2
$\begingroup$

Context

I am studying electromagnetic quantities through the framework of special relativity. I find it notation heavy and the issue of tensors is something I am still not fully grasping. The continuity equation is one of the first electromagnetic quantities in Zangwill's book [2], which I am studying from. I am not comprehending the matter, so I am asking for help.

Given the transformation $$ \begin{align} ct^\prime &= \gamma ( ct - \beta x_3 ) & ct &= \gamma ( ct' + \beta x_3^\prime) \\\\ {x^1}' &= x^1 & x^1 &= {x^1}' \\\\ {x^2}' &= x^2 & x^2 &= {x^2}' \\\\ {x^3}' &= \gamma (x^3 - \beta c t) & {x^3} &= \gamma ({x^3}' + \beta c t) \end{align} $$ I obtain that $$ \frac{\partial }{\partial \mathbf{r}'}=\begin{bmatrix} \gamma \left( \frac{\partial }{\partial ct} + \beta \frac{\partial }{\partial x^3} \right) & \frac{\partial }{\partial x^1} & \frac{\partial }{\partial x^2} & \gamma \left( \beta \frac{\partial }{\partial x^0} + \frac{\partial }{\partial x^3} \right) \end{bmatrix}. $$

Since four quantities $\frac{\partial}{\partial {ct} }, \frac{\partial}{\partial {x^1} }, \frac{\partial}{\partial {x^2} }, \frac{\partial}{\partial {x^3} }$ in a coordinate system $(ct, x^1, x^2, x^3)$ are related to four quantities $\frac{\partial}{\partial {ct'} }, \frac{\partial}{\partial {x'}^1 }, \frac{\partial}{\partial {x'}^2 }, \frac{\partial}{\partial {x'}^3 }$ in another coordinate system $(ct^\prime, {x^\prime}^1, {x^\prime}^2, {x^\prime}^3)$ by the transformation equations $${a'}_i=\frac{\partial{x}^j}{\partial{x'}^i}{a}_j$$ then they are components of a covariant four-vector. Therefore, I write the four-gradient covariant components in four-vector notation as $$ \partial_\mu=\left(\frac{\partial}{\partial{ct}},\boldsymbol{\nabla}\right). $$ This precisely conforms with what is found in [1]. Great.

Meanwhile, as in [2], it may be shown that the four quantities $c\rho,j^1,j^2,j^3$ in a coordinate system $(ct, x^1, x^2, x^3)$ are related to four quantities $c\rho',{j'}^1,{j'}^2,{j'}^3$ in another coordinate system $(ct^\prime, {x^\prime}^1, {x^\prime}^2, {x^\prime}^3)$ by the transformation equations $${j'}^i=\frac{\partial{x'}^i}{\partial{x}^k}{j}^k.$$ Therefore, I write the four-current contravariant components in four-vector notation as $$ j^\nu=\left(c\rho,\mathbf{j}\right).$$ Great, no problem here.

Now, I inquire whether the continuity equation in differential form, $$\frac{\partial \rho}{\partial t} + \boldsymbol{\nabla}\cdot\mathbf{j}=0,$$
can be written as the four-divergence. This inquiry is done in two attempts. The first attempt leads to an incorrect result. Reasoning is given to explain the incorrect result. The second attempt leads to the correct result.

Attempt 1(erroneous attempt)

If I assume---incorrectly---that I form the dot product $$ \vec{\nabla} \cdot \vec{j}, $$ and next apply the standard formulation (cf, [3,4]) using, for example, the $(+,-,-,-,)$ metric signature, then I obtain $$ \vec{\nabla} \cdot \vec{j} = \frac{\partial\rho}{\partial{ t}}- \boldsymbol{\nabla} \cdot\mathbf{j} . \tag{0} $$ Finally, if I were to set (0) equal to zero, then i would obtain---incorrectly---that $$ \frac{\partial\rho}{\partial{ t}}- \boldsymbol{\nabla} \cdot\mathbf{j} = 0 . \tag{1} $$ As pointed out by @Filippo, this is incorrect. The reason it is incorrect is because ``The Minkowski metric, the bilinear form, and the Minkowski inner product are all the same object; it is a bilinear function that accepts two (contravariant) vectors and returns a real number. [5]'' This quote is relevant to my question herein. Note that I do not have two contravariant vectors. Rather, one four-vector is contravariant, while the other in covariant.

Attempt 2 (correct)

If I assume that $$ \partial_\mu j^\mu = 0, $$ and next use Einstein summation (cf, examples in [6]), then I correctly obtain that $$ \partial_\mu j^\mu= \frac{\partial c\rho}{\partial{ct}} + \boldsymbol{\nabla} \cdot\mathbf{j} = 0 . $$

Question

The initial question, written after Attempt 1, but before Attempt 2 was,

``What am I misunderstanding and how can I correct it?''

The answers to the questions led to the correct solution which is given in Attempt 2.

Bibliography

[1] https://en.wikipedia.org/wiki/Four-gradient

[2] Zangwill, Modern Electrodynamics, pp 840-1.

[3] https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Essential_Graduate_Physics_-_Classical_Electrodynamics_(Likharev)/09%3A_Special_Relativity/9.05%3A_The_Maxwell_Equations_in_the_4-form

[4] https://en.wikipedia.org/wiki/Four-gradient#As_a_4-divergence_and_source_of_conservation_laws

[5] https://en.wikipedia.org/wiki/Minkowski_space#Minkowski_metric

[6] https://en.wikipedia.org/wiki/Tensor

$\endgroup$
0

2 Answers 2

4
$\begingroup$

What am I misunderstanding and how can I correct it?

Equation $(1)$ is wrong. With your definition of $j$ we clearly obtain $$\partial_\mu j^\mu= \frac{\partial\rho}{\partial{ t}}+ \boldsymbol{\nabla} \cdot\mathbf{j} $$ and hence the two continuity equations are equivalent.

$\endgroup$
2
  • $\begingroup$ What about the metric signature $(+,-,-,-)$? Do or don't we need to apply it here? Please advise how come $\endgroup$ Commented Nov 24, 2023 at 3:00
  • 1
    $\begingroup$ @MichaelLevy We do not need to apply it here. You have to apply it when the position of the indices is not how you would expect, e.g. if you are given a vector $X^\mu$, then you need to apply the metric to make sense of $X_\mu$ and conversely if you are given a 1-form $\alpha_\mu$, then you need the metric to make sense of $\alpha^\mu$. $\endgroup$
    – Filippo
    Commented Nov 24, 2023 at 3:07
1
$\begingroup$

You should use the Minkowski metric to give the sign you need. Zangwill does it a different way, using ict. The problem is that no one else does it that way, so you can't compare with other references.

$\endgroup$
1
  • $\begingroup$ I like Zangwill's book very much. However, his approach on special relativity is a bit obtuse $\endgroup$ Commented Nov 24, 2023 at 3:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.