I'm calculating spacetime intervals in standard setup with frames A and B
where t = t' = 0 and x = x' = 0, and all the motion is going only in x direction.

Objects A and B move in relative velocity v = 0.99c, (giving us $\gamma= 7.089$) for duration of what A perceives to be 100 seconds.

I need to calculate spacetime interval undergone by the objects A and B, from each of the frames. Meaning ie sB is spacetime interval of B as calculated from reference frame A and s'A is spacetime interval of A calculated from B ( the apostrophe signals from which frame we are calculating)

Now spacetime interval undergone by B as seen from reference frame A should be

$sB^2= (ct)^2 - x^2 = (100c)^2 - (99c)^2 = 199c^2$
time is 100 seconds as seen from A, in that time with speed 0.99c the B moves 99 light seconds

Now we know that what A perceives to be distance of 99 light seconds will B perceive to be contracted by factor of $\gamma$, so in their proper time they will traverse this distance in shorter time $t' = t /\gamma$

$s'B^2= (100c / \gamma)^2 = (14.1c)^2 = 199c^2$

so far so good, even though im not sure this is correctly reasoned about

now for the spacetime interval of point A, first from the reference frame of A :

$sA^2 = (100c)^2 = 10 000c^2$

and them from reference frame of B:

$s'A = (100c/\gamma)^2 - (99c/\gamma)^2 = 4c^2$

completely different number!

Shouldnt they be equal, given that spacetime interval is invariant?
I must have made an error in my reasoning but I dont see where, can someone help me?


1 Answer 1


Shouldnt they be equal, given that spacetime interval is invariant? I must have made an error in my reasoning but i dont see where, can someone help me?

Yes, they should be equal. The reason that they are not equal is because you incorrectly performed the Lorentz transform. I will show you where you made the mistake, but I will also show how I go about such problems to make sure that such mistakes are avoided as much as possible.

First, objects have worldlines, so they have an infinite number of events on each worldline. So there isn't just one "spacetime interval undergone by the objects". To reflect that, I like to write down the actual worldline for the objects whenever it is easy to do so. It looks like you are considering A's frame to be the unprimed frame, so in the unprimed frame A's worldline is $(c\ t, 0)$ where the first coordinate is $ct$ and the second coordinate is $x$. Similarly, B's worldline is $(c\ t,0.99 \ c\ t)$.

Now, we want to calculate the spacetime interval, so we need to select an event on each worldline. We will select the events at $t=100$ for each. We will label the events as $a$ and $b$ respectively using lower case to distinguish between the entire worldline for A and B and the specific events. So $a=(100 \ c,0)$ and $b=(100\ c,99\ c)$, again all in the unprimed (A) frame.

So, let's calculate the intervals. I will skip the details since you have that down correctly, and we get $sa^2=10000\ c^2$ and $sb^2=199\ c^2$ exactly as you had before.

At this point, we will transform from the unprimed frame to the primed frame. The problem where you made your mistake was that you did not use the full Lorentz transform. You tried to consider only time dilation and length contraction. By using the full Lorentz transform we get different values than what you had. Specifically, we get $a'=(708.9 \ c,-701.8 \ c)$ and $b'=(14.1 \ c,0)$. Note in particular, that these two events are not simultaneous in the unprimed frame where your calculation incorrectly shows that they are simultaneous. This error is due to using time dilation and length contraction rather than the Lorentz transform.

Finally, we can calculate the intervals as before and we get $s'a^2=10000\ c^2$ and $s'b^2=199\ c^2$, which agrees with our previous calculations as it should.

  • $\begingroup$ Thanks for your example of correct calculation. I was feeling that the last calculation has some trouble concerning time and simultaneity. I will accept your answer as solution to my problem. But I still do not understand which relativistic effect i should take into account when performing my step-by-step reasoning about events. I am trying to get a grasp of what is physically happening beyond the full lorentz transform, which, as you shown, yields the correct results $\endgroup$ Aug 9, 2019 at 15:27
  • $\begingroup$ You have to take all of the relativistic effects into account! That is the purpose for using the Lorentz transform. When I am discussing relativity with new students, in fact, I recommend not using the time dilation and length contraction formulas at all. It is far too easy to try to use them where they do not apply. If you use the full Lorentz transform in a situation where time dilation or length contraction is appropriate, then the Lorentz transform will automatically simplify correctly. You are better off getting an intuitive understanding of the Lorentz transform than a piecemeal approach $\endgroup$
    – Dale
    Aug 9, 2019 at 16:00
  • $\begingroup$ So you are saying a person should first learn the math before trying to understand why the math is in such way? I understand that this is the way chinese school of thought on learning goes. $\endgroup$ Aug 9, 2019 at 16:45
  • $\begingroup$ No, I am saying that you should be taught the right math first. Length contraction and time dilation are very error-prone and should be taught last. The Lorentz transform is, more general, more powerful, and less error-prone and should be taught first. Each concept should be taught so that it is understood why it is that way. $\endgroup$
    – Dale
    Aug 9, 2019 at 17:00

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