Special Relativity: What are the events in the question?

[AronYstad]

Homework Statement

A spacecraft is moving with a velocity of 85 km/s relative to Earth. How long does it take before an originally synchronized clock on the spacecraft falls 1.0 s behind a clock on Earth?

Relevant Equations

$$\Delta T = \gamma \Delta T_0$$
$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$

This was part of a test we did a while back, so I forgot how exactly I solved it, but I think I basically solved the question by putting the values into equations and hoping for the best, since I didn't have a good understanding back then. Since then, I have learnt that it's a good idea to break questions up into events, and that works for most questions. For example, if the question was about the time between the clock on the spacecraft showing 12:00 and the clock on the spacecraft showing 13:00, then those would be the two events, and $T_0$ would be the time measured on the spacecraft, since that reference frame observes the two events at the same spacial coordinate. However, in this question, I am having trouble breaking the problem up like that, since the question is about comparing two clocks, not just observing a single one.

 

[kuruman]

I would consider event A the departure of the spacecraft when the clocks are synchronized and the frames are coincident. Event B will be the arrival of the spacecraft at $x_{\!B}$ and time $t_{\!B}$ measured in the Earth frame and $x'_{\!B}$ and time $t'_{\!B}$ measured in the spacecraft frame. You have the Lorentz equations relating the primed and unprimed coordinates at event B plus the given condition that $t_{\!B}-t'_{\!B}=1~$s and that $x_{\!B}=v~t_{\!B}.$

 

[Chestermiller]

I would consider event A the departure of the spacecraft when the clocks are synchronized and the frames are coincident. Event B will be the arrival of the spacecraft at $x_{\!B}$ and time $t_{\!B}$ measured in the Earth frame and $x'_{\!B}$ and time $t'_{\!B}$ measured in the spacecraft frame. You have the Lorentz equations relating the primed and unprimed coordinates at event B plus the given condition that $t_{\!B}-t'_{\!B}=1~$s and that $x_{\!B}=v~t_{\!B}.$

In addition, I would take x’ = 0.

I look upon this as a spacecraft with x'=0 flying past a sequence of "stationary" clocks (earth frame) synchronized with the clock on earth (x = 0) and strung out in the x-direction. So the first event occurs at x = x' = 0, t = t' = 0. The 2nd event takes place at x' = 0 (in rocket), x = 85t (location of rocket in earth frame), sequence clock reading t, rocket clock reading t'.

 

[kuruman]

In addition, I would take x’ = 0.

Did you mean $x'_{\!B}=0$? That is not really needed for solving the problem because $$t'_{\!B}=\gamma\left(t_{\!B}-\frac{v~x_{\!B}}{c^2}\right)= \gamma\left(t_{\!B}-\frac{v^2~t_{\!B}}{c^2}\right).$$ With $t'_{\!B}=t_{\!B} - 1~\text{(s)}~$, one can find $t_{\!B}.$

 

[kuruman]

I look upon this as a spacecraft with x'=0 flying past a sequence of "stationary" clocks (earth frame) synchronized with the clock on earth (x = 0) and strung out in the x-direction. So the first event occurs at x = x' = 0, t = t' = 0. The 2nd event takes place at x' = 0 (in rocket), x = 85t (location of rocket in earth frame), sequence clock reading t, rocket clock reading t'.

I think it would be appropriate to use subscripts indicating specific coordinates attached to specific events in order to distinguish what's what when doing the algebra.

 

[vela]

Did you mean $x'_{\!B}=0$? That is not really needed for solving the problem because $$t'_{\!B}=\gamma\left(t_{\!B}-\frac{v~x_{\!B}}{c^2}\right)= \gamma\left(t_{\!B}-\frac{v^2~t_{\!B}}{c^2}\right).$$ With $t'_{\!B}=t_{\!B} - 1~\text{(s)}~$, one can find $t_{\!B}.$

But it does get you to the same intermediate result if you use $t_B = \gamma[t'_B + (v/c^2)x'_B]$ instead of the equation you chose, without the extra bit of algebra.

 

[kuruman]

But it does get you to the same intermediate result if you use $t_B = \gamma[t'_B + (v/c^2)x'_B]$ instead of the equation you chose, without the extra bit of algebra.

Ah yes, that it does.

 

[AronYstad]

You have the Lorentz equations relating the primed and unprimed coordinates at event B

I don't think I have learnt the Lorentz equations. I at least don't recall being taught anything with that name. Is there any other way to describe it?

 

[kuruman]

I don't think I have learnt the Lorentz equations. I at least don't recall being taught anything with that name. Is there any other way to describe it?

This is what I was referring to. It is possible that you may have seen them under a different name. They are a must if you wish to formulate relativistic problems in terms of events.

 

[AronYstad]

It is possible that you may have seen them under a different name. They are a must if you wish to formulate relativistic problems in terms of events.

Except for the Lorentz factor, I haven't seen those equations before. I have only ever seen the Lorentz factor used in the formula for time dilation that I put in the beginning of the thread, and the similar formula for length contraction. If I haven't been taught the Lorentz equations, then it can't be expected that I should use them on a test. So if they are required to formulate relativistic problems in terms of events, then what other ways are there to formulate the problem?

 

[jbriggs444]

So if they are required to formulate relativistic problems in terms of events, then what other ways are there to formulate the problem?

The problem is one where a simple calculation using time dilation alone is sufficient.

We are invited to adopt the [inertial] rest frame of the Earth. We are asked to consider a second clock which is moving uniformly relative to our chosen frame. We know that time dilation will cause us to measure that clock to be ticking slowly. How slowly? That is the Lorentz factor.

So we are left to do the algebra. If our clocks advance by $t$, the other clock advances by $\gamma t$. We want a difference of 1 second between our clock and the other clock: $1 = t - \gamma t$.

Solve for our clock time $t$ when this happens.

The trick to using time dilation properly (if you are not working with events and the full Lorentz transforms) is to figure out which clock is moving relative to the reference frame that you choose. That is the clock that will be ticking slowly.

The "ticking slowly" is the effect you get when comparing a moving clock (according to some frame) to a pair of clocks that are at rest (according to that same frame) and are synchronized (again according to that same frame).

So the heuristic that I use is to look for the single moving clock. That's the one that will be time dilated. I also look for the inertial frame against which that clock is seen to move. That is where the calculation of time dilation is valid.

If you want to cast this in terms of events then...

Event 1: Two clocks start ticking, both at zero. They are co-located. One clock moves off to the right at 85 km/s relative to the other. We will arbitrarily call it the "moving clock". The other will be the "rest clock".

Event 2: A third clock, at rest relative to the "rest clock" and some distance to the right begins ticking. at zero. This event is simultaneous (according to their shared inertial rest frame) with event 1. We will call this clock the "third clock".

Event 3: The "moving clock" passes the "third clock" we are told that the difference in their readings at this event is 1 second.

Event 4: [Irrelevant] The "rest clock" reaches the same reading as that of the "third clock" at event 3. But this event is pretty much irrelevant. The problem asks for the reading on the "rest clock" at this event. But, by construction, that reading is the same as the reading of the "third clock" at event 3.

We are asked for the reading on the "third clock" at event 3.

This is a perfect scenario for time dilation. One moving clock against two rest clocks, looking for elapsed time on the one inertial moving clock compared to that shared by two inertial and synchronized rest clocks.

 

[AronYstad]

The problem is one where a simple calculation using time dilation alone is sufficient.

We are invited to adopt the [inertial] rest frame of the Earth. We are asked to consider a second clock which is moving uniformly relative to our chosen frame. We know that time dilation will cause us to measure that clock to be ticking slowly. How slowly? That is the Lorentz factor.

So we are left to do the algebra. If our clocks advance by $t$, the other clock advances by $\gamma t$. We want a difference of 1 second between our clock and the other clock: $1 = t - \gamma t$.

Solve for our clock time $t$ when this happens.

The trick to using time dilation properly (if you are not working with events and the full Lorentz transforms) is to figure out which clock is moving relative to the reference frame that you choose. That is the clock that will be ticking slowly.

The "ticking slowly" is the effect you get when comparing a moving clock (according to some frame) to a pair of clocks that are at rest (according to that same frame) and are synchronized (again according to that same frame).

So the heuristic that I use is to look for the single moving clock. That's the one that will be time dilated. I also look for the inertial frame against which that clock is seen to move. That is where the calculation of time dilation is valid.

If you want to cast this in terms of events then...

Event 1: Two clocks start ticking, both at zero. They are co-located. One clock moves off to the right at 85 km/s relative to the other. We will arbitrarily call it the "moving clock". The other will be the "rest clock".

Event 2: A third clock, at rest relative to the "rest clock" and some distance to the right begins ticking. at zero. This event is simultaneous (according to their shared inertial rest frame) with event 1. We will call this clock the "third clock".

Event 3: The "moving clock" passes the "third clock" we are told that the difference in their readings at this event is 1 second.

Event 4: [Irrelevant] The "rest clock" reaches the same reading as that of the "third clock" at event 3. But this event is pretty much irrelevant. The problem asks for the reading on the "rest clock" at this event. But, by construction, that reading is the same as the reading of the "third clock" at event 3.

We are asked for the reading on the "third clock" at event 3.

This is a perfect scenario for time dilation. One moving clock against two rest clocks, looking for elapsed time on the one inertial moving clock compared to that shared by two inertial and synchronized rest clocks.

Thanks. That makes intuitive sense to me. However, when I asked about another question back in March about someone going from Earth to a dwarf planet in a spaceship, the answer I got was that the person in the spaceship measured the shorter time T₀, or what you just called t. This made me slightly confused, since I at first thought time dilation was symmetric, and since it appears to be the opposite case in this question. However, after a bit of thought, I think that maybe that other question had to do with the distance to the dwarf planet, and thus the measured time is lower on the spaceship because the distance shrinks with length contraction. But I don't think length contraction was ever brought up in that thread, so now I'm really unsure. Am I thinking somewhat correctly?

 

[kuruman]

It's probably easier to see what's going on if you consider the following (more realistic) situation instead of a spaceship. A pion ($\pi^+$) is a particle that has lifetime $\tau_0$ in its own frame and then decays to a muon ($\mu^+$) and a neutrino. If a pion is created in the lab at the origin O and travels with speed $v$, where must one place a detector D in order to detect the muon as soon as it is created? In other words, what should the distance OD be?

The events are A = creation of the pion and D = detection of the just-created muon. The key to answering the question is that if the detector detects the muon in the lab frame, it must also do so in the particle frame. If the event occurs in one frame, it must occur in the other. Also note that in both frames the pion travels at speed $v$.

In the lab frame
Let distance OD = $L_0$ and let $\tau_0=~$ the pion's lifetime in its own (proper) frame. The particle's lifetime in the lab frame is longer and given by $\tau={\gamma}~\tau_0$. It follows that for the pion to decay when it reaches D, it must have speed $$v=\frac{L_0}{\tau}=\frac{ L_0 }{\gamma~\tau_0}.$$In the particle frame
The particle lives for time $\tau_0$ and travels distance $L$ which is shorter than OD = $L_0$ by a factor of $\gamma$, i.e. $L=\frac{1}{\gamma} L_0$. It follows that for the pion to decay after covering distance $L$, it must have speed $$v=\frac{L}{\tau_0}=\frac{ L_0 }{\gamma~\tau_0}.$$The speeds match and the pictures are consistent.

 

[jbriggs444]

Thanks. That makes intuitive sense to me. However, when I asked about another question back in March about someone going from Earth to a dwarf planet in a spaceship, the answer I got was that the person in the spaceship measured the shorter time T₀, or what you just called t.

Yes, the person on the spaceship will measure the shorter time. His wristwatch is moving inertially. It is one clock. The single clock runs slow. [See the heuristic mentioned about midway in #11]. The time recorded by a single clock is "proper time".

By comparison, the time difference between the two clocks, one on Earth and one on the dwarf planet, synchronized according to an Earth standard will be the baseline. The time kept by an array of synchronized clocks is "coordinate time".

Proper time is dilated compared to coordinate time.

This made me slightly confused, since I at first thought time dilation was symmetric

Yes. Time dilation is symmetric.

If you compare a single Earth clock against a pair of co-moving rockets, the Earth clock will be dilated.

If you compare a single rocket clock against a clock on Earth and a co-moving clock on a dwarf planet then the rocket clock will be dilated.

Either way, proper time is slow compared against coordinate time. Equivalently, the delta on the one clock will be less than the delta between the two synchronized clocks.

However, after a bit of thought, I think that maybe that other question had to do with the distance to the dwarf planet, and thus the measured time is lower on the spaceship because the distance shrinks with length contraction. But I don't think length contraction was ever brought up in that thread, so now I'm really unsure. Am I thinking somewhat correctly?

Yes. That is correct.

Length contraction can explain why the rocket measures a reduced time. It travels what it measures as a reduced distance. [Or, to be picky, the rocket watches Earth and the dwarf planet whiz past with a reduced separation between them. If we say we are adopting a reference frame, we should actually do so]

If you turn around and ask how the rocket explains that the Earth and dwarf planet observers got a longer elapsed time according to their two clocks, the explanation will turn out to be the "relativity of simultaneity". The rocket observer will claim that the Earth clock and the dwarf planet clocks were not properly synchronized according to the rocket's standard of rest.

[Apologies if this got somewhat repetitive. I'm trying to say the same thing multiple times in multiple ways to get it to make sense and stick]

 

[AronYstad]

So it depends on what the question is asking for. If it's about time and velocity, but not about travelling a certain distance, then the question is almost always about what an observer sees in its own reference frame compared to what it sees in the other reference frame, and that's only time dilation. If it's about distance and velocity, and two observers measuring the time it takes to move a certain distance, then you need to take length contraction into account, meaning one of them will measure a shorter time, but both will still see the other one as slowed down. And in all possible questions, you need to take into account the formulation of the question in order to figure out what it is you need to calculate. Did I get that right?

 

[kuruman]

So it depends on what the question is asking for.

That is dangerous thinking and can lead you into pitfalls. There is no preferred frame. You can solve any problem in either frame. The only difference might be that solving in one frame may require a few more algebraic steps. The two frames are equivalent in that regard.

$\dots~$ then the question is almost always about what an observer sees in its own reference frame compared to what it sees in the other reference frame,$\dots$

No, no, no. Observer A sees an event in its own reference frame only and assigns coordinates $\{\vec r_A,t_A\}$ to it. That's all observer A sees. Likewise, observer B sees the same event in its own reference frame only and assigns coordinates $\{\vec r_B,t_B\}$ to it. When the two sets of coordinates are compared, they are related by the Lorentz transformation equations.

And in all possible questions, you need to take into account the formulation of the question in order to figure out what it is you need to calculate.

Of course. You always need the formulation of the problem to know what is given, what is not given and what you are asked to find.

 

[vela]

Thanks. That makes intuitive sense to me. However, when I asked about another question back in March about someone going from Earth to a dwarf planet in a spaceship, the answer I got was that the person in the spaceship measured the shorter time T₀, or what you just called t. This made me slightly confused, since I at first thought time dilation was symmetric, and since it appears to be the opposite case in this question. However, after a bit of thought, I think that maybe that other question had to do with the distance to the dwarf planet, and thus the measured time is lower on the spaceship because the distance shrinks with length contraction. But I don't think length contraction was ever brought up in that thread, so now I'm really unsure. Am I thinking somewhat correctly?

Time dilation is symmetric, but you need to recognize when the observers compare clocks, each is considering a different pair of events because of the relativity of simultaneity.

Say Alice is the observer on Earth, and Bob is the observer on the spaceship. Alice checks her clock when, according to her, Bob reaches the dwarf planet. She sees more time elapsed on her clock than Bob's. From Bob's perspective, however, Alice looked at her clock long after the dwarf planet reached him, so of course, she measured a longer time than what his clock read. If she had checked her clock when, according to Bob, the dwarf planet reached him, she would have seen less time had elapsed on her clock than on his. In other words, Alice looking her her clock when she says the trip ended and Alice looking at her clock when Bob says the trip ended are two different spacetime events.

 

[AronYstad]

No, no, no. Observer A sees an event in its own reference frame only and assigns coordinates {r→A,tA} to it. That's all observer A sees. Likewise, observer B sees the same event in its own reference frame only and assigns coordinates {r→B,tB} to it. When the two sets of coordinates are compared, they are related by the Lorentz transformation equations.

I was a bit unclear with the wording. What I meant was that the question would be about for example looking at your own clock, then looking at the clock on a spaceship and comparing them. I know that each observer only sees the event in its own reference frame. Also, as stated, I don't know the Lorentz equations.

 

[AronYstad]

That is dangerous thinking and can lead you into pitfalls. There is no preferred frame. You can solve any problem in either frame.

Is that true even when the problem says that the clock on the spaceship falls behind? Cause both observers will see the other observer's clock slowed down, right? So if the question says that the spaceship's clock falls behind, then doesn't that mean that that's what's observed from Earth?

 

[AronYstad]

Time dilation is symmetric, but you need to recognize when the observers compare clocks, each is considering a different pair of events because of the relativity of simultaneity.

Say Alice is the observer on Earth, and Bob is the observer on the spaceship. Alice checks her clock when, according to her, Bob reaches the dwarf planet. She sees more time elapsed on her clock than Bob's. From Bob's perspective, however, Alice looked at her clock long after the dwarf planet reached him, so of course, she measured a longer time than what his clock read. If she had checked her clock when, according to Bob, the dwarf planet reached him, she would have seen less time had elapsed on her clock than on his. In other words, Alice looking her her clock when she says the trip ended and Alice looking at her clock when Bob says the trip ended are two different spacetime events.

I got the right answer when I simply accounted for length contraction, and now you've confused me. Do I need to take all of that into account, or can I simply think of it as Bob measuring a shorter distance because of length contraction, and therefore a shorter time since both measure him to be travelling at the same velocity?

 

[kuruman]

I was a bit unclear with the wording. What I meant was that the question would be about for example looking at your own clock, then looking at the clock on a spaceship and comparing them.

To use @vela's example, Alice can look at her clock and take a picture of its reading at the time she has calculated that Bob has arrived. Bob takes a picture of his clock when the planet arrives to him. Then Alice and Bob exchange pictures using signals that travel at the speed of light and when these signals arrive, they compare readings. You cannot say that Alice looks at her clock and "then" at Bob's clock because that implies a simultaneity that just isn't there even after correcting for the signal transit times.

Here is yet another calculation. Let
$L_0=$ the Earth-planet distance measured by Alice.
$v=~$ the speed of the spaceship.
According to Alice:
Bob reaches the planet when her clock reads time $$t_A=\frac{L_0}{v}.$$According to Bob:
The distance to the planet is $L=\dfrac{L_0}{\gamma}$ so that when the planet reaches him, his clock reads $$t_B=\frac{L}{v}=\frac{L_0}{\gamma v}.$$When they compare pictures, they see that $$t_A=\gamma t_B.$$See @vela's explanation in post #17 about how Alice and Bob reconcile what's they see on the pictures.

 

[AronYstad]

To use @vela's example, Alice can look at her clock and take a picture of its reading at the time she has calculated that Bob has arrived. Bob takes a picture of his clock when the planet arrives to him. Then Alice and Bob exchange pictures using signals that travel at the speed of light and when these signals arrive, they compare readings. You cannot say that Alice looks at her clock and "then" at Bob's clock because that implies a simultaneity that just isn't there even after correcting for the signal transit times.

Here is yet another calculation. Let
$L_0=$ the Earth-planet distance measured by Alice.
$v=~$ the speed of the spaceship.
According to Alice:
Bob reaches the planet when her clock reads time $$t_A=\frac{L_0}{v}.$$According to Bob:
The distance to the planet is $L=\dfrac{L_0}{\gamma}$ so that when the planet reaches him, his clock reads $$t_B=\frac{L}{v}=\frac{L_0}{\gamma v}.$$When they compare pictures, they see that $$t_A=\gamma t_B.$$See @vela's explanation in post #17 about how Alice and Bob reconcile what's they see on the pictures.

Thanks for the explanation. My original statement wasn't about that type of question though. I said that for questions where it wasn't about distance, it would be about an observer looking at their own clock and the other clock and comparing them, for example in questions where you're only measuring how long it takes for a clock to fall behind with a certain amount. Can't you say it like that though? Comparing your clock with the other one?

 

[vela]

I got the right answer when I simply accounted for length contraction, and now you've confused me. Do I need to take all of that into account, or can I simply think of it as Bob measuring a shorter distance because of length contraction, and therefore a shorter time since both measure him to be travelling at the same velocity?

Your explanation reconciles Alice's and Bob's measurements about the same pair of events. Alice and Bob both agree that Alice's clock will say more time elapsed between the two events than Bob's did.

But time dilation is necessarily symmetric in accordance with the principle of relativity. From Bob's perspective, Alice's clock runs slow. When the dwarf planet reaches Bob, he will say Alice's clock shows less time has elapsed than his clock, which seems to contradict what I wrote in previous paragraph. My post was explaining how you reconcile this paradox.

 

[AronYstad]

Your explanation reconciles Alice's and Bob's measurements about the same pair of events. Alice and Bob both agree that Alice's clock will say more time elapsed between the two events than Bob's did.

But time dilation is necessarily symmetric in accordance with the principle of relativity. From Bob's perspective, Alice's clock runs slow. When the dwarf planet reaches Bob, he will say Alice's clock shows less time has elapsed than his clock, which seems to contradict what I wrote in previous paragraph. My post was explaining how you reconcile this paradox.

Sorry, English isn't my main language, so I don't really understand what you mean by "reconcile". However, are you saying that my calculations were correct, but to understand how they can be correct while keeping symmetry, you need to take into account that Bob sees Alice checking/stopping her clock some time after he has arrived, since their pairs of events are different?

 

[AronYstad]

Also, in my physics formula book, there is a note next to the formula for time dilation, which I find to be quite contradictory to the answers I have gotten in the two threads I have made here. Here is my best translation of the notes:

t is a time interval measured by a clock that's moving with the velocity v relative to the system where the event occurs

t₀ (proper time) is the corresponding time interval measured by a clock that's at rest relative to the system where the event occurs

The reason why I find this contradictory is both because from my current understanding, a time interval is measured between two events, not one, and also, I thought time dilation was about comparing two measurements made in systems with relative velocity to each other, from the perspective of one of them. The formulation there makes it seem like it's more like the Bob and Alice scenario mentioned in a post above, which from my current understanding is because of length contraction. Since the velocity remains the same, this means that the measured time will be longer or shorter by the gamma factor since the length is contracted by that same factor. Is this also called time dilation?

 

[malawi_glenn]

an event is a space + time coordinate.
A clock at rest in its own frame at say position $x$, if you have two different time readings, say $t_1$ and $t_2$, you have two events: $(x, t_1)$ and $(x, t_2)$.

 

[AronYstad]

an event is a space + time coordinate.
A clock at rest in its own frame at say position $x$, if you have two different time readings, say $t_1$ and $t_2$, you have two events: $(x, t_1)$ and $(x, t_2)$.

Yeah, I know that, and that's why I'm confused by the formulation in the formula book. It makes it seem like a time interval can be measured with only one event.

 

[malawi_glenn]

Do you have natur & kultur formula book?

 

[AronYstad]

No, I have "Formler & tabeller i fysik, matematik & kemi för gymnasieskolan" by Ekholm, Fraenkel and Hörbeck.

 

[malawi_glenn]

I have it here in front of me now. This is an old Swedish formulation that is very confusing. Even in the book Impuls 1 is incorrectly written see page 354 "where $t_0$ is the time for an event".

I have written to the authors many times over the years, never got any reply...

If you want to learn relativity, get rid of all gymnasium and "introductory modern physics" books. Study the real deal, like Susskinds book or Morins. One of these books are gonna serve you better than trying to cope the horrible swedish gymnasium books and asking questions here.

I could also send you my Fysik 3 book, which is written in swedish. Currently working on a new edition, correcting various typos and such. Just PM me.