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Lamp color before and after a phase of acceleration

  1. Jun 16, 2013 #1
    You have a yellow lamp in your spaceship. Now you accelerate your spaceship significantly. Then you stop accelerating. Does the lamp still have the same color? I would think it should, but I reckon differently as follows:

    Consider the situation from an outside observers point of view, resting in the frame of before the spaceship was accelerated.

    a) The light undergoes a Doppler shift when emitted from the lamp, but exactly the opposite Doppler shift applies in the retina of the moving observer, so this would not change the color.

    b) But, compared to the rest frame, time on the spaceship runs slower, meaning the moving observer's time runs slower by the Lorentz factor, allowing more light waves to hit his retina within his standard time interval, hence a higher frequency.

    Incidentally, the frequency is higher by the Lorentz factor, which does look right.

    To get rid of any strange effects going on in the light emitting process of the lamp, we could assume to have captured a beam of light in a mirrored box before the acceleration phase, letting it out afterwards. The arguments (a) and (b) still apply and the beam of light should now be blue shifted for the moving observer.

    Is it blue shifted or not :confused: ?
     
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  3. Jun 16, 2013 #2

    Nugatory

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    The observer is always at rest relative to himself, right?

    So we're always talking about the source moving relative to the observer... If the source is moving away from the observer, then the observer sees light that is is red-shifted relative to the frequency at the source; if the source is moving towards the observer the light is similarly blue-shifted.

    To reconcile this result with time dilation you have to remember that the light source is moving, so the distance that the light signal travels is continuously changing - the travel time for successive peaks and valleys is different so they take different times to reach the observer's retina.

    If you search in this forum for "Doppler" you'll find a ton of excellent space-time diagrams showing exactly how this all works out.
     
  4. Jun 16, 2013 #3
    This does not answer my question, so I try to be as short as possible.

    Does the person on the spaceship see a different color of light coming from a lamp in his spaceship before and after an acceleration phase?
     
  5. Jun 16, 2013 #4

    Nugatory

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    No. He's at rest relative to the lamp before and after.
     
  6. Jun 16, 2013 #5

    WannabeNewton

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    The lamp (source) will be at rest relative to him (the spaceship guy) before and after so why would he see any frequency shift before and after?

    EDIT: Looks like Nugatory answered a minute before me. My plans have been foiled again!
     
  7. Jun 16, 2013 #6

    ghwellsjr

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    No, but during, yes.
     
  8. Jun 16, 2013 #7

    Nugatory

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    Yes, but at least you read the question right the first time :smile:
     
  9. Jun 16, 2013 #8

    WannabeNewton

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    I was hoping I could get something clarified: so let's say we have the observer inside the spaceship and the lamp inside the spaceship is separated from the observer by some non-negligible distance. Now let's say the spaceship's rockets fire such that it is uniformly accelerating along the direction of separation between the observer and lamp. Would it be valid to explain the frequency shift of the light from the lamp during the acceleration phase, as measured by the observer, using the equivalence principle? That is, since the observer is in a Rindler frame (i.e. the frame of a uniformly accelerating observer), could he, instead of saying he is uniformly accelerating, say he is at rest in a uniform gravitational field and ascribe the frequency shift of the light reaching him from the lamp, which is some distance away inside the spaceship, to gravitational redshift (with the exact shift being given by the Rindler metric)?
     
  10. Jun 16, 2013 #9

    ghwellsjr

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    There is a transient at the beginning and at the end of the uniform time of acceleration which has not been specified but influences the answer.
     
  11. Jun 16, 2013 #10
    Well, because of the reasoning (a) and (b) in my original post.Both (a) and (b) are the reasoning of the guy who stayed at home or stayed in the rest frame of before the acceleration. Describing the situation from the outside should also come to the conclusion that the spaceship guy sees no change in color.

    Now that you support my suspicion that the color does not change, I wonder where (a) and (b) are wrong and/or incomplete.
     
  12. Jun 16, 2013 #11

    WannabeNewton

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    Right but excluding the contributions from that, could we ascribe the effects in between as outlined above is what I am asking.
     
  13. Jun 16, 2013 #12

    ghwellsjr

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    There are no transients in a Rindler frame since its acceleration has no beginning or end. Until the details of the acceleration, the materials and structure of the spaceship, etc are specified, there is no unique answer to the OP's question. Depending on these details, the initial transient could last longer than the uniform period of acceleration. But once the answer has been determined in one inertial frame, it can be transformed into any other inertial frame and it will yield identical results for what the observer actually sees, which I think is the issue he was asking about.
     
  14. Jun 16, 2013 #13

    WannabeNewton

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    I was just asking if, during a period of uniform acceleration, we can justify the observed redshifting using the equivalence principle. I wasn't trying to tie this in to the OPs question, I was just wondering if observed redshift in the frame of a uniformly accelerating observer can be translated over to a geometrical explanation using the EP. Imagine we are instead in a spaceship that has been uniformly accelerating from the start and has yet to stop.
     
  15. Jun 16, 2013 #14

    ghwellsjr

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    I think you mean uniformly accelerating without a start, but I don't see why it has to be a red shift. Seems to me that it could also be blue, depending on the configuration details.
     
  16. Jun 16, 2013 #15

    Vanadium 50

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    But we're here to answer the OPs question. Adding additional complications to a problem seldom makes things clearer.
     
  17. Jun 16, 2013 #16

    WannabeNewton

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    Hiya. I was using "red" to just refer to the phenomena as a whole, as it is usually done when speaking of gravitational redshift. Anyways, thanks for the responses George. According to wiki you are allowed to describe the effect using the EP when speaking of continuous uniform acceleration which is apparently what Einstein did except he used a box instead of a spaceship http://en.wikipedia.org/wiki/Gravitational_redshift#History. Anyways, I'll let you get back to the problem at hand :wink: I just wanted to have that clarified. Thanks again!

    Right sorry, I just didn't want to start a whole new thread for what I thought was a loosely related and (at the time) quick question.
     
    Last edited: Jun 16, 2013
  18. Jun 16, 2013 #17

    Dale

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    Time is also dilated for the moving lamp by the same factor. So this effect cancels out the retinal time dilation.
     
  19. Jun 16, 2013 #18
    (Many thanks for guiding the discussion back on track.)

    Yep, that make sense, but raises my next question. The observer who stayed at home will analyze the situation as you just described: lamp has time dilated emission, meaning it sends a lower frequency than before.

    Apart from just analyzing the situation, the observer left at home can also see (measure) the frequency. And what he sees must be the time dilated redshift combined with the redshift caused by the spaceship traveling away. Removing the motion-redshift computationally, the observer at home would now "see" the time dilation redshift. Is that true?
     
  20. Jun 17, 2013 #19

    Dale

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    Yes, that is correct. The relativistic Doppler shift is essentially the classical Doppler shift with time dilation on top. If you remove the classical Doppler, what you are left with is time dilation.

    In particular, when the emitter is moving at 90° to the detector there is no classical Doppler. This, so called, "transverse Doppler" is therefore equal to the time dilation and is a key difference between classical and relativistic physics.
     
  21. Jun 17, 2013 #20

    ghwellsjr

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    I have made a couple spacetime diagrams which might depict the scenario you have in mind. Here is the first one that shows the Inertial Reference Frame (IRF) in which the spaceship was at rest prior to its acceleration. I have shown the lamp as the thick red line with dots marking equal intervals of its Proper Time when it sends out light signals to you, shown in blue towards the front of the spaceship. The acceleration of the spaceship is unrealistic but intended only to show the Doppler shifts that you asked about.

    attachment.php?attachmentid=59629&stc=1&d=1371478902.png

    Prior to your acceleration, you receive the light signals at the same rate that they were sent and so there is no Doppler shift. Some time after your acceleration, you are also receiving the light signals at the same rate that they were sent even though both the lamp and you are time dilated in this IRF. Since I have set the final speed of the spaceship to be 0.6c in this IRF, the gamma factor is 1.25 and you can see that the Proper Time dots for both you and the lamp are spaced out by 1.25 so that in 5 nanoseconds of Coordinate Time, the Proper Time has advanced by only 4 nanoseconds.

    But during the acceleration transient, you are receiving the light signals Doppler shifted by a factor of 0.5. Even though the lamp accelerated prior to you (in this IRF), you don't see the effect of that until some time later but you do immediately see the effect of your own acceleration as soon as it happens.

    You also talked about an observer at rest in this IRF. In order to depict what he sees, we would have to drawn in light signals from each event of interest to wherever you want him to be (assuming that he is located along the same axis as the spaceship's acceleration). You have indicated that he sees the spaceship moving away which implies that he would be located at some negative Coordinate Location. You could have also placed him at some positive Coordinate Location and although he would see things differently, he could still reconstruct the same IRF (after it was all over and by using radar methods). But, in my opinion, it is far easier to forget about observers at rest in an IRF and just talk about the worldlines and events in the IRF in which case, we become the observers and we don't have to worry about light propagation time. Let me show you what I mean.

    Let's transform the entire scenario depicted in the above IRF into the IRF in which the spaceship is at rest after the acceleration:

    attachment.php?attachmentid=59630&stc=1&d=1371478902.png

    First, I want you to notice that there is no inertial observer that is at rest in this IRF. But we can observe this IRF with no problem.

    Next we want to note that in this IRF you accelerate before the lamp does and yet you continue to see the signals from the lamp exactly the same as you did in the first IRF, that is, you see them coming in with no Doppler shift prior to and after the acceleration transient but during the acceleration transient there is a Doppler shift of 0.5, even though in this IRF the Time Dilation factors for you and the lamp happen in the opposite order.

    Now if you want, you can add another observer at rest anywhere in this IRF and show how he can construct this IRF diagram using radar techniques. You can also show an inertial observer moving in this IRF diagram (such as the one who was at rest in the first IRF) and show how he can construct a diagram for his own rest frame. Furthermore, you can use the same radar technique to show how you, accelerating on the spaceship can construct a diagram for a non-inertial reference frame in which you remain at rest.
     

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