Explicitly, my question is: Suppose we have $f(x-x_p(t))$ in our unprimed frame. We now consider a primed frame moving along the x-axis, where $x' = \gamma(x-\beta ct)$ and $ct' =\gamma(ct-\beta x).$ How do we write $f(x-x_p(t))$ using the primed coordinates? Is this a well-phrased question? Here $x$ refers to the distance on an x-axis.

My question originates in the context of showing that the usual definition of a point charge $q$'s 4-current $J^\mu$ in the rest frame of the particle will net us $\int d^3x J^0=cq$ in every Lorentz frame; that is, $\int d^3x J^0=cq$ is a Lorentz scalar. I am very convinced of this fact. However, in giving a quick proof of this fact for a specific case, I have bumped up against how to find the Lorentz transformation of a 3-delta function when the argument of the delta function depends on coordinate time. I will give the context of my question below:

Suppose we have a particle lying still on the x-axis of an unprimed frame. Then our usual definition of the 4-current is $J^\mu = (cq\delta(x)\delta(y)\delta(x-x_p), 0,0,0)$. Clearly $\int d^3x J^0 = cq$.

On boosting to a frame moving with speed $c\beta'$ on the x-axis, we will have, via Lorentz transformations, $J'^\mu = cq\delta(y)\delta(z)\big(\gamma'\delta(x-x_p), -\gamma' \beta'\delta(x-x_p),0,0\big)$.

Using $x-x_p = \gamma(x'-x_p'(t'))$, $y'=y$, $z'=z$, we have

$J'^\mu = cq\delta(y')\delta(z')\big(\gamma'\delta(\gamma'(x'-x'_p(t'))), -\gamma' \beta'\delta(\gamma'(x'-x'_p(t'))),0,0\big)$.

Using $\delta(ax) = \frac{\delta(x)}{|a|}$, we have

$J'^\mu = cq\delta(y')\delta(z')\big(\delta(x'-x'_p(t')), -\beta'\delta(x'-x'_p(t')),0,0\big)$

We then see immediately that $\int d^3x' J'^0=cq$, which is exactly as I would anticipate.

However, now let's take one more boost in the x' direction to a double primed frame, which will be at velocity $c\beta''$ relative to the primed frame. Here is where my naive transformations in the style $x-x_p = \gamma'(x'-x_p'(t'))$ appear to break down.

We get

$J'^\mu = cq\delta(y')\delta(z')(\gamma''\delta(x'-x'_p(t'))(1+\beta''\beta'), -(\beta'+\beta'')\gamma''\delta(x'-x'_p(t')),0,0)$

Immediately, it is likely we need $\gamma''\delta(x'-x'_p(t')) = \frac{1}{1+\beta''\beta'},$ as this will get us the correct conserved charge and will also ensure the correct relativistic velocity addition in the spatial components.

Thus, naively saying $x'-x'_p(t') = \gamma'' (x''-x''_p(t''))$ doesn't cut the bill.

Thus, in general, when we have a spatial function of coordinate time, how do we write that function in the coordinates of some frame moving relative to our original one? I'm most interested in the case above, where we're considering some function of the spatial function of coordinate time, like $f(x-x_p(t))$.

Please let me know if you want any clarifications.


2 Answers 2


You're right, π‘₯β€²βˆ’π‘₯′𝑝(𝑑′)=𝛾″(π‘₯β€³βˆ’π‘₯″𝑝(𝑑″)) is the problem. It turns out to be right only if the particle is at rest in the primed frame -- that's why everything works for the first transformation.

The general case is that you have $F(x,t)$, a function of the coordinates of the spacetime event $(x,t)$ and you want to write it in terms of $x'$ and $t'$. In theory that's easy since you just use the (inverse) Lorentz transformation: $x=\gamma(x' + \beta t')$ and $t=\gamma(t' + \beta x')$. Just plug these in for $x$ and $t$ to write $F$ in terms of $x'$ and $t'$.

In your case you are looking at the function $F(x,t) \equiv \delta\left(x-x_p(t)\right)$, where $x_p(t)$ is itself a function that gives the x-coord of the particle at time $t$ in the unprimed frame. You want to write this as a function of $x'$ and $t'$ (the Lorentz transformation of $(x,t)$).

After we rewrite it as a function of $x'$ and $t'$ it will still be a delta function and your instinct of using the identity $\delta(f(x))=\frac{1}{|f'(x_{\rm root})|}\delta(x-x_{\rm root})$, (where $x_{\rm root}$ is a root of $f$) is a good one. I.e. we will have

$$ \begin{align} \delta\left(x-x_p(t)\right) &= \delta\left( f(x',t') \right), \end{align} $$ for some function $f(x',t')$. We will first want to find where this function is zero. Then we will find the derivative of this function with respect to $x'$ while holding $t'$ constant. We treat $t'$ as const because later you will want hold $t'$ constant and integrate the current density over all of space.

The root of $f(x',t')$ will occur when $x'=x'_p(t')$, i.e. when $x'$ equals the x-coord of the particle in the primed frame at the time $t'$ (in the primed frame). This is true because the zero of your original delta function occurs at $x=x_p(t)$, i.e. when the event $(x,t)$ is coincident with the particle. And $(x',t')$ is the same event as $(x,t)$. Therefore, $$\delta\left(x-x_p(t)\right) = \frac{1}{|df/dx(x'_{\rm root})|} \delta\left(x'-x'_p(t')\right).$$

To get the derivative of $f$ wrt $x'$ holding $t'$ fixed, just rewrite $x$ and $t$ in terms of $x'$ and $t'$ as above: $$ \begin{align} f(x,t) &= x-x_p(t) \\ &= \gamma(x' + \beta t') - x_p\left( \gamma(t'+\beta x') \right). \end{align} $$ Remember $x_p(t)$ is just some function of $t$ that gives the trajectory of the particle in the unprimed frame.

The deriv wrt $x'$ is $$ \begin{align} \left.\frac{\partial f}{\partial x'}\right|_{t'} &= \gamma - \frac{d x_p}{dt} \gamma\beta, \end{align} $$ where $dx_p/dt$ is the velocity of the particle in the unprimed frame at the time $t=\gamma(t' + \beta x')$.

Therefore the result is $$ \delta\left( x - x_p(t)\right) = \frac{1}{\gamma\left(1-\frac{dx_p}{dt} \beta\right)} \delta\left(x' - x'_p(t')\right). $$

For your first transformation the particle was at rest in the unprimed frame so $dx_p/dt=0$ and the jacobian is just a simple $\gamma$ which is what your transformation was predicting. For your second transformation (from primed to double primed) take the above equation and increase every symbol's superscript by one prime. In the primed frame the particle is moving at a velocity $-\beta'$ so $dx'_p/dt'=-\beta'$ and the jacobian is no longer just a factor of $\gamma$. I'm hoping that when you write it all out this time everything will work out. Let us know :)

  • $\begingroup$ Thank you for the clear answer. The fact that $x_p$ is just a function slipped my mind - rewriting the argument of $x_p$ in terms of the primed coordinates makes sense. It's also nice that you can argue by events coinciding that the root of $f$ in terms of the primed coordinates occurs at $x' = x'_p(t')$ in order to rewrite the delta function - without having to find $x_p(t)$ in terms of $x'_p(t')$. Everything works out well, including in the case where $x_p(t)$ is a more complicated function. $\endgroup$
    – user196574
    Nov 5, 2019 at 3:21

A general way of expressing a four-current due to a single charge is:

$J^\mu\left(x\right) =qc\int d\tau \, u^\mu \delta^{\left(4\right)}\left(x-\bar{x}\left(\tau\right)\right)$

Where $q$ is the charge, $c$ is the speed of light, $\delta^{\left(4\right)}\left(x-\bar{x}\left(\tau\right)\right)=\delta\left(c\left(t-\bar{t}\left(\tau\right)\right)\right)\delta^{\left(3\right)}\left(\mathbf{r}-\bar{\mathbf{r}}\left(\tau\right)\right)$ is the 4d delta function, $\tau$ is the proper time of the particle and $\bar{x}^\mu=\bar{x}^\mu\left(\tau\right)$ is the world line of the particle. This should be your starting point, IMHO.

Also, accrding to my calculations:

$\frac{1}{c^2}\int d^3 r\, \gamma\left(t\right) u_\mu\left(t\right) J^\mu\left(t,\mathbf{r}\right)=q$

where $\gamma$ is the Lorentz factor of the charge. So maybe this is what you were looking for.

Still, your question was about how to deal with $f=f\left(\mathbf{r}-\bar{\mathbf{r}}\left(t\right)\right)$ (I hope you don't mind me changing notation slightly). Now you want to consider some boosted frame with coordinates $\{ct',\,\mathbf{r}'\}$. Clearly you will have a procedure for changin the coordinates from $S'$ to $S$: $t=t\left(t',\,\mathbf{r}'\right)$ and $\mathbf{r}=\mathbf{r}\left(t',\mathbf{r}'\right)$. Essentially, the inverse of what you wrote. Then your function in $S'$ is simply:


However you have to be careful. Just because the function is not a vector it does not mean it is a true scalar. For example $\delta^{\left(3\right)}$, i.e. the 3d delta-function, is actually a density, hence when you change coordinate frame you will get Jacobians coming out (hello length contraction). Also, you have to be careful with integrals $\int d^3 r\dots$ is an integral over 3d space that is perpendicular to temporal axis in the current refrence frame. In a different reference frame this 3d space will no longer be perpendicular to temporal axis, it could be partially spatial and partially temporal.

Following from the comments

$\frac{1}{c^2}\int d^3 r\, \gamma\left(t\right) u_\mu\left(t\right) J^\mu\left(t,\mathbf{r}\right)=\frac{q}{c^2}\int d^3 r\, \gamma\left(t\right) u_\mu\left(t\right) \: \int cd\tau \: u^\mu\left(\bar{t}\left(\tau\right)\right) \delta^{\left(4\right)}\left(x-\bar{x}\left(\tau\right)\right))$

Expand the 4D delta function for the current observer. The world-line of the particle in the current frame is $\bar{t}=\bar{t}\left(\tau\right)$ and $\bar{\mathbf{r}}=\bar{\mathbf{r}}\left(\tau\right)$:

$\frac{1}{c^2}\int d^3 r\, \gamma\left(t\right) u_\mu\left(t\right) J^\mu\left(t,\mathbf{r}\right)=\frac{q}{c^2}\gamma\left(t\right) u_\mu\left(t\right) \int cd\tau \: \delta\left(c\left(t-\bar{t}\left(\tau\right)\right)\right) u^\mu\left(\bar{t}\left(\tau\right)\right) \int d^3 r\, \delta^{\left(3\right)} \left(\mathbf{r}-\bar{\mathbf{r}}\left(\tau\right)\right) $

The last integral over the 3d delta function trivially vanishes so

$\frac{1}{c^2}\int d^3 r\, \gamma\left(t\right) u_\mu\left(t\right) J^\mu\left(t,\mathbf{r}\right)=\frac{q}{c^2}\gamma\left(t\right) u_\mu\left(t\right) \int cd\tau \: \delta\left(c\left(t-\bar{t}\left(\tau\right)\right)\right) u^\mu\left(\bar{t}\left(\tau\right)\right)$

Next we change the integration variable: $\frac{d\bar{t}}{d\tau}=\gamma\left(\bar{t}\right)$. Clearly there is slight abuse of notation here, but since there is a bijective relationship between $\bar{t}$ and $\tau$, it is ok. So:

$\frac{1}{c^2}\int d^3 r\, \gamma\left(t\right) u_\mu\left(t\right) J^\mu\left(t,\mathbf{r}\right)=\frac{q}{c^2}\gamma\left(t\right) u_\mu\left(t\right) \int \frac{cd\bar{t}}{\gamma\left(\bar{t}\right)} \: \delta\left(c\left(t-\bar{t}\right)\right) u^\mu\left(\bar{t}\right)=\frac{q}{c^2}\gamma\left(t\right) u_\mu\left(t\right) \frac{1}{\gamma\left(t\right)} u^\mu\left(t\right)=\frac{q}{c^2}u_\mu\left(t\right) u^\mu\left(t\right)=q$

Since $u_\mu u^\mu = c^2$ at all times

I think a comprehensive answer can be reached using the language of differential forms.

One of the Maxwell's laws in differential forms is:

$d\star F=\mu_0 J$, where $J$ is the current density 3-form (of a charged particle in some motion) and $F=dA$ is the electromagnetic 2-form ($A$ is the vector potential 1-form).

Consider the integral of the left-hand-side over $\Omega$, the 3d sub-space of the space-time that is perpendicular the the four-velocity of the particle at some time in its history. At that time, therefore $\Omega$ is aligned with the rest-frame of the particle ($\tilde{S}$).

Then, using Stokes theorem:

$\int_\Omega d\star F = \oint_{\partial\Omega} \star F$

Now $\star F = -\frac{\tilde{\epsilon}_{ijk}}{2c} \tilde{E}^k d\tilde{x}^i\wedge d\tilde{x}^j + c g_{kk'}\tilde{B}^{k'}d\tilde{t}\wedge d\tilde{x}^k$

where $\tilde{E}$ and $\tilde{B}$ are the electric and magnetic fields in the rest-frame (for point-charge $\tilde{B}=0$) and latin indices run over spatial parts only. It follows that:

$ -\frac{1}{c} \int_{\Omega} J= \frac{1}{\mu_0 c^2} \oint_{\partial\Omega} \tilde{\epsilon}_{ijk}\tilde{E}^k d\tilde{x}^i\wedge d\tilde{x}^j= \epsilon_0 \oint_{\partial\Omega} \mathbf{\hat{n}}.\tilde{\mathbf{E}}\,d^2x = q$

Where I have used the integral form of the Gauss' law, and $\mathbf{\hat{n}}$ is the outward pointing normal vector.

Now, come back to the current 3-form. Explicitly this is given by:

$J=\epsilon_{\alpha\beta\sigma\mu} J^\mu dx^\alpha \wedge dx^\beta \wedge dx^\sigma$

Where $J^\mu$ is the usual current density. The integral over $\Omega$ may be now cosidered in terms of the current density:

$\int_\Omega J = \int_{\mathbb{R}^3} d^3 \breve{x} \, \frac{\partial x^\alpha}{\partial \breve{x}} \frac{\partial x^\beta}{\partial \breve{y}} \frac{\partial x^\sigma}{\partial \breve{z}} J_{\alpha\beta\sigma}\left(x\left(\breve{x}\right)\right) = -\frac{1}{c}\int_{\mathbb{R}^3} d^3 \breve{x} \, u_\mu J^\mu$

Where in the second step I used a pull-back to Eucledian 3d space (addressed by coordinated $\breve{x}, \breve{y}, \breve{z}$) in order to evaluate the differential form integral. The push-forward is

$ \begin{align} ct &= \gamma \frac{v}{c}\breve{x} + \gamma c\breve{t}_0 \\ x &= \gamma \breve{x} + \gamma v \breve{t}_0\\ y &= \breve{y}\\ z &= \breve{z} \end{align} $

Where $\breve{t_0}$ is constant. Basically $\breve{S} \cong \mathbb{R}^3$ is like the rest-frame of the charge at proper time $\tau = \breve{t}_0$

So (with some more work):

$q = \frac{1}{c^2} \int_{\mathbb{R}^3} d^3 \breve{x} \, u_\mu J^\mu$

It would seem therefore that the important part is the time at which you evaluate your integral. If you are evaluating at fixed lab time, which is like the algebraic derivation, I gave originally, then you need the Lorentz factor. If, on the other hand, you are evaluating at fixed proper time, like the differential forms approach above, then the Lorentz factor is not needed.

  • $\begingroup$ Thank you for the answer. Could you discuss how you calculated $\int d^3r \gamma(t) u_\mu(t) J^\mu (t, \vec{r})$ further? $\endgroup$
    – user196574
    Nov 5, 2019 at 3:44
  • $\begingroup$ Sure, see above $\endgroup$
    – Cryo
    Nov 5, 2019 at 9:33
  • $\begingroup$ That's interesting to me because I would have expected that $J^\mu = \rho_0 u^\mu$, with the $\rho_0$ the rest density of charge, as that's true in the rest frame of the density of interest. Then I would expect $J^\mu u_\mu = \rho_0 c^2$, so that $\frac{1}{c^2}\int d^3r J^\mu u_\mu = q$ if the density in the rest frame were to integrate to $q$, without any factors of $\gamma$. $\endgroup$
    – user196574
    Nov 5, 2019 at 18:27
  • $\begingroup$ @user196574. That does not work since $u^\mu$ is position independent, whereas current density of a point charge must be zero nearly everywhere, and can only be nonzero on the world-line of the charge. Anyway, I've added a proof that does not rely on explicit form of the charge density $\endgroup$
    – Cryo
    Nov 6, 2019 at 3:23

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