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What is wrong?

  1. Jun 13, 2013 #1
    What is wrong??

    Hi,

    I want to know where is the mistake in the statement below (I really don't know what is wrong, but it is obviously wrong)

    [itex] \frac{\partial ^{2}x}{\partial t^{2}} = \frac{\partial ^{2}x}{\partial x^{2}} (\frac{\partial x}{\partial t} )^{2} [/itex]

    But
    [itex] \frac{\partial ^{2}x}{\partial x^{2}} = \frac{\partial \frac{\partial x}{\partial x}}{\partial x} = \frac{\partial 1}{\partial x} = 0[/itex]

    This way:
    [itex] \frac{\partial ^{2}x}{\partial t^{2}} = 0[/itex]

    Thanks,
    John
     
  2. jcsd
  3. Jun 13, 2013 #2
    After so many posts, you surely know that you should post what is given and what you are supposed to find.

    Really.
     
  4. Jun 13, 2013 #3

    That first statement doesn't look true.
     
  5. Jun 13, 2013 #4

    Mark44

    Staff: Mentor

    What does the following simplify to?
    $$ \frac{\partial x}{\partial x}$$
     
  6. Jun 13, 2013 #5
    Sorry.

    Imagine a car traveling in the x axis. It's initial position is zero and its initial velocity ∂x/∂t is zero too. Although it has a variable acceleration a.

    [itex] a = \frac{\partial ^{2}x}{\partial t^{2}} [/itex]

    I've just multiplied by
    [itex]\frac{\partial x^{2}}{\partial x^{2}} [/itex]
    to get
    [itex] \frac{\partial ^{2}x}{\partial x^{2}} (\frac{\partial x}{\partial t} )^{2} [/itex]

    In my conception I just multiplied by 1 and got an absurd result. "a" is not always zero (I haven't even mentioned the function a(t))
     
  7. Jun 13, 2013 #6
    First of all, it is far from clear why you are using partial derivatives. You have functions of just one argument - the time - so ordinary derivatives would work fine.

    Second, ## a = \frac {d^2x} {dt^2} \ne (\frac {dx} {dt})^2 = v^2##.
     
  8. Jun 13, 2013 #7
    Yikes, I think you're having a problem with notation vs. concepts.

    First of all,

    [itex] \frac{\partial ^{2}x}{\partial x^{2}} (\frac{\partial x}{\partial t} )^{2}[/itex]

    is in no way, shape, or form, the same as

    [itex]\frac{\partial ^{2}x}{\partial t^{2}}[/itex]

    You did multiply by 1, but your mistake is thinking that the Leibniz notation is just like a simple fraction involving numbers being multiplied because it looks the same. It is a fraction, but ∂ is not a number. ∂/∂ is not 1, it's not anything, it's like saying +/+ or (grapefruit)/(I think Michael Shannon is a talented actor.)


    -edited out-

    I proved that it couldn't be true, but misread the equation myself in doing so. :/
     
    Last edited: Jun 13, 2013
  9. Jun 13, 2013 #8

    But why can I multiply dx/dt by dθ/dθ to get (dx/dθ)(dθ/dt) = w (dx/dθ).
    I've just multipied by one too, and only rearranged the terms, as I did in the initial derivative. What's the difference between them? Why in the first one I cannot do this and in the second one I can?
     
  10. Jun 13, 2013 #9
    Ok... this may be challenging to explain. What you've done is used symbols that look algebraic, who's values are not algebraic but defined, then did algebra with these symbols, assuming their values would be preserved.

    Put it this way, some people write arcsin(x) as sin^-1(x). But, that doesn't mean I can do algebra to both sides and get a true statement, because the symbol "sin^-1(x)" gets its value from definition.

    Here:



    See,

    [itex]\frac{\partial ^{2}x}{\partial x^{2}}[/itex]

    Is 0.

    This symbol, as a whole, means the acceleration of x with respect to itself. It will always be changing at a constant rate relative to itself (namely, 1) so you have made the claim:


    [itex] \frac{\partial ^{2}x}{\partial t^{2}} = 0(\frac{\partial x}{\partial t} )^{2} [/itex]

    [itex] \frac{\partial ^{2}x}{\partial t^{2}} = 0 [/itex]

    That's the claim, period. The fact that if you leave 0 as you wrote it and do algebra to get something that looks true means absolutely nothing because our "zero" is zero through conceptual definition and not algebraically so (IE, the numerator is not 0.)

    So it's like "changing your mind" mid way through.

    "This is zero because of what we say the symbols mean, not algebraically"
    to
    "Do normal algebra with those symbols"
    back to
    "Now let's return to what the symbols mean again and observe the result."

    Which will of course, end up as nonsense.
     
    Last edited: Jun 13, 2013
  11. Jun 13, 2013 #10

    Mark44

    Staff: Mentor

    Yes. That's where I was going when I asked this question.
    Minor quibble. It means the second derivative of x with respect to itself. Acceleration usually means the second derivative of position with respect to time.
     
  12. Jun 13, 2013 #11
    Well, of course I agree, but I wanted to stay in the OP's terms.
     
  13. Jun 13, 2013 #12
    In fact, the symbols were not used correctly even by the 17th century's standard. Mr Leibniz did a marvelous job to ensure that ## \frac {d} {dt} \frac {d} {dt} ## naturally ends up being ## \frac {d^2} {dt^2} ##, not ## (\frac {d \text{something}} {dt})^2 ##, simply because it has two d's and none of the 'something'.
     
  14. Jun 13, 2013 #13
    I think
    [itex] \frac{\partial ^{2}x}{\partial t^{2}} = \frac{\partial ^{2}x}{\partial x^{2}} (\frac{\partial x}{\partial t} )^{2} [/itex]

    was thought to be true because if you treat the RHS purely algebraically, you get


    [itex]\frac{\partial ^{2}x\partial ^{2}x^{2}}{\partial ^{2} x^{2} \partial ^{2} t ^{2}}[/itex]

    to which "cancelations" give

    [itex]\frac{\partial ^{2}x}{ (\partial t)^{2}}[/itex]


    Not due to an error involving

    Unless I'm missing it somewhere. :)
     
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