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My problem is really simple. I was reading the Strominger lectures where he defines the future charges $Q_\varepsilon^+$, and he does something that I don't understand. He says on the equation $(2.5.4)$ that without massive charges the following affirmation is true:

$$ I=\int_{\mathcal I_+^+} \varepsilon*F=0 $$

Where $F$ is the Electromagnetic Tensor, $\,\mathcal I^+$ is the Future Null Infinity, $\,\mathcal I^+_-$ is its past and $\,\mathcal I^+_+$ is its future. And the only condition for $\varepsilon$ is that it obeys $\varepsilon\big|_{\mathcal I^+_-}=\varepsilon\big|_{\mathcal I^-_+}$

The reason for this affirmation according to the book is that the electric field will vanish at $\mathcal I^+_+$. I don't know why these affirmations are true.


My attempts:

Attempt 1: If there is no charges, according to Maxwell equations:

$$ d*F=0\implies *F=d\omega $$

Where $\omega$ is a $1$-form. Then the integral is:

$$ I=\int_{\mathcal I_+^+} \varepsilon \,d\omega $$

Using $d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^p\alpha\wedge d\beta$ where $\alpha$ is a $p$-form:

$$ I=\int_{\mathcal I_+^+} \Big[ d(\varepsilon\omega) +\omega\wedge d\varepsilon \Big] $$

And I suppose I could do something with the generalized Stokes theorem or something like that. But I really have no clue.

Attempt 2: I kind of have an intuition, maybe it is wrong, $\mathcal I^+_+$ is like a $2$-manifold (on spacetime) of size going to zero, so all the integrals of the form $\int_{I^+_+}$ will be zero.

Any help or advice would be awesome.


References:

Lectures on the Infrared Structure of Gravity and Gauge Theory - Andrew Strominger

Lectures on Youtube

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Not 100% sure but here is my guess. Somehow the fact that the particle is massless must come into play. From equation 2.3.2

$$ F_{rt}\rvert_{\mathcal{I}^+} = \frac{e^2}{4 \pi r^2} \sum_{k=1}^n \frac{Q_k}{\gamma_k^2 (1 - \hat x \cdot \vec{\beta}_k )^2}. $$ When computing $\star F$ over an $S^2$ integral at $\mathcal{I}^+_+$, only the $\phi$ and $\theta$ components of $\star F$ come into play. This is the same as the $r$ and $t$ components of $F$. Note that $F_{rt}$ above goes as $1/r^2$ while our spherical integral will go as $r^2$, so a priori we will have a $\mathcal{O}(1)$ left over. So something about this expression must go to zero if the particles are massless. I think it is $\gamma_k$ of the particles. If $|\beta_k| = 1$ then $\gamma_k \to \infty$ and $F_{rt} \to 0$. (However, I think you also have to check that $(1 - \hat x \cdot \vec{\beta}_k)^2$ won't somehow mess this up. Also, would this mean that the expression is zero on the whole $\mathcal{I}^+$, save for maybe a discrete set of $\hat x$?)

Also, if you continually struggle to make sense of the book, I think some of the original papers go into more detail and may be a bit clearer on some technical points. A quick arxiv search reveals the following promising looking results.

https://arxiv.org/abs/1506.02906

https://arxiv.org/abs/1412.2763

https://arxiv.org/abs/1407.3814

https://arxiv.org/abs/1407.3789

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    $\begingroup$ You were right. The integral was (if I made the integral right): $\int_{\mathcal I_+^+}\varepsilon*F=-\sum_{k=1}^n\frac{1}{\gamma^2_k}\cdot\frac{Q_ke^2}{4\pi}\int \varepsilon\frac{\sin\theta \,d\theta \,d\phi}{(1-\vec\beta\cdot\hat r)^2}$ and when there are only massless charges $\gamma_k\to\infty$ and therefore $\int_{\mathcal I_+^+}\varepsilon*F=0$. I had to learn things that apparently I didn't know. Today I'm less noob than yesterday, so thank you very much for the advice. $\endgroup$ Jun 17, 2021 at 20:10
  • $\begingroup$ Typo: I had a wrong sign on $\int_{\mathcal I^+_+} \varepsilon * F$, I mean it must be: $\int_{\mathcal I^+_+} \varepsilon * F=\sum^n_{k=1}\frac{1}{\gamma_k^2}\cdot\frac{Q_k e^2}{4\pi}\int \varepsilon \frac{\sin\theta\,d\theta\,d\phi}{(1-\vec\beta\cdot\hat r)^2}$ $\endgroup$ Jul 6, 2021 at 23:20

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