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Susskind said that the square of a differential equal zero

  1. Apr 28, 2014 #1
    Hello,

    I watched a lecture by Leonard Susskind, in which he said that a differential is so short that when you square it, you get zero.

    What exactly could he mean by this?

    Thank you for your time.

    Kind regards,
    Marius
     
  2. jcsd
  3. Apr 28, 2014 #2

    jedishrfu

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    Perhaps you could post a link to the video and the time inside the video that he said this.

    He's probably referring to the size of the number for example if your differential was numerically evaluated to 10^(-8) then the square of it would be 10^(-16) which is a much smaller number and could be considered to be zero.

    The effect of this would allow you to drop higher order terms in a power series expansion of a function as an example to get a linear approximation of the function.
     
  4. Apr 28, 2014 #3

    maajdl

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    My guess:
    He might mean that the boundary of a boundary is empty.
    The differential operator he is talking about would be the "exterior derivative".
    It is this operator that -for example- allows you to replace a (closed) surface integration by an integration over the volume bounded by this surface.
    On the basis of this geometric interpretation, you can guess that the "derivative" of a "derivative" is alway zero! (exterior derivative is implied!). The boundary of the volume of a sphere is the surface of the sphere, which is closed. The boundary of the surface of a sphere is empty. Considering an infinitesimal volume in place of a sphere leads you to one special case of the rule d²=0 .

    http://en.wikipedia.org/wiki/Exterior_derivative
    http://en.wikipedia.org/wiki/Boundary_(topology)

    [​IMG]
     
    Last edited: Apr 28, 2014
  5. Apr 28, 2014 #4

    sophiecentaur

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    I came across this at school (I think it's what could be worrying you - ignore it if not), when being taught how to get to the rules for differentiating, from basics. It can be done quite easily for simple algebraic functions and for trig functions, using triangles. You arrive at an expression involving δx and (δx)squared and then found what the limit is as δx→0.
    (δx)squared will vanish (can be ignored), compared with δx. It's a common way to find how an expression will behave as one of the variables approaches zero. Your "differential" doesn't actually become Zero, it just is infinitely small compared with other things he's dealing with.
    This link goes through the basics of how differentiation works and, right at the end, show how you ignore terms with dx in.
     
  6. Apr 28, 2014 #5

    D H

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    A link to the video would be nice.

    I suspect it is related to his book The Theoretical Minimum: What You Need to Know to Start Doing Physics (co-authored with George Hrabovsky). It apparently uses "physics math", where one ignores the Δx2 terms (and higher) in calculating derivatives. It's much easier than the more rigorous epsilon delta approach to a limit.
     
  7. Apr 28, 2014 #6
    That's sometimes called an infinitesimal of higher order. Δx2 is and infinitesimal of higher order and can be dropped from calculations. The point is that there will be an infinitesimal of first order somewhere in the denominator of the expression that will cancel infinitesimals of first order in the numerator (or/and vice - verse). But infinitesimals of higher order do not get cancelled out and will eventually drop out when the limit is taken. That limit is often implicit but one needs to keep in mind that it's there.
     
  8. Apr 28, 2014 #7
    That is not often true, try calculating the infinitesimal volume of a spherical shell (r to r+dr).
     
  9. Apr 28, 2014 #8
    What's the problem with that? The infinitesimals of 1st order are the only ones that survive. What's your point?
     
  10. Apr 28, 2014 #9

    boneh3ad

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    Sounds to me a lot like what is commonly done in an order of magnitude analysis when linearizing an equation or making other similar approximations to make a problem more tractable.
     
  11. Apr 28, 2014 #10

    Nugatory

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    Not sure I understand what you mean... That volume is ##\frac{4\pi}{3}(r+dr)^3-\frac{4\pi}{3}r^3 = \frac{4\pi}{3}(3r^2dr+3r(dr)^2+(dr)^3)##; drop the terms of order 2 and higher and we get the expected ##4\pi{r}^2dr##.
     
  12. Apr 28, 2014 #11
    If you read the original post I quoted, he mentioned another infinitesimal as the denominator which cancels the non-first orders in the limit that they approach 0. I'm just trying to show that this is not always the case when we ignore higher order infinitesimals.
     
  13. Apr 28, 2014 #12
    wow Susskind is amazing, I was wondering too how small can you reduce a unit basis vector, maybe to the size of a differential?
    I guess very few people have the talents to become theorists?
     
  14. Apr 28, 2014 #13

    SteamKing

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    A unit basis vector always has a magnitude of 1, which is why they are called 'unit' vectors.
     
  15. Apr 28, 2014 #14
    well say you could create a vector particle whose integral from 0 to infinity would be 1 unit vector :)
    this seems to be a powerful technique for instance to discretize a space....
     
    Last edited: Apr 28, 2014
  16. Apr 28, 2014 #15

    SteamKing

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    I'm not sure what a 'vector particle' is. It's not clear how vectors could be used to discretize space, since space is not a vector quantity.
     
  17. Apr 29, 2014 #16
    You're dividing by 1/N where N is the total number of terms in the Rimanian sum that will eventually convert into an integral after the implicit limit is taken.
     
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