1. Mar 6, 2014

### nearc

Most relativity paradoxes are not true paradoxes they merely require some clarification about frame of references, stationary observer, etc… I hope some folk could help me clarify my “paradox” or point out where someone as already discussed this concept.

We have two ships on parallel vectors [see thumbnail, is there a better way to add images?], one ship [on the right] sends out a signal at step 0 [from the front tip of the ship] traveling the speed of light and that signal will eventually be picked up by the other ship [on the left] again the receiver is in the front tip. If the two ships are stationary then the signal will be received at step 4 denoted by the pair of white ships, however if the two ships are traveling at near light speed the receiving ship will not pick up the signal until step 9. If we say they are traveling near light speed but are not moving relative to anything then conventional wisdom dictates we treat them as stationary but this is where I am confused; how do we say the signal will be received in step 4 when it seems like it should be received in step 9?

Some notes about the diagram: Obviously using a band to show the propagation of a signal is not very accurate. Also I attempted to show the ships traveling slightly slower than the signal propagation. At step 0 the broadcasting ship front tip was in the middle of the band, whereas, in step 9 it was at the edge of the band. Hopefully the general idea can be gleamed.

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2. Mar 6, 2014

### Staff: Mentor

It's received at step 9 in the frame in which you drew your diagrams (in which the ships are moving); but it's received at step 4 in the frame in which the ships are at rest. Time is frame-dependent.

3. Mar 7, 2014

### ghwellsjr

I'll take a stab at pointing out what's wrong with your "paradox" and then show how to correctly depict your scenario.

After you upload your images, scroll down to where it says "Attach Files" and right-click on one of them and select "Copy Link Location" (or something equivalent). Then click in the edit window where you want the image to appear. Finally, click on the "Insert Image" at the top of the edit window and hit Ctrl-v (or something equivalent to paste in the Link Location) and click OK. Click on the Preview Post button to see how it looks before you Submit Reply. A word of caution: Make sure when you get all done that you continue to see the thumbnails at the bottom of the post. If they disappear (which happens to me all the time--I don't know why) then although you will continue to see the images embedded in your post, no one else will be able to see them and you will have to re-upload them and repeat the above steps.

At step 0, the tip of the first ship is not in the middle of the band, it is at the bottom edge of the top part of the band, just like it is in all the other steps. It just looks like the tip of the ship is in the middle because the band is just starting and the bottom part is still touching the upper part. This is the problem with your depiction, you should have been always referring to the bottom of the top part of the band in which case it would not be step 9 where the second ship detects the same portion of the band that the first ship is tracking. In fact, you should see that you would have to show more steps and if you think about it, the bottom edge of the top part of the band will never reach the second ship. Even though you say that you "attempted to show the ships traveling slightly slower than the signal propagation", what you are actually showing is that they are traveling at exactly the same speed as the signal propagation.

This is always a problem with scenarios that are attempting to be so close to the speed of light. Why do you want to be so close? Instead, let's do the same thing with the ships traveling at 80% of the speed of light. Then it will be clear what is happening. I have drawn a series of steps where the second ship is three units to the left of the first ship that emits the light. I show the light in the first step as a larger than normal red dot to make it a little more obvious but you should realize that it is really just a small dot not extending past the tip of the ship.

Step 0:

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

The minor grid lines help show the speeds of the ships at 80%c and you can see that it is at step 5 where the second ship detects the light. This is in accord with Special Relativity which says that time is dilated for moving objects by the factor gamma which is 1.667 at 0.8c. Instead of the light reaching the second ship at 3 units of time like it would in their mutual rest frame, it takes 1.667 times 3 or 5 units of time. Does this all make sense to you now?

EDIT: I have inserted the Proper Times at the tips of the two ships assuming that they have been previously synchronized to the Coordinate Time in their mutual rest frame. I'll post more about how I did this later.

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4. Mar 7, 2014

### Glurth

To expand a bit on what Peter said:

Your rings show a frame of reference that is stationary.
Your moving ships experience a different frame of reference, as they are moving so fast.

Since they are in difference frames of reference they cannot agree on what is simultaneous, i.e. they each experience a different amount of time pass between frames. The EVENT where the ship emits the light, and the EVENT where the light is received are fixed in space-time: however the experienced time between these events is what varies between frames of reference. (So the ships' passengers will think the light is recieved in (thier) time-frame 4, the stationary observer will see the light is recieved in (his/her) time-frame 9)

5. Mar 7, 2014

### ghwellsjr

Depicting a scenario in different frames of reference does not change what any ship or passenger or stationary observer experiences.

If the passengers in both ships had previously synchronized their clocks and the first ship emitted the light pulse at their mutual time 0, then in the OP's scenario, they both would see that the light was received when the other ship's clock reached 4 units of time. The stationary observer would also see the same thing. This can be shown in all reference frames.

6. Mar 7, 2014

### Glurth

Sorry, I absolutely should have qualified my comment with "I THINK".

Are you saying that between the two events in space-time, 4 time-frames units are measured to pass by the fast moving observers, and 4 frame-fame units are measured to pass by the stationary observer? Well that certainly throws me off. Why is there no time dilation between the fast and stationary observers?

7. Mar 7, 2014

### ghwellsjr

There is Time Dilation. I mentioned it in my first post in reference to my modified scenario since I consider the OP's diagrams useless. I said that the time when the second ship detects the light is at 5 units of time (the 3 units of time multiplied by gamma).

The frame steps represent Coordinate Time. In the mutual rest frame of the ships (which is not shown), the Coordinate Time matches the Proper Time of the ships' observers' clocks. Time Dilation is the ratio of the Coordinate Time shown for a moving clock (5 units of time) to the Proper Time of the moving clock (3 units of time).

When I have some more time, I'll draw some more diagrams illustrating what I am saying.

8. Mar 7, 2014

### Glurth

Thanks George, that matches what I was trying to say in #4, although a LOT more explicitly. Were you trying to correct/clarify my post with #5, or just expand it? If correction, what did I get wrong? (other than frame numbers: I used OP's.)

9. Mar 7, 2014

### ModestyKing

So what ghwelljr is saying is that both ships are experiencing time dilating, but not relative to each other, but relative to a stationary observer? So with ghwelljr's graph depictions, it takes 5 seconds for the signal to get from one ship to another (1.667 times 3) in the reference frames of both ships, but to someone observing, it'd have taken 3 units of time? (stationary).

10. Mar 8, 2014

### ghwellsjr

In order to determine the Proper Times located at the tips of the ships in post #3, I used the Lorentz Transformation process to convert events on the ship's worldline corresponding to increments of Coordinate Time into the IRF of the stationary ship. Since I already have a computer application that can do this for me, I let it do the number crunching. Here is a diagram showing the tip of either ship along the vertical axis of the diagrams in post #3. It is traveling at 0.8c for 5 units of Coordinate Time. The events of interest are shown as blue dots:

After transforming to 0.8c, we get this diagram:

I have marked in the Coordinate Times of the blue events which is also the Proper Times of those events in all frames and which I marked in the diagrams on post #3.

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11. Mar 8, 2014

### ghwellsjr

Here is a set of diagrams depicting the mutual rest frame of the two ships. I selected steps corresponding to the Proper Times that I determined in the previous post.

Step 0.0:

Step 0.6:

Step 1.2:

Step 1.8:

Step 2.4:

Step 3.0:

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12. Mar 8, 2014

### nearc

13. Mar 8, 2014

### ghwellsjr

You're welcome.

I'm wondering, too. Bell's paradox has to do with two spaceships in the same line, one in front of the other. Your scenario is with two spaceships side by side (with a great distance between them). Even with acceleration, as long as it's in the same direction that you showed your spaceships going, there would be no connection.

14. Mar 9, 2014

### ghwellsjr

Time for a clock/object/observer that is moving in an Inertial Reference Frame will be dilated. That means that it takes longer for the moving object to tick one unit of time than for the equivalent unit of Coordinate Time for the IRF. That is clearly stated. Saying that one object experiences time dilation relative to another object is meaningless. Each ship's Time Dilation is independent of any other ship or observer. It is only dependent on its speed in the IRF. You only need one object in one IRF to have Time Dilation. If you have multiple objects, you determine each ones speed and therefore each ones Time Dilation without regard to any other object.

You have this backwards. In the mutual rest frame of the two ships as depicted in post #11, it takes 3 units of Coordinate Time for the signal to propagate between the ships. In the original frame in which both ships are moving as depicted in post #3, it takes 5 units of Coordinate Time. People observing will not be able to see the Time Dilation. What they will see is that the time on the first ship was at 0 when the light was emitted and the time on the second ship was 3 when the light was detected and they will see both events delayed by whatever time it takes the light from those events to reach them. All observers will be able to see this too, but with possibly different delays.

15. Mar 9, 2014

### nearc

how do we explain other things like redshift/blueshift for example?

16. Mar 9, 2014

### ghwellsjr

We get redshift whenever the observer is moving away from the received signal and blueshift whenever the observer is moving toward the received signal. Look up Doppler.