Space-time interval invariance question

[bernhard.rothenstein]

Cinsider please the invariance of the space-time interval in an one space dimension approach
(x-0)²-c²(t-0)²=(x'-0)²-c²(t'-0)²
My question is: does it hold for arbitrary events (x,t) in I and (x',t') in I?
Does it hold only in the case when the events are genertated in I and I' by the same light signal (x=ct,t=x/c); (x'=ct',t'=x'/c) or in the case when the events are generated by the same tardyon moving with speed u in I and u' in I' i.e. (x=ut,t=x/u) and (x'=u't', t'=x'/u')?
Are x and x' the components of a "two" vector or only x=ct, x'=ct' and x=ut, x'=u't', u amd u' being related by the addition law of parallel speeds?
Thanks for your answer.

 

[Meir Achuz]

Cinsider please the invariance of the space-time interval in an one space dimension approach
(x-0)²-c²(t-0)²=(x'-0)²-c²(t'-0)²
My question is: does it hold for arbitrary events (x,t) in I and (x',t') in I?
QUOTE]
It holds for arbitrary x and t with x' and t' given by a LT from S to S'.

 

[Meir Achuz]

Does it hold only in the case when the events are genertated in I and I' by the same light signal (x=ct,t=x/c); (x'=ct',t'=x'/c) or in the case when the events are generated by the same tardyon moving with speed u in I and u' in I' i.e. (x=ut,t=x/u) and (x'=u't', t'=x'/u')?

It holds for any x and t. If x is written as x=ut, then x' will =u't', with u'given by the relativistic velocity addiltion.

 

[Meir Achuz]

Are x and x' the components of a "two" vector or only x=ct, x'=ct' and x=ut, x'=u't', u amd u' being related by the addition law of parallel speeds?

x and t are two components of a four-vector, as are x' and t'. Writing x=ut implies that a consstant velocity, which is not necessary for t^2-x^2 to be invariant.

 

[Fredrik]

The "interval" -(x^0)^2+\vec x^2 is invariant because it's the Minkowski space "scalar product" of a four-vector with itself. The "scalar product" (which isn't really a scalar product since the result can be negative) is defined by

\langle y,x \rangle=y^T\eta x=-y^0x^0+\vec y\cdot\vec x

This is invariant under Lorentz transformations because all Lorentz transformations satisfy the condition \Lambda^T\eta\Lambda=\eta.

\langle \Lambda y,\Lambda x\rangle=(\Lambda y)^T\eta (\Lambda x)=y^T\Lambda^T\eta\Lambda x=y^T\eta x=\langle y,x \rangle

 

[bernhard.rothenstein]

\Delta

x and t are two components of a four-vector, as are x' and t'. Writing x=ut implies that a consstant velocity, which is not necessary for t^2-x^2 to be invariant.

Thanks for your answer. Consider please the inertial reference frames I, I' and I" in the standard arrangement. I' moves with velocity V relative to I and I" moves with speed u relative to I and with speed u' relative to I' all speed showing in the positive direction of the overlapped x, x' and x" axes. A rod of proper length L(0) is located along the overlapped axes at rest relative to I". Observers from I measure its Lorentz contracted length
L=L(0)(1-u²/c²)^1/2. (1)
For observers from I' the length of the same rod is
L'=L(0)(1-u'²/c²)^1/2 (2)
Eliminating L(0) between (1) and (2) we obtain that the non-proper lengths are related by
L=L'(1-u²/c²)^1/2/(1-u'²)/c^21/2) (3)
Expressing the right side of (3) as a function of u' via the addition law of parallel speeds it becomes
L=L'(1-V²/c²)^1/2)/[1+Vu'/c²] (4)
resulting that L and L' do not transform via the Lorentz transformation. Under such conditions are L=Dx and L'Dx' the components of a "two vector? Equation (4) suggests that 1/L and 1/L' are. Is there some connection with the concept of wave vector?