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Astrophysics

  • one year ago

Observe only, please!

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  1. Astrophysics
    • one year ago
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    Evaluate the integral \[\int\limits \textbf A \times \textbf A'' dt\]

  2. Astrophysics
    • one year ago
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    @Empty

  3. Empty
    • one year ago
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    This one is one of my favorites. Hahaha

  4. Astrophysics
    • one year ago
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    Show me

  5. Empty
    • one year ago
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    Hint:\[ \textbf A' \times \textbf A' = 0\]

  6. Empty
    • one year ago
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    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.

  7. Astrophysics
    • one year ago
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    So can I integrate A and then differentiate A'' at the same time

  8. Empty
    • one year ago
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    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 \[\bar \tau = \bar r \times \bar F\]

  9. Astrophysics
    • one year ago
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    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.

  10. Empty
    • one year ago
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    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.

  11. Astrophysics
    • one year ago
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    \[\frac{ \vec dL }{ dt } = \frac{ \vec dr }{ dt } \times \frac{ \vec dp }{ dt }\] should I be doing this

  12. Astrophysics
    • one year ago
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    \[\vec r \times \vec F = \vec r \times \vec p\]

  13. Empty
    • one year ago
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    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. :)

  14. Empty
    • one year ago
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    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.

  15. Astrophysics
    • one year ago
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    \[\frac{ d }{ dt }(\vec r \times \vec p) = \vec r~' \times \vec p + \vec r \times \vec p ~ '\]

  16. Empty
    • one year ago
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    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

  17. Astrophysics
    • one year ago
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    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

  18. Empty
    • one year ago
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    Ok so now let's resolve these two terms, what do we have, can we rewrite these at all?

  19. Empty
    • one year ago
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    I think we can use that good thinking! Let's look at the first term and plug this in: \[\vec p = m \vec v\]

  20. Astrophysics
    • one year ago
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    Oh ok lol I tried it out, but wasn't sure it was the right way, and yeah haha that sounds good

  21. Empty
    • one year ago
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    So what does multiplying a vector quantity by a scalar quantity do to it? Describe it visually, not mathematically.

  22. Astrophysics
    • one year ago
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    Well if we multiply a scalar by a vector it gives us a scale factor of the vector without changing the direction

  23. Astrophysics
    • one year ago
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    So |dw:1436087673043:dw|

  24. Empty
    • one year ago
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    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.

  25. Astrophysics
    • one year ago
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    The noodle is gone, haha alright

  26. Astrophysics
    • one year ago
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    \[\vec r~ ' \times \vec p \implies \vec r~' \times (m \vec v) = m( \vec r~' \times \vec v) = (m \vec r~') \times \vec v\]

  27. Empty
    • one year ago
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    Good, wait what's \(\vec r \prime \)?

  28. Astrophysics
    • one year ago
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    I want to say velocity

  29. Astrophysics
    • one year ago
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    Since it's a position vector so the derivative would be velocity

  30. Empty
    • one year ago
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    You got it. :) Keep going see if you can discover anything else to evaluate this term.

  31. Empty
    • one year ago
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    Hint: use the fact that you mentioned earlier

  32. Astrophysics
    • one year ago
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    Kk, but should I leet r' = v?

  33. Empty
    • one year ago
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    You bet

  34. Empty
    • one year ago
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    Let it be what it is. :P

  35. Astrophysics
    • one year ago
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    \[\bar a \times \bar b = - \bar b \times \bar a\] \[(m \vec v) \times \vec v = - \vec v \times (m \vec v)\]

  36. Empty
    • one year ago
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    Awesome. :) Now play around with this thing you just found with that other thing you learned about the scalars.

  37. Astrophysics
    • one year ago
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    then let that = p

  38. Empty
    • one year ago
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    nooo

  39. Astrophysics
    • one year ago
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    lolol

  40. Astrophysics
    • one year ago
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    Yeah haha ok

  41. Empty
    • one year ago
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    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

  42. Astrophysics
    • one year ago
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    As in linear on one side

  43. Empty
    • one year ago
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    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

  44. Astrophysics
    • one year ago
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    Yeah I figured it out after I read it the second time, it's just the cross multiply that had me all woozy

  45. Astrophysics
    • one year ago
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    Ok so I use the same property for scalars again?

  46. Empty
    • one year ago
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    Yeah, so pull the mass scalar out of the equation

  47. Astrophysics
    • one year ago
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    \[(m \vec v) \times \vec v = - \vec v \times (m \vec v) \] \[(m \vec v) \times \vec v = -m \vec v \times \vec v \] so something like this

  48. Empty
    • one year ago
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    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.

  49. Astrophysics
    • one year ago
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    Right right

  50. Astrophysics
    • one year ago
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    \[m \vec v \times \vec v = -m \vec v \times \vec v\]

  51. Empty
    • one year ago
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    Ok I am going to stop hand holding now, this simplifies to something by normal algebra.

  52. Empty
    • one year ago
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    I'm gonna be a wrinkled old man before I get an answer out of you, is this how it's gonna be? :P

  53. Astrophysics
    • one year ago
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    Lol I'm using the draw tool to figure it out

  54. Empty
    • one year ago
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    Oh show me

  55. Astrophysics
    • one year ago
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    If it's angular momentum, I'm not getting that

  56. Empty
    • one year ago
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    I don't know, show me

  57. anonymous
    • one year ago
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    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.

  58. Empty
    • one year ago
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    smh @wio

  59. Astrophysics
    • one year ago
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    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

  60. anonymous
    • one year ago
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    Hmmm? I think in general \(\mathbf v \times \mathbf v = 0\), right? Because vectors are parallel to themselves.

  61. Empty
    • one year ago
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    Great, thanks a lot @wio for wasting our time. :/

  62. Astrophysics
    • one year ago
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    |dw:1436091897061:dw|

  63. Astrophysics
    • one year ago
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    No I was actually using this https://web.viu.ca/wattsv/math200/Overheads/section10.4/algcrossproduct.pdf to go over the properties

  64. Astrophysics
    • one year ago
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    Because I wasn't sure if you meant regular algebra or cross product algebra haha, so I was trying all these rules

  65. Astrophysics
    • one year ago
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    anyways let me work this out

  66. Astrophysics
    • one year ago
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    |dw:1436092214752:dw| I don't know if this is entirely right

  67. Empty
    • one year ago
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    You have written the right thing for the wrong reasons.

  68. perl
    • one year ago
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    .

  69. Empty
    • one year ago
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    You are not allowed to use the rule \[\vec v \times \vec v = 0\] Because you have not discovered it yet.

  70. Empty
    • one year ago
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    But this problem is where that rule comes from.

  71. Astrophysics
    • one year ago
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    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.

  72. Empty
    • one year ago
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    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?

  73. Empty
    • one year ago
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    Solve this algebra equation I am just making up right now: x=-x What's x?

  74. Astrophysics
    • one year ago
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    x=0

  75. Astrophysics
    • one year ago
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    omggg

  76. Astrophysics
    • one year ago
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    This is just like line integrals

  77. Astrophysics
    • one year ago
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    I was focusing so hard on the math didn't even pay attention to the concept, figures.

  78. Empty
    • one year ago
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    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 \[f(x,y)=-f(y,x)\]

  79. Astrophysics
    • one year ago
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    Ah interesting, so whenever for example lets just say QM concepts they talk about symmetries do they mean this?

  80. Empty
    • one year ago
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    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: \[f(x,y)=f(y,x)\]

  81. Astrophysics
    • one year ago
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    Fascinating

  82. Astrophysics
    • one year ago
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    Can we finish this problem though

  83. Astrophysics
    • one year ago
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    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

  84. Astrophysics
    • one year ago
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    Sorry not 0 = 0, the skew symmetry

  85. Empty
    • one year ago
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    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!

  86. Empty
    • one year ago
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    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?

  87. Astrophysics
    • one year ago
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    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

  88. Empty
    • one year ago
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    Yeah exactly, you use that to say the derivative of momentum is force. That completes the proof!

  89. Astrophysics
    • one year ago
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    Very cool haha, now how does this relate to the original problem xD

  90. Empty
    • one year ago
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    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

  91. Empty
    • one year ago
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    The next step to solving this problem is "fundamental". ;)

  92. Astrophysics
    • one year ago
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    Haha nice, now we apply FTC

  93. Astrophysics
    • one year ago
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    WOW, so what we actually just did is go backwards!

  94. Empty
    • one year ago
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    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

  95. Empty
    • one year ago
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    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

  96. Astrophysics
    • one year ago
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    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

  97. Empty
    • one year ago
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    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

  98. Astrophysics
    • one year ago
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    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.

  99. Astrophysics
    • one year ago
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    skew symmetry velocities*

  100. Empty
    • one year ago
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    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.

  101. Astrophysics
    • one year ago
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    I never fully understood cross product, so the cross product is always perpendicular of what's being "crossed"

  102. Astrophysics
    • one year ago
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    I guess that would make sense because a and b are in a plane where then your cross product is then in R^3

  103. Empty
    • one year ago
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    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|

  104. Astrophysics
    • one year ago
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    |dw:1436095034791:dw|

  105. Empty
    • one year ago
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    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.

  106. Empty
    • one year ago
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    Ahahahaha XD

  107. Empty
    • one year ago
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    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

  108. Astrophysics
    • one year ago
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    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.

  109. Astrophysics
    • one year ago
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    WHAT IS A TENSOR

  110. Astrophysics
    • one year ago
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    https://www.youtube.com/watch?v=f5liqUk0ZTw

  111. Empty
    • one year ago
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    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.

  112. Empty
    • one year ago
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    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.

  113. Astrophysics
    • one year ago
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    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.

  114. Astrophysics
    • one year ago
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    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 :))

  115. Empty
    • one year ago
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    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.

  116. Astrophysics
    • one year ago
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    Yeah definitely, computational physics and math is something I always wanted to learn.

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