| ▲ | How do you take a picture of a black hole?(quantamagazine.org) |
| 69 points by dverlaeckt80 11 hours ago | 58 comments |
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| ▲ | myrmidon 10 hours ago | parent | next [-] |
| What I find really interesting is that black holes don't really "require" infinite density; the mass density inside the event horizon never exceeds neutron star levels, and becomes in fact lower the heavier the black hole gets (=> event horizon enclosed volume grows faster than mass). |
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| ▲ | kakacik 37 minutes ago | parent | next [-] | | Some of.the biggest ones woild float on water... which tells you its not very useful to compare density based on Schwarzschild radius | |
| ▲ | Borg3 9 hours ago | parent | prev [-] | | For large blackholes of course. And this actually makes other interesting property. Our universe is in fact giant blackhole with our own event horizon outside it. This also support cyclic big bang theory.
Once entire universe will collapse to singularity, the smallest possible particles shatters, leaving pure energy. Pure energy doesnt have "static" mass so no gravity => kaboom, we have new iteration. Theory of multiverse (multiple black holes aka universes) is out of scope here because we will never be able to escape ours. | | |
| ▲ | raattgift 6 hours ago | parent | next [-] | | The observables in our sky do not match any interior metric for a black hole. In particular generically black hole metrics concentrate a central mass, and (from a Raychaudhuri equation perspective) the known sources of shears and pressures that work against this concentration are insufficiently strong to block ever greater collapse. Additionally, shears and pressures enter into the stress-energy tensor. That's why the centre of a BH can be a singularity. Observations of distant galaxies and the cosmic microwave background do match the (expanding vacuum) Lemaître-Robertson-Walker metric with the distribution of matter-density described by the Friedmann equations. Distant galaxies fill the sky and are receding from us, from our perspective; they are not concentrating in any sort of middle. There is insufficient evidence to conclude that a big crunch is in our future. "Pure energy" is represented in the stress-energy tensor just fine. General relativity's Einstein Field Equations exactly relates the stress-energy tensor to the curvature of spacetime. It gravitates, your argument just falls down. <https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor#/...> | | |
| ▲ | itsamario 26 minutes ago | parent | next [-] | | Science mapped out our solar system and developed base12 to account for rotation and orbit. What I haven't found is an accounting of our solar system's patterns in our galaxy. We know about axial precession but we don't know the official cause and some suspect it's related to a pull from another galaxy or a swayish wobble from our solar system's rotation in the galaxy. Same for a supposed magnetic flip in the earths core. They might be related but we seemingly only know of the precession measured as 1 degree per 72 years but the magnetic north pole doesn't align with any noted patterns nor cause. | |
| ▲ | thunderbong 6 hours ago | parent | prev [-] | | Thank you for your comment. I've heard the parent argument earlier as well. I wasn't aware of these points. |
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| ▲ | adrianN 8 hours ago | parent | prev | next [-] | | It is my understanding that energy produces a gravitational field regardless of the form it takes. | | |
| ▲ | Maxatar 3 hours ago | parent [-] | | That's correct, in fact almost nothing Borg3 has said is remotely accurate. One cool thought expressed by Penrose is the possibility that all matter will eventually decay or annihilate and all that's left will be massless radiation such as photons. If this happens then the concept of space and time itself will cease since from within the reference frame of a photon space and time are undefined. In this scenario the distinction between an infinitely expanded, cold, and dilute universe and an extremely small, hot, and dense universe becomes undifferentiable. Without any physical scale by which to distinguish the two, the far distant future of our universe where all matter has disappeared can be equivalently viewed as the big bang of a new universe. This is what Penrose calls conformal cyclic cosmology [1]. There was a time when physicists debated whether the universe would continue expanding forever or whether gravity would eventually halt its expansion and cause it to collapse back in on itself in a Big Crunch, potentially giving rise to another Big Bang. Conformal cyclic cosmology allows for an endless repeating cycle of universes even in the case where the universe expands forever, by identifying the infinitely distant future of one universe with the Big Bang of the next. This is highly speculative and almost certainly false, but it's at least grounded in actual physics and is kind of poetic. [1] https://en.wikipedia.org/wiki/Conformal_cyclic_cosmology | | |
| ▲ | sandworm101 2 hours ago | parent [-] | | >> all matter will eventually decay or annihilate and all that's left will be massless radiation such as photons. If this happens then the concept of space and time itself will cease since from within the reference frame of a photon space and time are undefined. So all matter is pushed out (dark energy) eventually reaching its own event horizon in the dimension we call time, then continues as photons (Hawking radiation). And eventually after all the matter has done so, everything collapses and this universe ends? Sounds like we are in an aging BH to me. |
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| ▲ | nkrisc 8 hours ago | parent | prev [-] | | What is “pure” energy? | | |
| ▲ | Borg3 7 hours ago | parent | next [-] | | Photons? They cannot be stopped, because they have no "static" mass?
As far as they move, they react to gravity, not sure if they can provide its own gravity field. Might be a case as far as they move. Not expert here :) | | |
| ▲ | raattgift 6 hours ago | parent | next [-] | | Photons have momentum. They can drive solar sails. They can kick electrons out of atomic orbitals. Lots of them streaming outwards hold up the outer layers of stars, keeping them from collapsing. Radiation pressure. Momentum-density is represented in the stress-energy tensor, and so is directly related to spacetime curvature. Light feels and generates gravity. A sibling comment points out the hypothetical kugelblitz as an extreme example. In cosmology textbooks you can read about the "radiation-dominated era" of the first roughly 50000 years of the universe where the energy density working strongest against the expansion was dense photons (and neutrinos). | | |
| ▲ | jjtheblunt 3 minutes ago | parent [-] | | I had a radiometer when i was a kid ... is that momentum what makes it spin, because i thought the light was heating the black surfaces more than the white surfaces, so the air molecules bounced more vigorously away from the black. i didn't think about photons having momentum when i was 8, though. https://a.co/d/00nIWNpB |
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| ▲ | pavel_lishin 6 hours ago | parent | prev [-] | | Apparently, they can: https://en.wikipedia.org/wiki/Kugelblitz_(astrophysics) |
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| ▲ | ButlerianJihad 7 hours ago | parent | prev [-] | | It is the subtitle of a #3 synth-pop hit by Information Society, which debuted only 16 months before the release of the unrelated Belgian techno anthem, "Pump Up the Jam". |
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| ▲ | ck2 35 minutes ago | parent | prev | next [-] |
| there's a serious group of astrophysicists that actually disagree the photo was assembled properly at all long paper made short I think their carefully demonstrated argument is it's a machine-learning hallucination to see what we want Dr. Becky has excellent coverage of it, will link it in a moment when I find it here it is, well worth watching: * https://www.youtube.com/watch?v=9U6bvR6SzMo Miyoshi, Kito & Makino (2024; claim the Sag A* image is wrong) * https://academic.oup.com/mnras/article/534/4/3237/7660988 |
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| ▲ | rustyhancock 10 hours ago | parent | prev | next [-] |
| One of the most remarkable aspects of physics is that Newtonian gravity calculations are quite spot on for all the parts of black holes layfolk consider important. 1. Both predict the same critical radius, now called the Schwarzschild radius. 2. Both predict a central singularity with formally infinite density in idealised collapse. 3. Both predict exactly the same finite time to fall into the singularity under equivalent idealised conditions. 4. Both give the same equation for radial free fall, using different definitions of time. What GR added was meaning to these things (and of course precise calculations outside of very idealised scenarios and symmetric conditions). |
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| ▲ | Maxatar 4 hours ago | parent | next [-] | | One significant difference in interpretation that layfolks do consider important is that it's possible to exit from within the event horizon of a black hole in Newtonian gravity (without exceeding the speed of light) whereas in general relativity that is not possible. In relativity the event horizon really is a point of no return, whereas in Newtonian physics it's the point at which escape velocity is equal to the speed of light. The escape velocity on the surface of the Earth is 11 km/s, but that doesn't mean that I as a human can't escape the surface of the Earth momentarily just by jumping and I certainly don't jump 11 km/s or even 1 km/s for that matter. Similarly in Newtonian gravity it would be possible for an object to enter the event horizon of a black hole, and then "jump" back out at a much lower speed than the speed of light and then get sucked back beyond the event horizon, and you can in principle do this over and over again so long as you had some sufficiently powerful source of thrust. In general relativity this is in principle not possible. | |
| ▲ | opus55_lover 7 hours ago | parent | prev | next [-] | | in what sense. the black hole is a result of the interaction of light with the curavture of space time. this is not modelled by newtonian gravity (or physics for that matter) at all. light is assumed to not interact with gravity in that framework in any way. | | |
| ▲ | Maxatar 3 hours ago | parent [-] | | >light is assumed to not interact with gravity in that framework in any way. Light absolutely interacts with gravity in Newtonian mechanics. One of the neat features of Newtonian gravity is that gravitational charge (which is mass) equals inertia so the two cancel each other out exactly in the equation of motion. No other force has this property. This means gravitational acceleration is independent of inertia, even in the case where inertia/mass approaches zero. |
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| ▲ | shiandow 7 hours ago | parent | prev | next [-] | | Could you provide a bit more detail? I don't quite see how one would define the event horizon for instance. | | |
| ▲ | post-it 7 hours ago | parent [-] | | The radius at which the escape velocity is the speed of light. |
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| ▲ | sandworm101 7 hours ago | parent | prev | next [-] | | >> Newtonian gravity calculations are quite spot on Which is why the concept existed long long before any talk of relativity or time dilation, back before anyone knew the name Einstein. (He was like 4yo when this came out) From American Physical Society:
"the true “father” of the black hole concept was a humble 18th century English rector named John Michell–a man so far ahead of his scientific contemporaries that his ideas languished in obscurity, until they were re-invented more than a century later." https://www.aps.org/apsnews/2009/11/1783-john-michell-black-... | |
| ▲ | ant6n 9 hours ago | parent | prev | next [-] | | Newtonian gravity gives infinite time to fall into black hole? I thought you need at least special relativity for time dilution. | | |
| ▲ | rustyhancock 9 hours ago | parent [-] | | Both have finite time falling in to the singularity. If you use a point test particle Newton and GR agree on that finite time. That's proper time I.e. the infalling particles wristwatch time. In GR, external observers would never see the infalling particle cross the event horizon but that's a coordinate artifact. If using our wrist watch time for that particle. | | |
| ▲ | gradus_ad 8 hours ago | parent | next [-] | | But observers can see a black hole grow over time, so does that mean they observe growth without consumption? | | |
| ▲ | rustyhancock 8 hours ago | parent | next [-] | | Were past what I could confidently answer (if I'm wrong hopefully someone corrects me) but I'm not sure I follow, if the BH grows it would encompass what we see as nearly smeared over it's horizon. There's also surprising results around what the external gravitation field appears to be for an object. I.e. no matter how mass is distributed behind the event horizon externally it looks like the spherically symmetric Schwartzchild metric. That has implications for what will emit gravitational waves. Eithway, smeared over the horizon but never crossed and crossed the horizon externally look the same from a gravity perspective. | | |
| ▲ | raattgift 7 hours ago | parent [-] | | The quanta article at the top is about BH shadows. The black hole shadow for Schwarzschild and for Kerr are very different. The latter, which we expect for spinning black holes, is asymmetric and distorted depending on the direction to the observer (maximal for an observer in the equatorial plane, minimal for an observer above a pole). Most astrophysical black holes have a lot of spin, mostly because their progenitors (stars, and smaller black holes formed by stellar collapse) have lots of spin. Infalling gas clouds tend to inspiral rather than fall straight in radially on one of the poles, and that also increases the BH's spin as it grows. So, no, the shadows of most astrophysical black holes will not look like Schwarzschild. You need a quadrupole for gravitational radiation. Blobs of gas and distant starlight generate a negligible quadrupole when they are near an isolated black hole. You get significant gravitational radiation when there's a quadrupole, which we can think of as a barbell: two heavy masses connected by an imaginary bar connecting them as they rotate around the barbell's centre of mass. (For extreme mass-ratios the centre of mass is inside one of the BHs, but for BHs of similar masses, it's near the middle of the "bar"). The gravitational radiation is because, if we think of the metric as reflecting a gravitational potential, that changes when the two items are eclipsing versus when they are spatially separated: you are attracted more strongly to the lined up configuration and will tend to fall more towards one or the other of the weights when they are not. Schwarzchild BHs are not barbell-like, and so don't emit gravitational waves. There is so much noise around real black holes (even ones with a vanishingly small accretion structure; there's lots of background objects in our sky!) that there is little hope of directly observing a horizon-crossing, even if the gravitational redshift didn't lead to dimming etc. However, we have seen bright flares that are good evidence for parts of things (stars, for example) falling in. Infallers are not smeared across the horizon. They are if anything asymptotically "frozen" in the distant observer's view. One could imagine a distant observer seeing a pile-up of objects, but the later ones will simply occlude the earlier infallers. I dunno where you get smearing from. Complementarity or holography? That's about where (for a particular mathematically-related pair of equally valid descriptions) information is encoded for outside observers to reconstruct the BH's history, and not about a physical infaller's apparent location. (An infaller doesn't experience any of this itself, and would tend to prefer a local description of its fate, but not for long). |
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| ▲ | raattgift 7 hours ago | parent | prev [-] | | Merging astrophysical ("real world") black holes are quite different from the idealization in the Schwarzschild setting. You need a gravitational quadrupole to emit gravitational radiation; in essence you need a "barbell" with an infinitesimally small "bar" connecting two large weights. (The weights can differ; the bar's length can change). Schwarzschild is characterized by a central mass, and so can emit no gravitational radiation. (One can do perturbative methods on Schwarzschild to introduce gravitatonal radiation; and in reverse the effective one-body formalism basically (over-simplified) turns a binary black hole with a large mass ratio into a test particle in a Schwarzschild background). Additionally, there is the Vaiyda null dust absorbing metric, where "null dust" is very much like light. In this metric, the central mass grows, and the apparent horizon (where outgoing null rays turn inwards) is r = 2m(v), where v is the advanced time. That locally-defined dynamical surface moves outwards. There is also an event horizon in Vaiyda, but it is neither locally detectable nor coincident with the apparent horizon (that global null surface lies well outside during accretion). That's the closest starting point to an answer for an isolated non-rotating black hole that grows by accretion; one can perturb Vaiyda with something more realistic than a constant "rain" of noninteracting massless (but not momentumless) particles. That's just for monopole growth though. Like with Schwarzschild, Vaiyda is non-radiating (and in V's case this is by construction). An astrophysical isolated single black hole will see infalling material that has angular momentum; anisotropic stresses (beyond the null direction) (radiation from matter-matter interactions); a gravitational quadrupole (emitting gravitational radiation); and for big lumps of infalling material, some self-gravity. In these cases, the black hole "shadow", the topic of the article at the top, reveals the details of the black hole, including growth. The shadow is fundamentally a gravitational lensing effect, and depends on the BH's mass, angular momentum, and charges. For a growing Vaidya black hole, the shadow is not static. The distant observer will see the shadow evolve in various ways. This is because it's a dark region whose boundary tracks the changing spacetime geometry. In the pure Schwarzschild case, the shadow is static, eternal, empty, and set by the boundary of the photon sphere at r = 3M rather (and rather bigger) than the event horizon at r = 2M. For an astrophysical black hole, the observer's view of an infaller will be largely washed away by various sources of noise and the gravitational redshift of the infaller. However projects like the GRAVITY collaboration (at ESO's Very Large Telescope) look for infrared flares around Sgr A*, and have spotted really interesting things like the "G2" gas cloud. This is a busy area of current research. TDECat is a catalogue of more than a hundred tidal disruption events, mostly from X-ray flares and delayed infrared emissions from various space instruments. A lot of this is stuff actually falling into the black hole, but "covered" in a huge burst of high-energy radiation from real matter colliding together faster and faster as it inspirals. Finally, we've seen the gravitational waves emitted by merging black holes of various mass-ratios. The result is a single bigger black hole and an enormous amount of gravitational radiation. The behaviour of matter near the merger and between the merger and us can corroborate that result. So we can see growth in that case, as well as where we can image a BH's shadow. These BHs are not very close to us. So clearly the idea that nobody can watch growth or consumption is wrong, although one might get nitpicky about which of these observables are indirect, but that's like saying that because Miss Marple sees only a bled-out corpse with lots of stab wounds there is no evidence that a murder has been committed. |
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| ▲ | raattgift 8 hours ago | parent | prev [-] | | > external observers would never Consider two strongly-interacting freefallers related by a boost colliding and sticking together (like a black tshirt and a floof of white pet fur) close to the apparent horizon. Consider a freefaller and an accelerated observer meeting close to the apparent horizon. The freefaller here could be distant starlight and the accelerated observer a powerful rocket. Are you saying these coincidences are forbidden in the vicinity of an apparent horizon (which is locally detectable), or are you distinguishing between the apparent horizon and the (global, but not locally detectable) event horizon? Where is the point where the t-shirt will remain unsullied by unremovable fur?
What light does an outwards-pointing camera on the spaceship see? Visser [2014] "Physical observability of horizons", https://link.aps.org/doi/10.1103/PhysRevD.90.127502 aka https://arxiv.org/abs/1407.7295 about the variety of horizons about which one should be careful. The equivalence principle demands that the local free-fall frames on the t-shirt and the fur are both smooth. At an enormous black hole, the curvature is also small (we can make the black hole so large that the curvature and tidal effects are comparable to the surface of the Earth). The result is the "no drama" conjecture, wherein nothing special happens immediately upon crossing the event horizon. Detecting the apparent horizon requires some cleverness to shear apart a cloud of coffee grinds or the like while the components are variously on hyperbolic trajectories (and will escape the region around black hole) and infalling trajectories (which will be stuck below a trapping surface). The shirt-and-fur will form a single extended object and continue to infall togther. A violation of this would be a violation of Local Lorentz Invariance, which would be beyond known physics, all of which enjoys LLI (and global Lorentz Invariance in the idealized flat spacetime of Special Relativity). This gets trickier just inside the horizon because of of causal disintegration: electromagnetic interactions from the inner part of an extended object can no longer reach up to the outer part of the extended object. For a time the outer part of the extended object can "catch up", but eventually intermolecular and other forces will fail, such that the tshirt will not experience being furry. Local physics is local. The tshirt-fur meeting above the apparent horizon may be too redshifted to be visible by various classes of observer above. The electromagnetic radiation from them meeting below the apparent horizon is stuck behind a trapping surface; observers above will not see a furry tshirt. Generically an accelerated freefaller shooting itself through a horizon will see freefallers as being fast (clocks running fast, high velocities). The effect increases as the horizon is approached, so a camera looking back at distant starlight sees a high blueshift and a focusing into a small region of the accelerated infaller's "sky". One can analyse this in several ways, but the acceleration/gravitation equivalence means the rocketing infaller is not much different from an observer standing still on a planet freely falling into the BH, or alternatively one can consider its suprageodesic in a convenient system of coordinates in a setting like Vaiyda's absorbing null dust. The inwards-accelerating rocket camera sees a blueshift from its own acceleration (because that is equivalent to gravitational blueshifting) and from the falling of the distant starlignt down to the black hole, which will be blueshifted in the backwards pointing camera. Finally on the inwards-accelerating rocket, we have Unruh/Rindler in play locally (and local is fairly large for an enormous black hole); the Rindler horizon is ahead of them, so the backwards view is not behind the Rindler horizon. They can still receive light from outside. You don't want to collide with anything on an intersecting geodesic if you are this type of infaller, and eventually that includes highly-blueshifted starlight. Sizzle. |
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| ▲ | 6 hours ago | parent | prev | next [-] | | [deleted] | |
| ▲ | ziiinq 10 hours ago | parent | prev [-] | | [dead] |
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| ▲ | JohnKemeny 10 hours ago | parent | prev | next [-] |
| Is it just me, or could they have added a, say, picture of a black hole? |
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| ▲ | cogogo 9 hours ago | parent | next [-] | | I thought I was the only one. But there is a really good video linked at the bottom of the page that has the two images and a summary of the work to build them. I also loved the analogies of reading the date of a coin in LA from NYC or seeing a tangerine on the moon. But they kept talking about how surprisingly small black holes are. I wanted that described in relative terms. Maybe I just did not read carefully enough but are they the size of a pinhead? The solar system? I know I can look this up. | | |
| ▲ | elvircrn 7 hours ago | parent [-] | | "bottom of the page that has the two images and a summary of the work to build them." A picture + explanation below it would carry more weight than this entire article tenfold. |
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| ▲ | bcraven 10 hours ago | parent | prev | next [-] | | It's the one in your head that you picture, as not many have been taken: https://www.si.edu/newsdesk/releases/astronomers-reveal-firs... | | |
| ▲ | somat 9 hours ago | parent | next [-] | | The EHT is a remarkable achievement in datagathering and collaboration. But I always sort of wonder, a bit of doubt if you will. seeing as the process(simplified nearly to the point of absurdity) is that of building a image generator based off what we think a black hole looks like then feeding it very vague spectral data and getting it to generate an image. isn't that a huge confirmation bias. That is. of course the black hole looks like simulated ones, that is how the generator was programmed to make them. | | | |
| ▲ | GreenLightGo 9 hours ago | parent | prev | next [-] | | Why does it look like it's moving to me? | | | |
| ▲ | lifestyleguru 9 hours ago | parent | prev [-] | | I remember times when the best available photo of Pluto looked like this. |
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| ▲ | 10 hours ago | parent | prev | next [-] | | [deleted] | |
| ▲ | ButlerianJihad 9 hours ago | parent | prev [-] | | This is a transcript of a 60-minute podcast. Would you like that TIFF in Zmodem format? Or they could read it out in base64? |
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| ▲ | tilt_error 6 hours ago | parent | prev | next [-] |
| Well, for starters you need a reasonably long exposure time. |
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| ▲ | spottedmarley 9 hours ago | parent | prev | next [-] |
| First you'll need a camera that can detect mathematical equations. |
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| ▲ | metalman 6 hours ago | parent | prev | next [-] |
| presumably the only way to get a good picture of a blackhole would be to get very lucky and catch a supernova going off next door, preferably at a slight oblique angle from one of the poles and be able to light the acretion disk and any dust
but I dont think we have the visible light gathering capacity to search at the distances required for even a slight chance of capturing the event, so the answer to the question is : you wait. |
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| ▲ | fuzzfactor 8 hours ago | parent | prev | next [-] |
| Nothing beats a close-up. |
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| ▲ | ahmedk29 an hour ago | parent | prev | next [-] |
| [flagged] |
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| ▲ | ethfan 7 hours ago | parent | prev | next [-] |
| [flagged] |
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| ▲ | jon128 7 hours ago | parent | prev | next [-] |
| They got an image of something, yeah. Or, did they? "So we had been perfecting these imaging algorithms on synthetic data. We’d fed these algorithms made-up data. So we would take a black hole image that we might be looking at. We’d corrupt it with noise, and we’d send it through these algorithms, and we’d refine these to the point where we were pretty confident in these algorithms."
Algorithms? For what? So they're not just taking a photo of something observed through a telescope.They're not really explaining this either. I'm trying to keep an open mind but I'm still not buying the whole black hole thing. There needs to be something more intellectually compelling than "we know it's a black hole because we know it's a black hole". Yes, you read that right. I am doubting this religion. "That was a moment where there was a lot of crying and screaming."
another odd quote from this article, provided one is assuming this is supposed to be around the scientific method. |
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| ▲ | atombender an hour ago | parent [-] | | It's not taking a photo. It's computation, based on a lot of real data, with a lot of inference and synthetic data. The idea is that it's broadly accurate but cannot capture fine detail. The approach has been calibrated against known objects like NRAO 530, a quasar near Sgr A* that has some known properties, and it's been used on both Sgr A* and M87. And the team has admitted there's a lot of uncertainty about exactly how accurate it is. Here [1] is a pretty good article about how it works. There's overwhelming scientific consensus that Sgr A* and M87* are black holes. It's not really provable at this point, but I don't think anyone doubts it. [1] https://www.discovermagazine.com/the-event-horizon-telescope... |
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| ▲ | monster_truck 10 hours ago | parent | prev | next [-] |
| I'd use a camera, personally |
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| ▲ | lopsotronic 6 hours ago | parent | prev [-] |
| Easy. Space isn't empty. Due to the insane distortion of "space"[1], every spare molecule in the vicinity is going to be lighting up at the speed of $E = mc^2$ . Which tends to be real visible. Black holes power some of the brightest things in the universe - light so powerful that it sometimes has its own gravity. A tame black hole is possibly the least fussy high-efficiency furnace imagineable. Of course, "tame black hole" possibly . . . isn't. [1] Whatever that happens to be, either bouncy rubber sheet or emergent property from deeper Hilbert Spaces or something beyond our imaginations. Or something that's just an illusion, an artifact of our specific perspective from our little spot in existence, like a rainbow. |
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| ▲ | encrypted_bird an hour ago | parent [-] | | Genuine question: why did you out $ signs next to the formula? I normally only see those put next to variable names. | | |
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