# B Second opinion needed

1. Mar 6, 2017

### 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 travelling 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

2. Mar 6, 2017

### Staff: Mentor

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.

3. Mar 6, 2017

### david316

Assume they are synchronized via some uncanny coincidence. If I was to violate all the laws of physic 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?

4. Mar 6, 2017

Nope.

5. Mar 6, 2017

### david316

Actually does this simplify it?

Two observers, A and B, are travelling 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?

6. Mar 6, 2017

### Orodruin

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

7. Mar 6, 2017

### david316

Better??

Two observers, A and B, are travelling 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?

8. Mar 6, 2017

### david316

Better still??

Two observers, A and B, are travelling 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?

9. Mar 6, 2017

### 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 $\sqrt{1 - (v/c)^2}$.

10. Mar 6, 2017

### david316

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 $\sqrt{1 - (v/c)^2}$.

11. Mar 6, 2017

### Orodruin

Staff Emeritus
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.

12. Mar 6, 2017

### Orodruin

Staff Emeritus
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.

13. Mar 6, 2017

### david316

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

14. Mar 6, 2017

### david316

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

15. Mar 6, 2017

### 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.

16. Mar 6, 2017

### pervect

Staff Emeritus
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.

17. Mar 6, 2017

### david316

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.

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.

18. Mar 6, 2017

### SiennaTheGr8

Which clocks?

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.

19. Mar 6, 2017

### Orodruin

Staff Emeritus
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.

20. Mar 6, 2017

### Staff: Mentor

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.