<p><a href="http://en.wikipedia.org/wiki/Negative_mass" rel="nofollow">Negative energy</a> should be gravitationally repulsive. But naturally occuring negative energy is too weak to be noticible on a large scale. Macroscopic known is the <a href="http://en.wikipedia.org/wiki/Casimir_effect" rel="nofollow">Casimir effect</a>.</p>

<p>From chemistry you'll certainly know <a href="http://en.wikipedia.org/wiki/Van_der_Waals_force" rel="nofollow">van-der-Waals forces</a>. There are several kinds of van-der-Waals forces. One of them are London-van der Waals forces, thought to be related to the Casimir effect.</p>

<p>But electromagnetism is too strong compared to gravity in theses cases for reasonable experiments to show gravitational repulsion.</p>

<p><a href="http://en.wikipedia.org/wiki/Dark_energy" rel="nofollow">Dark energy</a> might be interpreted as "gravitational repulsion"; but this won't be effective for the gravitational force between two objects, if it's distributed equally, as <a href="http://en.wikipedia.org/wiki/Lambda-CDM_model" rel="nofollow">current models</a> suggest.</p>

<p><a href="http://www.earthtech.org/publications/davis_STAIF_conference_1.pdf" rel="nofollow">Here</a> a paper about a couple of attempts to produce negative energy in the lab.</p>

CC BY-SA 3.0
2014-04-25T05:30:16.063Z
Hi Gerald. How do you relate the Casimir effect with the "negative energy"? I don't know too much about the latter (that is, however, just theoretical stuff, it seems) but I don't think these two are actually related.
Py-ser
CC BY-SA 3.0
2014-04-25T12:29:01.540Z
@Py-ser Take vacuum as space filled with the lowest possible energy, call this energy zero. By Heisenberg's uncertainty you'll get electromagnetic quantum fluctuations of any wavelength. Between two metallic plates longer wavelengths don't occur, reducing the energy contents below zero, the result is negative energy. Zero energy outside the plates excerts a pressure in comparison to the negative energy between the plates.
Gerald
CC BY-SA 3.0
2014-04-25T14:02:26.070Z
Related question on Physics SE: http://physics.stackexchange.com/questions/47922/can-negative-energy-be-created-by-the-casimir-effect
Gerald
CC BY-SA 3.0
2014-04-24T21:49:55.573Z
2014-04-25T13:59:48.347Z
2014-04-25T13:59:48.347Z
Gerald
<blockquote>
  <p>What I ask myself is: Couldn't there be "negative" bundles of mass just the other way that pushes matter away instead of invisible dark matter that pulls it?</p>
</blockquote>

<p>The galaxy rotation curve indicates a (positively massed) dark matter distribution that is close to spherically symmetric; cf. <a href="http://en.wikipedia.org/wiki/Dark_matter_halo">dark matter halo</a>. I take it that you are asking whether instead of modeling it as a positive-mass distribution around the galaxy, we can model it instead as a negative-mass distribution surrounding the galaxy pushing the stars inward.</p>

<p>No, that's not possible. Since the effect is spherically symmetric, the hypothetical negative-mass distribution would need to be spherically symmetric as well. However, Newton's shell theorem guarantees that <em>everywhere</em> inside a void surrounded by such a distribution with this symmetry, whatever the sign of the mass is, the gravitational force vanishes. So a negatively-massed dark matter distribution outside the galaxy would do nothing at all.</p>

<p>In general relativity, the analogous statement is guaranteed by Birkhoff's theorem, and although GTR is nonlinear (and there are stars in the way), in our galaxy gravity is weak enough to be handled by the linearized theory.</p>

<blockquote>
  <p>What is the proof for the fact "Gravity is always attractive" beside that fact that it seems to be? </p>
</blockquote>

<p>In general relativity, gravity doesn't have to be attractive. For example, inside a perfect fluid with density energy $\rho$ and pressure $p$, $\rho+3p&lt;0$ would imply that gravity is locally repulsive in the sense that a ball of test particles initially at rest expands rather than contracts under gravitational freefall. More generally, the <em>strong energy condition</em> characterizes whether or not gravity is attractive in this sense.</p>

<p>However, there is no known material that violates the strong energy condition, besides possibly dark energy, which is a cosmological constant in the standard ΛCDM model, as Gerald says.</p>

<blockquote>
  <p>Has someone tried to find something like that?</p>
</blockquote>

<p>There are many studies and simulations that consider dark matter with a nonzero pressure component, sometimes even anisotropic pressure components. For example, "hot" dark matter would have $p\sim\rho/3$, while "cold" dark matter would have $p\sim 0$, "warm" dark matter somewhere inbetween. The CDM in the ΛCDM model stands for "cold dark matter", naturally.</p>

<p>There are more exotic possibilities with <em>anisotropic</em> pressure components, etc. But I don't know of any paper that looked specifically for $\rho&lt;0$ dark matter, and for the above reasons I'd be surprised if one existed.</p>

CC BY-SA 3.0
2014-04-25T08:09:12.973Z
Thank you for the comprehensive answer and the description that there are dark matter effects that can not be explained with repulsion but only with attraction.
Jens
CC BY-SA 3.0
2014-04-25T04:46:45.510Z
2014-04-25T08:35:01.050Z
2014-04-25T08:35:01.050Z
Stan Liou
<p>I learned that gravity always pulls, it is always attractive. There is no antigravity, no "negative"  mass. I also learned that most matter is dark, invisible. We know that because we see visible matter moving towards invisible bundles of dark matter and other effects like accelaration or gravity lenses. </p>

<p>What I ask myself is: Couldn't there be "negative" bundles of mass just the other way that pushes matter away instead of invisible dark matter that pulls it? Has someone tried to find something like that? 
What is the proof for the fact "Gravity is always attractive" beside that fact that it seems to be? </p>

<p>Granted, it may be a stupid question, but I lack a bit in physics education that is not very related to chemistry.</p>

CC BY-SA 3.0
2014-04-25T17:45:22.380Z
This question appears to be off-topic because it is about it's about physics and not astronomy in particular
Eduardo Serra
CC BY-SA 3.0
2014-04-25T20:41:07.490Z
Since the question concerns the structure of the universe it is very well inside the domain of astronomy in my opinion. I would say most questions regarding astronomy are about physics, but not all physics questions are about astronomy.
Jens
CC BY-SA 3.0
2014-04-29T13:46:37.437Z
@EduardoSerra Dark matter is in the realm of cosmology and thus on topic here.
called2voyage
<gravity><dark-matter>
Has someone looked the other way?
Jens
# Has someone looked the other way?

(Asked by: username_0 on 2014-04-24)

I learned that gravity always pulls, it is always attractive. There is no antigravity, no "negative" mass. I also learned that most matter is dark, invisible. We know that because we see visible matter moving towards invisible bundles of dark matter and other effects like accelaration or gravity lenses.

What I ask myself is: Couldn't there be "negative" bundles of mass just the other way that pushes matter away instead of invisible dark matter that pulls it? Has someone tried to find something like that?
What is the proof for the fact "Gravity is always attractive" beside that fact that it seems to be?

Granted, it may be a stupid question, but I lack a bit in physics education that is not very related to chemistry.

**Tags:** gravity, dark-matter
> **Comments:**
> **username_1**: This question appears to be off-topic because it is about it's about physics and not astronomy in particular  
> **username_0**: Since the question concerns the structure of the universe it is very well inside the domain of astronomy in my opinion. I would say most questions regarding astronomy are about physics, but not all physics questions are about astronomy.  
> **username_2**: @EduardoSerra Dark matter is in the realm of cosmology and thus on topic here.  

---

## 2 answers

### Answer by username_5 on 2014-04-25. Score: 9 | Accepted answer

> What I ask myself is: Couldn't there be "negative" bundles of mass just the other way that pushes matter away instead of invisible dark matter that pulls it?

The galaxy rotation curve indicates a (positively massed) dark matter distribution that is close to spherically symmetric; cf. [dark matter halo](http://en.wikipedia.org/wiki/Dark_matter_halo). I take it that you are asking whether instead of modeling it as a positive-mass distribution around the galaxy, we can model it instead as a negative-mass distribution surrounding the galaxy pushing the stars inward.

No, that's not possible. Since the effect is spherically symmetric, the hypothetical negative-mass distribution would need to be spherically symmetric as well. However, Newton's shell theorem guarantees that *everywhere* inside a void surrounded by such a distribution with this symmetry, whatever the sign of the mass is, the gravitational force vanishes. So a negatively-massed dark matter distribution outside the galaxy would do nothing at all.

In general relativity, the analogous statement is guaranteed by Birkhoff's theorem, and although GTR is nonlinear (and there are stars in the way), in our galaxy gravity is weak enough to be handled by the linearized theory.

> What is the proof for the fact "Gravity is always attractive" beside that fact that it seems to be?

In general relativity, gravity doesn't have to be attractive. For example, inside a perfect fluid with density energy $\rho$ and pressure $p$, $\rho+3p<0$ would imply that gravity is locally repulsive in the sense that a ball of test particles initially at rest expands rather than contracts under gravitational freefall. More generally, the *strong energy condition* characterizes whether or not gravity is attractive in this sense.

However, there is no known material that violates the strong energy condition, besides possibly dark energy, which is a cosmological constant in the standard ΛCDM model, as Gerald says.

> Has someone tried to find something like that?

There are many studies and simulations that consider dark matter with a nonzero pressure component, sometimes even anisotropic pressure components. For example, "hot" dark matter would have $p\sim\rho/3$, while "cold" dark matter would have $p\sim 0$, "warm" dark matter somewhere inbetween. The CDM in the ΛCDM model stands for "cold dark matter", naturally.

There are more exotic possibilities with *anisotropic* pressure components, etc. But I don't know of any paper that looked specifically for $\rho<0$ dark matter, and for the above reasons I'd be surprised if one existed.

> **Comments:**
> **username_0**: Thank you for the comprehensive answer and the description that there are dark matter effects that can not be explained with repulsion but only with attraction.  

---

### Answer by username_3 on 2014-04-24. Score: 1

[Negative energy](http://en.wikipedia.org/wiki/Negative_mass) should be gravitationally repulsive. But naturally occuring negative energy is too weak to be noticible on a large scale. Macroscopic known is the [Casimir effect](http://en.wikipedia.org/wiki/Casimir_effect).

From chemistry you'll certainly know [van-der-Waals forces](http://en.wikipedia.org/wiki/Van_der_Waals_force). There are several kinds of van-der-Waals forces. One of them are London-van der Waals forces, thought to be related to the Casimir effect.

But electromagnetism is too strong compared to gravity in theses cases for reasonable experiments to show gravitational repulsion.

[Dark energy](http://en.wikipedia.org/wiki/Dark_energy) might be interpreted as "gravitational repulsion"; but this won't be effective for the gravitational force between two objects, if it's distributed equally, as [current models](http://en.wikipedia.org/wiki/Lambda-CDM_model) suggest.

[Here](http://www.earthtech.org/publications/davis_STAIF_conference_1.pdf) a paper about a couple of attempts to produce negative energy in the lab.

> **Comments:**
> **username_4**: Hi Gerald. How do you relate the Casimir effect with the "negative energy"? I don't know too much about the latter (that is, however, just theoretical stuff, it seems) but I don't think these two are actually related.  
> **username_3**: @Py-ser Take vacuum as space filled with the lowest possible energy, call this energy zero. By Heisenberg's uncertainty you'll get electromagnetic quantum fluctuations of any wavelength. Between two metallic plates longer wavelengths don't occur, reducing the energy contents below zero, the result is negative energy. Zero energy outside the plates excerts a pressure in comparison to the negative energy between the plates.  
> **username_3**: Related question on Physics SE: http://physics.stackexchange.com/questions/47922/can-negative-energy-be-created-by-the-casimir-effect  

---

