What kind of tensor is the electromagnetic field tensor?

Click For Summary
SUMMARY

The electromagnetic field tensor, denoted as F^{\mu \nu}, is confirmed to be a second-rank tensor that transforms under Lorentz transformations according to the relation F'^{\mu \nu} = \Lambda^\mu_{\ \alpha} \Lambda^{\ \nu}_{\beta} F^{\alpha \beta}. This transformation is derived from the covariant form of the Lorentz force law, m \ddot x^\mu = q F^{\mu \nu} \dot x_\nu. The discussion clarifies that while F^{\mu \nu} behaves like a mixed tensor, its transformation properties align with those of a contravariant tensor, leading to questions about the nature of its classification.

PREREQUISITES
  • Understanding of Lorentz transformations and their properties
  • Familiarity with tensor notation and types (covariant, contravariant, mixed)
  • Knowledge of the Lorentz force law in physics
  • Basic concepts of differential calculus, particularly the quotient rule
NEXT STEPS
  • Study the properties of second-rank tensors in the context of special relativity
  • Learn about the applications of the Lorentz transformation in physics
  • Explore the mathematical foundations of tensor calculus
  • Investigate the implications of mixed tensors in theoretical physics
USEFUL FOR

Physicists, particularly those specializing in electromagnetism and relativity, as well as students and researchers interested in advanced tensor analysis and its applications in theoretical frameworks.

center o bass
Messages
545
Reaction score
2
The covariant form of the Lorentz force can be written as

m \ddot x^\mu =q F^{\mu \nu} \dot x_\nu

and such a relation should prove by the quotient rule that F is indeed a tensor.
But what kind of tensor is it? One can show that it transforms from an unprimed
to a primed system like

F'^{\mu \nu} = \Lambda^\mu_{\ \alpha} \Lambda^{\ \nu}_{\beta} F^{\alpha \beta} = \frac{\partial x'^\mu}{\partial x^\alpha} \frac{\partial x^\nu}{\partial x'^\beta} F^{\alpha \beta},

where \Lambda^{\ \nu}_{\beta} is the inverse Lorentz transformation matrix. But what kind of tensors transforms like this? Does it have a name? I know about covariant, contravariant and mixed tensors. The closest I get is a mixed tensor but it would transform like

T^\mu_\nu = \frac{\partial x'^\mu}{\partial x^\rho }\frac{\partial x^\sigma}{\partial x'^\nu} T^\rho_\sigma.
 
Last edited:
Physics news on Phys.org
Hi Tim!

But then it should transform like

F'^{\mu\nu} = \frac{\partial x'^\mu}{\partial x^\alpha} \frac{\partial x'^\nu}{\partial x^\beta} F^{\alpha \beta}

and not as

F'^{\mu \nu} = \frac{\partial x'^\mu}{\partial x^\alpha} \frac{\partial x^\nu}{\partial x'^\beta} F^{\alpha \beta}

which arises from the fact that the inverse lorentz transformation gives the equality

\frac{\partial x^\mu}{\partial x'^\nu} = \Lambda_\nu^{\ \ \mu}

and that I have shown that

F'^{\mu \nu} = \Lambda^\mu_{\ \ \alpha} \Lambda^{\ \ \nu}_{\beta} F^{\alpha \beta}

from transforming the lorentz force law in covariant form.
 
Last edited:
Maybe there is a flaw in my proof? It goes as follows: Ignoring the constants m and q

\ddot x'^\mu = F'^{\mu \nu} \dot x'_\nu =\Lambda^\mu_{\ \alpha} (\dot x^\alpha) = \Lambda^\mu_{\ \alpha} (F^{\alpha \beta} \dot x_{\beta}) = \Lambda^\mu_{\ \alpha} F^{\alpha \beta} \Lambda_\beta^{\ \ \nu} \dot x'_\nu,
from which it should follow that

F&#039;^{\mu \nu} = \Lambda^\mu_{\ \alpha} \Lambda_\beta^{\ \ \nu} <br /> F^{\alpha \beta} = \frac{\partial x&#039;^\mu}{\partial x^\alpha} \frac{\partial x^\nu}{\partial x&#039;^\beta} F^{\alpha \beta}.

Here I have used the lorentz transformation and the inverse transformation
x&#039;^\mu = \Lambda^\mu_{\ \ \nu} x^\nu

x&#039;_\mu = \Lambda_\mu^{\ \ \nu} x_\nu

from which it follows that

\frac{\partial x^\mu}{\partial x&#039;^\nu} = \Lambda_\nu^{\ \ \mu}
and

\frac{\partial x&#039;^\mu}{\partial x^\nu} = \Lambda^\nu_{\ \ \mu}.
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

Similar threads

  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
932
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
673
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 59 ·
2
Replies
59
Views
5K