twin paradox

[Fred Champion]

How can you observe Doppler shift from something that is stationary with respect to you?

You are correct, the twin will not observe a doppler shift in the frequency of light from the clocks at a constant speed. I don't know why I wrote that last sentence of my previous post that way. Of course a doppler shift in light will be observed only when the observer's velocity is different from that of the source. I did a lousy job of expressing my thoughts. I'm really sorry for the confusion.

 

What I was trying to convey is that the "shift" in the light received from the clocks will still be present. It will not be recognized but will have to be "unwound" as the ship decelerates back to a rest state. The reason it will not be recognized is that the twin will not compare the number of cycles of the light he receives from each clock. I don't know how to say this any better than with an illustration.

 

Suppose the twin has set up his clocks so that the forward clock will transmit the video part of a movie and the rear clock will transmit the audio part. As he accelerates the video and audio signals the twin receives will not be in sync. When he decelerates to a rest state the video and audio will sync again.

 

The clocks are not moving relative to the twin.

You are correct, the clocks are not moving relative to the twin. However, the light emitted from the clocks has received no velocity component from the motion of the clocks. Light is emitted from a specific point in space and will propagate from that point in all directions at c.

 

An object thrown on a moving platform does have a velocity component from the motion of the platform, but this is not so for light.

[Delta1212]

You are correct, the twin will not observe a doppler shift in the frequency of light from the clocks at a constant speed. I don't know why I wrote that last sentence of my previous post that way. Of course a doppler shift in light will be observed only when the observer's velocity is different from that of the source. I did a lousy job of expressing my thoughts. I'm really sorry for the confusion.

 

What I was trying to convey is that the "shift" in the light received from the clocks will still be present. It will not be recognized but will have to be "unwound" as the ship decelerates back to a rest state. The reason it will not be recognized is that the twin will not compare the number of cycles of the light he receives from each clock. I don't know how to say this any better than with an illustration.

 

Suppose the twin has set up his clocks so that the forward clock will transmit the video part of a movie and the rear clock will transmit the audio part. As he accelerates the video and audio signals the twin receives will not be in sync. When he decelerates to a rest state the video and audio will sync again.

 

You are correct, the clocks are not moving relative to the twin. However, the light emitted from the clocks has received no velocity component from the motion of the clocks. Light is emitted from a specific point in space and will propagate from that point in all directions at c.

 

An object thrown on a moving platform does have a velocity component from the motion of the platform, but this is not so for light.

For the acceleration to cause Doppler shift, the clocks would have to be moving relative to the observer during the acceleration phase. If it's a persistent effect, the ship would either be crushed or ripped apart by continuos acceleration given enough time.

 

You would get some slight compression at the start equivalent to what you'd get in a gravitational field of equivalent strength, but it's not going to be enough of a difference to be measurable. You'd probably see more of a loss of synchronicity from relativistic effects if you picked one clock up and carried it across the room.

[md65536]

What I was trying to convey is that the "shift" in the light received from the clocks will still be present. It will not be recognized but will have to be "unwound" as the ship decelerates back to a rest state. The reason it will not be recognized is that the twin will not compare the number of cycles of the light he receives from each clock. I don't know how to say this any better than with an illustration.

 

Suppose the twin has set up his clocks so that the forward clock will transmit the video part of a movie and the rear clock will transmit the audio part. As he accelerates the video and audio signals the twin receives will not be in sync. When he decelerates to a rest state the video and audio will sync again.

This is generally true I think but it is avoidable.

 

See: http://www.mathpages.com/home/kmath422/kmath422.htm

This is mostly about Born rigid motion, where distance distortions are avoided, which partly supports your argument: "In other words, if we contrive to hold the spatial relations fixed during an acceleration, a phase shift is introduced between different parts of the object, just as, if the phase is held constant, there is spatial stretching."

 

The second part implies there are cases where it it possible to have no phase shift.

 

This all is only a problem if the accelerating twin has a non-negligible length, and in this case I think you need to specify the details of how it accelerates, or just how the different parts of the ship are coordinated. You can specify it some way so that there is a phase shift, or another way so that there is not.

 

(Unless you specifically set it up to avoid either spatial or temporal distortions, you'll get both, so I think you're generally right.)

Edited May 24, 2014 by md65536

[Delta1212]

This is generally true I think but it is avoidable.

 

See: http://www.mathpages.com/home/kmath422/kmath422.htm

This is mostly about Born rigid motion, where distance distortions are avoided, which partly supports your argument: "In other words, if we contrive to hold the spatial relations fixed during an acceleration, a phase shift is introduced between different parts of the object, just as, if the phase is held constant, there is spatial stretching."

 

The second part implies there are cases where it it possible to have no phase shift.

 

This all is only a problem if the accelerating twin has a non-negligible length, and in this case I think you need to specify the details of how it accelerates, or just how the different parts of the ship are coordinated. You can specify it some way so that there is a phase shift, or another way so that there is not.

 

(Unless you specifically set it up to avoid either spatial or temporal distortions, you'll get both, so I think you're generally right.)

For any realistic set up, wouldn't it be a negligible effect, or am I missing something?

[md65536]

For any realistic set up, wouldn't it be a negligible effect, or am I missing something?

Probably, but I don't know or plan to figure out the math to show that. Still, it *is* a real effect even if relatively insignificant. What is a realistic twin paradox setup anyway? Everything depends on the details. Since I have no idea how big the effect is I wouldn't say one way or the other without working through an example.

 

I'd rather avoid the issue by letting the ship length be negligible. Contrary to post 144, an observer and its clock are usually considered to have the same position, so that none of this matters. Not because the effect is small, but because the set up is described without the effect mattering. If someone else is interested in the effect, they could set it up differently, and calculate it.

[Fred Champion]

The shift during acceleration will be small. That's why I set the twin's clocks two light seconds apart. After reaching .5c the twin would see a shift of only .5 second in each clock. At any reasonable length for the ship the effect would be practically negligible.

 

However, I did point out that there would be apparent displacement of objects not along the ship's centerline. This will be significant. Example: Set the direction of "forward" at 0 degrees and "rearward" at 180 degrees. As the speed approaches c, light leaving an object 45 degrees from forward will reach the twin when he has traveled .707 times the distance to the object. In other words, the twin will "see" the object at 90 degrees from forward.

 

Note too that the profile (rate and duration) of the acceleration doesn't affect the final apparent shift, only the amount and duration of the doppler shift during acceleration. I have seen a couple of posts where there was some concern about how quickly a constant speed was reached. We all recognize the effect of acceleration is cumulative.

Edited May 25, 2014 by Fred Champion