Wikipedia:Reference desk/Archives/Science/2014 January 22

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January 22

Is this right? (Order of reactants)

The problem is Zero, First and Second order reaction rate... things (I'm so screwed I don't even know the correct way to say it. but anyway, I'm not sure if this is correct, but It's what I gleaned from what I've read online:

• A zero order reactant does not affect the rate of reaction if doubled
• a first order reactant makes the rate go two times as fast if doubled
• a second order reactant makes the rate go four times as fast if doubled.

Am I figuring this correctly?--Ye Olde Luke (talk) 04:04, 22 January 2014 (UTC)

According to our article rate equation. A zero-order reaction is at a constant rate with changing concentration of reactant, a first-order reaction is linearly proportional to the concentration of reactant, and a second-order reaction is proportional to the square of the concentration. So yes, the example numbers you cite are correct. SpinningSpark 16:29, 22 January 2014 (UTC)

Awesome, thanks! --Ye Olde Luke (talk) 01:06, 24 January 2014 (UTC)

User:Ye Olde Luke, be careful. Spinningspark's answer, and our article, say the reaction rate is proportional to the square of a second order reactant. Mathematically, we say x is proportional to y^2 if x=Cy^2, for some constant C. My understanding is that this constant would depend on the details of the chemical species involved, but I believe it is incorrect to say that doubling a second order reactant always means the reaction goes exactly 4 times as fast. But I'm speaking mathematically, not as a chemist; perhaps the meaning is a little different in chemistry. SemanticMantis (talk) 17:04, 24 January 2014 (UTC)

@SemanticMantis: If y is doubled in $x=Cy^{2}$ then x is quadrupled regardless of the value of C. SpinningSpark 01:46, 25 January 2014 (UTC)

Polar Bears

If a few polar bears were introduced to Antarctica, what would happen to all the penguins? — Preceding unsigned comment added by 216.114.215.239 (talk) 04:15, 22 January 2014 (UTC)

As long as those polar bears are all male the answer is not much. 220.239.51.150 (talk) 05:46, 22 January 2014 (UTC)

And if they are all female? --Bowlhover (talk) 05:54, 22 January 2014 (UTC)

Still not much. The IP editor is suggesting that they couldn't mate and thus would die off after a few years. Dismas|^(talk) 07:59, 22 January 2014 (UTC)

Polar bears live for around 25 years...if there is going to be ecological damage, I'd expect it to happen long before enough of the bears died out to prevent it. SteveBaker (talk) 16:54, 22 January 2014 (UTC)

It depends greatly how you intepret 'a few'. I would intepret it to be a fairly low number less than 20. 10 polar bears could definitely cause great ecological damage, heck even if only 5 die they could also cause great ecological damage if left unchecked. I mean a single rat, cat or dog can devaste an unadapted population [1]. That said, it's a lot less likely than if you import a breeding group. Nil Einne (talk) 00:14, 23 January 2014 (UTC)

The penguins would be fine. The polar bears couldn't get the wrappers off. Rojomoke (talk) 06:33, 22 January 2014 (UTC)

Some joking is fine, but I think the main focus should be answering the poster's question. That's part of civility.76.218.104.210 (talk) 12:28, 22 January 2014 (UTC)

The joke isn't even that unique [2] Nil Einne (talk) 00:14, 23 January 2014 (UTC)

Uniqueness is an absolute, so something either is unique or isn't. This info should allow you to supply perfecter responses in the future. :-) StuRat (talk) 00:26, 23 January 2014 (UTC)

[3], [4], [5], [6], [7], [8], [9], Comparison (grammar)#cite_note-4. Nil Einne (talk) 08:23, 23 January 2014 (UTC)

It's pretty difficult to predict the effect of introducing a non-native species; the oft-quoted examples of brown tree snakes, rabbits in Australia and, to a lesser extent gray squirrels, suggest it may not be good news for the penguins, but food webs are complicated and who knows what trophic cascade may follow. benmoore 13:10, 22 January 2014 (UTC)

Hear, hear! USA government actually promoted planting kudzu in the south, to disastrous effects. Experts usually know enough to know that they cannot predict the long-term outcomes of population dynamics of endemic species when exposed to introduced species. Really, we don't even know if polar bears are any good at catching penguins ;) SemanticMantis (talk) 15:12, 22 January 2014 (UTC)

It's important to note that many (most?) species of penguin don't live anywhere near the antarctic - so while it's possible, and even quite likely that some of them would suffer - it's undoubtedly true that not all of them would even notice. For the antarctic penguins, let's consider what polar bears eat - and how they hunt. Here are some quotes from our Polar bear article:

• "Mature bears tend to eat only the calorie-rich skin and blubber of the seal, whereas younger bears consume the protein-rich red meat. Studies have also photographed polar bears scaling near-vertical cliffs, to eat birds' chicks and eggs. For subadult bears which are independent of their mother but have not yet gained enough experience and body size to successfully hunt seals, scavenging the carcasses from other bears' kills is an important source of nutrition." -- There are plenty of seals in the Antarctic, although not the familiar species that the bear prey on. So we might expect the antarctic seals to have a very hard time of things - they aren't adapted to avoid land-predators, their biggest problem normally being killer whales. Once they get over the 6 months of "jet lag" due to mid-winter being in July, I think this would give the bears the diet they need without too much trouble. Since we know they will take birds eggs and chicks, they might well cause the penguins problems - we know that penguins are completely unafraid of humans - so very likely they'd just stand around looking stupid while the bears grab their eggs and chicks.
• "Most terrestrial animals in the Arctic can outrun the polar bear on land as polar bears overheat quickly, and most marine animals the bear encounters can outswim it. " - so the bears won't be able to catch penguins in the ocean. Whether a penguin can out-run a bear is hard to guess.

My best read on this is that if you could transfer the bears without screwing up their body clocks and have them hibernate at the right times - then they'd survive just fine on seal meat...possibly thriving because the seals don't expect attacks when on land. I think the bears might then not need to resort to hunting penguins - and when they do, it would be chicks and eggs that they'd grab. Since leopard seals are a major predator of penguins - and the seals would be in huge trouble from the bears - the penguins might actually do rather well out of the deal.

Bears prefer to hunt in sea-ice - which will constrain them to the oceans' edge - so some penguins that do their breeding inland might also be OK.

CONCLUSION: Both bears and penguins probably do just fine...very, very bad news for the seal population though. SteveBaker (talk) 16:32, 22 January 2014 (UTC)

See also polar bear which mentions difference between Artic and Antartic seals believed to be partially the result of polar bear predation. I wouldn't BTW underestimate the possible effect of polar bears on Antartic penguins. Seals in the Antartic and of course the terrain are somewhat different from those in the Artic. And more significantly, many examples have shown predatorsorganisms are nothing if not good at adapting to new preyfood species they encounter in a new environment. Several commentators seem to agree [10] [11] [12] [13] [14] [15] [16] [17]. (The first source BTW may be a little confusing, the paper it refers to [18] [19] is not about such long range translocations and in fact specifically mentions polar bears to Antartica as something they are not proposing.) And of course, if they are too successful at hunting seals and you're right there's a good chance to think they will be, they could easily quickly devaste the seal population even more so if you import a breeding group. And as they devaste the seal population and their plentyful food supply runs out, if they weren't already hunting penguins there's a good chance they will start. Nil Einne (talk) 00:14, 23 January 2014 (UTC)

I agree - this is a complicated scenario. Whether the bears would be able to make a significant dent in the size of the seal population is very hard to estimate...and for the penguins, that's a critical thing. If the bears can be satiated from the existing seal population - depressing their numbers, but not eradicating them - then the penguins win because the bears have full stomachs and there are fewer seals predating on them. If the bears wipe out the seals completely, then they'll have to start eating penguins to survive - so that's really bad news for the penguins. Somewhere there is a tipping point where bears are eating penguins - but the number of seal predators is depressed enough to make up for those losses. But an artificially-created scenario isn't likely to produce that result naturally - so the entire eco-system is more likely to go into frequent boom/bust cycles. The predation rates in the natural population have evolved to produce a stable balance - but it could take a hundred thousand years for evolution to catch up with a sudden man-made change like this, and that might be enough time to see the extinction of seals, then penguins and soon after, the bears. SteveBaker (talk) 16:48, 23 January 2014 (UTC)

You should also consider the seals that would have been eaten by the whales anyway that the polar bears got to first (that's a wash). So not as much attrition as originally thought.165.212.189.187 (talk) 21:19, 24 January 2014 (UTC)

I doubt it's "a wash" - if there were a one in fifty chance per month of a particular seal being eaten by a whale - then the one in fifty chance that the polar bear got it first is a pretty small percentage. Also, the whales aren't just going to go hungry if they fail to catch a seal - they'll keep looking until the get one. Of course since starving the whales would also cause them to go for more penguins, it's an even messier food mesh than I'd previously suggested. Bottom line here is *still* that this is a highly speculative question, and while we can come up with scenarios that apply pressure to one part or other of the population - we don't have the data to know which ones are the driving forces. 02:54, 25 January 2014 (UTC)

Elephants

Are elephants aware that the main reason they are being hunted by elephant poachers is for the ivory?76.218.104.210 (talk) 12:26, 22 January 2014 (UTC)

It seems like you're asking about awareness on a cognitive level, which seems unlikely. But on a related note, (and I'm not aware of the actual research the press was based on) a few years ago it was claimed (e.g.) that due to the artificial selection imposed on elephants, killing those with the largest tusks, the mean population tusk length was decreasing. I'm not sure how they accounted for survivorship bias and it seems unlikely such a fitness benefit could be realised in such a short time, but then again I think dramatic morphological changes were noted in Atlantic cod in the North Sea before strict fishing restrictions were put in place, so who knows. benmoore 13:03, 22 January 2014 (UTC)

tevil::There is simply no way to know what the elephants may or may not be "aware" of. We can't know what any other beings are thinking - I don't know what my next door neighbor is aware of without asking, and how would you ask an elephant? You might possibly be able to prove that they were *UN*aware (eg if the poachers removed the entire elephant carcass before removing the tusks - you'd be fairly sure that the elephants wouldn't know whether they were after the ivory or...the toenails or something) - but even if they saw the tusk removal happening, it's not clear what they would or wouldn't be able to deduce from that. Worse still, you're asking for us to know whether the elephant knows what the poacher is thinking - and that's two levels of unknowability. SteveBaker (talk) 16:13, 22 January 2014 (UTC)

This made me think of the bear that watched its parents get shot and killed and then when faced with the hunter rolled over on its back as if to say "dont shoot me; it seems that you like us to lay still so here you go, but dont kill me" anyone hear of this account or is it some "rural" legend?165.212.189.187 (talk) 16:33, 22 January 2014 (UTC)

I guess what I meant is that it seems the bear deduced (albeit somewhat incorrectly) that the hunter wanted all the bears to "lay still" (be dead).165.212.189.187 (talk) 18:27, 22 January 2014 (UTC)

Consider the human equivalent. If random passersby were targeted by a psycho killer with a particular fixation (say for left-handed people with big noses), would they be able to decipher the reason from a tiny sample set? No. Clarityfiend (talk) 01:41, 23 January 2014 (UTC)

Whoa, why consider this obtuse example?? Consider .... me on a toilet. So what? Instead, Clarity, please give us some clarity as to why the bear acted the way it did. 165.212.189.187 (talk) 17:38, 23 January 2014 (UTC)

I thought there might be some ways a very elephant-knowledgeable and imaginative human in elephant territory might discern such awareness from elephants' day to day behavior, or devise a clever test for that kind of awareness. Also, I respectfully disagree with Clarityfiend. I dont think the analogy Clarityfiend provided is very close to what's been happening in Chad, according to Wikipedia.--Richard Peterson76.218.104.210 (talk) 09:33, 23 January 2014 (UTC)

How not? Is an individual elephant likely to witness (and survive) more than one hunt? Drawing conclusions from one such incident or even two would be hard for people. Why would it be any easier for elephants? Clarityfiend (talk) 11:50, 23 January 2014 (UTC)

Well if there were an enormous , well over 50%, decrease of human population by widespread slaughter, by a lot of hunters, and our bodies were left to rot, and all that was taken from us were our upper front teeth, those people remaining who came upon our dead bodies could notice. Likewise, living elephants could come upon herds of murdered detusked elephants more than once.76.218.104.210 (talk) 08:31, 24 January 2014 (UTC)

I think you mean hard for unimaginative pessimists who can't ever draw any conclusions from anything ever, because we cant really "know" anything.165.212.189.187 (talk) 17:40, 23 January 2014 (UTC)

Could we please keep this civil, 165.212.189.187?76.218.104.210 (talk) 08:31, 24 January 2014 (UTC)

@User:165.212.189.187 - that's ridiculous. There are plenty of things that we can know and conclusions that we can draw about the world - we know that there are elephants, we know that they die and we know that people sometimes kill them - and often that is for their ivory. There are all sorts of conclusions that we can draw about the economics of the ivory trade and how scarcity pushes up prices and so forth. But we do not (as yet) have a way to know what another being is thinking unless they speak to us about it (and even then, we can't be certain). We can't possibly know whether the elephants are able to deduce the stated conclusion from the available evidence. I'm not sure who you're calling an "unimaginative pessimist" - but that is an extremely rude violation of Wikipedia's No Personal Attacks rule - and I suggest that you apologize. Furthermore, answering a scientific question "optimistically" is a very bad idea - a healthy dose of scepticism is required, and the problem that there isn't a way to read the minds of elephants is hardly a matter of "imagination". We really can't know what the elephants are thinking - so we don't know whether they are intellectually capable of deducing the conclusion from the available evidence - we don't even know whether they think in ways we do - they might not care to seek a cause for these deaths. We simply do not know - and with current technology, cannot possibly know - and that's undoubtedly the correct answer to this question. Here on the reference desk, attacking another editor for the answer they give is considered rude - you may feel free to come up with a counter-argument, and I'm sure it would be carefull considered on it's merits - but name-calling isn't going to get you anywhere. SteveBaker (talk) 14:25, 24 January 2014 (UTC)

Steve, thanks for supporting my point that I was trying to make with reverse psyc. obviously we can know things. And I was not calling anyone in particular an unimaginative pessimist. But those that are would probably be the only ones who agree w/ clarity.165.212.189.187 (talk) 14:56, 24 January 2014 (UTC)

The proper term for SteveBaker's position here is philosophical skepticism, specifically Pyrrhonism, the belief we can have no certain knowledge, and perhaps solecism, the belief only one's own mind is real, not pessimism. As for unimaginative, I think the WP:OR he has presented is quite imaginative, especially since the only source or article he has quoted in this entire thread is WP:NPA. μηδείς (talk) 04:52, 26 January 2014 (UTC)

You mean solipsism, I think. "Solecism" is a solecism. Tevildo (talk) 15:45, 26 January 2014 (UTC)

Yes, lol, the belief that other people are grammatical errors. μηδείς (talk) 17:06, 26 January 2014 (UTC)

Does the bird know why the snake creeps into its nest filled with eggs?165.212.189.187 (talk) 17:45, 23 January 2014 (UTC)

• The loss (not merely reduction) of tusks by elephants is very real: see [20]. There are also certain places where tusklessness is almost universal, at least in a sex [21] those these are likely genetic drift. (Pure speculation follows ::) Still, this blogger I ran across [22] makes an interesting point: now that poaching is illegal, tusklessness seems to be increasing faster than ever. Now I think if you work out the math of the proportional increase, while mindful of the very heavy proportional toll of illegal poaching on elephants, that this is explainable in terms of direct selection. Still, I would not altogether rule out the possibility of demonstrating that elephants could somehow be sexually-selecting for a trait of tusklessness out of an 'awareness' that the tusks lead to an early death. Note that 'awareness' is not defined, at all, in any organism, and can occur at any number of levels, which may or may not be relatable to human thought. Wnt (talk) 14:31, 24 January 2014 (UTC)

Note that the elephants can (and indeed must) be selecting for less impressive tusks even if they are completely unaware of the problem. If elephants with large tusks are killed off more frequently, and earlier in life that those with smaller or slower growing tusks - then those carrying the "large tusk" genes will reproduce at a lower rate than those with the "small tusk" or "no tusk" genes. Evolution does it's thing and the tusks get smaller. It most certainly does not require the girl elephants to spurn the large-tusked boy elephants (and vice-versa) because of an awareness of the likelyhood of her babies being killed sooner. Indeed, that would assume that elephants have at least a passing understanding of genetic inheritance - which is even less likely (IMHO) than that they understand that tusk size relates to the likelyhood of humans coming to kill them.

That said, it's amazing that such evolutionary pressure could come so rapidly. Elephants remain in breeding condition for around 40 years - and the problem of ivory hunting has only been a significant impact on their population during the last 100 or so years. So only a few generations of selective breeding have occurred during that time. It seems unlikely that evolution could act so quickly on such a long-lived animal. However, the facts are the facts! SteveBaker (talk) 15:18, 24 January 2014 (UTC)

Oh no, I wasn't saying it was required - as I indicated, all that was just speculation. It's just that if you could show such an effect it would provide evidence for the OP's hypothesis. Elephants notably respond to skeletal body parts of one another after death^[23], so it isn't inconceivable they could get a clue, but definitely not necessary. Wnt (talk) 18:09, 24 January 2014 (UTC)

Lorentz transformation and Ampere's law

I have been thinking about the derivation of Ampere's law by applying the Lorentz transformation to Coulomb's law. My undergraduate physics text only treats the case of two equal currents with equal charge velocity, but I am having trouble getting the right answer for the more general case. We have two relevant articles, Relativistic electromagnetism and Classical electromagnetism and special relativity, but neither of them directly solves my problem. Here is what I have got;

Consider two parallel lines of charge a distance r apart with a charge density of Q₁ and Q₂ coulombs/metre respectively. In a frame of reference at rest to the charges, the Coulomb force between them is given by,

$F={\frac {Q_{1}Q_{2}}{2\pi \epsilon _{0}r}}$

If the charges are in motion axially (ie, are actually currents) we must apply the Lorentz transformation so that,

$Q'={\frac {Q}{\sqrt {1-v^{2}/c^{2}}}}$

and

$F'={\frac {Q'_{1}Q'_{2}}{2\pi \epsilon _{0}r}}={\frac {Q_{1}Q_{2}}{2\pi \epsilon _{0}r{\sqrt {(1-v_{1}^{2}/c^{2})(1-v_{2}^{2}/c^{2})}}}}$

Note that the velocities can be different even if the two currents are the same. This would pertain, for instance, if one conductor had a much larger cross-section than the other. The amount by which F' exceeds F is the Ampere's force,

$F_{\mathrm {m} }=F'-F={\frac {Q_{1}Q_{2}}{2\pi \epsilon _{0}r}}\left[{\frac {1}{\sqrt {(1-v_{1}^{2}/c^{2})(1-v_{2}^{2}/c^{2})}}}-1\right]$

Making the substitutions

$I=Qv$

and

$c^{2}={\frac {1}{\mu _{0}\epsilon _{0}}}$

gives

$F_{\mathrm {m} }={\frac {\mu _{0}I_{1}I_{2}}{2\pi r}}\left[{\frac {c^{2}}{v_{1}v_{2}}}\left({\frac {1}{\sqrt {(1-v_{1}^{2}/c^{2})(1-v_{2}^{2}/c^{2})}}}-1\right)\right]$

which is Ampere's law if the part in square brackets approximates to unity for small (non-relativistic) velocities. Unfortunately, it doesn't. Applying the binomial approximation

$(1+x)^{n}\approx 1+nx,\quad x\ll 1$

to the part in round brackets (I'm not showing all the detail here to keep it short) results in,

$[...]\approx \left[{\frac {c^{2}}{v_{1}v_{2}}}\left({\frac {1}{2}}{\frac {v_{1}^{2}}{c^{2}}}+{\frac {1}{2}}{\frac {v_{2}^{2}}{c^{2}}}\right)\right]$

$={\frac {v_{1}^{2}+v_{2}^{2}}{2v_{1}v_{2}}}$

which is unity when the velocities are equal but is clearly wrong - the expression goes to infinity if one velocity goes to zero. I would like to think that I have invented a new kind of motor with infinite torque, but in reality I have probably just made a mistake. SpinningSpark 12:39, 22 January 2014 (UTC)

$I=Qv$ isn't correct! The proper formula for the drift velocity is $J=\rho v_{\it {avg}}$, where J is the current density (current per unit *area*) and $\rho$ is the charge density (charge per unit **volume**). Alternatively, $Qv={I \over nA}$ where n is the charge carrier density. Not sure how to work those into your expressions, my physics is a touch rusty. MChesterMC (talk) 14:20, 22 January 2014 (UTC)

Starting from,

$J=\rho v$

which we both agree to be correct, and multiplying by cross-sectional area s,

$Js=I=\rho sv$

Since

$\rho s={\frac {Q}{s}}s=Q$

where Q is to be read as charge per unit length (a common enough contrivance in elementary textbooks to avoid having to explicitly include terms for length of the conductors) we have

$I=Qv$

as claimed. SpinningSpark 16:00, 22 January 2014 (UTC)

Sorry, but you made a wrong assumption. You can transfer from one frame of reference to another one, but you can not transfer to two frames of reference simultaneously! If charges are moving with different speeds they are in two different reference frames. So, the textbook is absolutely correct—you can only obtain the Amprere's law by frame transformation if velocities are equal. Ruslik_Zero 19:40, 22 January 2014 (UTC)

So what is the correct approach? I surely need the force between the conductors as observed in the rest frame. SpinningSpark 22:46, 22 January 2014 (UTC)

Incorrect answer
The following discussion has been closed. Please do not modify it.
One approach that works is as follows: So that there isn't any Coulomb force between the wires as measured in the lab frame, assume that in the lab frame, there is no net charge density on either wire 1 or wire 2. I.e., assume that on wire 1, in addition to the linear charge density $Q_{1}$ of charge carriers traveling at speed $v_{1}$ in the lab frame, there is also a linear charge density $Q_{1}^{*}$ of charges that are stationary in the lab frame, such that in the lab frame, $Q_{1}+Q_{1}^{*}=0$. Similarly, on wire 2, in addition to the linear charge density $Q_{2}$ of charge carriers traveling at speed $v_{2}$ in the lab frame, there is also a linear charge density $Q_{2}^{*}$ of charges that are at rest in the lab frame, such that in the lab frame, $Q_{2}+Q_{2}^{*}=0$. The total force on wire 1 due to the charge densities on wire 2 is then a sum of four terms, one for each possible combination of a charge density on wire 1 and a charge density on wire 2. Each term is a Coulomb force, except that the charge density on wire 2 needs to be increased by a Lorentz factor to account for the length contraction within the charge density on wire 2 as observed in a frame in which the charge density on wire 1 is at rest. Thus, for each term, the Lorentz factor for the term depends on the relative velocity between the two charge densities involved. The total force per unit length on wire 1 as measured in the lab frame is thus: $F_{m}={\frac {1}{2\pi \epsilon _{0}r}}\left(Q_{1}\gamma _{12}Q_{2}+Q_{1}\gamma _{10}Q_{2}^{*}+Q_{1}^{*}\gamma _{02}Q_{2}+Q_{1}^{*}\gamma _{00}Q_{2}^{*}\right)\ ,$ where $\gamma _{ij}={\frac {1}{\sqrt {1-(v_{i}-v_{j})^{2}/c^{2}}}}\ ,$ and for convenience we're defining $v_{0}=0$ for the speed of the lab frame as measured in the lab frame. But since the expression for $F_{m}$ is in the lab frame, we can substitute $Q_{1}^{*}=-Q_{1}$ and $Q_{2}^{*}=-Q_{2}$ to get the simpler $F_{m}={\frac {Q_{1}Q_{2}}{2\pi \epsilon _{0}r}}\left(\gamma _{12}-\gamma _{10}-\gamma _{02}+\gamma _{00}\right)\ .$ Doing a Taylor expansion of $\gamma _{ij}$ and ignoring terms that are more than quadratic in the velocities gives $\gamma _{ij}\approx 1+{\frac {v_{i}^{2}}{2c^{2}}}+{\frac {v_{j}^{2}}{2c^{2}}}-{\frac {v_{i}v_{j}}{c^{2}}}\ .$ Substituting this low-speed approximation for $\gamma _{ij}$ in $F_{m}$ and simplifying gives $F_{m}\approx {\frac {Q_{1}Q_{2}}{2\pi \epsilon _{0}r}}\left(-{\frac {v_{1}v_{2}}{c^{2}}}\right)\ .$ Making the substitutions $I=Qv$ and $c^{2}={\frac {1}{\mu _{0}\epsilon _{0}}}$ then gives $F_{m}\approx -{\frac {\mu _{0}I_{1}I_{2}}{2\pi r}}\ .$ Red Act (talk) 07:02, 23 January 2014 (UTC)

Thanks for that, and indeed my Physics text also only treats the case of overall neutral conductors as well. It would appear that Ampere's law fails in a charged medium such as a plasma or electron beam. Is that true? The equations I derived (and yours if you put Q*₁ = Q*₂ = 0) above imply that there is a "surplus" force even if one current is zero (contrary to Ampere's law). SpinningSpark 10:04, 23 January 2014 (UTC)

Thinking about this some more, the infinities that come out of my equation are not so surprising—a simple algebraic feature. Setting a velocity to zero while requiring a finite current will result in an infinite charge density so no surprise an infinite force is needed to move them. More useful results are obtained if the expression is rearranged into the form;

$F=F_{\mathrm {e} }+F_{\mathrm {m} }+F_{\mathrm {d} }$

Where $F_{\mathrm {e} }$ is the expected Coulomb's law force, $F_{\mathrm {m} }$ is the expected Ampere's law force, and $F_{\mathrm {d} }$ is a term representing the unexplained discrepancy. I make,

$F_{\mathrm {d} }={\frac {\mu _{0}(I_{1}-I_{2})^{2}}{4\pi r}}$

for non-relativistic cases (preceded by a minus sign in the convention used by User:Red Act). This is most certainly not either a Coulomb force nor an Ampere force. Setting one current to go to zero results in a force that is dependant on neither the charge nor the velocity of the other conductor. It appears to be a self-induced circularly compressive force in the current carrying conductor and does not require another conductor for it to appear. Is there such a force? It would, for instance, tend to reduce the dispersion expected in an electron beam due to space charge effects. SpinningSpark 15:51, 23 January 2014 (UTC)

Oops, sorry! My first two equations for $F_{\mathrm {m} }$ above are incorrect, in that they're missing some Lorentz factors. The equations resulted in the correct answer, so I assumed the equations were correct, but in actuality the equations only produced the correct answer because the missing Lorentz factors happened to all balance out in the case of no net charge on the wires. I'll start from scratch with a new approach:

I'm going to use $\lambda$ for the linear charge densities instead of $Q$, because that's a lot more common notation, and will hopefully result in less reader confusion. Charge and charge density behave differently under a Lorentz transformation, so it's really an important conceptual distinction in this problem.

Your first post above can also lead to confusion in that it uses the same indicator, a prime, to indicate two different coordinate systems. Instead, I will use unprimed values to indicate the lab frame, one prime to indicate a coordinate system in which the first line of charges is at rest, and two primes to indicate a coordinate system in which the second line of charges is at rest.

As before, the notation I'll use for Lorentz factors is

$\gamma _{ij}={\frac {1}{\sqrt {1-(v_{i}-v_{j})^{2}/c^{2}}}}\ ,$

where $v_{1}$ is the speed of the first line of charges as measured in the lab frame, $v_{2}$ is the speed of the second line of charges as measured in the lab frame, and for convenience we also define $v_{0}=0$ for the speed of the lab frame as measured in the lab frame.

In any of the three frames of reference, a charge density as measured in that frame will be a Lorentz factor greater than the charge density as measured in the frame in which that charge density is at rest, due to length contraction. In general, charge densities don't transform so simply between frames, because the spatial charge density and the temporal charge density (i.e. current) intermix in a transformation. But in the frame in which the (spatial) charge density is at rest, the temporal charge density is zero, so there isn't an extra term in the transformation in that case to account for a temporal charge density in one frame contributing to the spatial charge density in the other frame. In equations, the three versions of this rule we will use are:

$\lambda _{1}=\gamma _{01}\lambda '_{1}$

$\lambda _{2}=\gamma _{02}\lambda ''_{2}$

$\lambda '_{2}=\gamma _{12}\lambda ''_{2}$

Force perpendicular to the boosts transforms the same as the unit lengths, so their ratio is the same in all frames. Thus, the total force per unit length between the lines of charge, which in either the primed or double primed frame is considered to be purely a Coulomb force per unit length, is

$F_{t}=F'_{t}={\frac {1}{2\pi \epsilon _{0}r}}\lambda '_{1}\lambda '_{2}={\frac {1}{2\pi \epsilon _{0}r}}\lambda '_{1}\gamma _{12}\lambda ''_{2}={\frac {\lambda _{1}\lambda _{2}}{2\pi \epsilon _{0}r}}{\frac {\gamma _{12}}{\gamma _{01}\gamma _{02}}}$

The "magnetic" force between the two lines of charge per unit length in the lab frame is that total force per unit length minus the Coulomb force per unit length as measured in the lab frame,

$F_{m}={\frac {\lambda _{1}\lambda _{2}}{2\pi \epsilon _{0}r}}\left({\frac {\gamma _{12}}{\gamma _{01}\gamma _{02}}}-1\right)$

Using the Taylor expansions

$\gamma _{ij}\approx 1+{\frac {v_{i}^{2}}{2c^{2}}}+{\frac {v_{j}^{2}}{2c^{2}}}-{\frac {v_{i}v_{j}}{c^{2}}}$

${\frac {1}{\gamma _{ij}}}\approx 1-{\frac {v_{i}^{2}}{2c^{2}}}-{\frac {v_{j}^{2}}{2c^{2}}}+{\frac {v_{i}v_{j}}{c^{2}}}$

and discarding terms that are more than quadratic in the velocities gives

$F_{m}\approx {\frac {\lambda _{1}\lambda _{2}}{2\pi \epsilon _{0}r}}\left(-{\frac {v_{1}v_{2}}{c^{2}}}\right)\ .$

Making the substitutions $I=\lambda v$ and $c^{2}={\frac {1}{\mu _{0}\epsilon _{0}}}$ then gives $F_{m}\approx -{\frac {\mu _{0}I_{1}I_{2}}{2\pi r}}\ .$

Red Act (talk) 04:45, 28 January 2014 (UTC)

Outstanding!! SpinningSpark 09:39, 28 January 2014 (UTC)

What is Abrogation?

I know the term abrogation in its colloquial and legal senses. But I see that it is widely used in scientific contexts, too, particularly in molecular biology, where it seems to refer to some kind of canceling or preventing of the expression of genes. I've Googled variously but been unable to find a decent definition or explanation of its technical nuances. Anyone resolve the mystery?—PaulTanenbaum (talk) 18:23, 22 January 2014 (UTC)

Don't have a detailed explanation, but Oxford English Dictionary (1997) gives for 'abrogate': "3. Immunol. To suppress or prevent (a physiological process)." Abrogation in this sense first attested in 1959 in Nature: "Abrogation by injected mouse blood of protective effect of foreign bone marrow in lethally X-irradiated mice.". - Lindert (talk) 20:44, 22 January 2014 (UTC)

You pretty much have it. It doesn't refer to any specific biological or experimental process in my experience, and simply refers to the state of some element, not necessarily genetic, not being present or functional. Also, as far as I am aware it is always used in practise to refer to an experimental manipulation, or a mutation, natural or otherwise, and never to a process natural to a wild type organism (for which you might use "repress" or "inhibit" instead). For example you could say "the expression of gene A was found to be abrogated in mutant B" (referring to the expression of a gene not being present any longer) or "inhibitor X was found to completely abrogate the function of protein Y" (referring to the function of a protein not being present any longer). Equisetum (talk | contributions) 11:37, 23 January 2014 (UTC)