July 28 Information

Can you plant and grow bing cherries or "sweet cherries" in the temperate region of the Northern hemisphere in your backyard, assuming that you own a property of a suburban land division? How long does it take to grow the cherry tree to its first fruit? Will it ever produce fruit by itself, or do you need two trees to produce offspring? How do you pollinate each flower? If it produces fruit through parthenocarpy, then would the offspring fruit be seedless like navel oranges? Sneazy ( talk) 00:50, 28 July 2013 (UTC) reply

I typed "How to grow bing cherries" into Google, and got a multitude of sites which seem to answer most of your questions. -- Jayron 32 01:14, 28 July 2013 (UTC) reply

The multitude of questions are really linked or are supposed to fall under one big inquiry - the parthenocarpy of cherry trees. Sneazy ( talk) 01:41, 28 July 2013 (UTC) reply

No, Parthenocarpy is something entirely different. What you seem to be asking about is Self-pollination. Cherries are self-pollinating, so you can grow a single tree and get fruit. Although parthenocarpy does occasionally occur in cherries, and you might find a seedless fruit or two on a tree, there are no cherry varieties that consistently produce only seedless fruit. A truly parthenocarpic variety like navel oranges doesn't have "offspring". They can only be propagated by cloning. The rest of your questions are specific to your geographical area and can be answered by Googling, or contacting your local extension agent. Dominus Vobisdu ( talk) 07:30, 28 July 2013 (UTC) reply

See Bing cherry if (like me) you've never heard of them. Alansplodge ( talk) 16:51, 29 July 2013 (UTC) reply

I have read that two vectors cannot be divided. Why is it so? Here is something contradictory - according to the Newton's second law of motion, force (a vector) divided by acceleration (a vector) gives mass (a scalar). What is the reason behind this? Publisher54321 ( talk) 03:23, 28 July 2013 (UTC) reply

In the case of Newton's 2nd law, you divide the vectors' magnitudes. However, given that vectors can be adequately represented as 1-row or 1-column matrices, which can't be divided, there really isn't such a thing as vector division. For example, I know of no such operation that would return a "quotient" in a way that reverses the vector cross-products or dot products. It seems like you can only do this to reverse multiplication of a scalar by a vector.-- Jasper Deng (talk) 03:48, 28 July 2013 (UTC) reply

As Jasper was saying, division is the inverse operation of multiplication. Cross product and dot product are two forms of multiplication for vectors, but neither has an inverse. Bubba73 ^{You talkin' to me?} 04:12, 28 July 2013 (UTC) reply

In general, vectors don't have a form of division. However, there are vector spaces that do admit division, like the complex numbers. Other examples of vector spaces that may have some form of division are: polynomials over a ring, complex valued functions from a topological space, formal power series, differentiable functions, analytic functions, many others. You may want to read our articles Vector_Space#Algebras_over_fields and Algebra over a field; granted, you aren't guaranteed division. If you are talking, strictly, about the geometric concept of vectors as arrows, then there is no natural, or obvious, division. It's important to remember that "multiplication" is a certain general type of combining objects and that "division" is the inverting of this, so there is no reason to expect that things called a "product" are going to have some nature of inverse operation. Phoenixia1177 ( talk) 04:19, 28 July 2013 (UTC) reply

With my vague understanding of set theory, the layman's answer is basically that much rests on the definition of "multiplication" used, and the more complex your data type gets, the harder it is to make a definition of multiplication and have a compatible definition of division.-- Jasper Deng (talk) 04:29, 28 July 2013 (UTC) reply

Yes and no. Division isn't really a thing; there may be multiplicative inverses, or multiplication may be cancellable, but there isn't a clear "division operation" in most cases. As for commonality, it depends on how far you cast your net. Every field is a vector space over itself and has division. More abstractly, if we can consider elements of a Module (mathematics) to be vectors, then every abelian group is a Z-module, so every field is a Z-algebra with multiplicative inverses. On the other hand, if we mean real vector in euclidean space considered as an arrow, then there isn't any obvious geometry based division. Finally, the obvious examples of vector structures that have some form of division, and aren't really abstract at all, would be the complex numbers; and slightly more abstract, the Quaternions and the space of continuous functions from the reals into the reals. Ultimately, the answer depends on just what you allow "vectors" to be; I would say the lay answer would be that vectors don't divide unless you provide additional structure. Phoenixia1177 ( talk) 06:09, 28 July 2013 (UTC) reply

Geometric algebra is one framework that systematically allows a vector to be divided by another vector, to produce a multivector that contains a scalar and a bivector component. One application is to define a geometric calculus that combines the vector-calculus operators div and curl into a single object, that works in arbitrary numbers of dimensions, and that can be inverted in ways that relate to Stokes' theorem and the Cauchy integral theorem. This derivative operator also allows Maxwell's equations of electromagnetism to be written in a particularly concise form. Complex-number division and quaternion division fall out as the specialisation of the system to 2 real dimensions and 3 real dimensions respectively. Jheald ( talk) 14:46, 28 July 2013 (UTC) reply

In Newton's 2nd Law, Force and Acceleration are not independent vectors. The law applies only to vector components that have the same direction. DreadRed ( talk) 13:41, 28 July 2013 (UTC) reply

There are some good-looking answers at [1] - apparently you can define a vector C = "A / B" such that B × C = A, but the problem is that there is no "1" such that "1" x A = A, which somehow degrades the system. I'm sure there are better things written about this! Wnt ( talk) 14:50, 28 July 2013 (UTC) reply

As indicated by Jheald above and Muphrid in Wnt's math.stackexchange link here, there is a natural manner in which vector division is defined. In effect, one must be a little more flexible about what our "numbers" are allowed to be (in addition to the already allowed scalars and vectors). The dot and cross products each discard necessary information, but together they constitute a complete product that can be inverted; in particular, left- and right-division by vectors are not only possible, but are standard operations. — Quondum 15:34, 28 July 2013 (UTC) reply

When you say "a complete product that can be inverted", does that mean there is an identity element? 170.170.59.138 ( talk) 23:29, 31 July 2013 (UTC) reply

When an Alzheimer's disease patient dies, how is it decided whether the disease caused the death? Or in other words, how is it decided whether someone died of Alzheimer's? Of course I understand that it's sometimes easy, e.g. the patient is killed in an automobile accident, but I'm addressing a situation in which the proximate cause of death is something physiological. Bringing this up because of Category:Deaths from Alzheimer's disease. Neither Douglas Engelbart nor Auguste Deter are in the category, even though Engelbart's "death came after a long battle with Alzheimer's disease", and Dr Alzheimer's description of the disease was based largely on his observations of its effects on Deter. Moreover, the category says "People who died with Alzheimer's disease" — with, not from, so it doesn't help me understand the definition of "dying from Alzheimer's". Nyttend ( talk) 11:42, 28 July 2013 (UTC) reply

Alzheimer's mostly kills by infections due to loss of basic motor and reasoning skills. Patients fall and suffer fractures, in a later stage they won't be able to walk, become incontinent, have difficulty eating and swallowing; these lead to bedsores, bladder infections, pneumonia caused by inhaled food. Treatment becomes more difficult because the patient can't cooperate. ( alzheimers.co.uk slate.com: how does alzheimers kill) Ssscienccce ( talk) 12:15, 28 July 2013 (UTC) reply

On Death and Dying; Causes of Death in Alzheimer's Disease Bus stop ( talk) 12:17, 28 July 2013 (UTC) reply

Yes, my grandfather had Alzheimer's, became bedridden and immobile, and died of pneumonia. μηδείς ( talk) 14:31, 28 July 2013 (UTC) reply

Can we get to the original question? I'm only asking about the medical classification of the deaths of Alzheimer's patients as being "caused by Alzheimer's" or "not caused by Alzheimer's", and seeking to apply it to our categorisation of people who had the disease before their deaths. Nyttend ( talk) 18:59, 28 July 2013 (UTC) reply

To clarify, although it's only anecdotal, the cause of death in my grandfather's case was indeed listed as pneumonia (he died after a week with a with a fever) although he had been declining for a decade. (I remember being quite surprised at the time that this, and not Alzheimer's was listed as the cause). In the case of my grandmother it was listed as stroke, even though in her case the immediate cause was aspiration pneumonia caused by an inattentive attendant. My suspicion is that just looking at death certificates or press reports you may not get a very accurate picture. If, as I understand, your concern is for wikipedia purposes, it might be better to classify as dying suffering the disease, rather than from it. Or perhaps look at AIDS and give death from complications of Alzheimer's. μηδείς ( talk) 01:27, 29 July 2013 (UTC) reply

"Cause of death" is, really, a legal definition, rather than a medical one. The purpose of that box on the death certificate is to make sure that the appropriate legal steps have been taken - if a homicide is suspected, that the police have been involved; if an accident took place, that the appropriate investigators have examined the circumstances; and if a disease was involved, that a report was made that to the public health authorities (who are interested in epidemic diseases like cholera and chronic ones like Alzheimer's). I apologise if that sounds like pedantry, but when you see "cause of death" statistics, for different jurisdictions and for different times, it's important to realise that many are happy to simply record "homicide / accident(misadventure) / disease / old-age" and leave it at that. For old age, and diseases related inextricably to it (of which various neuro-degenerative ailments like Alzheimer's are prime examples) there's a tricky legal and statistical problem - an 80 year old dies with a panoply of disease, a number of which may contribute to their death, but putting the blame on one is difficult. A typical 80 year old might have high blood pressure, cardiovascular and pulmonary dysfunctions and diseases, one or several latent carcinomas, and various organic or endocrine diseases or dysfunctions. If that person has Alhzeimer's too, and they die, how should the death certificate be written, and how should their death be recorded in the country's death statistics database? The UK's Alzheimer's Society says "Although dementia is a life-shortening illness, another condition or illness (such as pneumonia...) may actually cause a person's death. This other condition or illness will most likely be listed as the cause on the person's death certificate. Pneumonia is listed as the ultimate cause of death in up to two-thirds of people with dementia." The UK has undertaken a process to improve the determination listed on death certificates, as described in this document for England (similar changes were undertaken in Scotland and I think in Wales and Northern Ireland). Those changes try to capture, on the death certificate, more of the underlying cause, and not just to have an unhelpful determination like "old age" or "organ failure". They want doctors to list both the proximal cause (pneumonia, organ failure, heart failure, infection) and the underlying cause - and for someone elderly, that underlying cause might well be several serious complaints, occurring together, of which Alzheimer's is but one. For almost all elderly people in poor health, no detailed post-mortem pathological examination is performed - there's no legal basis for doing so (legally it doesn't matter which disease killed them) and medically little reason (because they've followed the downward path most people do). From a scientific perspective, if you were to do the most thorough examination of each person, it's still not clear what you should really record. Looking at the body of the deceased person, and running every scientific test available, you'd end up with a laundry list of contributory factors: the observable neural effects of Alzheimer's; the consequences of poor diet and hydration; general diseases often associated with age; and infections. Just as asking why the Roman Empire failed, it's a facile TL;DR to say "Visigoths" just as the death certificate has the doctor to write "pneumonia subsequent to Alzheimer's disease" - from a practical perspective, that's a sufficient cause of death. -- Finlay McWalterჷ Talk 22:33, 28 July 2013 (UTC) reply

• "The most frequent cause of death for people with AD is aspiration pneumonia. This type of pneumonia develops when a person is not able to swallow properly and takes food or liquids into the lungs instead of air." [2] -- Anthonyhcole ( talk · contribs · email) 00:05, 29 July 2013 (UTC) reply

100% percent of pneumonia deaths are caused, most immediately, by brains rotting due to lack of oxygen. Granted, 100% of all deaths are caused by brain damage, so that kind of specificity actually makes the matter vaguer. But, very basically, even those without Alzheimer's die from the same effects (though maybe not the same causes, and certainly quicker).

As Finlay McWalter said, it's not always clear where to draw the line between proximate and immediate causes, and no wide standard. Depending how you look at it, the most significant contributor to death is staying alive. InedibleHulk (talk) 03:33, 31 July 2013 (UTC) reply

Ah, the dangerous Life. Sexually transmitted and 100 % fatal. Sjö ( talk) 11:13, 1 August 2013 (UTC) reply

What would be the differences if earth was pyramid? Assume a pyramid with same volume. 201.78.161.229 ( talk) 18:39, 28 July 2013 (UTC) reply

It would collapse under its own weight and form a sphere.-- Gilderien Chat| List of good deeds 18:42, 28 July 2013 (UTC) reply

(... assuming it was made of stronger stuff ...) The moon would have an even more irregular orbit that is does with the current oblate spheroid shape. Dbfirs 19:46, 28 July 2013 (UTC) reply

Why would that be of the center of gravity does not change? Dominus Vobisdu ( talk) 19:57, 28 July 2013 (UTC) reply

Because the separation between the Earth and the Moon is not sufficent, compared to their radii, for them to be considered as point masses, so tidal forces become significant. These depend on the geometry of the orbiting bodies. Tevildo ( talk) 20:39, 28 July 2013 (UTC) reply

All the water and air would flow to the centre of the faces, so there would be circular shaped oceans surrounded by a ring of atmosphere, but away from there the elevation would be so high and atmosphere too thin. It would be impossible for living things to travel between the faces through the cold vacuum. Graeme Bartlett ( talk) 20:52, 28 July 2013 (UTC) reply

It would be part of some "some think its funny" nonsense comedy movie/novel like The Hitchhiker's Guide to the Galaxy (novel) instead of being real. -- Kharon ( talk) 23:22, 28 July 2013 (UTC) reply

Quite so Kharon. And we'd all have to eat Toblerone Martinevans123 ( talk) 23:38, 28 July 2013 (UTC) reply

And, anywhere on "land", you would literally fall sideways. Towards the water. - ¡Ouch! ( ^{hurt me} / _{more pain}) 09:19, 29 July 2013 (UTC) reply

Well, lets assume the Earth as a tetrahedron (it's got to be Platonic) with the same mass (or volume or surface area) as the real Earth, made from Unobtainium of uniform density and braced with polarity-reversed tachyon shielding to keep it from collapsing. What would the gravity field look like? If I start in the middle of one of the sides, and walk towards one of the edges, do I walk up the potential well, or down? Intuitively, the gravity field should be weakest at the 4 corners, or do I have that wrong? -- Stephan Schulz ( talk) 09:36, 29 July 2013 (UTC) reply

Stephan, to determine the gravitational force exerted by a tetrahedron shape of uniform density (or of any known density function), you have to solve Newton's law of gravitation in its integral form. This is a straightforward and classical mathematics problem; there is no "trick" or "gotcha"; it's just a volume integral of a vector function with somewhat ugly limits of integration to constrain the mass to a tetrahedron-shape. Depending how familiar you are with integral calculus, you should be able to solve (or make a computer solve this for you). Such integral field equation solution problems are typical "homework" questions in Marion & Thornton's classical physics text (they have a chapter explaining how to perform these types of calculations for gravity). In my younger years I could probably crunch the numbers on such a problem in ten or twenty minutes, or maybe faster; but now that I'm out of practice, it'd probably take a bit longer.

If I were forced to analyze the problem "intuitively" rather than analytically, I would try to reason about the magnitude of forces based on Gauss's law for gravity, and regarding the direction of forces, I would consider the symmetry of the tetrahedron. The force of gravity falls off as r², but the mass increases as r increases (following the integral of the intersection of a sphere and the tetrahedron - which is less than, but upper-bounded by, r³ - and therefore this is not trivial!) At the corners, your test-point is quite a bit farther away, but the gaussian pillbox it defines is only enclosing slightly more mass; and because a sphere is a convex hull around the tetrahedron, I intuitively suspect the increase in enclosed mass is small and negligible. So, the center-of-each-face ought to have a stronger field. I would defer to a complete analytic integral solution for the field and its potential-function, though; I'm having trouble "intuiting" the gradient of the magnitude of the integral of a three-dimensional geometric intersection. (I suspect most people also have trouble with that "intuition"). Nimur ( talk) 16:15, 29 July 2013 (UTC) reply

• To put it briefly, that the force of gravity would be less at the vertices would matter less than the fact the direction the gravity would be pulling you would not be "down" (normal to the face) but sideways, toward the center of the face. μηδείς ( talk) 16:26, 29 July 2013 (UTC) reply
  • Not towards the centre of the face, but at an angle to the face (maybe between 45 and 60 degrees in the "habitable zone", assuming the sea doesn't spread too far towards a vertex). Houses and roads would have to be built at an angle between 30 and 45 degrees to the face, but that's no worse than building on a steep hillside. (I haven't tried Nimur's integral, and I'm only guessing how the water would behave, so I'm happy to have my estimates corrected by a rigorous mathematical analysis) I'm struggling to estimate the 3-D shape of the four seas. Each would form some sort of dome. Dbfirs 16:59, 29 July 2013 (UTC) reply

The angle would differ depending on where you were, but the net drag would be toward the center of the face, in the same way that when falling down a mountain, one falls toward the foot of the mountain, although "straight" down would be into the mountain. One might as well view the tetrahedron as four mountains. The center of each face would be the nadir of a three-branched valley formed between any three of the mountains. μηδείς ( talk) 19:20, 29 July 2013 (UTC) reply

Yes, a component of gravity towards the centre of the face, but we wouldn't need to "fall down". I've climbed and traversed mountains steeper than that. I think the "mountains" would be too big to perceive them in the same way as we do Earth mountains, and the four oceans would hide the "valleys" from being noticed. Life would be lived on a steep slope, and the (imaginary) inhabitants would adapt. Dbfirs 07:50, 31 July 2013 (UTC) reply

In deference to better physicists than myself, that's not "Nimur's integral," insofar as anyone may "own" an integral, it's probably either Green's or Gauss's integral (or maybe Laplace... I'm struggling to recall my history of famous formulations of classical physics). Suffice to say, every student of physics and mathematics ought to have performed this integral, for at least the last hundred and fifty years of physics pedagogy. I do not recollect that Isaac Newton ever performed the integral to solve for the potential for an arbitrary mass-density distribution; he was far too practical to consider such nonsense.

Regarding calculating the height of the seas: I would first assume that the mass of the ocean is negligible compared to the mass of the planet; and then I would assume that sea level follows the gravitational equipotential; the solution for which we would obtain by solving the same integral mentioned above. Were we to get serious about things, we would have to assume the mass of the ocean is non-negligible and iterate the solution numerically. Nimur ( talk) 17:12, 29 July 2013 (UTC) reply

That's the approach I was thinking of, but it's too long ago since I did serious Maths or Physics. Dbfirs 07:50, 31 July 2013 (UTC) reply

Back to the "What would be different..." part, there would be more differences than just a crazy gravity field.

Total Lunar eclipses would be slightly more rare because of the shape of Earth's shadow, and solar eclipses would be slightly less rare. The additional eclipses would, however, only shadow the tips where there is no atmosphere anyway: a purely theoretic result.

Scratch the above. Both, lunar and solar eclipses would be significantly more common, because...

The Moon would spiral out from Earth even more slowly than it does now (because the tetrahedron (terrahedron?) is assumed to be rigid), and thus Earth would spin faster, and the Moon would be closer. On top of that, the Moon would have a shorter orbit, so there would be more chances at eclipses.

The seas would be so deep that the water near the bottom would be solid (a kind of ice different from the one we have on Earth).

Planet Earth would have ceased to be a planet in 2006, because planets must have assumed a hydrostatic equilibrium, and we have assumed that the terrahedron is strong enough to withstand gravity. Defeat by definition.

And finally, there would still have been believers in a flat Earth. Some things just don't change. - ¡Ouch! ( ^{hurt me} / _{more pain}) 05:35, 30 July 2013 (UTC) reply

Forgive me for my understanding of these concepts which I have picked up from popular science books. For the photons of light to travel at the speed of light, they have to have zero mass right? How then is a black hole able to bend light? Surely gravity can only pull or affect something that has mass and therefore weight? Sandman1142 ( talk) 22:06, 28 July 2013 (UTC) reply

It has zero rest mass, true, but since it travels at the speed of light, it has relativistic mass. Alternatively, think of a black hole curving space-time, and a photon travelling along a straight path across a curved surface.-- Gilderien Talk| List of good deeds 23:01, 28 July 2013 (UTC) reply

If you try to calculate the deflection of light due to a gravitating body by plugging the photon's relativistic mass into Newton's law of gravitation, you get an answer that's off by a factor of two. Newtonian gravity just becomes too inaccurate of a model to use when dealing with objects moving at or near the speed of light. To get the right answer when dealing with very fast particles, you need to use general relativity. And from the perspective of general relativity, whether a particle has a mass or not is irrelevant to the question of whether the particle is affected by gravity. A hypothetical particle that has even a zero relativistic mass would still have the same world line as a photon, as long as the hypothetical particle also travels at the speed of light of course, and as long as neither the photon nor the hypothetical particle undergo any proper acceleration (e.g., the photon doesn't go through a lens or something). What happens with either particle is the gravitating body curves spacetime, and the particle travels in a straight line along that curved spacetime. The particle's world line is determined entirely by the geometry of spacetime as affected by the gravitating body, and the particle's initial conditions of direction and speed. The particle's mass never enters into consideration in the least.

A particle that isn't affected by gravity wouldn't even be a meaningful concept in general relativity. In Newtonian gravity, it would mean a particle that moves in a straight line relative to a global inertial frame of reference. But a global inertial frame of reference is just a bit of fiction that works well as an approximation under appropriate circumstances. In reality, no such global inertial frame of reference exists, because spacetime isn't flat, due to the curvature of spacetime due to gravitating bodies.

It may help toward understanding this to point out that in the view of general relativity, gravity isn't a proper force. When gravity appears to act as a force, it's really just acting as a fictitious force due to using an accelerating frame of reference. E.g. when an apple falls to the earth, the apple isn't really accelerating toward the Earth, in that in the apple's proper frame, which is an inertial frame of reference, the apple is just standing still. It's really the frame of reference that's attached to the surface of the Earth that's accelerating upward, which is making the apple appear to be accelerating downward. All small objects appear to accelerate toward the earth at the same rate in a vacuum regardless of what the object's mass is or what it's made of, because the acceleration involved is really the acceleration of the reference frame attached to the Earth, not anything having to do with the small object. If you want, you can multiply that acceleration by the small object's mass to give a "force", but in determining the dynamics of the object you're just going to turn around and divide that "force" by the same mass of the object in order to determine the acceleration anyway, so you might as well not bother multiplying the acceleration by the small object's mass in the first place. The acceleration is really the more fundamental thing that's actually going on, not the fictitious "force". This is basically the equivalence principle of general relativity. Red Act ( talk) 06:57, 29 July 2013 (UTC) reply

Thank you for the answers and especially the link to the helpful relativistic mass article. Sandman1142 ( talk) 15:36, 29 July 2013 (UTC) reply

With Newton's second law and the work-energy principle, derivation for motion along a straight line is straightforward. But I ran into problems when trying to derive it more generally by taking the line integral (over the particle's path C):

${\displaystyle K=\int \limits _{C}\mathbf {F} \cdot d\mathbf {L} =\int \limits _{0}^{t}\mathbf {F} \cdot \mathbf {v} dt=m\int \limits _{0}^{v}({\frac {\mathbf {v} \cdot d\mathbf {v} }{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}+{\frac {v^{3}dv}{c^{2}({\sqrt {1-{\frac {v^{2}}{c^{2}}}}})^{3}}})}$

Now there's no question about the second term, which is purely a scalar function. But it seems like the correct formula ${\displaystyle K={\frac {mc^{2}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}-mc^{2}}$ is only obtained when ${\displaystyle \mathbf {v} }$ and ${\displaystyle d\mathbf {v} }$ are in the same direction, which is not necessarily true. I know I must be doing something wrong here, or otherwise I misunderstand the meaning of that dot product (which seems to be treated as the same as the scalar product ${\displaystyle vdv}$ in the article on kinetic energy). What is it?-- Jasper Deng (talk) 23:35, 28 July 2013 (UTC) reply

IIRC, "v squared" is a scalar value. One can not take a square root of a vector AFAIK. To get a particle to move in other than a straight line requires a force of some sort - such as gravity. Collect ( talk) 00:22, 29 July 2013 (UTC) reply

But that's not very helpful. When v is written without being boldened, we mean its magnitude only, without regarding its direction (hence why the integral cannot be evaluated by separating components of v, since the magnitude of the whole vector is in the denominator). In addition, when you take the dot product of a vector with itself, you get the same as its magnitude squared. Also, a derivation of kinetic energy has to be valid for any net force and path - Newton's second law and the definition of work make no distinction between different kinds of forces.-- Jasper Deng (talk) 00:24, 29 July 2013 (UTC) reply

Although it's counter-intuitive, the following might be a partial answer:

${\displaystyle m\int \limits _{0}^{v}{\frac {\mathbf {v} \cdot d\mathbf {v} }{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}={\frac {mv^{2}}{2{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}-{\frac {1}{2}}mv^{2}-\int \limits _{0}^{v}{\frac {mv^{3}dv}{2c^{2}({\sqrt {1-{\frac {v^{2}}{c^{2}}}}})^{3}}}}$

It seems like using the integration of parts eliminates the dependence on direction. But it still seems quite strange.-- Jasper Deng (talk) 03:15, 29 July 2013 (UTC) reply

I think the numerator of the second term should have been ${\displaystyle v^{2}(\mathbf {v} \cdot d\mathbf {v} )}$ instead of ${\displaystyle v^{3}dv}$. -- BenRG ( talk) 03:30, 29 July 2013 (UTC) reply

Well, I omitted the step where I differentiated the momentum expression with respect to time, using a product rule (differentiating the vector and scalar components separately). In the first term, I differentiated the vector v in the numerator with respect to time. In the second term, I differentiated the denominator with respect to time, so in that term the vector v in the numerator is unaltered. Taking its dot product with the vector v in the integrand produces a purely scalar-valued expression:

${\displaystyle {\frac {d\mathbf {p} }{dt}}\cdot \mathbf {v} =({\frac {d\mathbf {v} }{dt}}{\frac {m}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}+{\frac {mv\mathbf {v} }{c^{2}({\sqrt {1-{\frac {v^{2}}{c^{2}}}}})^{3}}}{\frac {dv}{dt}})\cdot \mathbf {v} }$

which, when the dot product is evaluated and multiplied by dt, produces the integrand.-- Jasper Deng (talk) 03:58, 29 July 2013 (UTC) reply

Now that I think about it, ${\displaystyle \mathbf {v} \cdot d\mathbf {v} =|\mathbf {v} |\,d|\mathbf {v} |}$, so your original integral is the same as ${\displaystyle m\int \limits _{0}^{v}({\frac {v\,dv}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}+{\frac {v^{3}dv}{c^{2}({\sqrt {1-{\frac {v^{2}}{c^{2}}}}})^{3}}})}$ and you must have made a mistake in evaluating it. -- BenRG ( talk) 05:13, 29 July 2013 (UTC) reply

No. It's not that. I tried to evaluate it last night and found that the correct formula comes if I can use your assumption. But I'm not sure if you can assume that. In an extreme case, what if the change in velocity were almost orthogonal to the velocity at that instant? Remember, in magnitude form, the dot product of two vectors is the product of their magnitudes and the cosine of the angle between them - the assumption seems to be only valid if that cosine is 1, which would essentially be the two vectors having the same direction.-- Jasper Deng (talk) 05:22, 29 July 2013 (UTC) reply

But BenRG isn't just making an assumption which may or may not be valid depending on the path. It's always true in general that ${\displaystyle \mathbf {v} \cdot d\mathbf {v} =|\mathbf {v} |d|\mathbf {v} |}$, because ${\displaystyle |\mathbf {v} |d|\mathbf {v} |=(\mathbf {v} \cdot \mathbf {v} )^{1/2}d(\mathbf {v} \cdot \mathbf {v} )^{1/2}=(\mathbf {v} \cdot \mathbf {v} )^{1/2}{\frac {1}{2}}(\mathbf {v} \cdot \mathbf {v} )^{-1/2}d(\mathbf {v} \cdot \mathbf {v} )={\frac {1}{2}}(\mathbf {v} \cdot d\mathbf {v} +d\mathbf {v} \cdot \mathbf {v} )=\mathbf {v} \cdot d\mathbf {v} }$. Red Act ( talk) 21:48, 29 July 2013 (UTC) reply

It can be proven much simpler if one notes that by definition ${\displaystyle \mathbf {v} ^{2}=|\mathbf {v} |^{2}}$. Differentiation of both side of this equality leads to ${\displaystyle \mathbf {v} \,d\mathbf {v} =|\mathbf {v} |\,d|\mathbf {v} |}$. Ruslik_ Zero 09:51, 30 July 2013 (UTC) reply