Question

interval
Show
- that the
s² = c²t² - x²-y²-z²
remains
invarient under
lorentz transformation​

Answer 1

Step-by-step explanation:

Lorentz transformations can be regarded as generalizations of spatial rotations to space-time. However, there are some differences between a three-dimensional axis rotation and a Lorentz transformation involving the time axis, because of differences in how the metric, or rule for measuring the displacements Δr and Δs , differ. Although Δr is invariant under spatial rotations and Δs is invariant also under Lorentz transformation, the Lorentz transformation involving the time axis does not preserve some features, such as the axes remaining perpendicular or the length scale along each axis remaining the same.

Note that the quantity Δs2 can have either sign, depending on the coordinates of the space-time events involved. For pairs of events that give it a negative sign, it is useful to define c2Δτ2 as −Δs2 . The significance of c2Δτ as just defined follows by noting that in a frame of reference where the two events occur at the same location, we have Δx=Δy=Δz=0 and therefore (from the equation for Δs2=−c2Δτ2 ):

c2Δτ2=−Δs2=(c2Δt)2.(5.6.26)

Therefore c2Δτ is the time interval c2Δt in the frame of reference where both events occur at the same location. It is the same interval of proper time discussed earlier. It also follows from the relation between Δs and that c2Δτ that because Δs is Lorentz invariant, the proper time is also Lorentz invariant. All observers in all inertial frames agree on the proper time intervals between the same two events.