<p>I think I lost a <span class="math-container">$c$</span> factor...</p>
<p>I try</p>
<p><span class="math-container">$$m&lt;\sqrt{2\kappa\rho}\dfrac{\hbar}{c^2}=10^{-34}\dfrac{\sqrt{16\pi \cdot 6.674\cdot 10^{-11}\cdot 10^{-28}}}{9\cdot 10^{16}}=0.64\cdot 10^{-69}kg=0.64\cdot 10^{-66}g$$</span></p>
<p>But yet, I am missing a 10 factor...Not a big thing like before...But...As well: The translated into English version seems to be bad in the units of <span class="math-container">$\rho_0$</span> and the final numbers, but also the Russian version! Or I am doing something wrong... Am I correct now or have I lost a 10 somewhere here?</p>
<p>Remark: the density should be <span class="math-container">$\rho_0=10^{-26}kg/m^3=10^{-29}g/cm^3\sim 6protons/m^3$</span> to get the result of the article.</p>
<p>Dimensional check:</p>
<p><span class="math-container">$$[m]=[\sqrt{2\kappa\rho}\dfrac{\hbar}{c^2}]=M=\dfrac{ML^2T^{-1}\sqrt{L^3T^{-2}M^{-1}ML^{-3}}}{L^2T^{-2}}$$</span></p>

CC BY-SA 4.0
2024-04-02T01:05:44.270Z
This does not seem like an answer, but something you should have edited into your original question.
StephenG - Help Ukraine
CC BY-SA 4.0
2024-04-03T17:36:54.200Z
@StephenG-HelpUkraine I think it IS an answer, since it remarks that the likely missing factor is in the mass/energy density (the number seems wrong by different reasons in both, the Russian and the English papers)
riemannium
CC BY-SA 4.0
2024-04-01T15:59:12.003Z
2024-04-01T16:20:38.790Z
2024-04-01T16:20:38.790Z
riemannium
<p>In the paper, <em>Relativistic theory of gravitation and the graviton rest mass. Theoretical and Mathematical Physics</em>, 65(1), 971–979. doi:10.1007/bf01028629, the mass bound</p>
<p><span class="math-container">$$m_g&lt;6.4\cdot 10^{-66}g=6.4\cdot 10^{-69}kg$$</span></p>
<p>is obtained. However, putting the numbers with <span class="math-container">$\rho=10^{-29}g/cm^3=10^{-26}kg/m^3$</span>   I get <span class="math-container">$m_g&lt; 6.4\cdot 10^{-60}kg=6.4\cdot 10^{-57}g$</span>, with <span class="math-container">$k=8\pi G_N$</span>. Where am I missing the power counting?</p>

CC BY-SA 4.0
2024-04-01T14:12:20.157Z
This?  https://inis.iaea.org/collection/NCLCollectionStore/_Public/17/084/17084017.pdf  Is an English language version available?
James K
CC BY-SA 4.0
2024-04-01T15:30:49.873Z
Yes: https://link.springer.com/article/10.1007/BF01028629 @JamesK
riemannium
CC BY-SA 4.0
2024-04-01T15:34:28.117Z
@JamesK Even if taking into account the values in the russian version, if you see the English variant seems wrong as well...I am puzzled because I am pretty sure I did the conversion factors right...and that is why I ask here...$\hbar=1.05\cdot 10^{-33}J/Hz$ and $c=3\cdot 10^8m/s$, from the English variant of the article you obtain $\sqrt{2\kappa\rho}\hbar/ c$. Oh, wait...I think I can see it now...Let me try a thing...
riemannium
CC BY-SA 4.0
2024-04-03T11:33:54.897Z
Python's `astropy.units` can help you converting between units.
pela
<general-relativity><graviton>
Cosmic massive graviton bounds
riemannium
# Cosmic massive graviton bounds

(Asked by: username_0 on 2024-04-01)

In the paper, *Relativistic theory of gravitation and the graviton rest mass. Theoretical and Mathematical Physics*, 65(1), 971–979. doi:10.1007/bf01028629, the mass bound

$$m\_g<6.4\cdot 10^{-66}g=6.4\cdot 10^{-69}kg$$

is obtained. However, putting the numbers with $\rho=10^{-29}g/cm^3=10^{-26}kg/m^3$ I get $m\_g< 6.4\cdot 10^{-60}kg=6.4\cdot 10^{-57}g$, with $k=8\pi G\_N$. Where am I missing the power counting?

**Tags:** general-relativity, graviton

> **Comments:**
> **username_1**: This?  https://inis.iaea.org/collection/NCLCollectionStore/_Public/17/084/17084017.pdf  Is an English language version available?  
> **username_0**: Yes: https://link.springer.com/article/10.1007/BF01028629 @username_1  
> **username_0**: @username_1 Even if taking into account the values in the russian version, if you see the English variant seems wrong as well...I am puzzled because I am pretty sure I did the conversion factors right...and that is why I ask here...$\hbar=1.05\cdot 10^{-33}J/Hz$ and $c=3\cdot 10^8m/s$, from the English variant of the article you obtain $\sqrt{2\kappa\rho}\hbar/ c$. Oh, wait...I think I can see it now...Let me try a thing...  
> **username_2**: Python's `astropy.units` can help you converting between units.  

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## 1 answers

### Answer by username_0 on 2024-04-01. Score: 1

I think I lost a $c$ factor...

I try

$$m<\sqrt{2\kappa\rho}\dfrac{\hbar}{c^2}=10^{-34}\dfrac{\sqrt{16\pi \cdot 6.674\cdot 10^{-11}\cdot 10^{-28}}}{9\cdot 10^{16}}=0.64\cdot 10^{-69}kg=0.64\cdot 10^{-66}g$$

But yet, I am missing a 10 factor...Not a big thing like before...But...As well: The translated into English version seems to be bad in the units of $\rho\_0$ and the final numbers, but also the Russian version! Or I am doing something wrong... Am I correct now or have I lost a 10 somewhere here?

Remark: the density should be $\rho\_0=10^{-26}kg/m^3=10^{-29}g/cm^3\sim 6protons/m^3$ to get the result of the article.

Dimensional check:

$$[m]=[\sqrt{2\kappa\rho}\dfrac{\hbar}{c^2}]=M=\dfrac{ML^2T^{-1}\sqrt{L^3T^{-2}M^{-1}ML^{-3}}}{L^2T^{-2}}$$

> **Comments:**
> **username_3**: This does not seem like an answer, but something you should have edited into your original question.  
> **username_0**: @username_3 I think it IS an answer, since it remarks that the likely missing factor is in the mass/energy density (the number seems wrong by different reasons in both, the Russian and the English papers)  

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