EM receivers are focused on absorption - of retarded waves. For advanced you need its time symmetric analog - receiver focused on emission, what requires its initial excitation ... usually avoided e.g. by cooling of radio telescope amplifier.

In contrast, LIGO just measures lengths - which are T/CPT invariant ... also for pulsar timing ( https://theconversation.com/to-map-the-vibration-of-the-universe-astronomers-built-a-detector-the-size-of-the-galaxy-244157 ) requiring orbiting SMBH (more than expected retarded alone) - imagining spacetime as 4D membrane minimizing action (tension), orbiting SMBH should perturb it in both directions.

Edited December 16Dec 16 by Duda Jarek

47 minutes ago, Duda Jarek said:

For advanced you need its time symmetric analog - receiver focused on emission, what requires its initial excitation ... usually avoided e.g. by cooling of radio telescope amplifier.

You're still stuffing photons from everywhere (including a hot radio telescope) back into the star and splitting helium. Metaphorically speaking. As KJW and Swanson keep pointing out, our causality is asymmetric.

My problem with trying to imagine some reverse causality with gravity is that it's like trying to have Ricci curvature "the other way." Scorched shuttle parts rising up out of the Australian Outback and going up into space. With something like beta decay it's much easier, with a distant neutrino and an electron from my Geiger counter converging on a more stable nucleus in some radioisotope lump. Still makes no sense thermodynamically.

There is Feynman diagram photon exchange between coupled e.g. two electrons - current cooled EM receivers are focused on retarded: such that photon was emitted by target, absorbed by sensor ... for advanced you need CPT version of this diagram - reversed coupling: photon emitted by receiver, absorbed by target.

8 minutes ago, Duda Jarek said:

There is Feynman diagram photon exchange between coupled e.g. two electrons - current cooled EM receivers are focused on retarded: such that photon was emitted by target, absorbed by sensor ... for advanced you need CPT version of this diagram - reversed coupling: photon emitted by receiver, absorbed by target.

I think the point being made is that you can't use Feynman diagrams to model thermodynamics. Where there are dissipative processes, you have intrinsic irreversibility.

Thermodynamics is for spontaneous emission, while here we are talking about stimulated.

It is valuable to think about hydrodynamical analog: wave behind marine propeller carries energy, momentum and angular momentum - like photon. We can practically apply time symmetry - just reversing its rotation: getting pulling photon, causing deexcitation of target.

Going to similar EM, we can analogously apply time symmetry to synchrotron radiation - just reversing electron trajectory, this way switching absorption and stimulated emission acting on target:

1 hour ago, Duda Jarek said:

Thermodynamics is for spontaneous emission, while here we are talking about stimulated.

Stimulated emission prevents one from inverting a population by simple absorption. It is very much a thermodynamics thing. For example, one of the reasons for using stronger magnets in NMR spectroscopy, and thus increasing the energy separation between the nuclear spin levels, is to reduce the thermal population of the higher energy level relative to the lower energy level, thereby improving signal strength against saturation resulting from stimulated emission.

Edited December 16Dec 16 by KJW

The boundary conditions it would require (not just for gravitational waves, but for any full-fledged macroscopic waves of any kind to exist) would be waves starting at spatial infinity in phase. They are unphysical.

And yes, the reason is entropic in nature. We need to solve the arrow-of-time problem before we can answer that question.

In the famous Wheeler-Feynman problem of electrodynamics as a half-advanced, half-retarded field, the boundary conditions were totally ad hoc, as the authors were keenly aware of, that there was a perfect EM absorber at spatial infinity.

It is my intuition that it would cause serious problems with causality too (what would have caused that identical physical condition infinitely far apart in the first place?), but that a story for another day.

All of this IIRC.

12 hours ago, Duda Jarek said:

As the main source of gravitational wave events is just orbiting of e.g. two black holes, and evolving toward minus time orbiting remains orbiting, so using Euler-Lagrange toward minus time (t -> -t), or the least action principle, there should be generated similar waves - for us being advanced of similar chirp shape as retarded. LIGO just measures lengths - invariant to time symmetry, so should see both retarded and advanced waves.

It should be noted that for the retarded solution, there is a concomitant decrease in the orbit when gravitational radiation is emitted. For the advanced solution, there is a concomitant increase in the orbit when gravitational radiation is absorbed. The difference between these two is that gravitational energy is naturally lost from the orbit, which is radiated away as gravitational radiation, whereas in the reverse, gravitational energy has to be supplied to the orbit by an input of gravitational radiation, which although possible, is unlikely to occur naturally.

On 12/7/2025 at 3:05 PM, Duda Jarek said:

  On 12/7/2025 at 6:12 AM, KJW said:

Although the equations are symmetric, that "running movie backward" is a solution that is not symmetric.

Sure, e.g. throwing a rock to a lake symmetric in equations, there appear asymmetries of solutions ... for physics we know many such asymmetries , like entropy gradient, emission asymmetry (e.g. circulating electron losing energy), tendency for black hole formation from direction of Big Bang.

But solving physics by the least action(GR)/Feynman ensemble(QFT) using e.g. boundary conditions in Big Bang and Big Crunch, they seem very similar just hot soups - symmetric ways of solving from symmetric boundary conditions, shouldn't the solution be also symmetric?

If the equations are time symmetric, this does not imply that the solutions are also time symmetric, though it does imply that a time reversal transformation of a solution is also a solution. But it should be noted that there is no special reason for such a solution to lack time symmetry. To attribute the lack of time symmetry of a solution of a time symmetric equation to something like the second laws of thermodynamics is wrong and a misunderstanding of the nature of symmetry in mathematics and the physics based on such mathematics. For example, consider the second-order ordinary differential equation:

[math]\dfrac{d^2 x}{d t^2} - k^2 x = 0[/math]

This is invariant under time reversal [math]t \Rightarrow -t[/math]. The general solution is:

[math]x = C_1\,e^{kt} + C_2\,e^{-kt}[/math]

where [math]C_1[/math] and [math]C_2[/math] are arbitrary constants of integration. For [math]C_1 = C_2[/math], the solution is symmetric about [math]t = 0[/math]. For [math]C_1 C_2 \gt 0[/math], but [math]C_1 \neq C_2[/math], the solution will be symmetric about some [math]t \neq 0[/math]. However, for [math]C_1 C_2 \lt 0[/math], the solution is unsymmetric about any [math]t[/math]. In this case:

[math]x = C_2\,e^{kt} + C_1\,e^{-kt}[/math]

is the time reversed solution about [math]t = 0[/math]. So, although this symmetric differential equation does have symmetric solutions, it also admits unsymmetric solutions. As the equations are purely mathematical, the reason for the existence of unsymmetric solutions is also purely mathematical.

Now consider the reversible reaction mentioned in an earlier post:

[math]\text{X} \rightleftharpoons \text{Y}[/math]

The rate equation for this reaction is the pair of first-order ordinary differential equations:

[math]\dfrac{d\text{[X]}_t}{dt} = -k\,(\text{[X]}_t - \text{[Y]}_t)\\\dfrac{d\text{[Y]}_t}{dt} = -k\,(\text{[Y]}_t - \text{[X]}_t)[/math]

where [math]\text{[X]}_t[/math] and [math]\text{[Y]}_t[/math] are the concentration (or number) of [math]\text{X}[/math] and [math]\text{Y}[/math] at time [math]t[/math], and [math]k[/math] is the rate constant. Under the time reversal [math]t \Rightarrow -t[/math], [math]\dfrac{d\text{[X]}_t}{dt} \Rightarrow -\dfrac{d\text{[X]}_t}{dt}[/math], [math]\dfrac{d\text{[Y]}_t}{dt} \Rightarrow -\dfrac{d\text{[Y]}_t}{dt}[/math], and therefore [math]k \Rightarrow -k[/math].

Thus, the rate equation is covariant rather than invariant. It is covariance that is required of the laws of physics, a weaker notion than invariance. However, the equation itself does remain unchanged under time reversal. Also, the equations are symmetric with respect to the interchange of [math]\text{X}[/math] and [math]\text{Y}[/math]. This rather than time reversibility is what the notion of reversibility of the reaction is referring to. The solution of the equation:

[math]\dfrac{d}{dt}(\text{[X]}_t + \text{[Y]}_t) = 0\\\text{[X]}_t + \text{[Y]}_t = \text{[X]}_0 + \text{[Y]}_0[/math]

[math]\dfrac{d}{dt}(\text{[X]}_t - \text{[Y]}_t) = -2k\,(\text{[X]}_t - \text{[Y]}_t)\\\text{[X]}_t - \text{[Y]}_t = (\text{[X]}_0 - \text{[Y]}_0)\,e^{-2kt}[/math]

[math]\text{[X]}_t = \text{[X]}_0\,\left(\dfrac{1}{2} + \dfrac{1}{2}\,e^{-2kt}\right) + \text{[Y]}_0\,\left(\dfrac{1}{2} - \dfrac{1}{2}\,e^{-2kt}\right)\\\text{[Y]}_t = \text{[X]}_0\,\left(\dfrac{1}{2} - \dfrac{1}{2}\,e^{-2kt}\right) + \text{[Y]}_0\,\left(\dfrac{1}{2} + \dfrac{1}{2}\,e^{-2kt}\right)[/math]

For [math]k \gt 0[/math], the usually forward time direction value, both [math]\text{[X]}_t[/math] and [math]\text{[Y]}_t[/math] approach [math]\dfrac{1}{2} (\text{[X]}_0 + \text{[Y]}_0)[/math] as [math]t \to \infty[/math]. By contrast, for [math]k \lt 0[/math], the reverse time direction value, provided [math]\text{[X]}_0 \neq \text{[Y]}_0[/math], both [math]\text{[X]}_{t'}[/math] and [math]\text{[Y]}_{t'}[/math] move away from [math]\dfrac{1}{2} (\text{[X]}_0 + \text{[Y]}_0)[/math] as [math]t' \to \infty[/math] (NB, [math]t'[/math] is reverse time and therefore [math]t' \gt 0[/math] even though [math]t \lt 0[/math]).

The difference between the first equation and the second set of equations is that the first equation is a second-order equation without any first-order derivatives, whereas the second set of equations are first-order equations. Unlike the second-order derivative, the first-order derivatives change sign under time reversal.

Considering the rate equation from a physical perspective, the rate constant is a transition probability per time (with dimension [math]\text{T}^{-1}[/math]}. Thus, it is naturally a positive number. Its change in sign under time reversal represents a time asymmetry in the laws of physics. Specifically, the notion of causality is time asymmetric with the effect in the future of the cause. This would appear to statistically eliminate time reversed solutions although microscopic reversibility still applies.

Edited December 16Dec 16 by KJW

6 hours ago, KJW said:

Stimulated emission prevents one from inverting a population by simple absorption. It is very much a thermodynamics thing.

Sure there is also statistics there, but stimulated - especially in superradiance, also in laser ... or wave behind marine propeller pushing or pulling energy from resonator is quite deterministic.

For example white hole would emit, causing excitation in sensor of telescope. Applying T/CPT symmetry to this scenario, shouldn't black hole cause deexcitation of telescope sensor if prepared as excited?

6 hours ago, joigus said:

And yes, the reason is entropic in nature. We need to solve the arrow-of-time problem before we can answer that question.

GR is solved by the least action principle - treating spacetime as 4D membrane minimizing tension as action - based on boundary conditions in both time directions.

If there are e.g. orbiting supermassive black holes there, shouldn't distortions they create propagate in both directions of this 4D membrane? (for https://theconversation.com/to-map-the-vibration-of-the-universe-astronomers-built-a-detector-the-size-of-the-galaxy-244157 )

Solving GR by least action, QFT by Feynman ensembles is CPT symmetric, requires Einstein's block universe philosophy of time - that we travel through already found 4D solution ... e.g. in S-matrix: <psi_f | U | psi_i> with one amplitude coming from the past, second from the future, we multiply them e.g. in Born rule - allowing for very nonintuitive Bell violation.

20 hours ago, Duda Jarek said:

LIGO just measures lengths - invariant to time symmetry, so should see both retarded and advanced waves.

Yes, as I said a few posts ago, LIGO would in principle see both types of waves. However, you need to remember that gravitational waves often aren’t the full story - some events such as BH/neutron-star mergers, or mergers of SMBH with accretion discs, will also produce an EM signature. If gravitational waves were advanced, we would see them passing LIGO long before the actual source event happens (as seen in other channels). Needless to say, so far at least nothing of the sort has ever been observed.

Sure it seems highly suspicious that, among ~300 GW events, there was observed only single EM counterpart - including advanced waves into considerations could help with.

The current EM counterparts are retarded - being certain their non-existence when required, should indicate it was advanced GW (?)

To have a chance to observe advanced EM counterparts, we would need telescopes with excited sensor - currently avoided by cooling.

Edited December 17Dec 17 by Duda Jarek

8 hours ago, Duda Jarek said:

  14 hours ago, KJW said:

Stimulated emission prevents one from inverting a population by simple absorption. It is very much a thermodynamics thing.

Sure there is also statistics there, but stimulated - especially in superradiance, also in laser ... or wave behind marine propeller pushing or pulling energy from resonator is quite deterministic.

For example white hole would emit, causing excitation in sensor of telescope. Applying T/CPT symmetry to this scenario, shouldn't black hole cause deexcitation of telescope sensor if prepared as excited?

Just like absorption, stimulated emission is not deterministic. In the case of absorption, a photon has a choice between two options: 1, excite the molecule by being absorbed; or 2, do nothing. Each option has a probability. In the case of stimulated emission, a photon also has a choice between two options: 1, de-excite the molecule by stimulating the emission of a photon; or 2, do nothing. Each of these options also has a probability. In fact, the probabilities for stimulated emission are identical to the probabilities for absorption. Thus, when the number of excited molecules equals the number of unexcited molecules, there is no longer any net absorption or net stimulated emission.

If [math]\text{X}[/math] and [math]\text{Y}[/math] from my previous post describe unexcited and excited states of a molecule, then the rate equation in my previous post describes the rate equation for absorption and stimulated emission, in which case the rate constant also depends on the illumination. It may come as a surprise that it is stimulated emission rather than spontaneous emission that is the reverse of absorption. However, the dependence of the rate constant on the illumination for both stimulated emission and absorption manifests the symmetry of the equations describing these two processes, unlike spontaneous emission.

Thus, both stimulated emission and absorption depend on the notion of causality and are therefore time asymmetric.

Edited December 17Dec 17 by KJW

For both marine propeller and shooting e.g. free electron laser, indeed we don't control the details of individual particles, only their statistical behavior ... but still allowing to cause excitation of the target - by retarded EM wave.

Applying T/CPT symmetry: reversing rotation of marine propeller, or reversing electron trajectory of free electron laser/synchrotron, the causation should reverse to causing deexcitation of target - by advanced EM wave ... with requirement that this target was initially excited, what is not true e.g. for most radio telescopes - preventing them from observation of advanced waves.

Applying time symmetry to synchrotron:

Moderator Note

We have clearly plunged into a previous closed topic, covering the same ground and having the OP blatantly disregard explanations again.

Ignoring explanations is not discussion, not does it show any effort to understand. It’s not an exchange made in good faith.