May 15, 2011 8:44 AM Subscribe

Sean Carroll: Distant time and the hint of a multiverse
posted by kliuless (52 comments total)
38 users marked this as a favorite

Also, what does TED have against transcripts? Listening to a talk is about the slowest way possible to learn something new.

Thinking > Reading > Listening

not that listening is necessarily a bad thing, it's just a slow way to learn stuff

posted by leotrotsky at 9:39 AM on May 15, 2011 [11 favorites]

Thinking > Reading > Listening

not that listening is necessarily a bad thing, it's just a slow way to learn stuff

posted by leotrotsky at 9:39 AM on May 15, 2011 [11 favorites]

Er, it's entirely possible I'm missing something, leotrotsky, but that gray box to the right has a link in it that says 'Open Interactive Transcript' that lets you read the entire lecture, and links each sentence to that point in the video. It's kind of rad, actually.

posted by carsonb at 10:11 AM on May 15, 2011 [12 favorites]

posted by carsonb at 10:11 AM on May 15, 2011 [12 favorites]

Oh my GOD, interactive transcript - I never noticed that before! That is unspeakably cool.

posted by fleetmouse at 10:15 AM on May 15, 2011

posted by fleetmouse at 10:15 AM on May 15, 2011

leotrotsky - No joke. I feel like TED is such a tease. Even if they just had a transcript with smaller clips embedded, that would be better for me. I don't have the time to do the whole thing, though always rationalize reading an article isn't me ditching work.

posted by scunning at 10:56 AM on May 15, 2011

posted by scunning at 10:56 AM on May 15, 2011

So this Multiverse Chicken that lays little universe eggs ... is it going to lay any more? Or has it laid all the universes it's going to, and everything will run down more or less together?

And since I totally love this stuff without understanding it in the least, what's the deal with dark energy? How can empty space contain energy, and why doesn't it get diluted as our universe expands?

posted by Quietgal at 11:01 AM on May 15, 2011

And since I totally love this stuff without understanding it in the least, what's the deal with dark energy? How can empty space contain energy, and why doesn't it get diluted as our universe expands?

posted by Quietgal at 11:01 AM on May 15, 2011

And what happens when the ultra-macro-multiverse-farmer decides to make an omelet?

posted by sammyo at 11:15 AM on May 15, 2011

posted by sammyo at 11:15 AM on May 15, 2011

leotrotsky: "*Thinking > Reading > Listening*"

Not for everyone.

posted by BuddhaInABucket at 11:18 AM on May 15, 2011

Not for everyone.

posted by BuddhaInABucket at 11:18 AM on May 15, 2011

I'm with BiddhaInBucket, to a certain extent. Good writing can be a better medium for ideas than spoken word, but a transcript of something originally intended to be spoken is almost necessarily NOT as good as hearing it presented out loud.

Which is why I love TED.

posted by es_de_bah at 11:27 AM on May 15, 2011

Which is why I love TED.

posted by es_de_bah at 11:27 AM on May 15, 2011

The short answer is that right now we don't have a good answer to this question, although there are various theories: the negative pressure may be a fundamental property of empty space which arises out of whatever spacetime is actually "made of" or it could be some kind of field ("quintessence") that fills empty space which may vary with time and place, but the field is so weak and the variance so slow that it's unlikely we'll ever be able to detect it directly. Or maybe it's something else...

Some fields increase in strength with distance (the strong nuclear force does this on very small scales). Fields are weird.

There are theories of dark energy suggest that it may increase over time until it tears the universe apart right down to atomic scales, perhaps birthing a whole panoply of baby universes in the process! Maybe Dark Energy is the cosmic chicken laying endless universe sized eggs...

posted by pharm at 11:30 AM on May 15, 2011

Quietgal, as pharm said, we don't really understand dark energy very well right now. If it is (as it appears to be) a cosmological constant, what that means is the "empty space" is DEFINED as having a certain amount of dark energy in it - perhaps there's a field with uniform value throughout space, but we don't know yet. Practically (which is a strange word to use in this context), the equations of general relativity allow for the vacuum to have an arbitrary positive or negative energy, and it turns out that we live in a Universe with a very, very small negative cosmological constant (before measuring it, our previous best guess for what it should be was 10^122 times larger than it is. So before it was discovered, we physicists just said "meh, it can't be that large, so it must be zero. This is how we think).

I realize it's an unsatisfying answer, but right now we don't have a better one. This is completely unknown territory, even for physicists. So it's more useful to think of what dark energy does. Imagine it this way: everything in the Universe is attempting to get to the lowest possible energy configuration available - a fact that I wish I was better at getting my undergrad students (when I have them) to understand. The negative cosmological constant in our Universe means that every cubic centimeter of space has a certain quantity of negative energy associated with it. If you doubled the volume, you'd make the energy value twice as low. Therefore, to get to the lowest possible energy, you just keep making the Universe bigger. (This is very handwavy, for a reason I'll get into below)

Now, you might say "ah, but as I double the volume, the energy density should decrease by half, so the energy will remain the same." This is the special thing about dark energy: it's a property of the metric, so if the volume doubles, the energy density remains the same. How is this possible, since we all know energy is conserved?

The first answer to that addresses the problem locally and enters into the territory of Noether's theorem, which states that every conserved property in the Universe comes from a symmetry of nature. Momentum is conserved because space is uniform: all directions are the same. Angular momentum is the same because the Universe is rotationally symmetric. So what's the symmetry associated with energy? Turns out it is time. But since the Universe is clearly changing in time, then energy does not have to be locally conserved - no symmetry, no conservation. Don't attempt to use this as a perpetual motion machine though, unless you have a much longer view of things than most people (or indeed, most things made out of protons).

In fact, even without dark energy, energy in the Universe isn't conserved. For example, photons from the Big Bang have a wavelength that stretches as the Universe expands. Longer wavelengths correspond to lower energies, so as the Universe gets bigger, each photon gets less energetic (this is redshifting, which allows us to measure the distance to a far away object, assuming you know the initial wavelength of light). However, this is in addition to the dilution of photons due to the expansion of space. So if we doubled the dimensions of the Universe, the density of photons would drop by a factor of 2^3 = 8, but their energy would also drop by a factor of 2. So the energy density goes to 1/16th of the initial value, and the energy goes to 8/16 = 1/2.

For those of you paying close attention, no this doesn't mean photons make the Universe expand. The energy goes down, but I'm neglecting an additional term that for relativistic particles serves to counteract the expansion - it's the "pressure," but I think that's a unhelpful way of thinking about it. For one thing, positive pressure makes the Universe slow down, negative pressure (like that of dark energy) makes it speed up. Which is the exact opposite of what you would have guessed from the word "pressure."

The second answer is that you could try define an energy in the gravitational field, and then as space expands, the contributions of gravity and the cosmological constant cancel each other out so you get global energy conservation. I don't think there's a completely consistent way to define the energy in this field though, but I'm not qualified to really go into it.

Finally, to expand on something pharm said, the strong force does increase in strength as you go further from a nuclear charge, but this is an effect on excitations of the field. The properties of dark energy are expected to be a consequence of the field itself, not the excitations of it. There's all sorts of amazingly interesting stuff about the strong nuclear force that come from the fact that it increases in strength over distance (and one of my favorite insider jokes from the first season of the Big Bang Theory), but it's a somewhat different issue than what we expect dark energy to be all about. Of course, we don't know yet, so maybe that's incorrect.

Isn't physics fun?

posted by physicsmatt at 12:29 PM on May 15, 2011 [16 favorites]

I realize it's an unsatisfying answer, but right now we don't have a better one. This is completely unknown territory, even for physicists. So it's more useful to think of what dark energy does. Imagine it this way: everything in the Universe is attempting to get to the lowest possible energy configuration available - a fact that I wish I was better at getting my undergrad students (when I have them) to understand. The negative cosmological constant in our Universe means that every cubic centimeter of space has a certain quantity of negative energy associated with it. If you doubled the volume, you'd make the energy value twice as low. Therefore, to get to the lowest possible energy, you just keep making the Universe bigger. (This is very handwavy, for a reason I'll get into below)

Now, you might say "ah, but as I double the volume, the energy density should decrease by half, so the energy will remain the same." This is the special thing about dark energy: it's a property of the metric, so if the volume doubles, the energy density remains the same. How is this possible, since we all know energy is conserved?

The first answer to that addresses the problem locally and enters into the territory of Noether's theorem, which states that every conserved property in the Universe comes from a symmetry of nature. Momentum is conserved because space is uniform: all directions are the same. Angular momentum is the same because the Universe is rotationally symmetric. So what's the symmetry associated with energy? Turns out it is time. But since the Universe is clearly changing in time, then energy does not have to be locally conserved - no symmetry, no conservation. Don't attempt to use this as a perpetual motion machine though, unless you have a much longer view of things than most people (or indeed, most things made out of protons).

In fact, even without dark energy, energy in the Universe isn't conserved. For example, photons from the Big Bang have a wavelength that stretches as the Universe expands. Longer wavelengths correspond to lower energies, so as the Universe gets bigger, each photon gets less energetic (this is redshifting, which allows us to measure the distance to a far away object, assuming you know the initial wavelength of light). However, this is in addition to the dilution of photons due to the expansion of space. So if we doubled the dimensions of the Universe, the density of photons would drop by a factor of 2^3 = 8, but their energy would also drop by a factor of 2. So the energy density goes to 1/16th of the initial value, and the energy goes to 8/16 = 1/2.

For those of you paying close attention, no this doesn't mean photons make the Universe expand. The energy goes down, but I'm neglecting an additional term that for relativistic particles serves to counteract the expansion - it's the "pressure," but I think that's a unhelpful way of thinking about it. For one thing, positive pressure makes the Universe slow down, negative pressure (like that of dark energy) makes it speed up. Which is the exact opposite of what you would have guessed from the word "pressure."

The second answer is that you could try define an energy in the gravitational field, and then as space expands, the contributions of gravity and the cosmological constant cancel each other out so you get global energy conservation. I don't think there's a completely consistent way to define the energy in this field though, but I'm not qualified to really go into it.

Finally, to expand on something pharm said, the strong force does increase in strength as you go further from a nuclear charge, but this is an effect on excitations of the field. The properties of dark energy are expected to be a consequence of the field itself, not the excitations of it. There's all sorts of amazingly interesting stuff about the strong nuclear force that come from the fact that it increases in strength over distance (and one of my favorite insider jokes from the first season of the Big Bang Theory), but it's a somewhat different issue than what we expect dark energy to be all about. Of course, we don't know yet, so maybe that's incorrect.

Isn't physics fun?

posted by physicsmatt at 12:29 PM on May 15, 2011 [16 favorites]

Nitpick: the value of the cosmological constant (or the energy density of whatever it is, if it exists) is positive, not negative.

posted by edd at 12:42 PM on May 15, 2011

posted by edd at 12:42 PM on May 15, 2011

Yeah, sorry. There's an extra minus sign in there that I absorbed into the definition of the cosmological constant which probably shouldn't have been.

Damn you minus signs.

posted by physicsmatt at 12:46 PM on May 15, 2011

Damn you minus signs.

posted by physicsmatt at 12:46 PM on May 15, 2011

To clarify, the positive cosmological constant implies a negative pressure, which causes an expansion of the Universe. This is why I hate the term pressure in this context; it completely screws with my mental map. Thanks edd, and people should probably ignore the 2nd paragraph in my long post above. It still sort of works, but the additional contortions you need to get it to completely agree with the math probably are more trouble than they're worth.

...and it was such a nice simple description too.

posted by physicsmatt at 12:51 PM on May 15, 2011 [1 favorite]

...and it was such a nice simple description too.

posted by physicsmatt at 12:51 PM on May 15, 2011 [1 favorite]

posted by Quietgal at 11:01 AM

Good question.

A possibly related question, one I have been waiting to see addressed for some time now, is what about the gravitational field of all that dark energy?

Other forms of energy, including photons, produce gravitational fields according to their energy as if they had mass as described by Einstein's E= mc^2.

Unless dark energy is an exception in this as it is in so much else, and since dark energy is supposedly the predominant form of energy in the universe, shouldn't we see large gravitational effects from it?

In a positively curved universe, such as the surface of a hypersphere, static or not, I could accept the possibility that if DE was perfectly evenly distributed, DE's gravitational field might be the same at every point and therefore hard to observe, but even then it would be a force working against the expansion of the universe, wouldn't it?

posted by jamjam at 12:58 PM on May 15, 2011

The gravitational field of dark energy is exactly what leads to its observable consequence of an accelerating universe. The pressure is not the source of the force that accelerate the universe - it is the gravity of dark energy that does.

More on this when I get off the iPhone to a keyboard perhaps...

posted by edd at 1:06 PM on May 15, 2011 [1 favorite]

More on this when I get off the iPhone to a keyboard perhaps...

posted by edd at 1:06 PM on May 15, 2011 [1 favorite]

jamjam, that depends for a start on whether dark energy is a field (that you might expect to exert a gravitational effect) or a fundamental property of spacetime itself,the cosmological constant then being a macroscopic reflection of this universal property.

Even if it is a field, I can't see any a priori reason why the gravitational effect of the energy in such a field would be expected to necessarily outweigh it's repulsive effects. Thoughts physicsmatt?

posted by pharm at 1:08 PM on May 15, 2011

Even if it is a field, I can't see any a priori reason why the gravitational effect of the energy in such a field would be expected to necessarily outweigh it's repulsive effects. Thoughts physicsmatt?

posted by pharm at 1:08 PM on May 15, 2011

To put it another way - dark energy is not a new force. It is a new source of gravity.

posted by edd at 1:10 PM on May 15, 2011

posted by edd at 1:10 PM on May 15, 2011

If I read enough of your posts, physicsmatt, can I have a degree?

posted by Thistledown at 1:26 PM on May 15, 2011

posted by Thistledown at 1:26 PM on May 15, 2011

jamjam, as I was screwing up in my attempt previously, the point is that the dark energy's effect on the expansion of the Universe is opposite normal matter: that's what negative pressure means. If there was a consistent way to define the energy of a gravitational field, you could interpret the positive energy of the field to be cancelled by the negative energy contribution of the pressure (thus preserving energy conservation), but again, this is a tricky thing to do consistently. I agree it is a very unsatisfying explanation, but in GR you have to include this additional effect beyond just the density, and we can write down theories that have this property of negative pressure. It's just hard to test which one we happen to be living in.

I'm not sure why you think the dark energy would cluster in the Universe we have, which has a perfectly flat curvature, but not in a positively curved Universe (another confusing point, most old cosmology texts were written when we thought the cosmological constant Lambda = 0, so they conflate curvature and whether the Universe is open or closed). We are seeing the large gravitational effects of it, but only on the expansion of the Universe. You have to differentiate between the effect of an energy source locally and on the metric. In the "rubber sheet" analogy of GR, dark energy doesn't cause the sheet to bend, but it does cause the sheet to get bigger. Regular matter would cause local deformations, and if enough was spread evenly around the sheet, it would also cause the sheet to get smaller (or, at least get bigger less quickly). I'm trying to think of whether a particle with negative pressure could be made to cluster, I'm thinking not, at least not on cosmological scales but I'm not 100% sure on that.

pharm, a true cosmological constant has equation of state (relation between pressure and density) w = -1, which is exactly what you need to never be diluted and so continually cause an expansion. It's an open question as to whether dark energy is exactly w = -1, or just some largish negative number (for some reason, I remember that we know w <~ -0.7, but I'm not sure if that's the best current value). Anything less than -1/3 will cause an expansion, but the density would change over time (photons have w = +1/3, normal matter has w = 0, and they get diluted away, for example). So I guess the answer to your question is "for some things with negative w, the pressure component wins and causes an expansion, for other things, the pressure is not sufficiently negative. We happen to live in a Universe where whatever dark energy is, has w sufficiently negative. Don't ask me why (instead, find out why, then tell me on your flight to Stockholm).

posted by physicsmatt at 1:41 PM on May 15, 2011

I'm not sure why you think the dark energy would cluster in the Universe we have, which has a perfectly flat curvature, but not in a positively curved Universe (another confusing point, most old cosmology texts were written when we thought the cosmological constant Lambda = 0, so they conflate curvature and whether the Universe is open or closed). We are seeing the large gravitational effects of it, but only on the expansion of the Universe. You have to differentiate between the effect of an energy source locally and on the metric. In the "rubber sheet" analogy of GR, dark energy doesn't cause the sheet to bend, but it does cause the sheet to get bigger. Regular matter would cause local deformations, and if enough was spread evenly around the sheet, it would also cause the sheet to get smaller (or, at least get bigger less quickly). I'm trying to think of whether a particle with negative pressure could be made to cluster, I'm thinking not, at least not on cosmological scales but I'm not 100% sure on that.

pharm, a true cosmological constant has equation of state (relation between pressure and density) w = -1, which is exactly what you need to never be diluted and so continually cause an expansion. It's an open question as to whether dark energy is exactly w = -1, or just some largish negative number (for some reason, I remember that we know w <~ -0.7, but I'm not sure if that's the best current value). Anything less than -1/3 will cause an expansion, but the density would change over time (photons have w = +1/3, normal matter has w = 0, and they get diluted away, for example). So I guess the answer to your question is "for some things with negative w, the pressure component wins and causes an expansion, for other things, the pressure is not sufficiently negative. We happen to live in a Universe where whatever dark energy is, has w sufficiently negative. Don't ask me why (instead, find out why, then tell me on your flight to Stockholm).

posted by physicsmatt at 1:41 PM on May 15, 2011

Thistledown: sadly no (and dear god, the hash I'm making of this explanation isn't helping anyone understand anything, I fear). To continue the joke though, sure: it'll take 5 years, there will be a test at the end, and you will have to spend that time in minor poverty with insanely high stress levels. I doubt my posts will be as fun as grad school though.

posted by physicsmatt at 1:44 PM on May 15, 2011

posted by physicsmatt at 1:44 PM on May 15, 2011

I'm going to petition mathowie to create a non-accredited MeFi physics degree for everyone who reads physicsmatt's comments for the next 5 years.

The poverty and stress levels are already a given when it comes to MetaFilter. :)

posted by hippybear at 1:52 PM on May 15, 2011

The poverty and stress levels are already a given when it comes to MetaFilter. :)

posted by hippybear at 1:52 PM on May 15, 2011

Hate to keep nitpicking, but curvature is about whether the universe is open or closed - those terms apply to the curvature, not the ultimate fate of the universe (which was confused in older textbooks and is what physicsmatt means)...

On the w constraints I think they might be a bit better now, but not better enough to get us any closer to figuring out what the heck is going on.

posted by edd at 1:54 PM on May 15, 2011

On the w constraints I think they might be a bit better now, but not better enough to get us any closer to figuring out what the heck is going on.

posted by edd at 1:54 PM on May 15, 2011

Damnit. Again. Thanks edd, keep keeping me honest. The worst part is that I know at least one of my colleagues reads metafilter. Hi guys!

posted by physicsmatt at 2:14 PM on May 15, 2011

posted by physicsmatt at 2:14 PM on May 15, 2011

In which I reveal how little of modern physics I understand: so if dark energy permeates empty space and there's no matter involved ('cuz it's empty), E = mc^2 doesn't apply here? Dark energy exists independent of mass? And it's never diluted (or used up or destroyed, presumably) either? This seems to violate the "no free lunch" principle, which is pretty much the level at which I grok physics.

Does dark energy have any properties resembling regular ol' everyday energy? Does it "run downhill" into heat, the way normal energy "dies"?

Thank you physicsmatt, pharm and edd for your patience and enthusiasm for Sunday Afternoon Special Topics in General Relativity.

We totally need to do a TEDxMetafilter series of talks.

posted by Quietgal at 2:53 PM on May 15, 2011

Does dark energy have any properties resembling regular ol' everyday energy? Does it "run downhill" into heat, the way normal energy "dies"?

Thank you physicsmatt, pharm and edd for your patience and enthusiasm for Sunday Afternoon Special Topics in General Relativity.

We totally need to do a TEDxMetafilter series of talks.

posted by Quietgal at 2:53 PM on May 15, 2011

Quietgal, mass is only one form that energy can take, so no, in this case E=mc^2 doesn't apply - though of course the cosmological constant has gravitational effects, but due to the pressure issue, it's complicated compared to what we normally consider. This energy is simply a property of the vacuum; it's the price you have to pay in order to have "space." Again, it's not clear why it has to be this way, Einstein realized it COULD be true way back when, but dismissed it as a blunder (as in, he said that the cosmological constant should be zero). Later physicists did a back-of-the-envelope calculation and got an answer so large that the Universe should have expanded so fast that every proton would be totally isolated - a rather boring existence that we clearly don't live in. So they said "well, it's not ~infinity, so it's zero." Then we found out it was nonzero but ridiculously tiny, and we have no idea why.

As for the no-free lunch principle, conservation of energy can certainly be violated in our Universe, as there is no time invariance over cosmological epochs. I don't think you can imagine using this as a way to get free energy though, as there's no way to "attach" your perpetual motion machine to the expanding space-time to extract energy. Also, even if you violate the 1st law of thermodynamics (energy conservation), the 3rd is still going strong (entropy increases), so don't worry, the Universe is still making sure that we'll all run out of useable energy at some point. That's your cheery thought for the day.

Dark energy can't "run downhill" because, as the energy of the vacuum, there's nowhere else for it to go. It's the minimum energy configuration of the Universe; everything else is approaching this minimum, but pure vacuum can't lose more energy as heat. It's possible that we live in a "false vacuum," with some lower energy configuration that we can eventually reach. If that's the case, eventually some region of the Universe would tunnel into this lower configuration, releasing a huge amount of energy as energetic particles (so energy in the form of heat and mass) in some volume that would expand at the speed of light. You do not want to be around if this happens. Previous eras of this could be similar to what drove the inflation that occurred "before" the Big Bang in some multiverse theories, so it's not all gloom and doom: you could get a new Universe out of the deal. You'll be vaporized of course, but eggs, omelets and all that.

posted by physicsmatt at 3:32 PM on May 15, 2011

As for the no-free lunch principle, conservation of energy can certainly be violated in our Universe, as there is no time invariance over cosmological epochs. I don't think you can imagine using this as a way to get free energy though, as there's no way to "attach" your perpetual motion machine to the expanding space-time to extract energy. Also, even if you violate the 1st law of thermodynamics (energy conservation), the 3rd is still going strong (entropy increases), so don't worry, the Universe is still making sure that we'll all run out of useable energy at some point. That's your cheery thought for the day.

Dark energy can't "run downhill" because, as the energy of the vacuum, there's nowhere else for it to go. It's the minimum energy configuration of the Universe; everything else is approaching this minimum, but pure vacuum can't lose more energy as heat. It's possible that we live in a "false vacuum," with some lower energy configuration that we can eventually reach. If that's the case, eventually some region of the Universe would tunnel into this lower configuration, releasing a huge amount of energy as energetic particles (so energy in the form of heat and mass) in some volume that would expand at the speed of light. You do not want to be around if this happens. Previous eras of this could be similar to what drove the inflation that occurred "before" the Big Bang in some multiverse theories, so it's not all gloom and doom: you could get a new Universe out of the deal. You'll be vaporized of course, but eggs, omelets and all that.

posted by physicsmatt at 3:32 PM on May 15, 2011

physicsmatt : I love it when you talk physics to me Baby.

posted by Poet_Lariat at 5:55 PM on May 15, 2011

posted by Poet_Lariat at 5:55 PM on May 15, 2011

Having watched the video and *payed attention this time*, I couldn't help but think of the 'box of gas' idea in terms of Conway's Game of Life. The Universe is a version of that game that's been set to work out every single configuration of some basic particles, ever. The configuration that prompted our iteration and all that's in it was very small, and smooth, and completely symmetrical, and it's progression was BANG→US→ETC. The next configuration might result in ETC.→BANG→US or it might just be one big green tea cupcake, but it won't happen until our progression has petered out. Would iterations be random one to the next, or progress relationally somehow?

Anyway, in this context I imagine Phil Connors was suddenly transported to another progression with an initial configuration like a self-replicator, but one that spat out Punxsutawney Feb. 2 over and over.

posted by carsonb at 8:15 PM on May 15, 2011

Anyway, in this context I imagine Phil Connors was suddenly transported to another progression with an initial configuration like a self-replicator, but one that spat out Punxsutawney Feb. 2 over and over.

posted by carsonb at 8:15 PM on May 15, 2011

Oh yeah, and I like the CGoL frame of reference because then you don't need a Universe-pooping Chicken leaving feathers everywhere.

posted by carsonb at 8:16 PM on May 15, 2011

posted by carsonb at 8:16 PM on May 15, 2011

OK, let me do this description correctly. As in, no more screwing around with analogies. I'm not sure it'll help make it clear to everyone, but I thought I'd try. It's a couple of long posts - I didn't know metafilter had a maximum length cut-off.

Hold on to your hats, there will be math.

posted by physicsmatt at 8:21 PM on May 15, 2011

Hold on to your hats, there will be math.

posted by physicsmatt at 8:21 PM on May 15, 2011

General Relativity comes down to Einstein's Equation:

R_{\mu\nu} - 1/2 R g_{\mu\nu} = 8 \pi G T_{\mu\nu}

G is the gravitational constant, which just tells you the strength of gravity. \mu and \nu are indices denoting space and time dimensions that run from 0 to 3 (the 0 is time, 1-3 are space). R_{\mu\nu} and T_{\mu\nu} are tensors, which you can think of here as matrices that have a 4x4 entries. g_{\mu\nu} is the metric. This is the thing that tells you how to measure distances in space and time (thus the name metric). In flat space, g_\mu\nu is (-1,+1,+1,+1) along the diagonal and zero everywhere else (particle physicists like myself like the opposite choices for signs, which is one of the reasons I kept fucking this description up in the previous posts. Don't ask, it's stupid). R_{\mu\nu} and R are the Ricci tensor and scalar, which are a really complicated collection of derivatives of g_\mu\nu (that is, how g changes when you move in any direction - including time). The left-hand combination is often just written as G_{\mu\nu}, the Einstein tensor. Don't worry about this too much; just think "left hand side is how space and time are bent."

T_\mu\nu is the stress-energy tensor. This is the thing that tells you how matter and energy is arranged at each point in space. So with the Einstein equation, you can tell how energy causes space to bend (which then tells you how the energy moves, which tells you how space bends again, and so on. It gets computationally difficult if the moving objects are massive enough to really matter).

What will be different for dark energy is T_\mu\nu, so let's understand what goes where. The T_00 entry is the energy density, \rho. The flow of energy across a surface of constant x-coordinate are the T_01=T_10 coordinates, across a surface of constant y is T_02 = T_20, and constant z is T_03=T_30. The T_11, T_22, and T_33 is the flux of momentum in the x,y,z direction across a surface of constant x,y, or z. This is called the pressure, which should make sense, as if momentum is flowing in an direction, you'll feel a push in that direction. If it flows in all directions, that's saying something is confined and trying to push out. The remaining T_12, T_13 etc entries are shear forces, and I'm going to ignore the hell out of them.

Now, the question is, what's T_\mu\nu for dark energy? For any equally distributed type of energy (dust, light, dark energy), you have no shears and no flow of energy, so the only form you have for T_\mu\nu is the diagonal (\rho, p, p, p), where \rho is the density (the same everywhere) and p is the pressure (equal in all directions. Everything else is zero. Now, if I know how to jump from something called the Lagrangian to T_\mu\nu, it'd be simple to prove to you that a field that has no kinetic energy term and a non-zero potential V (i.e. energy that the field has in space) would have T_\mu\nu = -V g_\mu\nu, so you would see that obviously, the pressure must be negative (i.e. T_\mu\nu = diagonal(\rho, -p,-p,-p) ). I'm sure that'll be an emotionally satisfying answer for everyone.

A slightly more satisfying answer is to imagine inflating a balloon. The outside air pressure (which is positive) is the thing that makes it hard to do that: it works against you, so you have to put energy into the system to get the volume to increase. Dark energy (by definition) makes it easier to create more space; creating more space not only costs you no energy, it GIVES you more energy (since at the end, you have more space with more energy density in it). So you do negative work "inflating" your balloon of dark energy. Thus, by the magic of handwaving, the pressure involved must be negative. Ta da!

You won't get the exact relation this way, and I'll just mention again that it's

\rho = - p.

The -1 is the "w" for dark energy, it's the "equation of state" for this particular type of energy.

Importantly, notice that, since T_\mu\nu for dark energy is just a constant times g_\mu\nu, I could have separated it out from normal matter and written Einstein's Equation as

R_{\mu\nu} - 1/2 R g_{\mu\nu} +\Lambda g_\mu\nu = 8 \pi G T_{\mu\nu}

here Lambda is a cosmological constant, and this is how Einstein first wrote his equation. Lambda is just a fudge factor, while everything else is set by the local conditions, \Lambda is just whatever it is. Turns out, it's small but not zero.

posted by physicsmatt at 8:22 PM on May 15, 2011 [1 favorite]

R_{\mu\nu} - 1/2 R g_{\mu\nu} = 8 \pi G T_{\mu\nu}

G is the gravitational constant, which just tells you the strength of gravity. \mu and \nu are indices denoting space and time dimensions that run from 0 to 3 (the 0 is time, 1-3 are space). R_{\mu\nu} and T_{\mu\nu} are tensors, which you can think of here as matrices that have a 4x4 entries. g_{\mu\nu} is the metric. This is the thing that tells you how to measure distances in space and time (thus the name metric). In flat space, g_\mu\nu is (-1,+1,+1,+1) along the diagonal and zero everywhere else (particle physicists like myself like the opposite choices for signs, which is one of the reasons I kept fucking this description up in the previous posts. Don't ask, it's stupid). R_{\mu\nu} and R are the Ricci tensor and scalar, which are a really complicated collection of derivatives of g_\mu\nu (that is, how g changes when you move in any direction - including time). The left-hand combination is often just written as G_{\mu\nu}, the Einstein tensor. Don't worry about this too much; just think "left hand side is how space and time are bent."

T_\mu\nu is the stress-energy tensor. This is the thing that tells you how matter and energy is arranged at each point in space. So with the Einstein equation, you can tell how energy causes space to bend (which then tells you how the energy moves, which tells you how space bends again, and so on. It gets computationally difficult if the moving objects are massive enough to really matter).

What will be different for dark energy is T_\mu\nu, so let's understand what goes where. The T_00 entry is the energy density, \rho. The flow of energy across a surface of constant x-coordinate are the T_01=T_10 coordinates, across a surface of constant y is T_02 = T_20, and constant z is T_03=T_30. The T_11, T_22, and T_33 is the flux of momentum in the x,y,z direction across a surface of constant x,y, or z. This is called the pressure, which should make sense, as if momentum is flowing in an direction, you'll feel a push in that direction. If it flows in all directions, that's saying something is confined and trying to push out. The remaining T_12, T_13 etc entries are shear forces, and I'm going to ignore the hell out of them.

Now, the question is, what's T_\mu\nu for dark energy? For any equally distributed type of energy (dust, light, dark energy), you have no shears and no flow of energy, so the only form you have for T_\mu\nu is the diagonal (\rho, p, p, p), where \rho is the density (the same everywhere) and p is the pressure (equal in all directions. Everything else is zero. Now, if I know how to jump from something called the Lagrangian to T_\mu\nu, it'd be simple to prove to you that a field that has no kinetic energy term and a non-zero potential V (i.e. energy that the field has in space) would have T_\mu\nu = -V g_\mu\nu, so you would see that obviously, the pressure must be negative (i.e. T_\mu\nu = diagonal(\rho, -p,-p,-p) ). I'm sure that'll be an emotionally satisfying answer for everyone.

A slightly more satisfying answer is to imagine inflating a balloon. The outside air pressure (which is positive) is the thing that makes it hard to do that: it works against you, so you have to put energy into the system to get the volume to increase. Dark energy (by definition) makes it easier to create more space; creating more space not only costs you no energy, it GIVES you more energy (since at the end, you have more space with more energy density in it). So you do negative work "inflating" your balloon of dark energy. Thus, by the magic of handwaving, the pressure involved must be negative. Ta da!

You won't get the exact relation this way, and I'll just mention again that it's

\rho = - p.

The -1 is the "w" for dark energy, it's the "equation of state" for this particular type of energy.

Importantly, notice that, since T_\mu\nu for dark energy is just a constant times g_\mu\nu, I could have separated it out from normal matter and written Einstein's Equation as

R_{\mu\nu} - 1/2 R g_{\mu\nu} +\Lambda g_\mu\nu = 8 \pi G T_{\mu\nu}

here Lambda is a cosmological constant, and this is how Einstein first wrote his equation. Lambda is just a fudge factor, while everything else is set by the local conditions, \Lambda is just whatever it is. Turns out, it's small but not zero.

posted by physicsmatt at 8:22 PM on May 15, 2011 [1 favorite]

Moving along, what does this all mean for the future of the Universe? This means applying Einstein's Equation to a particular situation: a expanse of spacetime large enough that the little perturbations of stars and galaxies don't matter. Turns out, on these scales, the Universe is homogenous and isotropic (the same in all directions). We can then simplify the metric g_\mu\nu to a nice form where we can set "comoving coordinates" in space that move apart in time by a scale factor a(t). This means that the physical distance between two comoving points today is a_0, but previously, was smaller, since a(t) < a_0 (since the Universe is getting bigger, not getting smaller). More specifically, the metric is now +1 in time and -a(t) in space (usually written in radial coordinates, but I am NOT doing that without LaTex handy).

Locally, since all the matter in the Universe is homogeneous, we can say that it can't be flowing anywhere (on the average, of course). This allows us to say some things about how the density is changing in time. In particular, after solving a simple differential equation (\del_\mu T^\mu_0 = 0 for those paying very close attention), you find that the density is proportional to

a(t)^{-3(1+w)}

which for w = -1 means that \rho is a constant over time. Wacky, huh?

Continuing along, in our simplified metric, assuming you know calculus and have the full definition of R_\mu\nu in front of you it's easy (but tedious) to rewrite Einstein's Equation into a new form. We want to know how the scale factor a(t) is changing in time. If the Universe has curvature, this is more complicated. Basically, curvature acts like a new contribution to T_\mu\nu that comes from the global nature of how the Universe is shaped. However, it's telling you how to measure distances on huge length scales, which is DIFFERENT from dark energy, which is a contribution to T_\mu\nu that's always proportional to the metric. You can have both, but our Universe doesn't so I'll ignore it.

posted by physicsmatt at 8:22 PM on May 15, 2011

Locally, since all the matter in the Universe is homogeneous, we can say that it can't be flowing anywhere (on the average, of course). This allows us to say some things about how the density is changing in time. In particular, after solving a simple differential equation (\del_\mu T^\mu_0 = 0 for those paying very close attention), you find that the density is proportional to

a(t)^{-3(1+w)}

which for w = -1 means that \rho is a constant over time. Wacky, huh?

Continuing along, in our simplified metric, assuming you know calculus and have the full definition of R_\mu\nu in front of you it's easy (but tedious) to rewrite Einstein's Equation into a new form. We want to know how the scale factor a(t) is changing in time. If the Universe has curvature, this is more complicated. Basically, curvature acts like a new contribution to T_\mu\nu that comes from the global nature of how the Universe is shaped. However, it's telling you how to measure distances on huge length scales, which is DIFFERENT from dark energy, which is a contribution to T_\mu\nu that's always proportional to the metric. You can have both, but our Universe doesn't so I'll ignore it.

posted by physicsmatt at 8:22 PM on May 15, 2011

With that simplification, the 00 entries from Einstein's Equation gives two new equations. The first says:

(\dot{a}/a)^2 = 8\pi G/3 \rho

where \dot{a} means the time derivative of a(t) - how the scale factor changes in time. \dot{a}/a is better known as H, the "Hubble constant," it tells you the rate at which the distance scales are changing at a particular time. If space was curved, there'd be a 2nd term on the right, which could allow H to be negative (a contracting Universe). Since we have a positive matter density and no curvature, we can't go from expansion into contraction. The smallest H could be would be zero (static Universe). However, this equation doesn't tell us how H changes in time (turns out that this constant isn't so constant).

The 2nd equation we get is

\ddot{a}/a = - 4\pi G/3 (\rho +3 p) = -4 \pi G/ 3 (1+3w) \rho

Here, \ddot{a} is the 2nd derivitive of a(t) (the change in the rate of expansion - an acceleration). Notice, if w > -1/3, the right-hand side is negative, and so the Universe is slowing down: H decreases with time. But, if the dark energy is the most important thing around, w = -1, and \ddot{a} > 0. This means that the rate at which the scale factor a(t) is changing is increasing. The Universe accelerates it's expansion.

So what are the take-aways? 1) dark energy has a very strange property, it’s “pressure” is negative, 2) this means that that it doesn’t dilute away and 3) this means it can cause the expansion of the Universe to speed up. 4) GR is a complete nightmare to write out using standard text characters.

This was written with consultation with Sean Carroll’s GR textbook. Get yours today (not paid for by Sean Carroll, but this thread is ostensibly about him).

posted by physicsmatt at 8:23 PM on May 15, 2011

(\dot{a}/a)^2 = 8\pi G/3 \rho

where \dot{a} means the time derivative of a(t) - how the scale factor changes in time. \dot{a}/a is better known as H, the "Hubble constant," it tells you the rate at which the distance scales are changing at a particular time. If space was curved, there'd be a 2nd term on the right, which could allow H to be negative (a contracting Universe). Since we have a positive matter density and no curvature, we can't go from expansion into contraction. The smallest H could be would be zero (static Universe). However, this equation doesn't tell us how H changes in time (turns out that this constant isn't so constant).

The 2nd equation we get is

\ddot{a}/a = - 4\pi G/3 (\rho +3 p) = -4 \pi G/ 3 (1+3w) \rho

Here, \ddot{a} is the 2nd derivitive of a(t) (the change in the rate of expansion - an acceleration). Notice, if w > -1/3, the right-hand side is negative, and so the Universe is slowing down: H decreases with time. But, if the dark energy is the most important thing around, w = -1, and \ddot{a} > 0. This means that the rate at which the scale factor a(t) is changing is increasing. The Universe accelerates it's expansion.

So what are the take-aways? 1) dark energy has a very strange property, it’s “pressure” is negative, 2) this means that that it doesn’t dilute away and 3) this means it can cause the expansion of the Universe to speed up. 4) GR is a complete nightmare to write out using standard text characters.

This was written with consultation with Sean Carroll’s GR textbook. Get yours today (not paid for by Sean Carroll, but this thread is ostensibly about him).

posted by physicsmatt at 8:23 PM on May 15, 2011

I'm totally citing this thread the next time I tilt at the windmill of LaTeX parsing on Metafilter. (As in, that pony would be one I rode against windmills all the day long.)

posted by kaibutsu at 9:56 PM on May 15, 2011

posted by kaibutsu at 9:56 PM on May 15, 2011

I think this is a bit misleading too. Things aren't generally trying to get to a low energy configuration - they're trying to get to a high entropy one. These generally end up being the same thing - if I have a collection of high energy state atoms, then if they spit that energy out as photons I get a collection of low energy state atoms plus a lot of photons, and the larger number of entities means the entropy as a whole will have gone up.

This is not the driving force behind the universe's expansion - cosmological expansion doesn't involve a change in entropy. Naturally the entropy of the universe is going up, but this is a consequence of what the stuff in the universe does to pass time, and the peculiar fact that entropy was low in the first place. This is what Sean Carroll was talking about, skimming the transcript - it's a very interesting question to ask why the universe began in such a state.

So the way the expansion rate evolves as a function of its constituents isn't to do with things trying to get to a low energy state, or even a high entropy one. It's a time-reversible consequence of gravitational forces. The universe is expanding now because it was expanding just a moment ago, modified by the gravitational forces within it - in the same way a baseball going across the field is moving because it was just a moment ago, and its movement has changed as a result of the forces upon it.

posted by edd at 11:44 PM on May 15, 2011

Actually I think you can. Most people think of pressure as force/area, but force=work done x distance. Substituting in, pressure = work done / (area x distance), so you get a new description of pressure as p=-dE/dV, since area x distance is the volume change.

Next take ρ=E/V. Differentiate with respect to V:

dρ/dV=(VdE/dV - E)/V

Set this to zero as we want something that stays constant energy densit despite volume changes and VdE/dV=E. Fiddle, setting E/V back to ρ and ρ=-p.

posted by edd at 12:34 AM on May 16, 2011

I was confused by this lecture. Let me see if I've got this straight.

Dark energy is present as a fundamental component of empty space.

And space is expanding.

Therefore the total amount of dark energy in the Universe is increasing.

As space expands at an increasingly fast rate, the amount of dark energy being created along with the empty space will also increase exponentially.

So according to this theory, the total energy of the Universe is increasing? And the cosmological constant driving this acceleration and resultant increase is dark energy itself? So the energy is driving the creation of itself out of nothingness? And as the Universe continues to expand infinitely the total energy will approach infinity?

This seems preposterous, so I suspect I'm misunderstanding something fundamental here...

posted by jet_manifesto at 3:19 AM on May 16, 2011

Dark energy is present as a fundamental component of empty space.

And space is expanding.

Therefore the total amount of dark energy in the Universe is increasing.

As space expands at an increasingly fast rate, the amount of dark energy being created along with the empty space will also increase exponentially.

So according to this theory, the total energy of the Universe is increasing? And the cosmological constant driving this acceleration and resultant increase is dark energy itself? So the energy is driving the creation of itself out of nothingness? And as the Universe continues to expand infinitely the total energy will approach infinity?

This seems preposterous, so I suspect I'm misunderstanding something fundamental here...

posted by jet_manifesto at 3:19 AM on May 16, 2011

Yeah, as physicsmatt pointed out, it's somewhat difficult to talk about conservation of energy in the context of some changing background spacetime. The Usenet Physics FAQ explains this to a greater or lesser extent. Sean Carroll also explains this.

Essentially, it's not at all as preposterous as it sounds.

posted by edd at 3:28 AM on May 16, 2011 [1 favorite]

Essentially, it's not at all as preposterous as it sounds.

posted by edd at 3:28 AM on May 16, 2011 [1 favorite]

(A commenter to the Sean Carroll blog post that edd links to adds: "The universe is not a steam engine". Leave your physical intuitions at the door in other words...)

posted by pharm at 3:54 AM on May 16, 2011

posted by pharm at 3:54 AM on May 16, 2011

Yeah, its not much of a stretch really - if space can increase, why can't energy? Where nothing was, something now is.

posted by jet_manifesto at 4:12 AM on May 16, 2011

posted by jet_manifesto at 4:12 AM on May 16, 2011

edd, yeah, probably should have said "lowest potential energy," which is true since there is a force pushing you "downhill." An object is quite happy to stay with a lot of kinetic energy.

I was also worried that the thermodynamic argument was circular, but looks like you can get w = -1 that way. Also, how are you getting those nice rhos?

Finally, since the expanding Universe has a horizon, and horizons have entropy (quite a lot of it, actually), it's quite possible that the 2nd law is driving the expansion. Doesn't explain why the Universe started in such a low entropy state though, which is what Sean is talking about.

posted by physicsmatt at 4:21 AM on May 16, 2011

I was also worried that the thermodynamic argument was circular, but looks like you can get w = -1 that way. Also, how are you getting those nice rhos?

Finally, since the expanding Universe has a horizon, and horizons have entropy (quite a lot of it, actually), it's quite possible that the 2nd law is driving the expansion. Doesn't explain why the Universe started in such a low entropy state though, which is what Sean is talking about.

posted by physicsmatt at 4:21 AM on May 16, 2011

I think there's a question about the balloon or piston analogy in that dark energy isn't obviously doing work in the way a gas might - what surface is it exerting pressure on when it expands? It's bothering me a little more since I posted anyway :)

It's easy enough to do Greek letters in html - http://www.escapecodes.info/ for example. MathJax is a nice way to do it if you're building your own pages incidentally.

posted by edd at 4:29 AM on May 16, 2011

It's easy enough to do Greek letters in html - http://www.escapecodes.info/ for example. MathJax is a nice way to do it if you're building your own pages incidentally.

posted by edd at 4:29 AM on May 16, 2011

If you're just writing simple terms, then you can get quite a long way with a bit of unicode & html:

R_{μν} - ½Rg_{μν} + Λg_{μν} = 8πGT_{μν}

posted by pharm at 4:35 AM on May 16, 2011

R

posted by pharm at 4:35 AM on May 16, 2011

...and I see that in my post at 5:32 pm yesterday, I wrote "3rd Law" when I meant to say "2nd." My kingdom for an edit button with a very generous window of action.

We'll just pretend that this all an exercise in elaborate performance art to demonstrate that having a Ph.D. doesn't mean you can't make silly mistakes in public.

posted by physicsmatt at 6:02 AM on May 16, 2011

We'll just pretend that this all an exercise in elaborate performance art to demonstrate that having a Ph.D. doesn't mean you can't make silly mistakes in public.

posted by physicsmatt at 6:02 AM on May 16, 2011

MetaFilter: this all an exercise in elaborate performance art

posted by hippybear at 6:49 AM on May 16, 2011

posted by hippybear at 6:49 AM on May 16, 2011

Ah, crap, you lost me at "tensor". Can anybody recommend a good discussion of General Relativity for someone whose math background ends at basic calculus? I can handle a few integrals and differential equations, but I never even tackled partial differentials. The Physics FAQ linked above and even Wikipedia on dark energy are already beyond me, so I think I need a pop-sci source.

*hangs head in shame, wanders off to AskMe to look for cooking questions to answer*

posted by Quietgal at 1:26 PM on May 16, 2011

*hangs head in shame, wanders off to AskMe to look for cooking questions to answer*

posted by Quietgal at 1:26 PM on May 16, 2011

Quietgal, for this purpose, "tensor" is just a grid with entries labeled by two numbers μ and ν both running from 0 to 3. The labels refer to things that have to do with the associated dimension (00 is "time-time", 01 is "time-x", 23 would be "y-x" and so on). To do this right, you have to worry about how to raise and lower indices, and contractions and the formal definition of a tensor and how to get R from g_μν, but don't try to dive into the deep end immediately. Sadly, I know of no good GR texts for non-physicists; I keep recommending Sean Carroll's from Eternity to Here, it doesn't spend a huge amount of time on GR, but he's the best physics writer for a general audience I know of right now. Also maybe Lisa Randall's Warped Passages, but that's more particle theory. It talks about extra dimensions, which requires some GR. I can't remember how much she goes into the basics though.

Thistledown, well then, what shall we talk about?

posted by physicsmatt at 8:59 PM on May 16, 2011

Thistledown, well then, what shall we talk about?

posted by physicsmatt at 8:59 PM on May 16, 2011

Hmm. Whilst that's a good description of how a tensor can be written, I'm not sure it's very helpful in described what they are!

Quietgal: As physicsmatt says, in terms of what you write down, a tensor is "just" a generalisation of scalars (1-dimensional values) & vectors (N-dimensional values). Sticking to a newtonian 3D metric for simplicity, a scalar valued field would be something that has a single value for each point in space: temperature say. A vector field encodes a direction and magnitude for each point in space, so that might be something like newtonian gravity. We usually write the value of a scalar as a single number (3), whilst a vector would usually be written as three numbers, representing the strength of the field along each of the dimensions: (0.5, 5.0, -2.3) say. Tensors generalise this to arbitrary n by m matrices.

Take a rank 2 tensor in 3D space. In physical terms such a tensor encodes a field which depends not only on where you are in space, but also how the values of the field change depending on which way you're pointing at the time. One example of such a field might be the strain tensor for deformation of 3D objects where Hooke's law applies: for each point in space, you have a "normal strain", ie the direction in which the object is being stretched or squashed, and a "shear strain", which is defined as the change in angle between originally perpendicular lines from the point in question. You can encode this information as a 3x3 tensor in 3D Newtonian space.

In physical terms, what a tensor "means" depends partly on the rank of the tensor, that is to say how many dimensions there are in the array needed to write it, partly on the physical theory being used which gives semantic meaning to the tensor components. So different tensors will have different "meanings". The mathematics for operating on them is the same however (which has given category theorists a field day) so they pop up in many fields of physics simply because they're the right tool for the job: if you need to work on fields that depend on more than just position and direction, or you're working in a metric space which isn't flat, then you're going to be working with tensors even if only implicitly & doing so explicitly lets you bring to bear a whole pile of proven mathematics to the job.

Higher dimensional tensors can encode ever more complicated relationships. Once you dive into special & general relativity you need to include a tensor that represents the very space you're working in: the metric tensor. (The metric tensor exists implicitly in Newtonian space, but the metric is flat, so you can just drop it from the calculations) At this point my physics comes to a grinding halt (since I never studied GR) and I defer to physicsmatt & edd.

(Writing this post has brought to mind the earworm from Alfred Bester's "The Stars my Destination". So now I'm going to have that running through my head all day. Meanwhile physicsmatt & edd can point out the special cases I've unknowingly skipped over and the inevitable egregious errors.)

posted by pharm at 3:11 AM on May 17, 2011

Quietgal: As physicsmatt says, in terms of what you write down, a tensor is "just" a generalisation of scalars (1-dimensional values) & vectors (N-dimensional values). Sticking to a newtonian 3D metric for simplicity, a scalar valued field would be something that has a single value for each point in space: temperature say. A vector field encodes a direction and magnitude for each point in space, so that might be something like newtonian gravity. We usually write the value of a scalar as a single number (3), whilst a vector would usually be written as three numbers, representing the strength of the field along each of the dimensions: (0.5, 5.0, -2.3) say. Tensors generalise this to arbitrary n by m matrices.

Take a rank 2 tensor in 3D space. In physical terms such a tensor encodes a field which depends not only on where you are in space, but also how the values of the field change depending on which way you're pointing at the time. One example of such a field might be the strain tensor for deformation of 3D objects where Hooke's law applies: for each point in space, you have a "normal strain", ie the direction in which the object is being stretched or squashed, and a "shear strain", which is defined as the change in angle between originally perpendicular lines from the point in question. You can encode this information as a 3x3 tensor in 3D Newtonian space.

In physical terms, what a tensor "means" depends partly on the rank of the tensor, that is to say how many dimensions there are in the array needed to write it, partly on the physical theory being used which gives semantic meaning to the tensor components. So different tensors will have different "meanings". The mathematics for operating on them is the same however (which has given category theorists a field day) so they pop up in many fields of physics simply because they're the right tool for the job: if you need to work on fields that depend on more than just position and direction, or you're working in a metric space which isn't flat, then you're going to be working with tensors even if only implicitly & doing so explicitly lets you bring to bear a whole pile of proven mathematics to the job.

Higher dimensional tensors can encode ever more complicated relationships. Once you dive into special & general relativity you need to include a tensor that represents the very space you're working in: the metric tensor. (The metric tensor exists implicitly in Newtonian space, but the metric is flat, so you can just drop it from the calculations) At this point my physics comes to a grinding halt (since I never studied GR) and I defer to physicsmatt & edd.

(Writing this post has brought to mind the earworm from Alfred Bester's "The Stars my Destination". So now I'm going to have that running through my head all day. Meanwhile physicsmatt & edd can point out the special cases I've unknowingly skipped over and the inevitable egregious errors.)

posted by pharm at 3:11 AM on May 17, 2011

Thanks, everyone, for the explanations and book recommendations. I've requested *From Eternity to Here* from the library already! Dunno how much will actually sink in, but it's always fun to read about the ineffable weirdness of the universe at the quantum level.

posted by Quietgal at 8:44 PM on May 17, 2011

posted by Quietgal at 8:44 PM on May 17, 2011

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The 7th Chevron doesn't appear to lock.

posted by leotrotsky at 9:33 AM on May 15, 2011 [7 favorites]