B Second Opinion Needed: Time of Light Pulse Arrival

[david316]

Hello,

My colleague and I can't agree on an answer to a hypothetical question so have come here for a independent opinion. Question is

Two observers, A and B, are traveling towards each other at a relative speed of 0.6c. Assume their clocks are synced at t = 0.

When A measures t = 0, he measures that B is 6 light-years away. We define this as the proper length.

When B measures t = 0, due to length contraction, he measures that A is 4.8 light-years away.

At the same time (t = 0 in B's frame), A sends a pulse of light towards B.

At what time does B measure the pulse arriving?

Thanks

 

[Dale]

Assume their clocks are synced at t = 0.

In which reference frame? Because of the relativity of simultaneity, when you say two distant clocks are synchronized you need to specify in which reference frame they are synchronized.

 

[david316]

In which reference frame? Because of the relativity of simultaneity, when you say two distant clocks are synchronized you need to specify in which reference frame they are synchronized.

Assume they are synchronized via some uncanny coincidence. If I was to violate all the laws of physics and pause time when the proper length apart was 6 light years, both the clocks would read the same thing (t = 0). Hence they are synced in both frames at the same instance in time. Does that make sense?

 

[SiennaTheGr8]

Does that make sense?

Nope.

 

[david316]

In which reference frame? Because of the relativity of simultaneity, when you say two distant clocks are synchronized you need to specify in which reference frame they are synchronized.

Actually does this simplify it?

Two observers, A and B, are traveling towards each other at a relative speed of 0.6c.

A measures that B is 6 light-years away. We define this as the proper length.

B due to length contraction, measures that A is 4.8 light-years away i.e 0.8 * the proper length

A sends a pulse of light towards B.

How long does it take until B receives the pulse?

 

[Orodruin]

Actually does this simplify it?

No. Again you posted a series of statements that are not defining the situation adequately in SR.

 

[david316]

No. Again you posted a series of statements that are not defining the situation adequately in SR.

Better??

Two observers, A and B, are traveling towards each other at a relative speed of 0.6c.

In A's frame of reference be measures that B is 6 light-years away. We define this as the proper length.

B due to length contraction, measures that A is 4.8 light-years away i.e, 0.8 * the proper length

A sends a pulse of light towards B.

In B's frame of reference how long does it take until B receives the pulse?

 

[david316]

Better still??

Two observers, A and B, are traveling towards each other at a relative speed of 0.6c.

In A's frame of reference B is 6 light-years away. We define this as the proper length.

In B's frame of reference due to length contraction, A is 4.8 light-years away.

A sends a pulse of light towards B.

In B's frame of reference how long does it take until B receives the pulse?

 

[SiennaTheGr8]

No, that doesn't work, @david316

If A and B are traveling toward each other inertially (i.e., no external force is acting on either of them), then their situations are entirely symmetrical.

From A's perspective, A is at rest and B is approaching at some speed $v$. From B's perspective, B is at rest and A is approaching at the same speed $v$.

From A's perspective, the elapsed time measured by B's wristwatch is dilated by a factor of $\frac{1}{\sqrt{1 - (v/c)^2}}$. From B's perspective, the elapsed time measured by A's wristwatch is dilated by a factor of $\frac{1}{\sqrt{1 - (v/c)^2}}$.

From A's perspective, the distances that B measures along the axis of their relative motion are contracted by a factor of $\sqrt{1 - (v/c)^2}$. From B's perspective, the distances that A measures along the axis of their relative motion are contracted by a factor of $\sqrt{1 - (v/c)^2}$.

 

[david316]

No, that doesn't work, @david316

If A and B are traveling toward each other inertially (i.e., no external force is acting on either of them), then their situations are entirely symmetrical.

From A's perspective, A is at rest and B is approaching at some speed $v$. From B's perspective, B is at rest and A is approaching at the same speed $v$.

From A's perspective, the elapsed time measured by B's wristwatch is dilated by a factor of $\frac{1}{\sqrt{1 - (v/c)^2}}$. From B's perspective, the elapsed time measured by A's wristwatch is dilated by a factor of $\frac{1}{\sqrt{1 - (v/c)^2}}$.

From A's perspective, the distances that B measures along the axis of their relative motion are contracted by a factor of $\sqrt{1 - (v/c)^2}$. From B's perspective, the distances that A measures along the axis of their relative motion are contracted by a factor of $\sqrt{1 - (v/c)^2}$.

Won't the distance between two stationary points in one observers frame be measured as contracted by an observer in a moving frame of reference? Hence if two points are 6 light years away in observers A frame of reference ( i.e. they are stationary in respect to him ) observer B will measure them as contracted by a factor of $\sqrt{1 - (v/c)^2}$.

 

[Orodruin]

Better??

No. You are still failing to specify what "at the same time" means as requested in #2. You need this to give the problem meaning.

You can do this in several ways, for example:

When the distance between A and B in A's reference frame is 6 light-years, A sends a signal to B. What is the time difference between this event and B receiving the signal in the reference frame of B?

Alternatively:

When the distance between A and B in A's reference frame is 6 light-years, A sends a signal to B. What is the time difference in the reference frame of B between the event on the world line of B that is simultaneous with the sending of the signal in A's rest frame and the event of B receiving the signal?

These are different questions with different answers.

Won't the distance between two stationary points in one observers frame be measured as contracted by an observer in a moving frame of reference?

There is no such thing as a "stationary" reference frame. Reference frames can only be moving or at rest relative to some object or other reference frame. This is true in classical mechanics as well as SR.

 

[Orodruin]

Won't the distance between two stationary points in one observers frame be measured as contracted by an observer in a moving frame of reference? Hence if two points are 6 light years away in observers A frame of reference ( i.e. they are stationary in respect to him ) observer B will measure them as contracted by a factor of $\sqrt{1 - (v/c)^2}$.

Also, length contraction requires these points to be measured at the same time in B's rest frame. The events satisfying this are not simultaneous in A's rest frame. Hence the need to specify what you mean.

If you tell us your answers we might be able to help you deduce what you are actually computing.

 

[david316]

No. You are still failing to specify what "at the same time" means as requested in #2. You need this to give the problem meaning.

You can do this in several ways, for example:

When the distance between A and B in A's reference frame is 6 light-years, A sends a signal to B. What is the time difference between this event and B receiving the signal in the reference frame of B?

Alternatively:

When the distance between A and B in A's reference frame is 6 light-years, A sends a signal to B. What is the time difference in the reference frame of B between the event on the world line of B that is simultaneous with the sending of the signal in A's rest frame and the event of B receiving the signal?

These are different questions with different answers.

There is no such thing as a "stationary" reference frame. Reference frames can only be moving or at rest relative to some object or other reference frame. This is true in classical mechanics as well as SR.

It would help me greatly if you could answer both the questions you have stated above. Thank you in advance.

 

[david316]

If you tell us your answers we might be able to help you deduce what you are actually computing.

I calculated that B would receive the pulse after 3 years have elapsed on his clock.

 

[SiennaTheGr8]

You're using "stationary" in a way that isn't helpful, I think. "Stationary" is only meaningful when specified relative to something.

From every inertial observer's perspective, points in space are stationary. Our spatial coordinate systems travel with us.

In the situation you've described, A's and B's situations are symmetrical. If, according to A, the distance between A and B is 6 light years exactly $t$ seconds before they crash into each other, then the same is true for B: according to B, the distance between B and A is 6 light years exactly $t$ seconds before they crash into each other.

 

[pervect]

Hello,

My colleague and I can't agree on an answer to a hypothetical question so have come here for a independent opinion. Question is

Two observers, A and B, are traveling towards each other at a relative speed of 0.6c. Assume their clocks are synced at t = 0.

As others have pointed out, this assumption is the root of the problem.

The meta-point here is that clock synchronization is frame dependent. See any of the almost innumberable threads on "Einstein's train" or "Relativity of Simultaneity", You could also read an exceprt from Einstien's book on Relativity about the "Relativity of SImultaneity" (ch 9), at http://www.bartleby.com/173/9.html, or perhaps the paper "The challenge of changing deeply held students beliefs about the relativity of simultaneity", at https://www.aapt.org/doorway/TGRU/articles/Vokos-Simultaneity.pdf

Another approach that might also help is to describe in detail the exchange of signals that lead you to believe the clocks are synchronized. For instance, A might send out a radar signal at some time t1 and is reflected back and received at time t2. Assume the reading on B's clock when the signal reaches B is tb. A is stationary in his own frame, so if t2-t1 is 12 years, A concludes from the constancy of the speed of light that B was 6 light years away at the time (t1+t2)/2 when the signal arrived. If t1=-6, t2=6, and tb=0, the clocks will be synchronized in A's frame. However, they won't be synchronized in B's frame. If B also sends out a radar signal in the above scenario, his clock readings will not be consistent with the notions that the clocks are synchronized in his frame.

 

[david316]

You're using "stationary" in a way that isn't helpful, I think. "Stationary" is only meaningful when specified relative to something.

From every inertial observer's perspective, points in space are stationary. Our spatial coordinate systems travel with us.

In the situation you've described, A's and B's situations are symmetrical. If, according to A, the distance between A and B is 6 light years exactly $t$ seconds before they crash into each other, then the same is true for B: according to B, the distance between B and A is 6 light years exactly $t$ seconds before they crash into each other.

I agree but that's not what I am saying. If there are 2 rulers that are 1 meter long in As frame (call them Ra) and two rulers that are one meter long in B frame (Rb), and both A and B have one of each ruler then A will measure Rb as 0.8 of a meter and B will measure Ra as 0.8 of a meter.

As others have pointed out, this assumption is the root of the problem.

The meta-point here is that clock synchronization is frame dependent. See any of the almost innumberable threads on "Einstein's train" or "Relativity of Simultaneity", You could also read an exceprt from Einstien's book on Relativity about the "Relativity of SImultaneity" (ch 9), at http://www.bartleby.com/173/9.html, or perhaps the paper "The challenge of changing deeply held students beliefs about the relativity of simultaneity", at https://www.aapt.org/doorway/TGRU/articles/Vokos-Simultaneity.pdf

Another approach that might also help is to describe in detail the exchange of signals that lead you to believe the clocks are synchronized. For instance, A might send out a radar signal at some time t1 and is reflected back and received at time t2. Assume the reading on B's clock when the signal reaches B is tb. A is stationary in his own frame, so if t2-t1 is 12 years, A concludes from the constancy of the speed of light that B was 6 light years away at the time (t1+t2)/2 when the signal arrived. If t1=-6, t2=6, and tb=0, the clocks will be synchronized in A's frame. However, they won't be synchronized in B's frame. If B also sends out a radar signal in the above scenario, his clock readings will not be consistent with the notions that the clocks are synchronized in his frame.

I have misused the word synchronised. The clocks just read the same thing when the two observers are 6 light years apart as defined in Observers A frame of reference. The can still record different changes in time relative to each other.

 

[SiennaTheGr8]

I have misused the word synchronised. The clocks just read the same thing when the two observers are 6 light years apart as defined in Observers A frame of reference. The can still record different changes in time relative to each other.

Which clocks?

Perhaps this will help you:

Each frame of reference consists of an imaginary latticework of rulers and clocks, all at rest relative to each other, pervading all of space. (We're ignoring gravity.)

From A's perspective, the clocks in A's frame of reference are all synchronized. From B's perspective, the clocks in B's frame of reference are all synchronized.

However:

From A's perspective, the clocks in B's frame of reference are NOT synchronized. From B's perspective, the clocks in A's frame of reference are NOT synchronized.

 

[Orodruin]

For the first case, "when the distance is 6 ly in A's rest frame" implies that the sending occurs when the distance to B is measured to be 6 ly "at the same in A's rest frame". Letting B move to the left in A's rest frame (unprimed), be at rest in the origin of its own (primed), and calling the time when A sends the signal in the unprimed frame t=0. The unprimed coordinates of the sending are t=0, x=6 (units years and ly throughout). Lorentz transformation gives $x'=\gamma x =1.25x = 7.5$. Since B is at x'=0, it takes the light 7.5 years to arrive.

For the second case: The gap between B and the light signal closes at a speed 1.6c in A's rest frame. In A's rest frame, it therefore takes the signal 6/1.6 years to reach B. B is time dilated relative to this and the time in B's rest frame between the events is therefore $6/(1.6 \gamma)=3$ years.

 

[Dale]

In B's frame of reference due to length contraction, A is 4.8 light-years away.

A sends a pulse of light towards B.

In B's frame of reference how long does it take until B receives the pulse?

4.8 years

If A is 4.8 ly away in B's frame at the moment that A emits a light pulse then by the second postulate it will take 4.8 y to arrive.

Notice the phrase "at the moment" above. That refers to simultaneity in B's frame. A will disagree.

Note also that the length contraction formula is not designed to be applied to this scenario. It assumes that the distance between the two end points is constant. Here you should use the Lorentz transform instead.

 

[david316]

4.8 years

If A is 4.8 ly away in B's frame at the moment that A emits a light pulse then by the second postulate it will take 4.8 y to arrive.

Notice the phrase "at the moment" above. That refers to simultaneity in B's frame. A will disagree.

Note also that the length contraction formula is not designed to be applied to this scenario. It assumes that the distance between the two end points is constant. Here you should use the Lorentz transform instead.

But when B receives the pulse of light won't he have moved a distance towards A? Hence he will be 4.8 years minus the distance traveled towards A?

 

[Orodruin]

4.8 years

If A is 4.8 ly away in B's frame at the moment that A emits a light pulse then by the second postulate it will take 4.8 y to arrive.

Notice the phrase "at the moment" above. That refers to simultaneity in B's frame. A will disagree.

Note also that the length contraction formula is not designed to be applied to this scenario. It assumes that the distance between the two end points is constant. Here you should use the Lorentz transform instead.

The 4.8 ly is incompatible with the first statement that the distance between A and B at the time of emission in A's rest frame is 6 ly. The length contraction should be taken in the other direction if you consider how it is derived from the Lorentz transform. The result consistent with the first statement is 7.5 years. See #19.

You can also see this as follows: Let A carry 6 ly long stick ahead of him. The light signal is emitted when the other end touches B. While the stick in B's rest frame has length 4.8 ly, the events of the signal emission and the stick touching B are not simultaneous (they were simultaneous for A!)

Now, let B carry a 7.5 ly stick. Its length is 6 ly in A's rest frame and therefore A will emit the signal when he touches the other end at a distance of 7.5 ly in B's rest frame. It therefore takes the light 7.5 years to arrive.

 

[Orodruin]

But when B receives the pulse of light won't he have moved a distance towards A? Hence he will be 4.8 years minus the distance traveled towards A?

See #22. Also, B does not move in B's rest frame by definition of "B's rest frame".

 

[david316]

For the first case, "when the distance is 6 ly in A's rest frame" implies that the sending occurs when the distance to B is measured to be 6 ly "at the same in A's rest frame". Letting B move to the left in A's rest frame (unprimed), be at rest in the origin of its own (primed), and calling the time when A sends the signal in the unprimed frame t=0. The unprimed coordinates of the sending are t=0, x=6 (units years and ly throughout). Lorentz transformation gives $x'=\gamma x =1.25x = 7.5$. Since B is at x'=0, it takes the light 7.5 years to arrive.

I don't understand how something traveling the speed of light can take longer the 6 years to travel 6 light years. I calculate the it will take 3.75 years to arrive as measured in Observers A frame. 7.5 years total for it to get there and come back.

 

[david316]

See #22. Also, B does not move in B's rest frame by definition of "B's rest frame".

My bad. B will measure A as have moved closer.

 

[Orodruin]

I don't understand how something traveling the speed of light can take longer the 6 years to travel 6 light years. I calculate the it will take 3.75 years to arrive as measured in Observers A frame. 7.5 years for it to come to get there and come back.

It is not traveling 6 ly, it is traveling 7.5 ly in the rest frame of B. The travel distance is frame-dependent.

My bad. B will measure A as have moved closer.

Which is irrelevant to how the light moves in B's rest frame.

 

[Herman Trivilino]

Two observers, A and B, are traveling towards each other at a relative speed of 0.6c. Assume their clocks are synced at t = 0.

They would need to share the same location to synchronize their clocks at $t = 0$. Either that, or specify how they are to synchronize their clocks if they're separated by some distance.

When A measures t = 0, he measures that B is 6 light-years away. We define this as the proper length.

If it's a proper length of 6 light-years then it's the distance between two objects located a distance of 6 light-years apart, but those two objects would have to be at rest relative to each other. One of those objects could be A, and the other say a space buoy that's at rest relative to A and located 6 light years away. Moreover, A could synchronize his clock with a clock on the buoy, and set things up so they both read $t_A=0$ just as B flies past the buoy. And B agrees to also set his clock to $t_B=0$ just as he passes the buoy.

When B measures t = 0, due to length contraction, he measures that A is 4.8 light-years away.

That works.

At the same time (t = 0 in B's frame), A sends a pulse of light towards B.

Now you encounter the problem the others are pointing out. B will not agree that the clock at A was properly synchronized with the one on the buoy, that is, B will not agree that the clock at A was properly synchronized with the clock at B. Thus the notion of "when" A sends the pulse is ambiguous.

(My apologies if you've since sorted this part out, but while I was writing others were answering.)

 

[david316]

The 4.8 ly is incompatible with the first statement that the distance between A and B at the time of emission in A's rest frame is 6 ly. The length contraction should be taken in the other direction if you consider how it is derived from the Lorentz transform. The result consistent with the first statement is 7.5 years. See #19.

You can also see this as follows: Let A carry 6 ly long stick ahead of him. The light signal is emitted when the other end touches B. While the stick in B's rest frame has length 4.8 ly, the events of the signal emission and the stick touching B are not simultaneous (they were simultaneous for A!)

Now, let B carry a 7.5 ly stick. Its length is 6 ly in A's rest frame and therefore A will emit the signal when he touches the other end at a distance of 7.5 ly in B's rest frame. It therefore takes the light 7.5 years to arrive.

Got it. Makes sense. Thanks. Now I need to work out what question I was actually asking!

 

[Orodruin]

That works.

For some definition of "works". It works in the sense that in your setup with the forward buoy, A will be 4.8 ly away in B's frame when the buoy passes B. It is still not uniquely defining which emission moment is intended. A's clock will not show t=0 at this time in B's rest frame. At the time A's clock shows t=0, A is 7.5 ly away in B's frame.

 

[Dale]

But when B receives the pulse of light won't he have moved a distance towards A?

Not in B's frame. B is stationary in B's frame.

 

[david316]

The 4.8 ly is incompatible with the first statement that the distance between A and B at the time of emission in A's rest frame is 6 ly. The length contraction should be taken in the other direction if you consider how it is derived from the Lorentz transform. The result consistent with the first statement is 7.5 years. See #19.

You can also see this as follows: Let A carry 6 ly long stick ahead of him. The light signal is emitted when the other end touches B. While the stick in B's rest frame has length 4.8 ly, the events of the signal emission and the stick touching B are not simultaneous (they were simultaneous for A!)

Now, let B carry a 7.5 ly stick. Its length is 6 ly in A's rest frame and therefore A will emit the signal when he touches the other end at a distance of 7.5 ly in B's rest frame. It therefore takes the light 7.5 years to arrive.

In the explanation above... If A was carry a 6 ly long stick ahead of him and a light signal was emitted from B to A when the stick touched B, it would take 6 years to reach A?

 

[Orodruin]

In the explanation above... If A was carry a 6 ly long stick ahead of him and a light signal was emitted from B to A when the stick touched B, it would take 6 years to reach A?

According to whom and based on what simultaneity convention?

 

[Jeronimus]

It would help if you would start with a less complicated scenario first.

Imagine a 3rd observer C.

He observes both A and B moving towards each other with a relative speed of 0.6c. Their clocks are synced and will remain synced from in his frame of reference. When they are at a distance of 10 lightseconds, their clocks both display 0s and they release a lightbeam at the same time from C's perspective.
C draws a spacetime diagram where he places himself at x=0, t=0. A at x= -5ls, t=0 and B at x= 5ls, t=0

Observer C will measure the lightbeams to take 7.69...s to reach A and B respectively.

Since the situation is completely symmetric. A will measure the same time for his lightbeam to reach B as B will measure for his beam to reach A. They will also both measure the same distance between each other.
That is, the _instance_ of A with a clock count of 0s and the _instance_ of B with a clock count of 0s.

However, space separated events which are simultaneous in a given inertial reference frame, are NEVER simultaneous in another inertial reference frame. Observer A at x= -5ls, t=0 and a clock count of 0s is an event within C's reference frame.
Observer B at x= 5ls, t=0 is another event.
Those events are simultaneous for C.

For A, those events cannot be simultaneous, nor can they for B.

So when A checks his clock and registers 0 seconds, the instance of B he measures at a distance will not be an instance with a clock count of 0 seconds, but a higher value. Otherwise said: From A's perspective, when he registers 0 seconds on his clock, B's clock will not be a 0s but higher.

The same is true the other way around. When B registers 0s on his clock, A's clock will not be at 0s but higher for exactly the same value.You can always find an observer C which observes both A and B to be moving at the same speed relative to him. He can send a lightsignal to both A&B such that their clocks are in sync from his perspective and will remain in sync since he calculated the same time dilation for both.
Doing this, A and B will get the same results for their measurements for same clock counts on their respective clocks.

Syncing the clocks from any other inertial frame of reference in which A and B are observed to move at different speeds, would lead to different results for both. And of course, the clocks would not remain in sync either.

This is why above posters insisted on you defining the reference frame the clocks would be synced from.

If for example, A observes his clock count to be at 0s and at the same time he observes/measures B's clock count to be at 0s, B will never observe his clock count to be at 0s and A's at 0s as well, because of the difference of reference frames.
So it makes no sense for you to just say that they are in sync at 0s. They can only be in sync from A's perspective OR from B's, but not both.

 

[david316]

According to whom and based on what simultaneity convention?

What is the time difference in the reference frame of A between the event on the world line of A that is simultaneous with the sending of the signal in B's rest frame and the event of A receiving the signal?

 

[Orodruin]

It would help if you would start with a less complicated scenario first.

Imagine a 3rd observer C.

I fail to see how the introduction of yet another observer makes the scenario "simpler".

Observer C will measure the lightbeams to take 10s to reach A and B respectively.

This is incorrect, A and B are still moving.

They will also both measure the same distance between each other

It is unclear what you mean by this. The distance between A and B is changing in all frames and measuring "the same distance" is therefore inherently referring to "at the same time", but this has different meanings in the frames of A and B.

 

[Orodruin]

What is the time difference in the reference frame of A between the event on the world line of A that is simultaneous with the sending of the signal in B's rest frame and the event of A receiving the signal?

In B's frame, the signal is sent when the front of A's rod arrives. Since the rod rest length was 6 ly, its length in B's frame is 4.8 ly and at the same point in time in B's frame, A is therefore 4.8 ly away. The separation speed of the signal and A in B's frame is -1.6c and so it takes the signal 4.8/1.6 = 3 years to arrive in B's frame. The proper time elapsed for A between these events is time dilated and therefore equal to 3/1.25 = 2.4 years.

The time elapsed between the sending of the signal and the arrival of the signal based on A's simultaneity convention (ie, not B's) is 6 years.

 

[Jeronimus]

I fail to see how the introduction of yet another observer makes the scenario "simpler".

This is incorrect, A and B are still moving.It is unclear what you mean by this. The distance between A and B is changing in all frames and measuring "the same distance" is therefore inherently referring to "at the same time", but this has different meanings in the frames of A and B.

Introducing a 3rd observer does make this a lot simpler because it makes it easier to see the symmetry of the situation and understand that both A and B will have to arrive at the same measurements when they both do their measurements at same clock counts.

You were right about the 10s value however. That was a mistake.

I clarified what i meant by them measuring the same distance. When syncing their clocks the way i described it, using a 3rd observer which sees them both incoming at the same speed, then when they do measurements on the distance, they get the same results for same values on their clock count.

 

[david316]

The 4.8 ly is incompatible with the first statement that the distance between A and B at the time of emission in A's rest frame is 6 ly. The length contraction should be taken in the other direction if you consider how it is derived from the Lorentz transform. The result consistent with the first statement is 7.5 years. See #19.

You can also see this as follows: Let A carry 6 ly long stick ahead of him. The light signal is emitted when the other end touches B. While the stick in B's rest frame has length 4.8 ly, the events of the signal emission and the stick touching B are not simultaneous (they were simultaneous for A!)

Now, let B carry a 7.5 ly stick. Its length is 6 ly in A's rest frame and therefore A will emit the signal when he touches the other end at a distance of 7.5 ly in B's rest frame. It therefore takes the light 7.5 years to arrive.

How do I interpret this in terms of the distance being both 4.8 light years and 7.5 light years in B's rest frame?

 

[Orodruin]

How do I interpret this in terms of the distance being both 4.8 light years and 7.5 light years in B's rest frame?

Just as you need to define what you mean by time you need to define what you mean by "distance". This is intimately related to the definition of what "simultaneous" means.

 

[Dale]

The 4.8 ly is incompatible with the first statement that the distance between A and B at the time of emission in A's rest frame is 6 ly

I don't think that you and I disagree, but I think that we are making different assumptions about a poorly specified problem. From the revised description in the post I was responding to https://www.physicsforums.com/threads/second-opinion-needed.906688/#post-5710231 , the time and distance of emission is specified in B's frame.

In B's frame the worldline of B is $(t,0)$ and the worldline of A is $(t,0.6t-4.8)$ and a pulse of light emitted by A at $(0,-4.8)$ arrives at B at $(4.8,0)$.

Taking the Lorentz transform, in A's frame the worldline of B is $(t',-0.6t')$ and the worldline of A is $(t',-6.0)$. So at t'=0 they are indeed 6 ly away in A's frame.

However, A does not emit the pulse at t'=0 (A's frame). Instead, A emits the pulse of light at $(3.6,-6.0)$ in A's frame. This is what I was referring to about the disagreement.

 

[Orodruin]

Indeed, it is also the setting for OP's last problem statement for which I concluded 2.4 and 6 years, depending on the simultaneity convention.

 

[Jeronimus]

No matter how hard you try, this non-problem cannot be solved until we are told the clock count of B, A will measure when his clock count is at 0, or alternatively, the clock count of A, B will measure when his clock count is at 0.
Hence, whatever results are given here, can but only be false

I proposed a third observer C to do the syncing in such a way, it would allow to shed more light onto the core of the problem.

 

[russel_]

In the interests of clarifying things, and saving my sanity, I thought I'd offer some background on how this problem came about and hopefully clarify where the ambiguities may have been introduced.

It all started when this Scientific American article (https://www.scientificamerican.com/article/time-and-the-twin-paradox-2006-02/) came up in discussion - it claims to "explain" the twin paradox in terms of the time it takes for light to reach each observer. Now I claimed that this is misleading, because while the time delays might be interesting to consider in terms of what you'd actually "see", they are not relevant to the effects derived in Special Relativity. That is, time dilation and other effects are "real" and not just tricks due to how long it takes light to reach observers from distance events.

In order to demonstrate this, I pointed out that Einstein's original derivation let's observers have multiple clocks sitting in their frame of reference, all carefully synchronized with a clock next to them, and in this way the time it takes for light to reach each observer is eliminated (i.e., they measure the time of events in their frame by looking at their local clocks next to the event, no delay due to light propagation, or at least it's "cancelled").

This led to a number of progressively simpler thought experiments based on the twin paradox, trying to demonstrate this and other things that emerged in the process. Part of this involved having a fixed distance in A's frame (A was on Earth and the distance was to some star). Eventually we wound up with a situation like this:

A and B are two observers moving relative to each other at 0.6c
Two markers, 1 and 2, are stationary relative to A
As measured by A, the two markers are 6 light-years apart
When B's clock measures t = 0, B measures A as being next to marker 1 and himself as next to marker 2
When B's clock measures t = 0, B observers a pulse of light depart from A directed toward him

And the sticking point was how long it takes the light to reach B (3 years or 4.8 years?). I think that B would measure the distance between the markers as 4.8 light-years (the markers are stationary in A's rest frame, thus moving relative to B, so the 6 light-years apart markers are observed by B as length-contracted), but even if this is wrong, that doesn't matter for the next point: assuming he does measure the distance as 4.8 light-years, he will always measure the time it takes the light to reach him as 4.8 years by definition (or rather, by the two postulates of SR), regardless of the motion of A, or the motion of the markers, or anything else in the question - if the light starts 4.8 light-years away as measured his frame, it takes 4.8 years to reach him.

Now hopefully you can see why the 6 light-years was called a "proper distance". The distance originally referred to actual things at rest relative to A. Unfortunately I made the colossal blunder of suggesting that we remove the markers from the description (oops) because I wanted to whittle thing down to core ideas, in particular the one above: that if B measures the light starting 4.8 light-years away then it will always take 4.8 light-years to reach him, regardless of how we got to that distance and the motion of B relative to A or the markers.

The other ambiguity of "setting A's clock to t = 0"... I wasn't really paying attention to :). I don't think it's relevant to the point above, which I what I really care about, so I'm happy to let A set his clock to t = 0 whenever he likes, and whatever that means, and whether A and B see whatever event was used to set t = 0 on their clocks as being simultaneous, doesn't matter - if we just care about knowing when B measures the pulse of light departing and we have the specified positions above. Nonetheless I can see how this lax phrasing would further confuse things.

But at the end of the day the thing I care about most is that the statement above hold regardless of anything else in the problem: "if the light starts 4.8 light-years away as measured his frame, it takes 4.8 years to reach him"?

 

[Jeronimus]

Here is the initial post in the form of two x/t diagrams, as i would imagine you attempted to describe it, ignoring the inconsistencies with an attempt to point at what i believe you have not fully understood yet.

The magenta/light blue filled circle is Observer B. "He measures t=0" i assume to mean that he draws an x-t diagram where he places himself at x=0, t=0 and then goes on to map all other events within his reference frame onto the x-t diagram.

As you can see, observer B who in the left x-t diagram is at x=0, t=0 has a clock count of 10s currently.
He measures observer A to be at 6 lightseconds away of him, at x=6ls, t=0 with a clock count of 8s. Observer A is the white filled circle.

Observer A however, CANNOT measure the distance to the _instance_ of observer B with a clock count of 10s. As you can see from the right x-t diagram, which is A's reference frame, observer B with a clock count of 10s is not on the simultaneous axis of A.
That instance of observer B with a clock count of 10s is at about x'~ -7.5ls, t' ~ -4.5s (won't calculate the exact values), where the magenta/light blue filled circle is on the right diagram.

Instead, the instance of B which has a clock count of 13.6s(deep blue filled circle), located at the spacetime position of x'=4.8ls, t'=0 is simultaneous to the instance of A with a clock count of 8 seconds when observed from within A's reference frame.

And once again, the instance of observer B with a clock count of 13.6s, won't be able to measure the distance to observer A with a clock count of 8 seconds, because that instance of A is NOT simultaneous to this instance of B.
Instead it will be the instance of A which i marked as a pink filled circle which the instance of observer B with a clock count of 13.6s will observer as simultaneous to himself...

I hope you get the idea now.

As for the question for how many seconds it will take for the light beam to reach B when shot by A at a distance of 4.8ly, Dale already answered this correctly. 4.8 years of course, but i doubt that this was what you were trying to _really_ figure out.

That light beam in the diagrams would be the orange line. I used lightseconds instead of years because it would require too much editing.Those spacetime diagrams are taken from a screenshot of my twin paradox simulation. You might want to check it out here and download the code if you want to run it yourself (link is in the description).



maybe this will lift some of the confusion, hopefully not add to it :D

 

[russel_]

As for the question for how many seconds it will take for the light beam to reach B when shot by A at a distance of 4.8ly, Dale already answered this correctly. 4.8 years of course, but i doubt that this was what you were trying to _really_ figure out.

I'm pretty sure it is :)

Of course there are other things around it but the originating dispute was over - assuming B measures a 4.8 light-year distance & light departing at t = 0 - whether 3 years could possibly be correct. That's why the last three statements in the original problem explicitly gave t = 0 and 4.8 light-years as the distance for B's measurements. I couldn't think of a better explanation of why the 4.8 years answer immediately follows from this other than "Because the speed of light is constant and B is stationary in his own frame", and I was simply hoping someone would either have a more convincing explanation or at least back that up (i.e., Dale's response).

All the other issues raised here have resulted from removing the markers at the last minute so that the distance became ambiguous, coupled with the extraneous specification of t = 0 on A's clock.

 

[Ibix]

@david316 @russel_: The problem with your thought experiment is that the relativity of simultaneity means that A and B don't agree on when the experiment starts for the other person. If A calculates the time at which B is six light years away and sends a signal to B, then B looks at his clock when he receives the signal and subtracts six years he will not agree that A was six light years away then.

This is a general problem: there is no unique way to define what "at the same time" means for separated observers. Talking about "proper distance" doesn't save you because the proper distance is just the distance between two objects in their rest frame - and A and B are moving with respect to each other, so they don't have a rest frame (A has one, B has one, but they're not the same frame) so proper distance is not defined here. So your thought experiment is ambiguous. And you haven't realized that, so your thinking is subtly self-contradictory, and that is why you can't resolve it.

The twin paradox is a way to avoid the ambiguities by starting and finishing with both observers in the same place. The Scientific American article you linked therefore doesn't go into the ambiguities you introduced by trying to simplify things. It's not saying that relativistic effects are caused by lightspeed delays. Rather, it's using lightspeed delays to show that "acceleration causes the age difference" is not correct (or not complete, at least). The observations each twin makes of the other are different in many ways, not just that "one of them turns round".

In many ways the "exchanging light signals" explanation is very good because it deals solely in what each twin sees, not in how they interpret what they see. But I find the geometric explanation to be better. Follow the "Insights author" link on Orodruin's profile page to read his article on that.

 

[Ibix]

All the other issues raised here have resulted from removing the markers at the last minute so that the distance became ambiguous, coupled with the extraneous specification of t = 0 on A's clock.

All of the issues stem from "synced at t=0" which is an ambiguous statement unless the clocks are at the same place at t=0, which they aren't.

Re-reading the initial problem statement I suspect it is possible to resolve it if you set t=0 (or some other agreed value) when A and B meet.

 

[Jeronimus]

I'm pretty sure it is :)

Of course there are other things around it but the originating dispute was over - assuming B measures a 4.8 light-year distance & light departing at t = 0 - whether 3 years could possibly be correct. That's why the last three statements in the original problem explicitly gave t = 0 and 4.8 light-years as the distance for B's measurements. I couldn't think of a better explanation of why the 4.8 years answer immediately follows from this other than "Because the speed of light is constant and B is stationary in his own frame", and I was simply hoping someone would either have a more convincing explanation or at least back that up (i.e., Dale's response).

All the other issues raised here have resulted from removing the markers at the last minute so that the distance became ambiguous, coupled with the extraneous specification of t = 0 on A's clock.

What could be more convincing than using the most fundamental explanation? One of the two postulates SR is grounded upon.

If someone at 4.8 lightyears away of you shoots a lightbeam towards you, then the light coming towards you will be traveling at c ~ 300000km/s. If it did not do that, then we could throw SR into the trash bin because all you need to derive the formulas of SR mathematically is the postulate of light always traveling at 300000km/s independent of which inertial frame of reference you observe it from.

That and the first postulate, stating that the laws of physics are the same in every inertial frame of reference. Basically, that no IFR is special compared to another.

So of course it will be 4.8 years. Can you imagine a scenario where it would take longer or less than 4.8 years without violating one of the most fundamental rules of SR - light always traveling at c in a vacuum absent of gravity?

 

[russel_]

All of the issues stem from "synced at t=0" which is an ambiguous statement unless the clocks are at the same place at t=0, which they aren't.

Re-reading the initial problem statement I suspect it is possible to resolve it if you set t=0 (or some other agreed value) when A and B meet.

Well, the "synced at t=0" statement wasn't in the original (pre-forum) problem statement, so in this regard I claim innocence :).

 

[Jeronimus]

Well, the "synced at t=0" statement wasn't in the original (pre-forum) problem statement, so in this regard I claim innocence :).

You are guilty of adding too much detail and wording it badly however. The problem could be stated as follows:There are two observers A & B traveling at a relative speed of v= 0.6c towards each other.

At a distance of 6 lightyears, there is an object O which is at rest, relative to observer A.

Observer B measures the distance between O and A to be 4.8 lightyears (length contraction).

Observer B measures/calculates A shooting a lightbeam towards him, just as he passes by object O (simultaneously as measured/calculated by observer B).

how long will it take for the lightbeam to reach observer B? 4.8y of course.

How long would it take for the lightbeam to reach observer B, from observer's A perspective? This is a much more difficult question!(and much closer to the value you hoped for)

Never mind, it does not seem to be a secret anymore. I just calculated it to be 2.4 years, a value my eye already caught in some post earlier i did not fully read.