Charge invariance with Heaviside's function

[Frostman]

Homework Statement

Prove the invariance of the charge in any inertial frame of reference

Relevant Equations

Heaviside's function
Four-current

I followed a demonstration in one of my electromagnetism books, but it is not clear to me.
My problem is at the starting point.
The book begins by considering the office defined in the following way:

$$Q=\int d^4xJ^\alpha(x)\partial_\alpha\theta(\eta_\beta x^\beta)$$

where $\eta_\beta=(1,0,0,0)$

How do I relate to the definition I know?
$$Q=\int d^3x J^0(x)$$

I know that the derivative of the Heaviside function is the Dirac delta, but evaluated where? Are the following accounts I have done correct?
$$Q=\int d^4xJ^\alpha(x)\delta(t)$$

 

[TSny]

Your result $Q=\int d^4xJ^\alpha(x)\delta(t)$ can't be correct. In the original expression for $Q$, $\alpha$ is a dummy summation index. So $\alpha$ cannot appear as a lone index in the answer.

It might help to write out the given expression for $Q$ as $$Q=\int d^4x \left[J^0(x)\partial_0 +J^1(x)\partial_1 + J^2(x)\partial_2 + J^3(x)\partial_3 \right]\theta(\eta_\beta x^\beta)$$
What does $\eta_\beta x^\beta$ simplify to?

 

[Frostman]

Your result $Q=\int d^4xJ^\alpha(x)\delta(t)$ can't be correct. In the original expression for $Q$, $\alpha$ is a dummy summation index. So $\alpha$ cannot appear as a lone index in the answer.

It might help to write out the given expression for $Q$ as $$Q=\int d^4x \left[J^0(x)\partial_0 +J^1(x)\partial_1 + J^2(x)\partial_2 + J^3(x)\partial_3 \right]\theta(\eta_\beta x^\beta)$$
What does $\eta_\beta x^\beta$ simplify to?

I hope I understand the definition of the Heaviside function well

For $\beta=0$, $\theta(\eta_\beta x^\beta)=1$
For $\beta\neq0$, $\theta(\eta_\beta x^\beta)=0$

Right?

 

[TSny]

For $\beta=0$, $\theta(\eta_\beta x^\beta)=1$
For $\beta\neq0$, $\theta(\eta_\beta x^\beta)=0$

It doesn't make sense to say "For $\beta = 0$, $\theta(\eta_\beta x^\beta)=1$".

In the expression $\eta_\beta x^\beta$, the Einstein summation convention implies that $\beta$ is a summation index. So, $$\eta_\beta x^\beta = \eta_0x^0 + \eta_1x^1 + \eta_2x^2 +\eta_3x^3$$
What does this reduce to using the given values of $\eta_0, \eta_1, \eta_2$ and $\eta_3$?

Therefore, what does $\theta(\eta_\beta x^\beta)$ reduce to?

 

[Frostman]

It doesn't make sense to say "For $\beta = 0$, $\theta(\eta_\beta x^\beta)=1$".

In the expression $\eta_\beta x^\beta$, the Einstein summation convention implies that $\beta$ is a summation index. So, $$\eta_\beta x^\beta = \eta_0x^0 + \eta_1x^1 + \eta_2x^2 +\eta_3x^3$$
What does this reduce to using the given values of $\eta_0, \eta_1, \eta_2$ and $\eta_3$?

Therefore, what does $\theta(\eta_\beta x^\beta)$ reduce to?

The only term that remains is $x^0$ because $\eta_\beta$ is defined as $\eta_\beta=(1,0,0,0)$
So $\theta(\eta_\beta x^\beta)=\theta(\eta_0x^0)=\theta(x^0)$

 

[TSny]

The only term that remains is $x^0$ because $\eta_\beta$ is defined as $\eta_\beta=(1,0,0,0)$
So $\theta(\eta_\beta x^\beta)=\theta(\eta_0x^0)=\theta(x^0)$

Yes. Does this help to simplify the expression $$Q=\int d^4x \left[J^0(x)\partial_0 +J^1(x)\partial_1 + J^2(x)\partial_2 + J^3(x)\partial_3 \right]\theta(\eta_\beta x^\beta)$$

 

[Frostman]

Yes. Does this help to simplify the expression $$Q=\int d^4x \left[J^0(x)\partial_0 +J^1(x)\partial_1 + J^2(x)\partial_2 + J^3(x)\partial_3 \right]\theta(\eta_\beta x^\beta)$$

The only derivative $\partial_\alpha$ that can act is only this one $\partial_0$. So it remains
$$Q=\int d^4x J^0(x)\partial_0\theta(x^0)$$

 

[TSny]

The only derivative $\partial_\alpha$ that can act is only this one $\partial_0$. So it remains
$$Q=\int d^4x J^0(x)\partial_0\theta(x^0)$$

Yes

 

[Frostman]

Can I proceed like this to reduce it to the 3D relationship?
$$Q=\int dx^0d^3x J^0(x)\partial_0\theta(x^0)=\int d^3x J^0(x)dx^0\partial_0\theta(x^0)=\int d^3x J^0(x)$$

 

[TSny]

Can I proceed like this to reduce it to the 3D relationship?
$$Q=\int dx^0d^3x J^0(x)\partial_\theta(x^0)=\int d^3x J^0(x)dx^0\partial_\theta(x^0)=\int d^3x J^0(x)$$

This looks a bit sketchy to me. Note that $\partial_\theta(x^0)$ should be $\partial_0\theta(x^0)$.

I might start by writing $J(x)$ as $J(x^0, \vec x)$. So $$Q=\int dx^0d^3x J^0(x)\partial_0\theta(x^0) = \int dx^0d^3x J^0(x^0, \vec x) \partial_0\theta(x^0) = \int d^3x \int dx^0 J^0(x^0, \vec x) \partial_0\theta(x^0)$$ Then proceed to rewrite $\partial_0\theta(x^0)$ in terms of a Dirac delta function.

 

[Frostman]

This looks a bit sketchy to me. Note that $\partial_\theta(x^0)$ should be $\partial_0\theta(x^0)$.

I might start by writing $J(x)$ as $J(x^0, \vec x)$. So $$Q=\int dx^0d^3x J^0(x)\partial_0\theta(x^0) = \int dx^0d^3x J^0(x^0, \vec x) \partial_0\theta(x^0) = \int d^3x \int dx^0 J^0(x^0, \vec x) \partial_0\theta(x^0)$$ Then proceed to rewrite $\partial_0\theta(x^0)$ in terms of a Dirac delta function.

Yes, I corrected, I forgot a zero. Excuse me, but this is my first time working with the Heaviside and Dirac delta function together.

$$\partial_0\theta(x^0)=\delta(x^0)$$

Is it correct?

$$\int d^3x \int dx^0J^0(x^0, \vec x) \delta(x^0)$$

Or can I suppose that $\delta(x^0) = \delta(x^0-0)$ so I can rewrite as

$$\int d^3x \int dx^0J^0(x^0, \vec x) \delta(x^0-0)=\int d^3x J^0(0, \vec x)$$

 

[TSny]

Yes, that all looks good.

 

[Frostman]

Perfect, one last step and I believe I am there. Given this definition of charge Q we have that

The effect of a Lorentz transformation on Q is then evidently simply to change $\eta$:

$$Q'=\int d^4x J^\alpha(x)\partial_\alpha \theta(\eta_\beta'x^\beta)$$
For $J_\alpha\partial_\alpha$ I understood because the product of a covariant and contravariant four-vector is a relativistic invariant, but for the argument of the Heaviside function I should not also have a $x'^\beta$?

 

[TSny]

A Lorentz transformation will change at least some of the components of each of the 4-vectors $J^\mu$, $\partial_\mu$, $\eta_\mu$, and $x^\mu$. So it's not just the components $\eta_\beta$ that will change. So, $Q'$ needs to be written as $$Q'=\int d^4x' J'^\alpha(x)\partial'{_\alpha} \theta(\eta'_\beta x'^\beta)$$ [The above expression is wrong. See post #16 below for corrections.]

But for any two 4-vectors, $w$ and $v$, $w^\alpha v_{\alpha}$ is invariant under a Lorentz transformation. That is, $w^{\alpha}v_{\alpha} = w'^{\alpha}v'_{\alpha}$. Note that one of the $\alpha$'s must be "up" and the other $\alpha$ "down".

So, $J^{\alpha} \partial_\alpha = J'^{\alpha} \partial'_\alpha$. Similarly, $\eta_\beta x^\beta = \eta'_\beta x'^\beta$.

What can you say about $d^4x$? How is it related to $d^4x'$?

 

[Frostman]

A Lorentz transformation will change at least some of the components of each of the 4-vectors $J^\mu$, $\partial_\mu$, $\eta_\mu$, and $x^\mu$. So it's not just the components $\eta_\beta$ that will change. So, $Q'$ needs to be written as $$Q'=\int d^4x' J'^\alpha(x)\partial'{_\alpha} \theta(\eta'_\beta x'^\beta)$$

But for any two 4-vectors, $w$ and $v$, $w^\alpha v_{\alpha}$ is invariant under a Lorentz transformation. That is, $w^{\alpha}v_{\alpha} = w'^{\alpha}v'_{\alpha}$. Note that one of the $\alpha$'s must be "up" and the other $\alpha$ "down".

So, $J^{\alpha} \partial_\alpha = J'^{\alpha} \partial'_\alpha$. Similarly, $\eta_\beta x^\beta = \eta'_\beta x'^\beta$.

What can you say about $d^4x$? How is it related to $d^4x'$?

$$d^4x'=\bigg|\frac{\partial x'}{\partial x}\bigg|d^4x$$
Where $\bigg|\frac{\partial x'}{\partial x}\bigg|$ is the determinant of jacobian of Lorentz-transformation, for the Lorentz proper transformation $det\Lambda^\mu{}_\nu=1$.

The demonstration I am following is that of Weinberg chapter 7 page 40-41. And it uses what I quoted...

 

[TSny]

Thanks for the reference to Weinberg's text. After looking at pages 40-41, I realize that I made a mistake when I wrote

$$Q'=\int d^4x' J'^\alpha(x)\partial'{_\alpha} \theta(\eta'_\beta x'^\beta)$$

The ${\eta'}_\beta$ here should be the unprimed $\eta_\beta$, where $\eta_\beta \equiv (1, 0, 0, 0)$. Thus, we should write $$Q'=\int d^4x' J'^\alpha(x')\partial'{_\alpha} \theta(\eta_\beta \, x'^\beta)$$ Only with the unprimed $\eta$ does the integral reduce to the required $\int d^3x' {J'}^0(0, \vec x') \,\,$ (using the same manipulations as you did for $Q$).

Using the invariance of $d^4x$ and $J^\alpha \partial_\alpha$, you can write $$Q'=\int d^4x J^\alpha(x)\partial{_\alpha} \theta(\eta_\gamma x'^\gamma)$$ For convenience I replaced the dummy index $\beta$ by $\gamma$.

Since $x^\mu$ is a 4-vector, we can replace ${x'}^\gamma$ by $\Lambda^{\gamma}\!\:_\beta x^\beta$. So, $\eta_\gamma{x'}^\gamma = \eta_\gamma \Lambda^{\gamma}\!\:_\beta x^\beta = \left(\Lambda^{\gamma}\!\:_\beta \eta_\gamma \right) x^\beta = \eta'_\beta x^\beta$, where in the last step we follow Weinberg and define $\eta'_\beta \equiv \Lambda^{\gamma}\!\:_\beta \, \eta_\gamma$.

I think this is how you can get to Weinberg's expression for $Q'$ near the top of page 41. (Maybe there's an easier way.)

 

[Frostman]

In the first step you indicated all the primate terms, except $\eta_\beta$ because I must be able to reduce to the definition of charge as a three-dimensional integral of $J'^0$. I then observe that in reality the other primate quantities are Lorentz invariants. When I exchange the transformation between $x'^\beta$ and $\eta_\beta$ I don't have the problem again that I have to arrive at the definition of the charge as a three-dimensional integral of $J^0$?
I am not very convinced that the moment I introduce a Lorentz transformation $\eta_\beta$ is the only unprimed term ...

 

[TSny]

In the first step you indicated all the primate terms, except $\eta_\beta$ because I must be able to reduce to the definition of charge as a three-dimensional integral of $J'^0$.

Yes

I then observe that in reality the other primate quantities are Lorentz invariants. When I exchange the transformation between $x'^\beta$ and $\eta_\beta$ I don't have the problem again that I have to arrive at the definition of the charge as a three-dimensional integral of $J^0$?
I am not very convinced that the moment I introduce a Lorentz transformation $\eta_\beta$ is the only unprimed term ...

Weinberg's calculation is tricky in my opinion. I think it's important to realize that Weinberg does not treat $\eta_\beta$ as a 4-vector. The quantities $(\eta_0, \eta_1, \eta_2, \eta_3)$ are just 4 numerical quantities defined to have the values $(\eta_0, \eta_1, \eta_2, \eta_3) = (1, 0, 0, 0)$. These quantities are used in both the unprimed and primed frames to set up integrals representing the total charge in the two frames. $$Q=\int d^4x J^\alpha(x)\partial{_\alpha} \theta(\eta_\beta \, x^\beta)$$ $$Q'=\int d^4x' J'^\alpha(x')\partial'{_\alpha} \theta(\eta_\beta \, x'^\beta)$$
As outlined in post #16, the integral for $Q'$ can be manipulated into Weinberg's form $$Q'=\int d^4x J^\alpha(x)\partial{_\alpha} \theta(\eta'_\beta \, x^\beta)$$ where the primed quantities $\eta'_\beta$ are defined as $$\eta'_\beta \equiv \Lambda^\gamma{}_\beta \, \eta_\gamma \;\;\;\; \rm {Definition}$$ This equation for $\eta'_\beta$ does not represent a Lorentz transformation from the unprimed to the primed coordinates. The correct way to transform a covariant vector $U_\beta$ is $$U'_\beta = \Lambda _\beta{} ^\gamma \, U_\gamma \;\;\;\; \rm {Lorentz \; transformation \;\;[see \; Weinberg's \; equation \; (2.5.3)]}$$ Note how the $\Lambda$ matrix in the definition of $\eta'_\beta$ differs from the $\Lambda$ matrix in the Lorentz transformation.