix of the textbook of Group Theory in Physics by Wu-Ki Tung, the transpose of a matrix is defined as the following, Eq.(I.3-1)
$${{A^T}_i}^j~=~{A^j}_i.$$
This is extremely confusing for ...
- 962