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koustav
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are spacelike and timelike orthogonal?what is the mathematical proof
This is incorrect. Let ##V = (0,1,0,0)## and ##W = (0,0,1,0)##. Then ##V\cdot W = 0## and both ##V## and ##W## are space-like.SiennaTheGr8 said:but any four-vector that's orthogonal to a spacelike vector must be timelike, and vice versa
Any vector orthogonal to a time-like vector is space-like. Any vector orthogonal to a light-like vector is either proportional to the light-like vector itself or space-like. A vector orthogonal to a space-like vector may be time-like, space-like, or light-like.SiennaTheGr8 said:Retracted!
I need to rethink the conditions/exceptions for that "rule."
can you help me with the mathematical proofOrodruin said:Any vector orthogonal to a time-like vector is space-like. Any vector orthogonal to a light-like vector is either proportional to the light-like vector itself or space-like. A vector orthogonal to a space-like vector may be time-like, space-like, or light-like.
SiennaTheGr8 said:Not exactly, no, but any four-vector that's orthogonal to a spacelike vector must be timelike, and vice versa.
Spacelike just means that the magnitude of the three-vector spatial component is greater than the time component. Timelike means the opposite. And two vectors are orthogonal if their dot product is zero. So let's say we have the following two four-vectors:
##\mathbf{A} = (A_t, \mathbf{a}) \qquad \mathbf{B} = (B_t, \mathbf{b})##.
If they are orthogonal, then their dot product is zero:
##\mathbf{A} \cdot \mathbf{B} = (A_t)(B_t) - (a)(b) = 0##,
which means that:
##\dfrac{A_t}{a} = \dfrac{b}{B_t}##.
Assuming that ##\mathbf{A}## and ##\mathbf{B}## aren't lightlike (magnitude of zero), then this result can only occur if one of them is spacelike and the other is timelike. In other words, it must be the case that if ##A_t > a##, then ##b > B_t## (and vice versa).
In the case of a time-like vector, go to a reference frame where it is proportional to (1,0,0,0). Do the corresponding thing for space-like and null vectors.koustav said:can you help me with the mathematical proof
This should beSiennaTheGr8 said:If they are orthogonal, then their dot product is zero:
##\mathbf{A} \cdot \mathbf{B} = (A_t)(B_t) - (a)(b) = 0##,
Spacelike and timelike orthogonal are two types of orthogonal relationships in spacetime. Spacelike orthogonal means that the two vectors are perpendicular in space, while timelike orthogonal means that they are perpendicular in time.
The mathematical proof for spacelike and timelike orthogonal is derived from the Minkowski metric, which describes the geometry of spacetime. It involves solving for the dot product of two vectors and determining whether the result is positive, negative, or zero.
When two vectors are orthogonal in spacetime, it means that they are perpendicular to each other in both space and time. This is a fundamental concept in special relativity and plays a crucial role in understanding the behavior of objects in motion.
Yes, understanding spacelike and timelike orthogonal is important in various fields such as physics, astronomy, and engineering. It helps in predicting and analyzing the behavior of objects moving at high speeds, as well as in designing space-time diagrams and spacetime intervals.
The mathematical proof for spacelike and timelike orthogonal can be complex for those who are not familiar with advanced mathematics and concepts in special relativity. However, with some basic understanding and guidance, it can be comprehensible to most people.