Astrophysics
  • Astrophysics
Observe only, please!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Astrophysics
  • Astrophysics
Evaluate the integral \[\int\limits \textbf A \times \textbf A'' dt\]
Astrophysics
  • Astrophysics
@Empty
Empty
  • Empty
This one is one of my favorites. Hahaha

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Astrophysics
  • Astrophysics
Show me
Empty
  • Empty
Hint:\[ \textbf A' \times \textbf A' = 0\]
Empty
  • Empty
This is a fun problem, I actually just found this in a book and was stumped for a whole day last week solving it, so this is a nice coincidence.
Astrophysics
  • Astrophysics
So can I integrate A and then differentiate A'' at the same time
Empty
  • Empty
Hahaha I guess I came from this when considering a physics problem: we can calculate torque as the derivative of angular momentum. Therefore the integral of torque is angular momentum. So from this definition of angular momentum: \[\bar L = \bar r \times \bar p\] Take the derivative of \(\bar L\) to show that \[\frac{d \bar L}{dt} = \bar \tau\] where \[\bar \tau = \bar r \times \bar F\]
Astrophysics
  • Astrophysics
Using that differentiating trick you once posted, where you can differentiate in the integral, and yeah haha it's in my physics textbook, but we never did this chapter, it's one of the most difficult I read a bit of it and did not understand a thing, this is where I learnt vectors are transformations and stuff...haha.
Empty
  • Empty
That's ok, just calculate this derivative of this function, and we'll go from there: \[\bar L = \bar r \times \bar p\] L is the angular momentum, r is the position, p is the momentum.
Astrophysics
  • Astrophysics
\[\frac{ \vec dL }{ dt } = \frac{ \vec dr }{ dt } \times \frac{ \vec dp }{ dt }\] should I be doing this
Astrophysics
  • Astrophysics
\[\vec r \times \vec F = \vec r \times \vec p\]
Empty
  • Empty
You're on the right path, now you just need to separate it out like you would the product rule: \[(fg)' = f'g+fg'\] Similarly we have \[(\bar a \times \bar b)' = \bar a ' \times \bar b + \bar a \times \bar b '\] Also we can sorta just see this is true by remembering: \[\bar a \times \bar b = | a| |b| \sin \theta \] If we interchange the order of a and b, then we will be changing the orientation of the angle between them to be in the opposite direction: \[\bar b \times \bar a = | a| |b| \sin -\theta = -| a| |b| \sin \theta \] so we can sorta see that we have discovered \[\bar a \times \bar b = - \bar b \times \bar a \] Which is the "right hand rule", a result of seeing that it is dependent on an odd function like this. Sorta weird but this is also the same reason interchanging rows of a matrix's determinant causes the sign to flip as well. The determinant is just like a more generalized cross product... Not to overwhelm you or anything! But you often see determinants used for calculating cross products normally, so this shouldn't be too much of a stretch and the linear algebra knowlege will come eventually. :)
Empty
  • Empty
I see what you're doing, and it looks like you're sorta skipping ahead so let's try to focus on just this single step: \[\frac{d}{dt} (\vec r \times \vec p) = \] Apply the product rule here, you will get a sum of two terms.
Astrophysics
  • Astrophysics
\[\frac{ d }{ dt }(\vec r \times \vec p) = \vec r~' \times \vec p + \vec r \times \vec p ~ '\]
Empty
  • Empty
PERFECT!!! I think the main problem is you know what the result should be so you're frustrated because what you're getting right now isn't that because there are some tricky steps. I remember us having this problem before, you're anticipating what it should be when you should just let yourself pretend you're dumb temporarily and don't know what's coming haha XD
Astrophysics
  • Astrophysics
Lol I'm actually not, pretty happy about this, I don't mind making mistakes in this kind of stuff, it's pretty hard and the only way I'll learn it is through the struggle :P
Empty
  • Empty
Ok so now let's resolve these two terms, what do we have, can we rewrite these at all?
Empty
  • Empty
I think we can use that good thinking! Let's look at the first term and plug this in: \[\vec p = m \vec v\]
Astrophysics
  • Astrophysics
Oh ok lol I tried it out, but wasn't sure it was the right way, and yeah haha that sounds good
Empty
  • Empty
So what does multiplying a vector quantity by a scalar quantity do to it? Describe it visually, not mathematically.
Astrophysics
  • Astrophysics
Well if we multiply a scalar by a vector it gives us a scale factor of the vector without changing the direction
Astrophysics
  • Astrophysics
So |dw:1436087673043:dw|
Empty
  • Empty
Yeah, definitely, so this gives us the special property that we can pull it out of the cross product. The cross product is the area of a parallelogram and scaling the vector is the same thing as scaling the entire area: |dw:1436087867828:dw| Scaling the vector from the larger vector to the smaller vector scales the area of the cross product by the same proportion. And since we can consider this as simply just scaling the proportion it's not actually scaling dependent upon the particular vector it scaled. The simplest example would be looking at length times width of a rectangle. if you scale the length or scale the width by the same number you'll get the same area: (s*x)*y = x*(s*y) = sxy So in the terms of cross products: \[\vec a \times (s \vec b) = s(\vec a \times \vec b) = (s \vec a) \times \vec b \] We will use and that other property you mentioned earlier axb=-bxa to simplify this left hand term: \[\vec r' \times \vec p \] Try to see if you can figure it out.
Astrophysics
  • Astrophysics
The noodle is gone, haha alright
Astrophysics
  • Astrophysics
\[\vec r~ ' \times \vec p \implies \vec r~' \times (m \vec v) = m( \vec r~' \times \vec v) = (m \vec r~') \times \vec v\]
Empty
  • Empty
Good, wait what's \(\vec r \prime \)?
Astrophysics
  • Astrophysics
I want to say velocity
Astrophysics
  • Astrophysics
Since it's a position vector so the derivative would be velocity
Empty
  • Empty
You got it. :) Keep going see if you can discover anything else to evaluate this term.
Empty
  • Empty
Hint: use the fact that you mentioned earlier
Astrophysics
  • Astrophysics
Kk, but should I leet r' = v?
Empty
  • Empty
You bet
Empty
  • Empty
Let it be what it is. :P
Astrophysics
  • Astrophysics
\[\bar a \times \bar b = - \bar b \times \bar a\] \[(m \vec v) \times \vec v = - \vec v \times (m \vec v)\]
Empty
  • Empty
Awesome. :) Now play around with this thing you just found with that other thing you learned about the scalars.
Astrophysics
  • Astrophysics
then let that = p
Empty
  • Empty
nooo
Astrophysics
  • Astrophysics
lolol
Astrophysics
  • Astrophysics
Yeah haha ok
Empty
  • Empty
Really the main point of this: \[\vec a \times (s \vec b) = s(\vec a \times \vec b) = (s \vec a) \times \vec b \] is that we can reduce to this form in the middle from either one of the end terms: \[s(\vec a \times \vec b) = s \ \vec a \times \vec b \] That way we think of only the vector part as being connected, while the scalar can freely multiply either if we like, but we can drop the parenthesis while doing algebra because we understand it has this nice property of just scaling the entire area and doesn't really matter the vector it came from. This is called "bilinearity" but don't worry about it. :P
Astrophysics
  • Astrophysics
As in linear on one side
Empty
  • Empty
nooo linear on BOTH sides, just like bicycle has two tires, bilinear is linear on both terms. when I write: \[s(\vec a \times \vec b) = s \ \vec a \times \vec b \] I don't associate s with the vector a, I just am writing it as a coefficient on the object \[\vec a \times \vec b \] which is the area. It's just commonly written that coefficients go on the left, like you write: \[ax^2+bx+c\] instead of \[x^2a+xb+c\] They're equivalent! You just think one is more awkward than the other by the same reasoning. :P
Astrophysics
  • Astrophysics
Yeah I figured it out after I read it the second time, it's just the cross multiply that had me all woozy
Astrophysics
  • Astrophysics
Ok so I use the same property for scalars again?
Empty
  • Empty
Yeah, so pull the mass scalar out of the equation
Astrophysics
  • Astrophysics
\[(m \vec v) \times \vec v = - \vec v \times (m \vec v) \] \[(m \vec v) \times \vec v = -m \vec v \times \vec v \] so something like this
Empty
  • Empty
Something like that, you should end up with something that has no parenthesis. So that second thing you wrote looks good, but throw away the parenthesis on the left side of the equation too.
Astrophysics
  • Astrophysics
Right right
Astrophysics
  • Astrophysics
\[m \vec v \times \vec v = -m \vec v \times \vec v\]
Empty
  • Empty
Ok I am going to stop hand holding now, this simplifies to something by normal algebra.
Empty
  • Empty
I'm gonna be a wrinkled old man before I get an answer out of you, is this how it's gonna be? :P
Astrophysics
  • Astrophysics
Lol I'm using the draw tool to figure it out
Empty
  • Empty
Oh show me
Astrophysics
  • Astrophysics
If it's angular momentum, I'm not getting that
Empty
  • Empty
I don't know, show me
anonymous
  • anonymous
Hmmm, well if \((A\times B)' = A'\times B+A\times B'\), then I guess\[ (A\times A')' = A'\times A' + A\times A'' \]So if you integrate both sides: \[ \int A\times A''~dt = A\times A' - \int A'\times A' ~dt \]It's like integration by parts I guess.
Empty
  • Empty
smh @wio
Astrophysics
  • Astrophysics
Nooo, I don't even know if I can do the regular algebra with the cross product so I've been trying to apply the algebraic properties of cross product
anonymous
  • anonymous
Hmmm? I think in general \(\mathbf v \times \mathbf v = 0\), right? Because vectors are parallel to themselves.
Empty
  • Empty
Great, thanks a lot @wio for wasting our time. :/
Astrophysics
  • Astrophysics
|dw:1436091897061:dw|
Astrophysics
  • Astrophysics
No I was actually using this https://web.viu.ca/wattsv/math200/Overheads/section10.4/algcrossproduct.pdf to go over the properties
Astrophysics
  • Astrophysics
Because I wasn't sure if you meant regular algebra or cross product algebra haha, so I was trying all these rules
Astrophysics
  • Astrophysics
anyways let me work this out
Astrophysics
  • Astrophysics
|dw:1436092214752:dw| I don't know if this is entirely right
Empty
  • Empty
You have written the right thing for the wrong reasons.
perl
  • perl
.
Empty
  • Empty
You are not allowed to use the rule \[\vec v \times \vec v = 0\] Because you have not discovered it yet.
Empty
  • Empty
But this problem is where that rule comes from.
Astrophysics
  • Astrophysics
Yeah, I was struggling when I was just using the other two rules so I had to look up rest of the properties, of course I remember them from calculus haha but I wanted to know all the properties, maybe I shouldn't have skipped ahead.
Empty
  • Empty
You have already shown that: \[m \vec v \times \vec v = - m \vec v \times \vec v\] Take your mind out of the details and just think practically about what this must mean. What does it mean for something to be equal to the negative of itself?
Empty
  • Empty
Solve this algebra equation I am just making up right now: x=-x What's x?
Astrophysics
  • Astrophysics
x=0
Astrophysics
  • Astrophysics
omggg
Astrophysics
  • Astrophysics
This is just like line integrals
Astrophysics
  • Astrophysics
I was focusing so hard on the math didn't even pay attention to the concept, figures.
Empty
  • Empty
They both have the same property, called skew symmetry. Skew symmetry just means when you interchange two quantities the new thing you get is negative. For instance \[f(x,y)=x^2y^3-x^3y^2\] is a skew symmetric function in the variables x and y, you should show that \[f(x,y)=-f(y,x)\]
Astrophysics
  • Astrophysics
Ah interesting, so whenever for example lets just say QM concepts they talk about symmetries do they mean this?
Empty
  • Empty
Yes, in some cases this is part of it. Specifically fermions have this skew symmetric property when you interchange them while bosons are symmetric. You can think of symmetric as just like skew symmetric, only you don't get a negative sign: \[f(x,y)=xy\] So the symmetric property is satisfied by: \[f(x,y)=f(y,x)\]
Astrophysics
  • Astrophysics
Fascinating
Astrophysics
  • Astrophysics
Can we finish this problem though
Astrophysics
  • Astrophysics
I want to see where the torque and what not comes in where you were going with it and what exactly this 0 = 0 has to do with this problem
Astrophysics
  • Astrophysics
Sorry not 0 = 0, the skew symmetry
Empty
  • Empty
Ok so back to the original place we were at. I'll recap what we've done since this sorta was a long path here!
Empty
  • Empty
I started by saying the angular momentum was the cross product of position and linear momentum. \[\vec L = \vec r \times \vec p\] By the product rule we got: \[\frac{d \vec L}{dt} = \vec r \prime \times \vec p + \vec r \times \vec p \prime \] Now we've reasoned out that the first term is zero! \[\frac{d \vec L}{dt} = \vec r \times \vec p \prime \] Now I claimed the derivative of angular momentum was torque! I'll remind you the definition of torque is: \[ \vec \tau = \vec r \times \vec F\] So we are almost there it appears to the definition, what is left to acknowledge before we can say for certain that \[\frac{d \vec L}{dt} = \vec \tau \] is true?
Astrophysics
  • Astrophysics
I think we should figure out \[\frac{d \vec L}{dt} = \vec r \times \vec p \prime\] or we can use \[\vec p~' = \frac{ d \vec p }{ dt }\] which equals the force
Empty
  • Empty
Yeah exactly, you use that to say the derivative of momentum is force. That completes the proof!
Astrophysics
  • Astrophysics
Very cool haha, now how does this relate to the original problem xD
Empty
  • Empty
So why bother with all this? Because we have shown this to be true we can write: \[\vec \tau = \vec r \times \vec F\] using the simple identity force is mass times acceleration. \[\vec F = m \vec r '' \] Remembering that scalars come out of cross products due to our bilinearity (fancy word for saying that we can multiply a number by a length or width of an area and it scales the area by the same amount) we talked about earlier: \[\vec \tau = m \vec r \times \vec r ''\] Since we just showed this was equal to the derivative of angular momentum: \[\vec L ' = m \vec r \times \vec r ''\] If we wanted to integrate torque it would look similar to our problem we began with, don't you think? :D
Empty
  • Empty
The next step to solving this problem is "fundamental". ;)
Astrophysics
  • Astrophysics
Haha nice, now we apply FTC
Astrophysics
  • Astrophysics
WOW, so what we actually just did is go backwards!
Empty
  • Empty
So you have more than just solved your problem, you have now been shown a physical application of this specific integral which means this is more than just a mathematical curiosity! Lucky! You will have to spend some time wracking your mind on this to really understand it because you don't have a real grasp of cross products yet. But you're getting there and this is a great step towards understanding. :D
Empty
  • Empty
By the way here's a tiny snippet from a quantum mechanics pdf I just randomly found about the Pauli exclusion principle describing symmetry like we discussed earlier in case you were curious: http://prntscr.com/7p0mrq
Astrophysics
  • Astrophysics
Yeah, this is a very nice haha, I'll try doing the original problem on my own now and going over everything we did. Very cool
Empty
  • Empty
There's more than one way to solve this problem too, different ways to think about it. But yeah this is a fun problem definitely, good title. ;P
Astrophysics
  • Astrophysics
I think we can also use this skew symmetry with classical energy and the one that Dirac used (relativistic) which would give us the skew velocities if we solved for the square root, neat stuff.
Astrophysics
  • Astrophysics
skew symmetry velocities*
Empty
  • Empty
You shouldn't really have to memorize any of the identities, you should have the most basic concept that the cross product gives you exactly this: |dw:1436094560604:dw| It gives you a vector with magnitude that is the area of the parallelogram made by the vectors and in a direction perpendicular to them in a right handed way. So if we think of bxa instead of axb it will have exactly the same area but pointing "upside down" relatively speaking, which is what the negative sign is all about.
Astrophysics
  • Astrophysics
I never fully understood cross product, so the cross product is always perpendicular of what's being "crossed"
Astrophysics
  • Astrophysics
I guess that would make sense because a and b are in a plane where then your cross product is then in R^3
Empty
  • Empty
Well there's a bit of a lie in all this. A cross product is something that strictly exists in 3D space and in an ugly sense it exists in 2D space as well. It's actually a pseudovector, something that's the "dual" of a vector. Don't worry about this. Just remember the right hand rule as this for now and think of the sort of rotational aspect of this: |dw:1436094887352:dw|
Astrophysics
  • Astrophysics
|dw:1436095034791:dw|
Empty
  • Empty
Maybe because it's bothersome to leave you hanging on something so interesting, I'll just say this: The cross product has the magnitude of an area because there are things that aren't scalars and vectors. There are other objects. A scalar is a 0 dimensional object, a point. A vector is a 1 dimensional object, a line segment. A bivector is a 2 dimensional object, a patch of area. So these all have magnitudes, but an area isn't quite a vector. It just turns out our three vectors x, y, and z can all be cleanly mapped to its "dual" which represent the basis bivectors which are just unit areas of the xy, xz, and yz plane. So we map the xy one to a z unit vector, etc... like this. This is also why if you take the curl of a 2D vector field you get 2 components that are 0 and one component in the "z" direction pointing out of the plane. It's really just a pseudoscalar value because there is only one area element in 2D. Don't worry about this too much, I am only saying this for the fact that I find it fun and interesting.
Empty
  • Empty
Ahahahaha XD
Empty
  • Empty
You really wanna burn your noodle, check this out. He starts out with some good stuff. You know Maxwell's equations? Well it turns out if you represent them in tensors or clifford algebra, which are part of what's called differential geometry, it turns out that Maxwell's equation is really just a single equation. http://www.av8n.com/physics/clifford-intro.htm
Astrophysics
  • Astrophysics
That's a nice way to put it, so when we are making traces for example for triple integrals or what ever, we're actually mapping it using basis bivectors...hope I used the right terminology haha.
Astrophysics
  • Astrophysics
WHAT IS A TENSOR
Astrophysics
  • Astrophysics
https://www.youtube.com/watch?v=f5liqUk0ZTw
Empty
  • Empty
Something you are working towards. Imagine using matrices to do calculus but the matrices are all not just squares, they could be cubes of numbers or higher.
Empty
  • Empty
Yeah I remember watching that. That's kinda what tensors are, but he misses some important stuff about what really makes a tensor a tensor. There's a special quality they have such that they represent invariant objects in space independent of coordinate systems. Moral of the story is you need to understand linear algebra first. ;) I don't want to overwhelm you, only inspire you. Tensors are the way in which modern physics is truly done, so you have that to look forward to. I'm not there yet either, but it's exciting and understanding relativity and quantum mechanics are two things I plan on understanding before I die, it's what matters to me.
Astrophysics
  • Astrophysics
It does seem very fascinating, and I'll have to review linear algebra for sure again, it's a very powerful tool like geometry, like I think even the first things you learn in LA is insane, when you realize you can figure out like hundred, million, what ever unknown variables in a snap, even that had me interested haha. Maybe when I review LA I'll post some questions to and you can OBSERVE ONLY.
Astrophysics
  • Astrophysics
Thanks again for everything, it's super late and I think my noodles are burnt enough and what I'm saying now probably doesn't make much sense, thanks again, and good night :))
Empty
  • Empty
Hahaha yeah I think it might be best to get into learning how to use MatLab, which is basically just a matrix calculator with other nice math functions and programming to go with it. It will broaden your horizons greatly to utilize computers for doing math and physics. In the past, they were damned to using only what they could easily evaluate with pen and paper and their minds. Computers and programming allow us to evaluate a wider range of problems and is a new level of mathematical thinking to complement your current knowledge. I can help you with some of this sorta stuff later, I am a long shot from being considered an expert in any of this, but I can definitely be a messenger for it haha.
Astrophysics
  • Astrophysics
Yeah definitely, computational physics and math is something I always wanted to learn.

Looking for something else?

Not the answer you are looking for? Search for more explanations.