Sign convention in the space-time 4-vector

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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?
 

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  • #2
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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.
 
  • #3
jtbell
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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.
 
  • #4
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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.
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 =)
 
  • #5
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My real question is the following: why we don't use the standard convention for scalar products of vectors (+,+,+,+) in GR?
 
  • #6
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My real question is the following: why we don't use the standard convention for scalar products of vectors (+,+,+,+) in GR?
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.
 
  • #7
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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).
 
  • #8
robphy
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My real question is the following: why we don't use the standard convention for scalar products of vectors (+,+,+,+) in GR?
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].
 
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  • #9
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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.
 
  • #10
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My real question is the following: why we don't use the standard convention for scalar products of vectors (+,+,+,+) in GR?
Because it would require complex coordinates, which is pretty irritating! :smile:

Pete
 

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