New answers tagged

0

Let me denote the inflation field energy density w.r.t to the critical density as $\Omega_{\rm f}$. We can write the Hubble function as $$H(z) = H_0\sqrt{\Omega_{\rm m,0}(1+z)^3 + \Omega_{\rm r,0} (1+z)^4 + \Omega_{\Lambda} + \Omega_{f}(z)}$$ Before the inflation, the universe was radiation dominated. This implies that $$H(z) \approx H_0\sqrt{\Omega_{r,0}(1+...


1

Here is a timeline of the universe in the current Big Bang model. In this list one sees that $ 1 degree kelvin= 8.621738 X10^{-5} eV$ As chemical potentials are of the order of eV , $10^5$ in the y axis is rather late. By that time there will be proto-stars formed and the chemical potential in the particular system might not be negligible. But for most ...


1

Note the CMB does not look the same in every direction. There's a very noticeable dipole. Adjusting to remove this dipole does let us say something about our relative velocity to the frame in which the CMB is isotropic, which is as "special" a frame as any. Our position, velocity, etc will all vary constantly though, since we are rotating around ...


1

Special relativity, flat spacetime, has no way to define a rest frame. On cosmological scales, spacetime isn't flat, and we have the CMB. The Milky Way is moving with respect to it at around 600 km/s (0.2% c), so yes, you can define an preferred frame with respect to it. But there is a problem with calling it "The Rest Frame". It is position ...


0

I think you have to look at the whole picture to tease out the relative odds. With an extremely low entropy Big Bang in the past, and the Second Law increasing the entropy ever since, and treating the universe as an isolated system (this part may be controversial), the Fluctuation Theorem says "as the time or system size increases, the probability of ...


0

Your idea about a possible resolution doesn’t work, because of conservation of information. First, let me define some terms. By "brain like yours" I will mean the exact current configuration of your brain. By "good"/"bad" states of the universe, I will mean states of the universe with an evolved/Boltzmann brain like yours. Your ...


0

As I said before in response to another question, the notion of complex curvature does exist: see Y. Martinez-Maure, Real and complex hedgehogs, their symplectic area,curvature and evolutes. The author says the paper is to appear in the Journal of Symplectic Geometry. For instance, a complex circle with (complex) radius R has a (complex) radius of curvature ...


0

I am answering with respect to a finite universe only. There will come a time when the radius $R_{ou}$ of the observable universe will be $$R_{ou}\ =\ \pi\ R_{curv}.$$ This is as far as you can see any emitted photons. If $$R_{ou}\ < t_{CMB}$$ where $t_{CMB}$ is the time it would take for a CMB photon to reach the observer in a non-expanding universe, ...


0

Rather than worry about the exact definition of 'each' and the synchronization issues, perhaps it is simpler, to say that the time coordinate is something like "the proper time for the fluid element since the big bang", or "since radiation decoupled from matter", which ties the synchronization point to something physical. As far as the ...


5

To understand this, keep in mind that the galaxies (or, to be more accurate, clusters of galaxies) are in freefall motion. In a Newtonian picture, this means that there is no force on them other than gravity, and in the GR picture their worldlines are timelike geodesics. Meanwhile if we ignore or "switch off" the dark energy component, then the ...


1

$$dt = \frac{dt}{da}da \frac{a}{a}$$ and we know that $1/a = 1+z$ and $da = -(1+z)^{-2}dz$ and $\frac{da/dt}{a} = H_0E(z)$ so we obtain $$dt = \frac{1}{H_0E(z)} \times \frac{-dz}{(1+z)^2}\times(1+z)$$ $$dt = -\frac{dz}{(1+z)H_0E(z)} \equiv -\frac{dz}{(1+z)H(z)} $$


0

A pattern of Fraunhofer lines can identify atoms. The ratio of frequencies for pairs of lines from a pattern for a particular atom is independent of redshift. The ratio of a frequency of a line from a distance source compared with the known laboratory frequency is the observed redshift which has three components: (a) Doppler, (b) gravitational, and (c) ...


0

It is the nature of understanding the cosmology of the observable universe that we assume as an axiom, regarding a cosmological model, that at a large scale it is homogeneous. It is impossible to know what is actually going on in that part of the universe which is not part of our observable universe. However, this fact has no relevance to whether or not a ...


1

An infinite universe which is approximately homogeneous should not flatten out. Flattening out is a local process, but on the large scale of homogeneity the GR gravity dynamics overwhelm the local process. The scale of flattening is generally at the galaxy level, possibly also regarding galaxy clusters. I am curious about the quote in your question, and I ...


1

Do the measurements from SH0ES and Planck cause the Hubble tension to alter this table? No, it does not. The problem with the $H_0$ tension is most likely due to the $\Lambda$CDM model (e.g., we need an another model) The Hubble tension is about the disagreement in the value of the $H_0$. If you try to measure the Planck constant at the moon in the far ...


4

Age of the universe is $$t = H_0^{-1}\int_0^{\infty}\frac{dz}{(1+z)E(z)}~~(1)$$ where $E(z) = \sqrt{\Omega_{\rm m,0}(1+z)^3 + \Omega_{\rm r,0} (1+z)^4 + \Omega_{\rm \Lambda,0}+ \Omega_{\rm k,0}(1+z)^2}$ For instance if you assume a flat-matter dominated universe ($\Omega_{\rm m,0}, \Omega_{\rm \Lambda,0}, \Omega_{\rm k,0}$), the age of the universe becomes $$...


0

I like physics. And I just wanted to comment that coldness is observed in temperature. While heat is a type of energy. There is no relation I know of with heat and temperature except the one with "Specific heat" which is, Energy = mass * Specific Heat * Temperature $$H = mS\theta$$ where $\theta$ is change in temperature. So energy sorts of flow ...


1

A neutrino condensate would involve lowering the energy of a pair of neutrinos. The only particles that could participate in such interactions would be those that are already at the top of a degenerate "Fermi sea" of neutrinos. In other words the formation of pairs would have the effect of "squaring off" the occupation index distribution ...


2

I don't have a definite answer for you. But I don't believe the existing answers have given this an appropriate treatment and I would like to leave an extended comment. You can extract useful work from a temperature gradient-- e.g. a different in temperature. Many thermodynamic processes will only care about the difference in temperature rather than absolute ...


0

If you have a barrel with nothing but air in it, you can’t put it to much use. But if someone helpfully liquefies the air for you, suddenly you can use it to power a mechanism, despite the fact that all the helpful person did was take away energy. One way of thinking about it is that all machines run on entropy, not energy. If you have access to a very cold ...


2

Edit- here is the link to article that made me think about this. somewhere in middle it is written that scientists can harness the cold energy using some active input method. The following statement from the article is poorly worded: “Essentially, a sky-facing surface passes its heat to the atmosphere as thermal radiation, losing some of its heat to space ...


5

There's a more recent scientific study about this topic, which was then covered in articles like this and this. I also asked a similar question a few months ago. Basically, this is called "thermoradiative photovoltaics" and involves generating energy by emitting heat (as infrared rays) to a heat sink. The proposed technology would use the Earth as ...


2

If you want to transfer heat from a cold environment to a warm environment, you need a heat pump but then part of the provided heat will come from the work done by the heat pump, so it comes from the fuel that drives the heat pump. To get all the heat from only the cold environment, you need to use a heat pump that exploits temperature differences in the ...


3

Heat is not “converted to heat energy”. Heat is the transfer of energy due solely to a temperature difference. Without a temperature difference there can be no heat. The consequences of that transfer can, but doesn’t necessarily, result in work. In the case of the thermoelectric effect, heat can generate a voltage which in turn can produce electrical work. ...


40

Strictly speaking, heat is not converted into energy - instead heat is energy. A thermoelectric generator is sometimes loosely described as turning heat into energy, but what actually happens is that a temperature difference between a heat source and a cold sink (usually the surrounding environment) causes heat/energy to flow between them and some part of ...


18

You need a difference in temperature between two places to generate useful energy. It is possible to use the coldness of space to generate energy, but only if you also have something warm nearby. Some satellites and space probes carry radioisotopes to be the warm thing. https://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generator .


0

The CKM matrix is for quarks and the PMNS matrix is for leptons. In a broken SN symmetry model the strength of CPV is unfixed. An Improved Standard Model Comes with Explicit CPV and Productive of BAU. JMP, 11, 1157-1169. The CPV strength characterized by Jarlskog invariant is much stronger than the value given by present standard model. Thus, there could ...


0

If the universe were spatially of finite and not too large volume then we could, in principle, discover this by observations in the future. For example, there might be evidence of light setting off in opposite directions eventually arriving at the same point after travelling around the universe, or something like that. This can happen even if the average ...


3

In the full gravitational path integral, you would integrate over all components $g_{\mu\nu}$ of the metric. You would also need some way to account for diffeomorphism invariance. The minisuperspace is an approximation which consists of only integrating over the lapse $N(t)$ and the scale factor $a(t)$, assuming both are functions of time only and not space ...


-1

surely if we do an experiment and if the consequence comes out that the energy is constant We do experiments in the laboratory, which can be a few meters long or even two laboratories thousand of kilometers away from each other, but still the dimensions are the dimensions of the earth. With these experiments conservation of energy is always shown to be ...


0

"Then why does thermodynamics and conservation laws states that energy of our universe remains constant?" Because this refers only to any local region of the universe for a period of time small relative to the age of the universe, rather than to the universe as a whole during its entire lifetime. The dynamics of the expanding universe prevents ...


1

A simple analogy. Consider a still pool of water. If it is still then there are no vortices and no ripples. If we reserve the word "thing" to mean "vortex, ripple, stuff like that" then when the pool is still, entirely unmoving, then there is "no-thing" there. When the pool is disturbed there will be ripples and vortices, and ...


3

You can define quite a few "universe temperatures": Cosmic microwave background (2.7K as of now) Cosmic neutrino background (theoretical, but lower than CMB, probably like 2K) Cosmic gravitational waves background (theoretical, but more or less obvious, lower than the above) Temperature of the baryon matter in some region of the universe (way ...


16

This is a helpful review: A new study by an international team of researchers, including members of the Kavli Institute for the Physics and Mathematics of the Universe (Kavli IPMU), suggests that the mean temperature of gas in large structures of the Universe has increased about 3 times in the last 8 billion years, to reach about two million Kelvin today. ...


21

There is a difference between the "temperature of the universe" and the temperature of the cosmic microwave background radiation (CMBR). The former can be changed by physical processes going on in the universe and for example, the conversion of gravitational potential energy, or the release of nuclear binding energy, into the thermal energy of ...


0

This is not an answer, but questions perhaps you might find related to your issue. The question you asks does not belong to physics, but to philosophy. Physics has nothing to say about the subjective experience (qualia) because it has no effect on the behavior of matter. Some, including Daniel Dennett even claim that qualia is an illusion and actually does ...


0

When you ask about a "theory" it is important that you not confuse it with a "speculation". The following is a very good definition for "theory". From https://en.wikipedia.org/wiki/Theory In modern science, the term "theory" refers to scientific theories, a well-confirmed type of explanation of nature, made in a way ...


4

We presume that the rod is rigid You cannot presume that the rod is rigid. Rigid objects are incompatible with relativity and the known laws of physics. Upon removing the un-physical presumption, what happens is simply that the rod stretches and possibly breaks, depending on the material characteristics.


2

There are many theories about the nature of Time. Whether the Universe is "open" with Time indefinitely long or "closed" with a finite Time ending in a Big Crunch is a matter of ongoing debate, although the standard model of cosmology has no mechanism for ending Time. Some cosmologists propose a "Big Rip", in which the expansion ...


3

From your question, I reckon you don't want a too-in-depth answer. Please correct me if you want some equations to make sense of what I write. This is a diagram widely used in cosmology. Roughly, you can see $\Omega_M$ as the matter density, and $\Omega_\Lambda$ as the Dark Energy density. There are various theories that try to explain what the universe ...


8

A description in which spacetime comes back to a situation like the south pole in your question is the possibility in general relativity called "Big Crunch". We don't think the universe will do this (the matter density is not high enough) but you asked if there were models which allowed that sort of thing. The Big Crunch is in some respects like ...


0

You are thinking of change in far too restrictive of a sense. Consider a static cone. Even though the cone is static we can perfectly validly say that the radius of the cone cross section changes from apex to base. The radius is small near the apex and it gets large near the base. This would be described by $\frac{d}{dx}r=k$. Mathematically, there is no ...


1

Look at a painting. The left is different from the right, the top is different from the bottom, and yet (if the painting is of something coherent) every little bit of the image is related in some way to the little bits on either side of it. Look at a movie. The beginning is different from the end, but (almost) every frame is related to the frames before and ...


0

As user253751 says in his comment, different points of the block can have different time values. So you need a direction for time and some kind of wave of time, that progresses in that direction. Basically the fully determined block universe is being read by the passing time, and we feel this as change.


2

If there were any regions in the observable universe where antimatter was more abundant than matter then we would be able to observe the gamma radiation produced by the annihilation of matter and antimatter at the boundaries of these regions. Even if the boundaries were in intergalactic space, the gamma radiation would still be detectable. But no such gamma ...


0

The simulation is not correct. In your code, if targets[i].x * H > c then that target moves faster than the photon, i.e., outside of the light cone. In reality, all locations in the universe are equivalent and there is no location at which galaxies outrun light. Another issue with the simulation is that $H$ is constant, which is accurate in the large-time ...


1

At the time of writing, the US particle physics community is trying to address this and similar questions in a large community study. Dark matter being the elephant in the room that it is, it seems clear that we need all possible experiments and detectors and observatories we can get our hands on in order to make progress towards understanding the true ...


0

Sometimes it's useful to make a distinction between vectors and pseudovectors (sometimes called polar vectors and axial vectors, respectively). Both objects transform the same way under rotations of the basis vectors, but pseudovectors get an additional minus sign under inversions which change the handedness of the basis. As an example, consider two ...


0

Physics studies how the universe behaves. Why it does so is outside the scope of physics. Never the less, there are a couple of things that can be said. Consider a pot of water. It just sits there. Raise the temperature. Now it boils. As long as the temperature (and other properties) stays the same, the water stay the same. You can say the same thing about ...


Top 50 recent answers are included