# Why $\int_{\mathcal I_+^+} \varepsilon*F=0$ for any $\varepsilon$ when there are no massive charges?

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

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

• 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. Jun 17, 2021 at 20:10
• Typo: I had a wrong sign on $\int_{\mathcal I^+_+} \varepsilon * F$, I mean it must be: $\int_{\mathcal I^+_+} \varepsilon * F=\sum^n_{k=1}\frac{1}{\gamma_k^2}\cdot\frac{Q_k e^2}{4\pi}\int \varepsilon \frac{\sin\theta\,d\theta\,d\phi}{(1-\vec\beta\cdot\hat r)^2}$ Jul 6, 2021 at 23:20