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

Sign convention in the space-time 4-vector

  1. Apr 30, 2008 #1
    What is the rationale for the sign convention in the space-time 4-vector? How is it related to the sign convention in the energy-momentum 4-vector, if at all?
  2. jcsd
  3. Apr 30, 2008 #2
    It's all related to the norm or dot product of 4 vectors. The Lorentz group keeps the inner product [itex]v_0 w_0 - v_1 w_1 -v_2 w_2 ...[/itex] invariant. So start with a vector with components
    [itex]v_\mu = (v_0, v_1,v_2, ... v_3)[/itex] and when you raise the index with the metric, you get [itex]v^\mu = (v_0, -v_1, -v_2, ... -v_3)[/itex].

    Hope that helps.
  4. Apr 30, 2008 #3


    User Avatar

    Staff: Mentor

    I assume you mean, why do we write [itex]ds^2 = (cdt)^2 - dx^2 - dy^2 - dz^2[/itex] rather than [itex]ds^2 = dx^2 + dy^2 + dz^2 - (cdt)^2[/itex]?

    For one thing, with this choice, we get a positive number for a "causally connectable" spacetime interval. Also, if we use the same convention for all four-vectors, then for energy-momentum [itex]E^2 - (p_x c)^2 - (p_y c)^2 - (p_z c)^2 = (m_0 c^2)^2[/itex] which is also a positive number.

    We could do it the other way, but it seems to me that this way, we get fewer minus signs associated with the invariant quantities that we're usually interested in.
  5. Apr 30, 2008 #4
    Relativists seem to prefer the latter, while field theorists seem to prefer the former. Which sucks when you're both a relativist and a field theorist =)
  6. Apr 30, 2008 #5
    My real question is the following: why we don't use the standard convention for scalar products of vectors (+,+,+,+) in GR?
  7. Apr 30, 2008 #6
    Because that's not what nature chose to give us. Also, you'll notice that there's no cross-over with that metric. In other words, there is no limiting speed like the speed of light.
  8. Apr 30, 2008 #7
    Have you ever learnt about tensor calculus? If you have brief understanding of tensor, the answer is obvious. This is because the metric ds squared equals to (ct) squared minus the sum of squares of (dx), (dy) and (dz).
  9. Apr 30, 2008 #8


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That's not actually a "convention".
    The signature arises from the geometrical structure of the space you are dealing with.
    Often one deals with a Euclidean space... which is often implicit... but it's there.

    In relativity, as others have pointed out, the geometrical structure of spacetime has a Lorentzian-signature metric tensor... with one sign different from the rest.
    Whether it's (-,+,+,+) or (+,-,-,-) or (+,+,+,-) or (-,-,-,+) is the choice of convention. (It's not just the signs... it's also the use of x0 or x4. These days x0 is preferred to allow consideration of more or fewer spatial dimensions.) Pick your favorite, make it known, and use it consistently [and be prepared to translate if necessary].
    Last edited: Apr 30, 2008
  10. May 1, 2008 #9
    Here is a book I would like to introduce you if you does not have strong enough background on tensor calculus. Schaum's outline series, theory and problems of tensor calculus by David C. Kay. This book is easy to follow and should be able to learn all the materials in it within a few weeks normally especially if you already have some knowledge of relativity.
  11. May 1, 2008 #10
    Because it would require complex coordinates, which is pretty irritating! :smile:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Sign convention in the space-time 4-vector