Science desk
< June 24	<< May \| June \| Jul >>	June 26 >

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

June 25

Species in which females are aggressive and males passive

Are there any species in which females are aggressive while males are passive?Uncle dan is home (talk) 04:49, 25 June 2017 (UTC)[reply]

The spotted hyena is a species of hyena that is very unique in the class Mammalia in that the females of the species have fewer nipples , lack an external vaginal opening, and possess a pseudo-penis. (Essentially they possess a very large clit; keep in mind the anatomical and development similarities between the clitoris and the penis.) The species operates in a matriarchal social system; the females are larger than the males and dominate over them. Even the lowest ranking females of the group have dominion over the highest ranking males. The female spotted hyenas tend to stick together over a large period of time while the male does not often form strong social ties to the den and is more migratory. A mother hyena is extremely protective of her cubs especially towards male hyenas. Penile erection is a sign of submission the male displays to a female hyena. Additionally, the spotted hyena is considered to be one of the most intelligent species of mammal in the world, ranking as smart or smarter than many species of primate. UltravioletAlien (talk) 06:18, 25 June 2017 (UTC)[reply]

Homo sapiens? Greglocock (talk) 06:45, 25 June 2017 (UTC)[reply]

Many species of hymenoptera. In bees, for example, the males (drones) only exist to mate, then they die. The female queen and workers run the place. StuRat (talk) 14:36, 25 June 2017 (UTC)[reply]

Popular culture is well acquainted with the black widow; among insects I think the preying mantis also. But the anglerfish is stranger than them all. Wnt (talk) 16:36, 25 June 2017 (UTC)[reply]

Aren't there some ~~grasshoppers~~ where the female decapitates the male after mating ? "Would you care to come in for a decap... uh, nightcap, I mean." StuRat (talk) 05:40, 26 June 2017 (UTC)[reply]

Actually Preying mantids, although subsequent research showed that this happens far more frequently when they're in captivity than in the wild, as the article discusses. {The poster formerly known as 87.81.230.195} 2.221.82.167 (talk) 11:44, 26 June 2017 (UTC)[reply]

Thanks, that's what I was thinking of. StuRat (talk) 05:43, 27 June 2017 (UTC)[reply]

Physics: Can a changing spacetime be used to generate unlimited energy?

I discovered that energy is not conserved in general relativity after reading this by Sean Carroll. He seems to argue that energy is not conserved if spacetime is changing.

The work done on a mass by a force F over a distance L is

$\int _{0}^{L}F\cdot dx$

If energy is not conserved in general relativity due to a changing spacetime, I wondered if it could be exploited to generate unlimited energy. The first example I considered was Hubble's law. Hubble's law says

${\dot {x}}=Hx$ .

Taking the derivative of both sides gives

${\ddot {x}}=H{\dot {x}}=H^{2}x$ .

Now, by Newton's second law and special relativity

$F={\frac {m}{\sqrt {1-{\frac {{\dot {x}}^{2}}{c^{2}}}}}}{\ddot {x}}$ .

But work can only be recovered up to the Hubble radius, so

$L={\frac {c}{H}}$

Substituting all of that and evaluating the integral gives

$\int _{0}^{\frac {c}{H}}{\frac {m}{\sqrt {1-{\frac {(Hx)^{2}}{c^{2}}}}}}H^{2}x\cdot dx=mc^{2}$

according to WolframAloha.

The energy generated is exactly the energy content of the mass used to generate that energy!

In fact, if you use different rules for the expansion of space (e.g. ${\dot {x}}=Hx^{2}$ ), you get the same answer ( $mc^{2}$ ). Here's the WolframAlpha for the example ${\dot {x}}=Hx^{2}$ .

This means you can't really use the expansion of space to generate energy, as the energy content of the mass used to generate that energy is the same as the amount of energy generated (and that mass disappears over the horizon once it reaches c/H).

That made me wonder if mass really IS conserved in a sense in general relativity, or if there is some other law preventing the generation of unlimited energy.

I know only the basics of physics. If you are an expert on physics (or if you know someone who is an expert on physics), can you let me know if a changing spacetime can or can't be used to generate unlimited energy? Perhaps a generalization of my reasoning above?

PeterPresent (talk) 09:41, 25 June 2017 (UTC)[reply]

The blog piece you quote seems very misleading. I mean, directly contrary to what he says, there is a difference between saying that energy is not conserved, versus saying that energy is conserved when the negative energy of the gravity field is taken into account. The difference is that you understand right off with the second option that if you make any circular set of changes to the gravity field, if in the end you have the same field you started with, you have the same energy you started with.

But the key issue here is what he says near the end:

First, unlike with ordinary matter fields, there is no such thing as the density of gravitational energy. The thing you would like to define as the energy associated with the curvature of spacetime is not uniquely defined at every point in space. So the best you can rigorously do is define the energy of the whole universe all at once, rather than talking about the energy of each separate piece. (You can sometimes talk approximately about the energy of different pieces, by imagining that they are isolated from the rest of the universe.) Even if you can define such a quantity, it’s much less useful than the notion of energy we have for matter fields.

If that is true, then it certainly seems like a worthy challenge for theoreticians. Surely there must be some way to divide up this energy neatly, without making any simplifying assumptions ... mustn't there? And if you have that math worked out, it seems like there ought to be other uses for it. Wnt (talk) 16:49, 25 June 2017 (UTC)[reply]

law of conservation of energy only applies to isolated system. A system in a changing gravity field (i.e., a changing spacetime, in relativity framework) is NOT isolated, and can gain (or lose) energy (for instance : pendulum energy is mgh, if g increases, so does energy). No big deal, even in a newtonian framework.
This law of conservation itself derives from the principle of least action, that general relativity also respects (ie: Einstein had not to get rid of this principle, he only had to adopt Einstein–Hilbert action instead of previous action (physics)).

So you wont generate energy through changing spacetime: you need as much energy to change space-time as it would get through the change. No free lunch (according to current theory, that seems to work pretty well, so we have no reason to change).

Gem fr (talk) 14:21, 26 June 2017 (UTC)[reply]

Feynman Lectures. Exercises PDF. Exercise 7-4 JPG

copyvio, see talk

The following discussion has been closed. Please do not modify it.

. .

...
DELETED We now turn to the case of elliptic orbits. This is essentially about three ellipses: both bodies move along ellipses (light-heavy, heavy-on-small) and, in addition, the relative motion of bodies also occurs along an ellipse (see Lectures, Chapter 7 ). All three ellipses are similar to each other, that is, they share the same eccentricity. If we also take into account that the center of mass of the system remains stationary (it lies in the common focus of the orbits of both bodies), and the distances from the center of mass to the bodies are inversely proportional to their masses, we come to the conclusion that the arrangement of bodies and their orbits is as in figure.

Denoting by v_1 and v_2 the velocities of the bodies M_1 and M_2 at the time when they are at apogee. As can be seen from the figure,

${\tfrac {v_{1}}{v_{2}}}={\tfrac {a_{1}+c_{1}}{a_{2}+c_{2}}}={\tfrac {a_{1}(1+e)}{a_{2}(1+e)}}={\tfrac {a_{1}}{a_{2}}}$
(Indices 1 and 2 denote quantities related to ellipses along which the masses M_1 and M_2 move).

To get the same expressions for elliptical orbits as for circular ones, remember that an ellipse can be imagined as a circle visible at some angle to its plane or (which is the same thing) as the projection of a circle onto an inclined plane. In other words, an ellipse can be obtained from a circle if you change the scale along one of the coordinate axes. Acceleration of the body while moving it along the circumference was calculated in "Lectures" (fig. 7.4). To obtain the acceleration of the body (for example, M_1) in the case of interest to us, imagine that its orbit is obtained from a circular increase in the scale in the "vertical direction" a_1 / b_1 times. The value of x will not change, and s will increase and become equal to s₁ = (a₁ / b₁) s.
Substituting x² = 2Rs (valid for the circle) their values after increasing the scale x₁ = x, s₁ = (a₁ / b₁) s and R = b₁ (The "horizontal" dimensions have not changed, so the small semiaxis of the ellipse is equal to the radius of the original circle), we obtain

$x^{2}=2{\tfrac {b_{1}^{2}}{a_{1}}}s_{1}$

Thus, the radius of curvature of the ellipse at the point of intersection with the semimajor axis is ${\tfrac {b_{1}^{2}}{a_{1}}}$ . Assuming that for a very small period of time the first body moves along a circular orbit of this radius, we can write

${\tfrac {v_{1}^{2}a_{1}}{b_{1}^{2}}}={\tfrac {GM_{2}}{(a+c)^{2}}}={\tfrac {GM_{2}}{a^{2}(1+e)^{2}}}$

(Here a and c are the orbital parameters of the relative motion of the bodies: a = a_1 + a_2, c = c_1 + c_2). Similarly for the second body

${\tfrac {v_{2}^{2}a_{2}}{b_{2}^{2}}}={\tfrac {GM_{1}}{a^{2}(1+e)^{2}}}$

Adding the last two equalities and expressing v_2 in terms of v_1, we obtain

${\tfrac {v_{1}^{2}(1+e)}{a_{1}^{2}(1-e)}}={\tfrac {G(M_{1}+M_{2})}{a^{3}}}$
It remains only to find out what relation the quantity on the left-hand side of this equation has to the period of revolution. First of all, note that the area that the radius vector of the body M_1 (drawn from point 0) "sweeps out" per unit time is 1/2 v_1 (a_1 + c_1) = 1/2 v_1a_1 (1 + e). Although in fact we calculated the rate of change of the "swept out" area for the moment when the body M1 is at apogee, according to Kepler's second law, this speed does not change when the body moves in orbit. Therefore, the value 1/2 v_1a_1 (1 + e) T (here T is the period of revolution) is equal to the area of the orbit of the body M_1. The area of the ellipse is easy to calculate if you realize that when you zoom in on one of the axes, the area of the figure increases by the same factor as the scale. Therefore, the area of the ellipse is
$\pi b_{1}^{2}{\tfrac {a_{1}}{b_{1}}}=\pi a_{1}b_{1}=\pi a_{1}^{2}{\sqrt {1-e^{2}}}$

It is now easy to see that
$T={\tfrac {2\pi a_{1}}{v_{1}}}{\sqrt {\tfrac {1-e}{1+e}}}$ , and $T^{2}={\tfrac {4\pi ^{2}a^{3}}{G(M_{1}+M_{2})}}$

— MEPhI , Solutions (Google Translate)

Apparent WP:COPYVIO, see talk μηδείς (talk) 23:27, 25 June 2017 (UTC)[reply]

Here is my solution png. The question is can we use 3rd Kepler's law for relative circular orbit and relative elliptical orbit? Or the 3rd Kepler's law is just approximation.

Username160611000000 (talk) 12:46, 25 June 2017 (UTC)[reply]

See Kepler's third law -- it is an approximation. Wnt (talk) 16:53, 25 June 2017 (UTC)[reply]

Hm, they uses 2nd Kepler's law during the proof. So we cannot derive the exact formula for circular orbit and then extrapolate the approximation formula on case of elliptical orbits, can we? Username160611000000 (talk) 09:46, 26 June 2017 (UTC)[reply]

The second law should not be an approximation so long as the right center is used - the actual barycenter in the case of a two-body problem, rather than the center of the larger body. This is because of conservation of angular momentum. Naturally, any torque from additional masses would tend to throw it off. Wnt (talk) 17:12, 26 June 2017 (UTC)[reply]

Looking at the problem myself, you have a force of GmM/D^2 pulling inward. Note I use D for the distance between them which is only approximately r (the radius of the circle) for the smaller mass. The centripetal acceleration needed is -w^2*r, where w is angular velocity (I'll be lazy and avoid the omega). The second is a vector; the first is a magnitude, so I'll wave my hands and make the "-" go away, then convert acceleration to force by multiplying by the mass, leaving me with GmM/D^2 = mw^2*r for the little mass = and GmM/D^2 = Mw^2*r for the big one. But what is r? Well, it's a (literally) weighted average of the full distance D for one mass and 0 for the other, i.e. (0*m + D*M)/(m+M) or (D*m + 0*M)/(m+M). The first (DM/m+M) defines r=0 at the small mass and the second (Dm/m+M) puts r=0 at the large. Either way we flip that and get GmM/D^2 = mw^2*DM/(m+M) = Mw^2*Dm/(m+M), depending on which mass you consider. So G(m+M)/D^3 = w^2. The period is 2 pi w, so this is a period inversely proportional to D^(3/2) and directly proportional to the square root of m+M.

Note that we use the inaccurate Kepler's third law version in the first section of orbital period, without warning people it is wrong, even saying that mu depends on the 'more massive body' rather than 'total mass'... then get out the right version in the third section down. I think I actually tried to fix that once and got rebuffed, but there is an article that looks way more mysterious than it ought to if anyone wants to hack at it. Wnt (talk) 17:44, 26 June 2017 (UTC)[reply]

Bringing home the process of making industrial nutritional yeast - how to do it?

I know very little about the industrial process of making nutritional yeast other than the reading on Wikipedia. Some kind of unnamed strain is used to culture the yeast cells on top of the agar plate, I presume. The agar plate probably contains glucose from molasses or sugarcane, which is food for the nutritional yeast. Then, the yeast cells are heat-deactivated, harvested, washed, dried, packaged, and sold in grocery stores. In a home kitchen, an agar plate can probably be replaced by a tomato sauce jar lid. Maybe the agar can be replaced by gelatin (may be easier to buy than agar). The person may want to heat-sterilize whatever metal (has to be metal, as plastic will just melt) instrument he's using and layer the Baker's yeast (same species, but may be a different strain) on the culture plate with the sterilized rod. The culture plate (lid) may be screwed back on the tomato sauce jar container, but the container is placed upside-down, because of the yeast cells. Incubate (in the conventional oven?) for 24 hours or until the culture plate seems to be covered by yeast. Then, the person should select for "desirable" phenotypic characteristics. But, what is considered desirable? Maybe it's best to harvest everything? For the washing process, a cheesecloth, perhaps, may be used to filter the yeast, but that assumes that the yeast will not fall through the holes of the cheesecloth. For drying, the yeast cells may be gathered into a jar. The jar is mostly sealed except for a space to allow a blow-dryer in and dry the yeast cells. Afterwards, the yeast may be used as topping on foods.

I have a number of concerns. (1) Does the strain of yeast really matter? Can Baker's yeast strain replace the Nutritional yeast strain? (2) Is the umami flavor of Nutritional yeast caused by the glutamate release of the cells? (3) Will the protocol above actually work as planned? Are there any holes or false assumptions (such as the cheesecloth which may have holes that are too large for the yeast cells)? (4) What "desirable" characteristics are people selecting for when they make nutritional yeast on an industrial scale? 50.4.236.254 (talk) 19:09, 25 June 2017 (UTC)[reply]

As a side question, what is the formal job title of the person who cultures the plates industrially? How can one apply to such a position at a company? 50.4.236.254 (talk) 19:21, 25 June 2017 (UTC)[reply]

I'm sure, without even looking, that industrial yeast is raised in a broth and filtered out. I'll take a look a bit later. Abductive (reasoning) 21:27, 25 June 2017 (UTC)[reply]

As you know there are different strains of yeast. So the starting culture has its growing medium, tailored to suit the type of yeast that one wants to be in abundance ( by adjusting the mediums pH and things), (which insistently, did not require modern science – just experience gained over many generations). By this process, one can produce starting cultures selected for either bread making, wine, beer, larger, etc. This was achieved long before the invention of the microscope. Aspro (talk) 23:34, 25 June 2017 (UTC)[reply]

Just so you know, I don't think that's an answer to my question. My question is not about the different strains of yeast. I already know that. Nor is it about the process of making different strains of yeast in a pre-modern fashion. It is, at the most basic level, about whether deactivated baker's yeast/brewer's yeast/wild yeast can replace "nutritional" yeast and still have that umami-rich flavor. 50.4.236.254 (talk) 11:27, 26 June 2017 (UTC)[reply]

[Adding the 'close small' coding that 50.4.236.254 forgot, thus rendering the entirety of the following queries small. {The poster formerly known as 87.81.230.195} 2.221.82.167 (talk) 11:39, 26 June 2017 (UTC)][reply]

A liquid growing medium should be far more productive that a plate. Grow the yeast in a jar full of a sugar solution with added micronutrients. Adapt methods used for making ginger beer to optimize yeast growth. The yeast will settle in the bottom as brown "sludge". Roger (Dodger67) (talk) 08:12, 27 June 2017 (UTC)[reply]