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I What is a "free vector"?

  1. Aug 11, 2016 #1
    Hello there, I had a look at the notes you shared and I have a question
    what does "Vectors in spacetime are always fixed at an event; there is no such thing as a “free vector” that can move from place to place" mean?
    so okay, an event is when you fix the coordinates of time and space , so of course this comes with a vector of 4-coordinates, but what would a free vector be?
     
  2. jcsd
  3. Aug 11, 2016 #2

    ShayanJ

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    Its not a technical term that has a physical\mathematical meaning, actually its supposed to be the familiar notion everyone has about vectors from the Euclidean case. In the Euclidean case, you can compare vectors in different points. For example you can compare angular momenta of two particles in two different positions. As if you can move one of them to the location of the other and see which is longer. The vectors are free in this way. I should emphasize that this is not technical term.
     
  4. Aug 11, 2016 #3
    So you cannot "move" them around mathematically for comparison etc because they have meaning only when connected with an event? Like they don't have much info or meaning except when connected strictly with some very specified coordinates?
     
  5. Aug 11, 2016 #4

    Nugatory

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    You're surely comfortable with vectors written in Cartesian coordinates. Let's take an example in ordinary two dimensional space, as on the surface of a sheet of paper. We'd write a vector ##V## of length 2 and pointing towards the top of the sheet (in the positive y direction) as ##\vec{V}=0\hat{x}+2\hat{y}##, or more concisely as ##(0,2)##. In the notation of the notes that started this thread, we'd say ##V^0=0## and ##V^1=2##. As long as we're using cartesian coordinates, that representation works anywhere; a vector of length two pointing straight up everywhere will be ##(0,2)## everywhere. This is a free vector.

    But suppose that we were using polar coordinates.... ##\vec{V}## still is a vector of length two and pointing straight up, but it will have different components at different places. At the point corresponding to cartesian coordinates ##(x=1,y=0)## we have ##\vec{V}=0\hat{r}+2\hat{\theta}=(0,2)##. At the point corresponding to cartesian coordinates ##(x=0,y=1)## we have ##\vec{V}=2\hat{r}+0\hat{\theta}=(2,0)##, and at the point corresponding to cartesian coordinates ##(x=-1,y=0)## we have ##\vec{V}=0\hat{r}-2\hat{\theta}=(0,-2)##. Now when we say ##(0,2)##, we also have to specify the point that we're talking about, because ##(0,2)## means different things at different places. That's a vector that isn't free.
     
    Last edited: Aug 11, 2016
  6. Aug 12, 2016 #5
    thank you so much, that was very helpful! So a non free vector is one that could be at any place , since many functions lead to these points, so we need to specify which point of spacetime.
    But, for the Cartesian case, it wouldn't work everywhere, but only in the point of x=0 so we could move the vector facing upwards, up and down the y-axis, but not on the x-axis, since x is zero.
    Another point is that, if I get this right, 0 corresponds to time coordinate whereas 1,2,3 are for spacial coordinates, do I get this right? I guess you used 0 for x for simplicity.
     
  7. Aug 12, 2016 #6

    Nugatory

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    It works everywhere. The vector is ##0\hat{x}+2\hat{y}## where ##\hat{x}## and ##\hat{y}## are unit vectors in the x and y directions. No matter where you are on the plane, that vector will point up, in the positive y direction.
    right.
     
  8. Aug 12, 2016 #7

    sophiecentaur

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    I wonder what you actually mean by a vector that can move about. A vector involves magnitude and direction and not an 'origin'. You can have a Vector Field which describes the values of a vector quantity over a range of locations. There is an idea of 'freedom' there(?).
     
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