# Really easy question from a neophyte

## Main Question or Discussion Point

OK, here we go. This will sound very dumb, and I accept that fully.

Three objects are arranged as follows:
We have a star or other heavenly body capable of emitting a single pulse of light on cue. I'll call it emitter.
We next have two other objects 1 light year from emitter. These other two objects, A and B, begin our hypothetical by being almost exactly next to one another. A will remain in its position throughout, 1 LY from emitter, but on cue, B will begin a frantic race at one-half the speed of light, in a direction directly away from emitter and A. Thus, they will all be in a straight line, EAB, as the motion of B continues, once it starts.
Now, I have a starter pistol, which I can magically get them all to hear at the same instant. When I fire the pistol, the emitter will emit a single burst of light, aimed at A and B. At the same instant that this occurs, B will begin moving away from the other two (on that line) at 1/2 the speed of light.
Now, after one light year, the light will reach B, who has remained in position relative to emitter. Meanwhile, B will have moved .5 light years on down the line. When will the light reach A, and how long will it appear to A that the time from B receiving the light to A receiving the light?
Thanks a million.
CH

## Answers and Replies

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Could you check your A's and B's?

ghwellsjr
Science Advisor
Gold Member
OK, here we go. This will sound very dumb, and I accept that fully.

Three objects are arranged as follows:
We have a star or other heavenly body capable of emitting a single pulse of light on cue. I'll call it emitter.
We next have two other objects 1 light year from emitter. These other two objects, A and B, begin our hypothetical by being almost exactly next to one another. A will remain in its position throughout, 1 LY from emitter, but on cue, B will begin a frantic race at one-half the speed of light, in a direction directly away from emitter and A. Thus, they will all be in a straight line, EAB, as the motion of B continues, once it starts.
Now, I have a starter pistol, which I can magically get them all to hear at the same instant. When I fire the pistol, the emitter will emit a single burst of light, aimed at A and B. At the same instant that this occurs, B will begin moving away from the other two (on that line) at 1/2 the speed of light.
Now, after one light year, the light will reach B, who has remained in position relative to emitter. Meanwhile, B will have moved .5 light years on down the line. When will the light reach A, and how long will it appear to A that the time from B receiving the light to A receiving the light?
Thanks a million.
CH
You've got your questions all mixed up. I think you mean:

Now, after one light year, the light will reach A, who has remained in position relative to emitter. Meanwhile, B will have moved .5 light years on down the line. When will the light reach B, and how long will it appear to A that the time from A receiving the light to B receiving the light?

Is that correct?

Thanks buddy. You are correct.

ghwellsjr
Science Advisor
Gold Member
Ok, well here's a spacetime diagram to illustrate your scenario. The emitter is shown as the thick blue line and sends out a pulse of light shown as a thin blue line. It arrives at A shown in red and continues on to B shown in black at the coordinate time of 24 months or two years but A doesn't see this happening until the time of 36 months (or 3 years). The dots indicate the passage of one-month's intervals for each object. Note that B sees the light after about 20.8 months.

I'm just curious: why was this question of interest to you?

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Bill_K
Science Advisor
Choran,

The way you have this problem stated, it does not even involve relativity, since it all takes place in A's rest frame. It's just a standard algebra "word problem" involving trains, planes and automobiles.

The light ray starts from the origin and travels at velocity c. Therefore its path is x = ct.

Observer B starts at x = 1 and travels at velocity 0.5 c. Its path is therefore x = 1 + 0.5 ct.

They meet where these two paths intersect, namely ct = 1 + 0.5 ct. The solution is ct = 2, or t = two years.

George, for some reason the diagram won't open and won't download on my Mac. I asked the question because I assumed that the answer was as set forth by Bill K, but wanted to check. It didn't seem to me that it had any relativistic aspects--that was my point in asking. Told you it was a dumb question!

ghwellsjr
Science Advisor
Gold Member
I re-uploaded the diagram, can you see it now?

Yes George, I can see it now. I would have thought that A would see the light reach B one year after it arrived at his (A's) location. It passed him, and caught B a year later. Why three years?
Would the answer be the same if instead of the beam continuing on to B, A relayed the beam to B, who was proceeding away from him at .5c? In other words, a relay throw rather than one straight from the outfield emitter.
Thanks.

George, I get the "three years". Where does the 20.8 months come from? Thanks again, buddy.

... Where does the 20.8 months come from? ...
B sees the light after 24 months have passed according to clocks that remain stationary relative to E (blue line) and A (red line). According to a clock carried by B the elapsed time is 20.8 months due to the time dilation effect, because B is moving relative to E and A. (Count the dots along the black line).

Thanks, Yulos, got it. Another dumb question, since I've never worked with these diagrams before. How is the scale prepared? This isnt a physics questions, I just want to know how the diagrams are prepared--it is a program that builds in the time dilation scale along whichever line you wish for whatever velocity you choose, or...?
Thanks all.

Let me clarify: Talking about the marks along the black line.
Thanks again.

... Another dumb question, since I've never worked with these diagrams before. How is the scale prepared? This isnt a physics questions, I just want to know how the diagrams are prepared--it is a program that builds in the time dilation scale along whichever line you wish for whatever velocity you choose, or...?
Thanks all.
It usually takes a lot of calculations, but I think ghwellsjr has written his own software to produce his nicely presented drawings... not sure, but I have not seen it commercially available anywhere. If I recall correctly, mentz114 has a time space program you can download from his blog here at phsicsforums.

ghwellsjr
Science Advisor
Gold Member
Thanks, Yulos, got it. Another dumb question, since I've never worked with these diagrams before. How is the scale prepared? This isnt a physics questions, I just want to know how the diagrams are prepared--it is a program that builds in the time dilation scale along whichever line you wish for whatever velocity you choose, or...?
Thanks all.
Yes, my diagrams are produced by an application I wrote using LabVIEW software. I specify the coordinates of the starting event of each worldline and its speed for so many ticks of unit time, then I can extend each worldline with a different speed and another number of ticks, repeating as many times as I want. The Time Dilations are applied by the program at this initial stage based on the speeds. I can also show the propagation of individual photons. Then I can transform the entire diagram at any speed (short of ±c) with respect to the first one. A this point, all the Time Dilations are determined automatically by the Lorentz Transformation process. Computers are great at doing lots of repetitive computations such as the Lorentz Transformation on a whole lot of events.

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ghwellsjr
Science Advisor
Gold Member
I would have thought that A would see the light reach B one year after it arrived at his (A's) location. It passed him, and caught B a year later. Why three years?
You originally asked about the appearance to A of the light reaching B and now you're asking about what A would see. Those questions are frame invariant.

The question of when the light reaches B in the rest frame of E and A applies only to that one frame and can be easily seen (by us, not by A) in the diagram above. Now after A receives the image coming from B, he can use the time on his clock when the light pulse passed him (1 year) and the time on his clock when he saw the reflection (3 years) and assume that the light took the same time to get from him to B as it did to get from B back to him and conclude (by averaging) that the light arrived at B at his own 2-year mark.

Now I'm going to use my software to transform the diagram for the rest frame of E and A into the rest frame of B (after he departs from A):

Note that in this frame, the arrival of the light at B occurs at A's time of 18 months instead of 24 months as it did in the previous frame. However, B still sees the light arriving at his location when his clock is at 20.8 months (count the dots) and A still sees the reflection at his time of 36 months, even though in this frame it is A that is experiencing Time Dilation instead of B as in the original frame. And since A's observations are the same in this frame as in the original frame, he can still make the same assumption about the time it takes for light to propagate each way, even though there is a 3-to-1 difference in this frame and he can arrive at the same conclusion that the light arrived at his 24-month mark instead of his 18-month mark.All the observers are completely oblivious to the coordinates specified in the different frames that we use to describe scenarios.

Now let's take this one step further. Let's assume that when A receives the reflection of the light pulse off of B at his own time of 36 months, he also observes that B's clock was reading 20.8 months. Now since A has determined that the reflection happened at his own time of 24 months, A can calculate the ratio of the rate of his own clock to B's clock. He gets 24/20.8 = 1.154. This, by the way, is the Lorentz factor for the speed of 0.5c. Keep this in mind for later.

Would the answer be the same if instead of the beam continuing on to B, A relayed the beam to B, who was proceeding away from him at .5c? In other words, a relay throw rather than one straight from the outfield emitter.
Thanks.
Yes, it would be the same.

George, I get the "three years". Where does the 20.8 months come from? Thanks again, buddy.
I think the answer is obvious in the above diagram.

Now let's do something more with the above diagram. Let's assume that B sent out a burst of light at his time of 7 months aimed toward A. You can mentally draw this into the above diagram and you will see that the burst arrives at A at his time of 12 months but B won't see this until his own time of 20.8 months. Now he can do the same thing that A did to determine, based on the light taking just as long to get to A as it takes to get back, that his own clock was at the average of 7 and 20.8 or 13.9 when A's clock was at 12. And you can see in the above diagram that that is the time at which a light pulse sent at month 7 would arrive at A. B can also calculate the Time Dilation of A's clock which is 13.9/12 = 1.158. Note how this is essentially the same value the A got for B's clock, as it should be.

Finally, I want to show you another frame in which all parties, E, A, and B are all traveling at the same speed, 0.268c.

In this frame, all parties are experiencing the same Time Dilation with respect to the coordinate time and yet all parties make the same observations as they did in the other two frames. This should be telling--different frames have no bearing on what any observer actually sees.

And in this frame, A's clock is also at 20.8 years when the light gets to B but A still continues to determine that it arrived at 24 months. He also continues to determine that the Time Dilation of B's clock is 1.15 and if B sends out a burst of light at his time of seven months, B determines that the Time Dilation of A's clock is 1.15.

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Thank you very much. Great idea with the software!