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I read a problem from this page here: https://byjus.com/physics/lorentz-transformations/#:~:text=What%20is%20the%20purpose%20of,another%20at%20a%20constant%20velocity.

Basically in summary the question poses this question: Spacecraft S’ is crossed by another spacecraft S at a speed of c/2 on their way to Alpha Centauri. When spacecraft S crosses S’, the captain of S’ sends a signal that lasts for 1.2s. Calculate the time interval of the signal using Lorentz Transformation in the perspective of spacecraft S.

After going through the Lorentz Transformation I get the answer 1.6s which is also the answer they give. I think this is the time interval of the signal as observed by the spacecraft travelling at c/2.

Although this doesn't make sense to me. The faster you go the faster time goes so a short time for you is a long time for someone not travelling as fast. So if a the signal is 1.2s shouldn't the signal as observed by you be shorter because time goes faster for you.

I feel like I am missing something trivial or misinterpreting the Lorentz Transformation. Can you please explain why how I am thinking is incorrect?

Thank you so much

Also if the link does not work for you then search up byjus lorentz transformation, then click the site with the title Lorentz Transformation - Definition, Equations, Formula ... which is usually the first result. That one is the site that I have linked. If you scroll down till you get to Q1, you should get the problem I stated above.

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  • $\begingroup$ Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Commented Feb 23, 2023 at 0:32
  • $\begingroup$ The link says: Access denied. You do not have access to byjus.com. $\endgroup$
    – Qmechanic
    Commented Feb 23, 2023 at 0:33
  • $\begingroup$ any problem that introduces the primed frame before the unprimed frame is sus . $\endgroup$
    – JEB
    Commented Feb 23, 2023 at 4:12

4 Answers 4

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The faster you go the faster time goes so a short time for you is a long time for someone not travelling as fast.

This understanding violates the principle of relativity.

In special relativity, everyone sees their own clocks tick at one second per second regardless of their motion through the vacuum, so long as it is inertial.

In both special relativity and classical mechanics, if you see a clock moving towards you, it appears to tick faster than one second per second, and if you see a clock moving away from you it appears to tick slower than one second per second. In classical mechanics this is a well known effect with well established formulas, called the Doppler effect, and those formulas require an underlying carrier medium which we would call “the lumineiferous ether”. We would say that if you are stationary relative to the ether and a clock is moving at speed $v$ towards you, then you see it tick at $c/(c-v)$ ticks per second. For small velocities this approximates as $\approx 1 + \frac v c + \left(\frac v c\right)^2$.

Special relativity, on the other hand, has the temerity to suggest that there is no luminiferous ether, so that everyone should just believe that they are at rest relative to the medium of wave propagation! Moreover the formula is slightly different, $$ \frac f {f_0}=\sqrt{c+v\over c-v}\approx 1+\frac vc + \frac12 \left(\frac v c\right)^2.$$ The same to first order, but just the tiniest bit slower in that second order term.

So if you take the relativistic formula and try to correct for the classical Doppler effect, then you will decide that the clock is “apparently” ticking at a slower frequency $f_0/\gamma$ as if their seconds were longer “dilated” by this factor $$\gamma = \frac1{\sqrt{1-(v/c)^2}}.$$

Of course that is not what they think, if that clock is inside a spaceship and someone inside that spaceship is looking at your clocks, doing the same argument, then they think their clocks are ticking at one second per second and your clock is ticking slower.

This leads to an interesting conundrum: if they think that my clocks are ticking slow, and I think that their clocks are ticking slow, can't I just call them up on the phone and we will find out who is right? Surely on the phone one of us will appear to be chattering too fast and the other will be sluggish and then we will know!

Except, they are at a distance from me, my phone works by sending microwave radiation which is a sort of light, that takes a certain time, and so there's a propagation delay. If they are approaching me, I say something into my phone, and they wait a few seconds, and then they hear me speaking very fast. Then they say something into their phone and I wait a few seconds and I hear them speaking very fast. The contradiction is resolved by this time delay. (And the argument was always specious, it required using a formula for the Doppler effect in classical mechanics and correcting for it in relativistic mechanics. If you want I could slow down their speech with a Doppler approximation and they could slow down mine and we would agree that the other person seems sluggish. If we did this with real phones, they probably send datagrams and we wouldn't hear anything different, they would just start “buffering” if we were moving away from each other and they couldn't get enough data fast enough to fill their buffers.)

If information could propagate instantaneously, no time delay, nowhere for the contradiction to hide, and yes, Nature would have to anoint one of us as right and the other as wrong. But as far as we can tell nothing goes faster than the speed of light and the speed of light is not fast enough to decide who is right.

So everybody sees everyone else's clocks moving slowly, except that this is based on a classical reconstruction because the effect is second order and is overwhelmed by the first order effect of the Doppler shift.

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  • $\begingroup$ I don't understand again. You are saying a clock moving away from you would tick slower. However I thought Lorentz Transformation finds what is observed by the other observer(in this case spaceship S'). So spaceship S' observes 1.2 sec. Spaceship S apparently records 1.6 sec. Spaceship S' and S, I think, are looking at clocks within their own spaceships. I don't see when they would be looking at the clock in the other spaceship and when the effect you are talking about would play a part. $\endgroup$ Commented Feb 24, 2023 at 2:56
  • $\begingroup$ The Lorentz transformation does not tell you what is literally observed, it tells you what you would classically reconstruct in your coordinate system based on your observations. For example: you might believe that if you saw a ball moving past you at relativistic speeds, it would appear to be a “pancake” due to length contraction. This is incorrect, it is a theorem of special relativity that anything which presents a circular outline will continue to do so under Lorentz boosts; instead it will appear strangely rotated to your eye (an effect called Terrell rotation). $\endgroup$
    – CR Drost
    Commented Feb 24, 2023 at 18:42
  • $\begingroup$ Please can you explain why the effect you talk about would even arise in the first place. Because I can clearly classically reconstruct that the S is not recording time with S' clock. Sorry if I am sounding mad. No one here is explaining why spaceship S clock is greater in the first place. $\endgroup$ Commented Feb 26, 2023 at 4:18
  • $\begingroup$ So “why does it even arise in the first place” is a tricky question and my answer is just that we happen to live in a world where, whenever you accelerate in some direction $\hat x$ with acceleration $a~\hat x$, then as far as you can tell when propagation delays and classical Doppler is corrected for, any clock ahead of you by coordinate $x$ ticking at rate $f$ appears to have ticked anomalously at rate $f~(1+a~x/c^2).$ Time dilation and length contraction are second order effects of this basic effect, this is the most important thing to understand about the Lorentz transformation. $\endgroup$
    – CR Drost
    Commented Feb 26, 2023 at 5:21
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You cannot tell which is moving faster, only which appears to be moving faster to the observer. Spaceship S sees S' as moving at c/2 while S is not moving. Spaceship s' sees the reverse: S is moving at c/2 while S' is not moving.

Spaceship S' emits a signal that lasts for 1.2 seconds on the S' clock. Spaceship S receives the signal that lasts 1.6 seconds on the S clock. The term "proper time" is the time an event takes when the entire event occurs at the same location. S' observes the proper time. The device emitting the signal does not change for observers on spaceship S'. Proper time is always the shortest time.

The time duration of the signal would change even without relativity. The signal must travel from S' to S. They are together when the signal begins. They are far apart when the signal ends. The end of the signal must travel quite a distance before spaceship S receives it. This is part of the extended time for S. Shifting of distance and time due to special relativity also contribute.

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  • $\begingroup$ That last paragraph you wrote really made sense to me. So I did the math. The spacecraft moving at half the speed of light would have the end of the signal catch up to it in 0.6 seconds. Plus the 1.2 seconds of the length of the signal. You get 1.8. Did the extra 0.2 seconds get subtracted because of time dilation? Also this all is obviously assuming the signal travels at the speed of light, but it doesn't state that in the problem. $\endgroup$ Commented Feb 24, 2023 at 2:36
  • $\begingroup$ You are correct about the 0.2sec loss. Consider a rod in the S' (sender) frame of reference. The rod extends from S' to where S is when the end of the signal reaches S. S' says the end of the signal should reach S 1.8sec after S' receives the start of the signal. The rod looks shorter to S because the rod is moving for S. The shortened distance rduces the time to less than 1.8sec. $\endgroup$ Commented Feb 27, 2023 at 18:18
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I think it is easier to see these kinds of things by explicitly showing the Lorentz transformations in a diagram. These look as follows in frame $S$ and $S'$ respectively. The main point is that how fast time is ticking depends on the frame of reference.

enter image description here

enter image description here

Some explanations for these diagrams. Point $B$ is located at $t=1.2$ seconds$^*$ in frame $S$. For spacecraft $S'$ this event does not occur after 1.2 seconds. How many seconds have passed for spacecraft $S'$? We can reason this by performing a Lorentz transformation that places $B$ at the origin. The dashed line shows how $B$ would be transformed for different Lorentz transformations. So we can read off the time at point $A$, which turns out to be ~1.04 seconds. Every point on the dashed line has the same proper time. This means that if you sent out a bunch of constant velocity observers from the origin, they will see 1.04 seconds on their personal clock when they cross the dashed line.

An important point to notice is that in frame $S$, point A occurs after 1.04 seconds and point B occurs after 1.2 seconds. In frame $S'$, point A occurs after 1.2 seconds and point B occurs after 1.04 seconds. So the order of events changes depending on reference frame. Likewise, point C occurs after 1.6 seconds in frame $S$, which is the answer to your question, but in frame $S'$ it occurs after about 1.89 seconds.

$*$ Note: the way I drew these diagrams the time axis is not in seconds but in $c\times$ time, or in lightseconds. This should not change the reasoning because it is just a scale factor.

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One cause of confusion in this type of problems is to consider only spaceships instead of frames.

So, spaceship S' send a signal during $1.2 s$. The frame S is passing by with velocity $0.5 c$. Suppose there are a row of ships in this frame S, all with their clocks synchronized.

And when the signal starts, both clocks of S' and of the ship nearby (from the row of S) show zero.

After $1.2 s$ in S' frame, looking to a clock of the spaceship of the S-row passing by in that moment, (not the original one that is now too far), it will show $1.6 s$. That is the meaning of the Lorentz transformation.

The same experiment could be done by one of the spaceships from S, and a row of spaceships of S'. The time of S' would be greater in this case.

Note that when S' arrives in Alpha Centauri, the ship's clock will show less time than the local clock there, if we suppose that the clock in the star is syncronized with the Earth. In this case, Earth and Alpha Centauri are like 2 spaceships with their clocks synchronized, passing by S' .

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  • $\begingroup$ Why is the time in the other spaceship greater though? $\endgroup$ Commented Feb 24, 2023 at 2:40
  • $\begingroup$ The Lorentz transformation of time is $t' = \gamma(t - \frac{vx}{c^2})$. It is necessary that $x = 0$ for the relation $t$ ans $t'$ be only $\gamma$. That is why the spaceship compares its clock with the clock of the other frame that is close to it. $\endgroup$ Commented Feb 24, 2023 at 11:18
  • $\begingroup$ Why is the time in spaceship S greater? That is my question. You just say it is. I don't understand how when S is moving when spaceship S' looks over it sees a greater time, but when S stops the Alpha Centauri time is greater. $\endgroup$ Commented Feb 26, 2023 at 4:21
  • $\begingroup$ Time dilation means: the clocks of a frame that is moving in relation to me records greater time intervals than mine. But for $x=0$, so we must compare clocks of that frame that are momentarily at my location. $\endgroup$ Commented Feb 26, 2023 at 13:12

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