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Is there any simple way to intuitively understand spacetime interval, proper time and proper length?

Qmechanic
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Toneri Otsutsuki
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    What is wrong with the literal way? It is simple and intuitive. Any other way is bound to be more confusing. – m4r35n357 Jan 23 '21 at 09:27
  • what is the literal way? – Toneri Otsutsuki Jan 23 '21 at 09:33
  • @m4r35n357 that depends. The question might be trivial for the site but there is nothing much intuitive in defining (delta s)^2. I guess OP is searching for an answer that gives the taste of it. – Alchimista Jan 23 '21 at 11:00
  • @ToneriOtsutsuki it just is the difference of two numbers, and represents either proper time or proper distance depending on the sign of that difference. – m4r35n357 Jan 23 '21 at 18:31

1 Answers1

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Let's start with proper length and proper time because they are probably the easiest to understand.

Let us have to inertial reference frames $S$ and $S'$ where $S'$ is moving relative to $S$. Thus, we define $S$ as "stationary" (because we are in this frame). Now a "proper" measurement on an object or event is any measurement made by an observer which is stationary with respect to that object or event.

This means that if we measure the length of an object stationary in $S$, we measure proper length. However, if we measure the length of an object in $S'$, this is not proper length because it will be contracted due to length contraction. An observer in $S'$ could measure proper length of an object in $S'$ because as seen from his frame, the object is stationary.

Now on to the invariant spacetime interval. You are probably familiar with the pythagorean theorem:

$$(\Delta {d})^{2}=(\Delta {x})^{2}+(\Delta {y})^{2}+(\Delta {z})^{2}$$

which gives the same distance for every observer in euclidean geometry. However, Minkowski spacetime is not euclidean and thus obervers at different velocities measure different distances due to length contraction. The invariant spacetime interval, however, is defined in such a way that it gives the same result for every observer. It can be written as either

$$(\Delta s)^{2}=(\Delta ct)^{2}-(\Delta x)^{2}-(\Delta y)^{2}-(\Delta z)^{2}$$ $$(\Delta s)^{2}=-(\Delta ct)^{2}+(\Delta x)^{2}+(\Delta y)^{2}+(\Delta z)^{2}$$

For more information, you may want to read Spacetime interval on Wikipedia.

jng224
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