Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

If it works in 3D, then it works in 4D

  1. Oct 15, 2012 #1
    "If it works in 3D, then it works in 4D"

    I've been watching Leonard Susskind's lectures on special relativity here. At time 0:59:40 he says, "If the vectors are Lorentz invariant, then if the first 3 components of a 4vectors are equal to the first 3 components of another vector (equation), then the 4th components will be equal as well". He uses this to justify using the same equations that work for the space components to write the same equations with 4-vector notation. That's a neat trick. I think I've heard this other places. But I wonder if anyone could actually prove this to me. I'd like to see the proof worked out somewhere. Thanks.
  2. jcsd
  3. Oct 15, 2012 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: "If it works in 3D, then it works in 4D"

    If two vectors have the same spacelike components but different timelike components, then a Lorentz boost will always be able to make their spacelike components unequal. So if you have a relation that holds identically between the spacelike components, in all frames, the timelike components must also be equal.

    I think you have to be careful not to apply this inappropriately. For example, we have lots of Newtonian relationships between spacelike components of vectors, but these relationships are only approximate, so it doesn't follow that they hold for four-vectors. As an example, four-velocities don't add the way three-velocities do.
  4. Oct 15, 2012 #3

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: "If it works in 3D, then it works in 4D"

    Another way to look ar:

    Suppose [itex]a^0 = b^0[/itex], [itex]a^1 = b^1[/itex], and [itex]a^2 = b^2[/itex]. Lorentz invariance gives

    [tex]\left(a^0\right)^2 - \left(a^1\right)^2 - \left(a^2\right)^2 - \left(a^3\right)^2 = \left(b^0\right)^2 - \left(b^1\right)^2 - \left(b^2\right)^2 - \left(b^3\right)^2 .[/tex]

    What can you coclude from the above and that a and b are related by a restricted Lorentz transformation?
  5. Oct 15, 2012 #4
    Re: "If it works in 3D, then it works in 4D"

    Thank you. That was pretty clear. But then the question becomes, how does this differ if the signature of the metric were (+,+,+,+)? What then does "Lorentz transformations" have to do with this?
  6. Oct 17, 2012 #5
    Re: "If it works in 3D, then it works in 4D"

    I'm not sure we're addressing the issue. Susskind may have misspoke. He was trying to justify extending a 3D equation to a (3+1)D equation if it is Lorentz invariant. I wonder if there is a theorem on that. I don't think this is the same question as to if the 4th components are equal when the first three components are equal. The question is what justifies extending and equation to make it apply to more dimensions. This would be like taking the distance formula
    [tex]d = \sqrt {{x^2} + {y^2}} [/tex]
    and extending to include another dimension to get
    [tex]d = \sqrt {{x^2} + {y^2} + {z^2}} [/tex]
    if d is invariant.

    What is the criterion to do this extention? Just because we know it works in 2D doesn't tell us that it will work in 3D? Does it work simply because we requrie that the extra z dimension be orthogonal to the other dimensions? Or is it that we can get the same answer if we permutate the variables and get the same answer (This would be a rotations, right)? Or is it that you can set the new dimension to zero and get the old result? Or maybe it's a combination of these, or something else entirely. I wonder.

    He's turning [itex]F(x,y,z)[/itex] into [itex]F(t,x,y,z)[/itex], and I don't understand when we can do this.
    Last edited: Oct 17, 2012
  7. Oct 17, 2012 #6
    Re: "If it works in 3D, then it works in 4D"

    let me just ask 1 question then: what's the curl operator in 4-D?
  8. Oct 17, 2012 #7
    Re: "If it works in 3D, then it works in 4D"

    The curl is generalized by the exterior derivative, or in general the cross product is generalized by the wedge, ##\wedge##.

    The wedge product of two vectors (and hence, the exterior derivative of a vector field) is no longer a vector but a two-index antisymmetric tensor called a bivector (or a bivector field). You can use wedges instead of cross products in 3d as well, and this makes a lot of results easily generalize to other spaces.

    A quick summary of vector algebra and calculus using wedges:

    $$a \wedge b = - b \wedge a$$

    Generalized dot product of a vector and a bivector (BAC-CAB rule)--note the result is a vector.
    $$a \cdot (b \wedge c) \equiv (b \cdot a)c - (c \cdot a)b$$

    Equality of mixed partial derivatives:
    $$\nabla \wedge \nabla \wedge A = 0$$

    Laplacian of a vector field:
    $$\nabla^2 A = \nabla \cdot (\nabla \wedge A) + \nabla \wedge (\nabla \cdot A)$$

    Product rule for scalar field ##\phi## and vector field ##F##:
    $$\nabla \wedge (\phi F) = (\nabla \phi) \wedge F + \phi \nabla \wedge F$$
  9. Oct 18, 2012 #8
    Re: "If it works in 3D, then it works in 4D"

    I can understand when you're adding another space dimension. I'm not so sure about extentions that add a time dimension. But perhaps these both are of the same form. They both seem to be of the more general form [itex]d{s^2} = {g_{ij}}(\vec x)d{x^i}d{x^j}[/itex], which seems to be valid for all dimensions. Is that the trick? Express it in terms of tensor notation. And if the formula has the same geometrical meaning in all dimensions, then it is valid to extend it? What about when you're not sure what the 4 dimensional version would mean?
  10. Oct 18, 2012 #9
    Re: "If it works in 3D, then it works in 4D"

    I think that's along the right track. It's a basic trick that if you can write an expression in terms of Lorentz covariant objects, then the result must also be Lorentz covariant. It's done, for example, in finding the EM field of a moving point charge. You can start with just the equation of a stationary point charge and steadily find expressions for the coordinate distance ##r## and such in terms of covariant quantities to get something that is overall covariant.
  11. Oct 20, 2012 #10
    Re: "If it works in 3D, then it works in 4D"

    I'm thinking about another requirement. Is it the case that whatever scalars are consturcted by adding one more dimension, then any minimization of the action in the higher dimension that uses the scalar must also be a minimum of action in the lower dimensional version? I'm thinking that the only reason to add dimension would be for physical reasons in which case the minimal action principle becomes relevant. And I'm thinking the lower dimensional action is achieved by setting the added dimension to zero so that the lower dimensional action would still need to be minimized. Or maybe the action goes along with having the same geometric meaning.
    Last edited: Oct 20, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook