<p>I will answer with what I now believe to be the correct answer. We see some gamma-ray burst emission hit our satellite which carries an amount of energy <span class="math-container">$E_{obs}$</span>. If we say the burst was emitted equally in all directions (isotropic) then we derive a quantity <span class="math-container">$E_{iso}$</span>. If we say that the burst was actually emitted in a tight beam then we derive <span class="math-container">$E_\theta$</span> where <span class="math-container">$\theta$</span> is the opening angle of the burst and we will find <span class="math-container">$E_{iso} &gt; E_{\theta}$</span>. <span class="math-container">$E_{iso}$</span> is used to compare burst energies in the literature without needing to find the opening angle.</p>

CC BY-SA 4.0
2020-09-01T11:23:38.610Z
2020-09-01T11:23:38.610Z
GLD
<p>In the literature on gamma ray bursts I see several references to <span class="math-container">$E_\gamma$</span> and <span class="math-container">$E_{\gamma,iso}$</span> for example as in <a href="https://arxiv.org/abs/astro-ph/0405503" rel="nofollow noreferrer">Piran (2005)</a></p>
<blockquote>
<p><span class="math-container">$E_{k,iso,52}$</span> is “isotropic equivalent” kinetic energy,
discussed below, in units of 10<sup>52</sup> ergs, while <span class="math-container">$E_{k,θ,52}$</span> is the real kinetic energy in the jet i.e:
<span class="math-container">$E_{k,θ,52} = (θ^2/2)$$E_{k,iso,52}$</span>. One has to be careful which of the two energies one discusses. In
the following I will usually consider, unless specifically mentioned differently, <span class="math-container">$E_{k,iso,52}$</span>, which
is also related to the energy per unit solid angle as: <span class="math-container">$E_{k,iso,52}/4π$</span></p>
</blockquote>
<p>My interpretation is that when an observation of a GRB is made, an energy <span class="math-container">$E$</span> is recorded. <span class="math-container">$E_{\gamma,iso}$</span> is the total energy that would have been emitted by the explosion if the energy was released equally in all directions and by considering our distance away from it. However, because we know that GRBs are tightly beamed, <span class="math-container">$E_\gamma$</span> is the corrected energy by taking <span class="math-container">$E_{\gamma,iso}$</span> and considering only a small fraction of it in a tight beam. However, this all seems a tad confusing and I can't see the point in talking about <span class="math-container">$E_{\gamma,iso}$</span> when we think they are very tightly beamed.</p>
<p>I would appreciate some clarification on what exactly <span class="math-container">$E_\gamma$</span> and <span class="math-container">$E_{\gamma, iso}$</span> are and where they come from.</p>
<p>Thanks</p>

<gamma-ray-bursts>
What is the difference between $E_\gamma$ and $E_{\gamma,iso}$ in gamma ray bursts?
GLD
# What is the difference between $E_\gamma$ and $E_{\gamma,iso}$ in gamma ray bursts?

(Asked by: username_0 on 2020-06-25)

In the literature on gamma ray bursts I see several references to $E\_\gamma$ and $E\_{\gamma,iso}$ for example as in [Piran (2005)](https://arxiv.org/abs/astro-ph/0405503)

> $E\_{k,iso,52}$ is “isotropic equivalent” kinetic energy,
> discussed below, in units of 1052 ergs, while $E\_{k,θ,52}$ is the real kinetic energy in the jet i.e:
> $E\_{k,θ,52} = (θ^2/2)$$E\_{k,iso,52}$. One has to be careful which of the two energies one discusses. In
> the following I will usually consider, unless specifically mentioned differently, $E\_{k,iso,52}$, which
> is also related to the energy per unit solid angle as: $E\_{k,iso,52}/4π$

My interpretation is that when an observation of a GRB is made, an energy $E$ is recorded. $E\_{\gamma,iso}$ is the total energy that would have been emitted by the explosion if the energy was released equally in all directions and by considering our distance away from it. However, because we know that GRBs are tightly beamed, $E\_\gamma$ is the corrected energy by taking $E\_{\gamma,iso}$ and considering only a small fraction of it in a tight beam. However, this all seems a tad confusing and I can't see the point in talking about $E\_{\gamma,iso}$ when we think they are very tightly beamed.

I would appreciate some clarification on what exactly $E\_\gamma$ and $E\_{\gamma, iso}$ are and where they come from.

Thanks

**Tags:** gamma-ray-bursts


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## 1 answers

### Answer by username_0 on 2020-09-01. Score: 2

I will answer with what I now believe to be the correct answer. We see some gamma-ray burst emission hit our satellite which carries an amount of energy $E\_{obs}$. If we say the burst was emitted equally in all directions (isotropic) then we derive a quantity $E\_{iso}$. If we say that the burst was actually emitted in a tight beam then we derive $E\_\theta$ where $\theta$ is the opening angle of the burst and we will find $E\_{iso} > E\_{\theta}$. $E\_{iso}$ is used to compare burst energies in the literature without needing to find the opening angle.


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