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Rewriting a Vector in Terms of Two Arbitrary Vectors

  1. Aug 30, 2012 #1
    I just started grad school from an engineering background. My research has led me to need more physics & math which is why I'm in the class, but I don't have a very broad Math background. This problem given is exactly as it is typed above. The text we are using is Arfken: Math Methods for Physicists (7th).

    1. The problem statement, all variables and given/known data
    Separate [itex]\vec{v}[/itex] into [itex]\vec{g_{1}}[/itex] and [itex]\vec{g_{2}}[/itex].

    (It has a figure with the 3 vectors listed above all starting from the same point. No angles or magnitudes are given.

    2. Relevant equations
    Hint:use [itex]\vec{v}[/itex]=[itex]\vec{v^{i}}[/itex][itex]\vec{g_{i}}[/itex], apply from the right the vector product with [itex]\vec{g_{2}}[/itex] and use the dot product to obtain [itex]\vec{g^{1}}[/itex]; [itex]\vec{g^{2}}[/itex] analogous

    Result: [itex]\vec{v}=\vec{g_{2}}\frac{(\vec{g_{1}} x\vec{g_{2}})\ldot(\vec{g_{1}} x \vec{v})}{(\vec{g_{1}} x \vec{g_{2}})^{2}}-\vec{g_{1}}\frac{(\vec{g_{1}} x\vec{g_{2}})\ldot(\vec{g_{2}} x \vec{v})}{(\vec{g_{1}} x \vec{g_{2}})^{2}}[/itex]


    3. The attempt at a solution
    I know that I am suppose to use the curl to find [itex]\vec{v^{1}}[/itex] and somehow there is abstract algebra involved as I've asked for help and that is what he told me. The weird thing is that I don't see how I work what the hint says in with the curl advice. I can do the arithmetic involved (ie I can take vector & scalar products, etc), but I'm just not even sure where to start to end up with the "Result". This is due tonight (I'm not asking for the answer here, just to be clear) so I will update the post as/if I figure more out. At this point I am just looking for advice in general as I am clueless as to what the "path" is if that makes sense.
     
    Last edited: Aug 30, 2012
  2. jcsd
  3. Aug 30, 2012 #2
    Also I have no idea how you square a cross product? I don't think that it means the magnitude or anything.
     
  4. Aug 30, 2012 #3

    gabbagabbahey

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    There is no need to calculate the curl of any vectors for this problem. Just use dot (\cdot is the [itex]\LaTeX[/itex] code) products and cross products (\times is the [itex]\LaTeX[/itex] code).

    Have you tried using the hint they gave you?

    What you want to do is find scalars [itex]v_1[/itex] and [itex]v_2[/itex] such that [itex]\mathbf{v}=v_1\mathbf{g}_1 + v_2\mathbf{g}_2=v_i\mathbf{g}_i[/itex] (note that I use only subscripts here for indices just so you don't confuse them with exponents).

    What do you get when you use this expression to calculate [itex]\mathbf{v} \times \mathbf{g}_i[/itex]?

    Some authors use [itex]\mathbf{x}^2[/itex] and [itex]|\mathbf{x}|^2[/itex] interchangeably for the square of the magnitude of [itex]\mathbf{x}[/itex]. In this case, [itex](\mathbf{g}_1 \times \mathbf{g}_2)^2 = |\mathbf{g}_1 \times \mathbf{g}_2|^2 \equiv (\mathbf{g}_1 \times \mathbf{g}_2) \cdot (\mathbf{g}_1 \times \mathbf{g}_2)[/itex]
     
  5. Aug 30, 2012 #4
    Thanks gabbagabbahey, so I see how the problem works. My problem is that I didn't understand the notation given in the hint. The vector equation for [itex]\vec{v}[/itex] will be[itex] \vec{v}_{1}\vec{g}_{1} + \vec{v}_{2}\vec{g}_{2}[/itex] Thanks to you, I understand that part and I get:

    [itex]
    v_{1}g_{1}\vec{g}_{2}-v_{2}g_{2}\vec{g}_{1} [/itex]

    I also know that the dot product will be of the form:

    [itex]\vec{a} \cdot\frac{\vec{b}}{||\vec{b}||}[/itex]

    Why is it that for [itex]\vec{a}[/itex] and [itex]\vec{b}[/itex] the vectors represented by the corss products were used? In other words, why did they cross [itex]\vec{g}_{1}\times\vec{g}_{2} and \vec{g}_{1} \times \vec{v}[/itex] to get to [itex]g_{1}[/itex]
    and [itex] v_{1}?[/itex]

    If I can figure that out then I feel like I will understand this problem a lot better. I have looked up projections, but that hasn't proven beneficial as of yet. Maybe I'm looking in the wrong topics/source?

    Thanks Again!
     
  6. Aug 30, 2012 #5

    gabbagabbahey

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    Be careful here. You are looking to express [itex]\mathbf{v}[/itex] as a linear combination of the vectors [itex]\mathbf{g_1}[/itex] and [itex]\mathbf{g_2}[/itex] as

    [tex]\mathbf{v}=v_1\mathbf{g}_1 + v_2\mathbf{g}_2[/tex]

    [itex]v_1[/itex] and [itex]v_2[/itex] are scalars in this equation, so it makes no sense for you to write them as vectors in your first equation (in other words, you shouldn't have the arrows over them).

    Also, I'm not sure where you got your second equation from... what are [itex]g_1[/itex] and [itex]g_2[/itex] doing in there? where did the negative sign come from?:confused:

    I'm not sure exactly what you mean here, but I can guess that you are assuming something like if [itex]\mathbf{v} =v_1\mathbf{g}_1 + v_2\mathbf{g}_2[/itex], then [itex]\mathbf{v} \cdot \mathbf{g_1} = v_1[/itex], but this is only true if [itex]\mathbf{g}_1[/itex] and [itex]\mathbf{g}_2[/itex] are orthonormal (like [itex]\mathbf{i}[/itex] and [itex]\mathbf{j}[/itex].

    In general, [itex]\mathbf{g}_1[/itex] and [itex]\mathbf{g}_2[/itex] won't be orthogonal, or normalised (have a magnitude of 1), so all taking the dot product gives you is [itex]\mathbf{v} \cdot \mathbf{g_1} = v_1 |\mathbf{g}_1|^2 +v_2 \mathbf{g}_1 \cdot \mathbf{g}_2[/itex], which isn't all that useful here.

    Instead, do exactly what they say to in the Hint. Start by taking the vector (cross) product of [itex]\mathbf{g}_i[/itex] with [itex]\mathbf{v}[/itex] from the right. That is, start by computing [itex]\mathbf{v} \times \mathbf{g}_i[/itex]... what do you get when you do that?
     
  7. Aug 30, 2012 #6
    I'm sorry, I had a typo in that first line. This is really my 2nd time using LaTEX. What I meant to say was:

    The vector equation for [itex]\vec{v}[/itex] will be[itex] v_{1}\vec{g}_{1} + v_{2}\vec{g}_{2}[/itex] Thanks to you, I understand that part and I get:

    I worked out the cross product and got [itex] v_{1}\vec{g}_{1}\vec{g}_{2} - v_{2}\vec{g}_{2}\vec{g}_{1}[/itex] which seems like it could be correct. If I didn't get that part right time then I don't feel like I fully understand what is meant by [itex]\vec{g}_{i}[/itex]. When I computed the cross product I wrote it out as [itex]\vec{g}_{1}[/itex] for the g1 direction and [itex]\vec{g}_{2}[/itex] for the g2 direction.

    If my result is still wrong hopefully you can at least see where the flaw is in my thought process. For now I'm going to operate under the premise that I'm correct and now I need to figure out how to get [itex] v_{1} [/itex] and [itex] v_{2} [/itex]. As I'm still lost I want to make sure that I am at least reading something that I will need. Any advice on this and I'd appreciate it. I'm not sure why I'm having so much trouble with this, it seems like it should be much easier than what it has turned out to be. I'm still waiting for that aha! moment I guess...
     
    Last edited: Aug 30, 2012
  8. Aug 30, 2012 #7

    gabbagabbahey

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    I think you may be having trouble with indices and einstein's summation convention. [itex]v_i\mathbf{g_i}[/itex] is just a shorthand way of writing [itex]\sum_{i=1}^2 v_i\mathbf{g}_i = v_1\mathbf{g}_1 + v_2\mathbf{g}_2[/itex]. Whenever a term has and index that is repeated exactly once, a summation over that index is implied. Whenever a term has an index that is repeated more than once, you've royally messed up somewhere. This last part is very important to remember when working with indices, using the Einstein summation convention.

    That said, if you are say that since [itex]\mathbf{v}=v_i\mathbf{g}_i[/itex], the cross product is given by [itex]\mathbf{v} \times \mathbf{g}_i = v_i\mathbf{g}_i \times \mathbf{g}_i[/itex], then you are claiming something that is nonsense since the index [itex]i[/itex] occurs more than twice. The proper thing to do is to notice that since [itex]i[/itex] is summed over in the equation [itex]\mathbf{v}=v_i\mathbf{g}_i[/itex], it is
    just a "dummy" index that you can replace with any other letter that is not used elsewhere in the expression, so replace it with [itex]j[/itex] and then you have [itex]\mathbf{v} \times \mathbf{g}_i = v_j\mathbf{g}_j \times \mathbf{g}_i[/itex].

    To see what this means, consider [itex]i=1[/itex], for example:

    [tex]\mathbf{v} \times \mathbf{g}_1 = v_j\mathbf{g}_j \times \mathbf{g}_1 = \left( v_1\mathbf{g}_1 + v_2\mathbf{g}_2\right) \times \mathbf{g}_1 = v_2\mathbf{g}_2 \times \mathbf{g}_1 [/tex].

    since [itex]\mathbf{g}_1 \times \mathbf{g}_1=0[/itex]

    So, hopefully now you can see why taking this cross-product is useful. For [itex]i=1[/itex], it allows you to get an expression for [itex]v_2[/itex] that doesn't involve [itex]v_1[/itex], and vice versa for [itex]i=2[/itex].

    How about the next step in the hint? What do you get when you take the dot product of this result with [itex](\mathbf{g}_1 \times \mathbf{g}_2)[/itex]?
     
  9. Aug 30, 2012 #8
    A basic idea of what's going on here is this: any vector can be decomposed in terms of other vectors like so.

    [tex]v = (v \cdot g^1) g_1 + (v \cdot g^2) g_2[/tex]

    The vectors denoted by down indices are "tangent" vectors (they're tangent to a coordinate axis), and the up indices denote "cotangent" vectors (they're cotangent, or perpendicular, to planes spanned by coordinate axes). This is the general rule for extracting components of vectors with respect to a particular basis. The difference between [itex]g_1[/itex] and [itex]g^1[/itex] doesn't exist in Cartesian coordinates, but in more general coordinates it does exist. If you're familiar with this concept, then so far so good. The only question, then, is how do we figure out the [itex]g^1[/itex] and [itex]g^2[/itex] vectors?

    Your book should tell you how to find these vectors, though, and this is definitely what the hint is trying to tell you to do.
     
  10. Aug 30, 2012 #9
    I really appreciate both of the effort you guys put into helping me tonight. The deadline just passed and I didn't get this one, but 6/7 isn't so bad. I am going to work on this more tomorrow morning. Put way more hours into the HW than I feel like I should have, for now I think I need sleep.

    @Murphrid
    I completely agree that it is probably in my book, but so far I haven't been able to bring it all together. Any advice on what I might be able to look up?
     
  11. Aug 30, 2012 #10
    These vectors go by various names. Not being familiar with the particular book, they could go by any of the following names: covariant vectors, cotangent vectors, one-forms, dual vectors, reciprocal vectors, and so on. In particular, you're looking for a basis, so I would look for dual basis or reciprocal basis.

    The 3D case is well-studied, but the 2D case is actually pretty rare to see. I know how to do it with geometric algebra, but to do it with cross products and such seems like a pain.

    Edit: the hint does try to guide you to a way of isolating the components in a simple way, though. What I described is a bit more of a turn-crank method. The process gabba described in post #7 should do fine for pointing you in the right direction.
     
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