# Space-time interval invariance question

• bernhard.rothenstein
In summary, the invariance of the space-time interval in a one space dimension approach holds for arbitrary events (x,t) in I and (x',t') in I, and is not limited to events generated by the same light signal or by the same tardyon moving with speed u in I and u' in I'. The components x and x' are part of a four-vector, as are t and t'. The length measurements of a rod in different reference frames can be related through a non-Lorentz transformation equation, suggesting a possible connection to the concept of a wave vector.
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?

bernhard.rothenstein said:
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'.

bernhard.rothenstein said:
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.

bernhard.rothenstein said:
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.

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 $\Lambda^T\eta\Lambda=\eta$.

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

$$\Delta$$
clem said:
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)/c21/2) (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?

## 1. What is the concept of space-time interval invariance?

The concept of space-time interval invariance refers to the idea that the measurement of time and distance between two events in space-time is independent of the observer's reference frame. This means that different observers, regardless of their relative motion, will measure the same space-time interval between two events.

## 2. How is the space-time interval calculated?

The space-time interval is calculated using the Minkowski metric, which takes into account both time and distance. It is represented by the equation: Δs² = c²Δt² - Δx² - Δy² - Δz², where c is the speed of light and Δt, Δx, Δy, and Δz represent the differences in time and space between two events.

## 3. Why is space-time interval invariance important in physics?

Space-time interval invariance is important in physics because it is a fundamental principle of special relativity. It helps to explain the connection between space and time, and allows for a consistent understanding of how the laws of physics apply in different reference frames.

## 4. Is the space-time interval always the same for all observers?

Yes, the space-time interval is always the same for all observers. This is a fundamental principle of special relativity and has been confirmed by numerous experiments and observations. However, the perception of time and distance may differ for different observers due to the effects of relative motion and gravitational forces.

## 5. How does the concept of space-time interval invariance relate to the speed of light?

The concept of space-time interval invariance is closely related to the speed of light. In fact, the constant speed of light is what allows for the space-time interval to be the same for all observers. This is because the Minkowski metric includes the speed of light, and any changes in the relative speed of observers must be accounted for in the differences in time and space.

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