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Gradient four-vector

  1. Apr 7, 2013 #1
    Hi,
    1. The problem statement, all variables and given/known data
    The gradient ∇3 can be generalized for spacetime as:
    4 =(∇3 ,d/dct)=(d/dx,d/dy,d/dz,d/dct)
    Show that ∇4 is a four-vector.

    2. Relevant equations



    3. The attempt at a solution
    I just have to write that :
    d/dx'=γ(d/dx-βd/dct)
    d/dy'=d/dy
    d/dz'=d/dz
    d/dct'=γ(d/dct-βd/dx)
    And
    4 =(d/dx,d/dy,d/dz,d/dct)=(∇3 ,d/dct) ..?

    Thanks
     
  2. jcsd
  3. Apr 7, 2013 #2

    haruspex

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    I don't understand the question. Of course it's a 4-vector. (a, b, c, d) is a 4-vector. Is there some wording missing?
     
  4. Apr 7, 2013 #3
    Hi,
    No, this is the complete wording.
     
  5. Apr 7, 2013 #4

    WannabeNewton

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    What? The 4-gradient ##\nabla_{a}## is a derivative operator. It isn't a 4-vector of any kind. I don't think you are phrasing it correctly.
     
  6. Apr 7, 2013 #5

    haruspex

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    A vector operator is still a vector. You can add them, and multiply them by scalar values.
     
  7. Apr 8, 2013 #6

    WannabeNewton

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    An operator acting on a vector isn't an element of the vector space that the vector belongs to. ##\nabla_{a}## is certainly not a 4-vector just because it is linear (it isn't a linear functional - it's just linear). How are you claiming that ##\nabla_{a}\in \mathbb{R}^{4}## (I used ##\mathbb{R}^{4}## because the OP is working in minkowski space-time). That doesn't make any sense.

    In fact what is true is that ##\frac{\partial }{\partial x}|_{p},\frac{\partial }{\partial y}|_{p},\frac{\partial }{\partial z}|_{p},\frac{\partial }{\partial t}|_{p}## are 4-vectors in ##\mathbb{R}^{4}##.
     
    Last edited: Apr 8, 2013
  8. Apr 8, 2013 #7

    micromass

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    The OP is asking about 4-vectors. So saying that ∇ is a 4-vector is the same as saying that ∇ is an element of ##\mathbb{R}^4##. This is obviously nonsense.

    Of course, we can add ∇ and multiply it by scalars. But that means that it is a vector in a vector space. But since the OP asks about 4-vectors, he's specifically asking about the vector space ##\mathbb{R}^4##. At least, that's how I interpret the question.
     
  9. Apr 8, 2013 #8

    vela

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    The OP is supposed to show that ##\nabla## transforms like a four-vector. In other words, he or she needs to show that it makes sense to say that ##\partial'_\mu = \Lambda_\mu{}^\nu \partial_\nu##.
     
    Last edited: Apr 8, 2013
  10. Apr 8, 2013 #9
    Actually, strictly speaking, shouldn't it transform like a 1-form?
     
  11. Apr 8, 2013 #10

    WannabeNewton

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    The coordinate vector fields ##\partial_{\mu}## do indeed transform like one-forms (hence why they are usually written with a lower index) but they are still 4-vectors as they form a basis for the tangent space when evaluated at each point. Also it doesn't make any sense to say ##\nabla## transforms like a 4-vector because it is not a vector in minkowski space-time, it is map on the ring of smooth functions on minkowski space-time. This bad notation is the kind you would see in Srednicki's QFT text.
     
  12. Apr 8, 2013 #11

    haruspex

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    I gather from various posts that, within relativity theory at least, "4-vector" has a more specific meaning than I was aware of. It's not clear to me whether this is a matter of definition or just common usage.
     
  13. Apr 8, 2013 #12

    micromass

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    Ah, maybe you're right. I don't know any relativity. I just looked up 4-vector in wikipedia.

    To avoid confusion, perhaps we should let the OP tell us how he defined "4-vector" in his course.
     
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