<p>No, because the strain amplitude from this binary is vanishingly small compared with variations in atmospheric pressure due to a variety of mechanisms at frequencies similar to the expected frequency of the gravitational waves from both the short-period orbit and the longer period modulation; and because air pressure is probably not sensitive to the passage of gravitational waves in any case, since they do not alter <em>volumes</em> in space.</p>
<p><strong>Details</strong></p>
<p>The quadrupole gravitational wave strain expected from a binary in a circular orbit is given by
<span class="math-container">$$h =  -\frac{4G \omega_{\phi}^2 \mu a^2}{rc^4} \left[\frac{(1 + \cos^2 i)}{2}   \cos(2\omega_{\phi} t)\ {\mathbf e_+}  
       + \cos(i)\, \sin(2 \omega_{\phi} t)\ {\mathbf e_{\times}} \right]\ , $$</span>
where <span class="math-container">${\mathbf e_+}$</span> and <span class="math-container">${\mathbf e_\times}$</span> are unit 4x4 matrices representing the two polarisations of gravitational wave radiation, <span class="math-container">$\omega_\phi$</span> and <span class="math-container">$a$</span> are the angular frequency and semi-major axis of the binary, <span class="math-container">$\mu= m_1m_2/(m_1+m_2)$</span> is the reduced mass of the binary,<span class="math-container">$i$</span> is the inclination of the orbital plane to the line of sight and <span class="math-container">$r$</span> is the distance to the binary.</p>
<p>The numbers for this system can be obtained from <a href="https://arxiv.org/abs/1401.0535" rel="noreferrer">Ransom et al. (2014)</a>:<br />
<br />
<span class="math-container">$a = 5.76\times10^8\ $</span> m<br />
<span class="math-container">$\omega_\phi = 4.46\times 10^{-5}\ $</span>rad/s <br />
<span class="math-container">$m_1 = 1.438M_\odot ,\ m_2 = 0.198M_\odot ,\ \mu = 0.174M_\odot $</span><br />
<span class="math-container">$i = 39.2^\circ$</span> <br />
<span class="math-container">$r = 1300\ $</span>pc</p>
<p>Thus
<span class="math-container">$$ h = 1.5\times10^{-25}\, \cos(2\omega_{\phi} t)\ {\mathbf e_+} + 1.5\times10^{-25}\, \sin(2 \omega_{\phi} t) {\mathbf e_\times} $$</span>
i.e. Almost circularly polarised gravitational wave radiation, with a dimensionless strain amplitude of <span class="math-container">$1.5\times 10^{-25}$</span>. This is about 4 orders of magnitude smaller than the gravitational waves detected by LIGO and would cause changes in spatial dimension of order 1 part in <span class="math-container">$10^{25}$</span>.</p>
<p>Even if this did cause changes in pressure by virtue of changes in volume (but see below), these changes would be of order a few parts in <span class="math-container">$10^{25}$</span>. Is that how precisely you are getting measurements of the air pressure?</p>
<p>Contrary to your statement, atmospheric pressure <em>is</em> sensitive to seismic activity and it is affected by all sorts of weather-related phenomena and tides at a wide variety of frequencies (e.g., <a href="https://link.springer.com/article/10.1134/S1069351313010138" rel="noreferrer">Shven et al. 2013</a>) from hours to months. These variations are measurable at amplitudes that are vastly bigger than the gravitational wave signal you want to detect. There is no possibility of detection against this noise background.</p>
<p>In fact, the passage of a transverse gravitational wave through space <em>would not</em> change the volume. A compression in one direction is balanced by a rarefaction on the orthogonal direction - e.g. see <a href="https://physics.stackexchange.com/questions/341988/can-a-gravitational-wave-stretch-a-volume-equally-in-3-dimensions">https://physics.stackexchange.com/questions/341988/can-a-gravitational-wave-stretch-a-volume-equally-in-3-dimensions</a> . So it is difficult to see why gravitational waves would change the pressure of a gas at all.</p>

CC BY-SA 4.0
2023-11-06T12:26:04.060Z
Cancel culture? It is always bad when measurement results contradict theory.
9herbert9
CC BY-SA 4.0
2023-11-06T14:37:42.883Z
@9herbert9 In what way have you been "cancelled"? You asked "Could this be a GW?" and I've given an answer. Are you saying that you asked in bad faith and only answers that match your pre-conceived ideas are acceptable?   You are absolutely free to contribute your own answer. Possibly you are referring to comments added underneath my question - removal of those has nothing to do with me, they possibly contravened site policies on comments? Note, this is a Q+A site, not a discussion forum. Comments clarifying your question should be edited into your question, not left as comments on answers.
ProfRob
CC BY-SA 4.0
2023-11-06T15:21:25.500Z
Sorry, I didn't refer to your post above. I just don't understand why the moderator deleted the additions below. Regardless: I didn't know that Stackexchange is one-sided and doesn't allow fruitful discussions. Not even any questions. A pity.
9herbert9
CC BY-SA 4.0
2023-11-06T16:29:13.620Z
@9herbert9 This isn't a discussion site - particularly to discuss unpublished results or personal theories. I can't now recall whether your comments were actually attached to my question, however I see that you had a (non) answer deleted. You were informed beforehand that this was likely to happen and asked to clarify your question or, if you had additional questions, to post these as new questions. This is not "cancel culture" it is following the rules of this Q+A site. Other discussion forums are available I believe.
ProfRob
CC BY-SA 4.0
2023-11-12T15:43:37.410Z
Is a surprising measurement result only credible if an explanation is immediately provided? Wilhelm Roentgen received a Nobel Prize for his accidental discovery of X-rays, although no one knew why they were created and what effects they had. My suggestion: check for yourself! It's very easy if you know reception technology.
9herbert9
CC BY-SA 4.0
2023-11-05T15:33:43.557Z
2023-11-05T16:53:47.150Z
2023-11-05T16:53:47.150Z
ProfRob
<p>After the stellar triple system PSR J0337+1715 was discovered a few years ago, I was interested in whether the GW could be detected here. From the known orbit times it follows that the GW frequency 14.2054e-6 Hz would have to be <em>phase</em> modulated with f_mod=1/(327 days). It's easy to build a software defined receiver for these frequency, the problem is the antenna. The LIGO interferometers cannot receive such low frequencies; LISA does not yet exist. So I used air pressure as an &quot;antenna&quot; because earthquakes don't interfere. Surprisingly, I discovered a weak signal with the characteristics described above exactly at the target frequency. Additionally it is <em>phase</em>-modulated with f_2=1/(365 days). Could this be a GW? It fits the triple system.</p>

CC BY-SA 4.0
2023-11-05T14:28:58.777Z
Why would the fact that air pressure is modulated on periods of 1 year have anything to do with a binary that has a period of 327 days? And why do you think air pressure should be modulated by gravitational waves? And where is your high precision air pressure data coming from?
ProfRob
CC BY-SA 4.0
2023-11-05T19:01:21.517Z
Gravitational waves are exceedingly faint and don’t cause large shifts in things. If you read a bit on the subject, you’ll soon find that when a gravitational wave crosses the solar system, it changes the Sun–Saturn distance by the size of about one proton. So, no, there is no way in hell that you can detect gravitational waves on such a small scale as with an air-pressure “antenna.” If what you found is really with the same period as that pulsar system (or anything else, for that matter), then it’s simply a coincidence.
Pierre Paquette
<gravitational-waves>
A mysterious signal at 14.2054 µHz
9herbert9
# A mysterious signal at 14.2054 µHz

(Asked by: username_0 on 2023-11-05)

After the stellar triple system PSR J0337+1715 was discovered a few years ago, I was interested in whether the GW could be detected here. From the known orbit times it follows that the GW frequency 14.2054e-6 Hz would have to be *phase* modulated with f\_mod=1/(327 days). It's easy to build a software defined receiver for these frequency, the problem is the antenna. The LIGO interferometers cannot receive such low frequencies; LISA does not yet exist. So I used air pressure as an "antenna" because earthquakes don't interfere. Surprisingly, I discovered a weak signal with the characteristics described above exactly at the target frequency. Additionally it is *phase*-modulated with f\_2=1/(365 days). Could this be a GW? It fits the triple system.

**Tags:** gravitational-waves

> **Comments:**
> **username_1**: Why would the fact that air pressure is modulated on periods of 1 year have anything to do with a binary that has a period of 327 days? And why do you think air pressure should be modulated by gravitational waves? And where is your high precision air pressure data coming from?  
> **username_2**: Gravitational waves are exceedingly faint and don’t cause large shifts in things. If you read a bit on the subject, you’ll soon find that when a gravitational wave crosses the solar system, it changes the Sun–Saturn distance by the size of about one proton. So, no, there is no way in hell that you can detect gravitational waves on such a small scale as with an air-pressure “antenna.” If what you found is really with the same period as that pulsar system (or anything else, for that matter), then it’s simply a coincidence.  

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## 1 answers

### Answer by username_1 on 2023-11-05. Score: 10

No, because the strain amplitude from this binary is vanishingly small compared with variations in atmospheric pressure due to a variety of mechanisms at frequencies similar to the expected frequency of the gravitational waves from both the short-period orbit and the longer period modulation; and because air pressure is probably not sensitive to the passage of gravitational waves in any case, since they do not alter *volumes* in space.

**Details**

The quadrupole gravitational wave strain expected from a binary in a circular orbit is given by
$$h = -\frac{4G \omega\_{\phi}^2 \mu a^2}{rc^4} \left[\frac{(1 + \cos^2 i)}{2} \cos(2\omega\_{\phi} t)\ {\mathbf e\_+}
+ \cos(i)\, \sin(2 \omega\_{\phi} t)\ {\mathbf e\_{\times}} \right]\ , $$
where ${\mathbf e\_+}$ and ${\mathbf e\_\times}$ are unit 4x4 matrices representing the two polarisations of gravitational wave radiation, $\omega\_\phi$ and $a$ are the angular frequency and semi-major axis of the binary, $\mu= m\_1m\_2/(m\_1+m\_2)$ is the reduced mass of the binary,$i$ is the inclination of the orbital plane to the line of sight and $r$ is the distance to the binary.

The numbers for this system can be obtained from [Ransom et al. (2014)](https://arxiv.org/abs/1401.0535):  
  
$a = 5.76\times10^8\ $ m  
$\omega\_\phi = 4.46\times 10^{-5}\ $rad/s   
$m\_1 = 1.438M\_\odot ,\ m\_2 = 0.198M\_\odot ,\ \mu = 0.174M\_\odot $  
$i = 39.2^\circ$   
$r = 1300\ $pc

Thus
$$ h = 1.5\times10^{-25}\, \cos(2\omega\_{\phi} t)\ {\mathbf e\_+} + 1.5\times10^{-25}\, \sin(2 \omega\_{\phi} t) {\mathbf e\_\times} $$
i.e. Almost circularly polarised gravitational wave radiation, with a dimensionless strain amplitude of $1.5\times 10^{-25}$. This is about 4 orders of magnitude smaller than the gravitational waves detected by LIGO and would cause changes in spatial dimension of order 1 part in $10^{25}$.

Even if this did cause changes in pressure by virtue of changes in volume (but see below), these changes would be of order a few parts in $10^{25}$. Is that how precisely you are getting measurements of the air pressure?

Contrary to your statement, atmospheric pressure *is* sensitive to seismic activity and it is affected by all sorts of weather-related phenomena and tides at a wide variety of frequencies (e.g., [Shven et al. 2013](https://link.springer.com/article/10.1134/S1069351313010138)) from hours to months. These variations are measurable at amplitudes that are vastly bigger than the gravitational wave signal you want to detect. There is no possibility of detection against this noise background.

In fact, the passage of a transverse gravitational wave through space *would not* change the volume. A compression in one direction is balanced by a rarefaction on the orthogonal direction - e.g. see <https://physics.stackexchange.com/questions/341988/can-a-gravitational-wave-stretch-a-volume-equally-in-3-dimensions> . So it is difficult to see why gravitational waves would change the pressure of a gas at all.

> **Comments:**
> **username_0**: Cancel culture? It is always bad when measurement results contradict theory.  
> **username_1**: @username_0 In what way have you been "cancelled"? You asked "Could this be a GW?" and I've given an answer. Are you saying that you asked in bad faith and only answers that match your pre-conceived ideas are acceptable?   You are absolutely free to contribute your own answer. Possibly you are referring to comments added underneath my question - removal of those has nothing to do with me, they possibly contravened site policies on comments? Note, this is a Q+A site, not a discussion forum. Comments clarifying your question should be edited into your question, not left as comments on answers.  
> **username_0**: Sorry, I didn't refer to your post above. I just don't understand why the moderator deleted the additions below. Regardless: I didn't know that Stackexchange is one-sided and doesn't allow fruitful discussions. Not even any questions. A pity.  
> **username_1**: @username_0 This isn't a discussion site - particularly to discuss unpublished results or personal theories. I can't now recall whether your comments were actually attached to my question, however I see that you had a (non) answer deleted. You were informed beforehand that this was likely to happen and asked to clarify your question or, if you had additional questions, to post these as new questions. This is not "cancel culture" it is following the rules of this Q+A site. Other discussion forums are available I believe.  
> **username_0**: Is a surprising measurement result only credible if an explanation is immediately provided? Wilhelm Roentgen received a Nobel Prize for his accidental discovery of X-rays, although no one knew why they were created and what effects they had. My suggestion: check for yourself! It's very easy if you know reception technology.  

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