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Assume that a far star sends light toward a receiver. If we move this antenna such that it accelerate first for a moment, and then it moves with constant speed, we can see that the frequency of the received light will be shifted instantly in antenna's frame(toward blue or red, doesn't matter). The question is, in antenna's frame after acceleration, (which is indeed inertia) how can the frequency of the light still be shifted? After all, the light transmitter is far like 100 light years away, so when antenna moves, in antenna's frame we would expect some kind of delay to see a movement for the the far star (transmitter) and ofcourse the Doppler shift effect (antenna is at the rest in its frame, so the only reason for receiving doppler shifted frequency would be the movement of transmitter itself in this frame). But there is no such delay in formulas at least.

If you don't get what i am trying to say note that in every frame, Doppler shift happens because of transmitter movements (every frame consider itself at the rest!). If transmitter starts its movement while it's 100 light years away, we will see transmitter movement and Doppler shift effect 100 years later. However, when antenna moves, there is no such delay which is very strange. It is as if transmitter doesn't move, but frequency has been changed out of thin air.

I understand that in accelerated systems laws of physics changes, metric is different and etc. However, even by knowing accelerated frames metrics (like Rindler and such), i can't show that there is indeed a solution for this problem. Because after all, in reality antenna will recieve Doppler shifted light even after it maintains its speed and becomes inertia

I won't accept an answer without math even though i might upvote one. Everybody can say that accelerated systems are different, i need a thorough solution.

Update: you can generalize this question to every effects in SR, like Lorentz contraction between two points in a far away space.

Update 2: Thanks to the help of all users, i am convinced that in accelerated frames there is nothing wrong with none local effects, for example if i rotate my head, i will see movement of stars immediately, hence @ThePhoton 's answer is acceptable for me now.

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    $\begingroup$ Your use of the word "dilation" doesn't make sense to me. Do you perhaps mean "delay"? $\endgroup$ – Acccumulation Jun 18 at 20:16
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    $\begingroup$ @Acccumulation My first language is not english, you are completely right. Thanks. $\endgroup$ – Paradoxy Jun 18 at 20:32
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    $\begingroup$ Just in case please note that a receiver can also consider himself moving in a frame of the stationary source. At least it is exactly what is going on in the Einstein’s 1905 paper (§7 Theory of Doppler principle and aberration) hermes.ffn.ub.es/luisnavarro/nuevo_maletin/… Mr. Einstein clearly says that “an observer is moving with velocity $v$ relatively to an infinitely distant source of light of frequency $\nu$,“ and „the velocity of the observer referred to a system of co-ordinates which is at rest relatively to the source of light“ $\endgroup$ – Albert Jun 18 at 21:29
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    $\begingroup$ @Albert is right. In the frame of an inertial observer that observer is at rest. But relativity does not say that the observer is ignorant of any motion relative to other frames. $\endgroup$ – Aaron Stevens Jun 18 at 21:34
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    $\begingroup$ @RBarryYoung, Unlike what most of people thinks, you can work on accelerated frames in SR as well! You just need a flat spacetime which is the case here. But even if we claim that GR will solve this question, still i'd like to see how, because i've read GR as well so i can understand an answer from it if someone finds it. $\endgroup$ – Paradoxy Jun 19 at 13:07
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The question is, in antenna's frame after acceleration, (which is indeed inertia) how can the frequency of the light still be shifted?

The light frequency was always shifted in this frame. (i.e. the frequency was always different in this frame than in the frame where the receiver is intially not moving)

You have two inertial frames, we can call them "A" and "B". Initially the receiver is at rest in frame A, and then it accelerates until it is at rest in frame B. But frame B wasn't created by this action. Frame B always existed. The only difference is that initially the receiver is at rest in frame A, so it measures the source frequency in frame A. Later the receiver is at rest in frame B, so it measures the source frequency in frame B.

But the frequency of the light in the frame B never changed, so there's no reason to worry about how it could have changed instantaneously.

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    $\begingroup$ There is no "antenna's frame". There is frame A and frame B. The antenna is initially at rest in frame A, then later it is at rest in frame B. When the antenna changes velocity to be at rest in frame B, it doesn't change anything about the star in frame B. The star is still moving in frame B the way it always has been. $\endgroup$ – The Photon Jun 18 at 22:28
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    $\begingroup$ RE "there is no 'antenna's frame'", of course what I mean is the antenna's frame is not an inertial frame. Trying to analyze what happens in this non-inertial frame is beyond my knowledge. But you shouldn't expect it to behave like an inertial frame, because it isn't one. $\endgroup$ – The Photon Jun 18 at 22:42
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    $\begingroup$ The antenna doesn't "change frames". It exists in both frames at all times. Sometimes it's at rest in one frame and moving in the other. Other times the other way around. The frame, and the behavior of the star in each frame, and the behavior of the waves coming from the star (as measured in each frame) don't depend on what the antenna does. The antenna just lets some operator measure the waves, with the result depending on how the antenna is moving relative to the source. $\endgroup$ – The Photon Jun 19 at 0:59
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    $\begingroup$ This comment series reminds me of one of the more difficult things when it comes to working with frames. Frames are 100% abstract concepts, divorced from any physical object. They are simply a way of measuring real life effects in terms of vectors. The calculations (such as those for Doppler shift) happen regardless of objects in the world -- they happen purely based on transforms between frames. However, if you choose the "right" frame, it becomes easy to convert the vectors measured in that frame to meaningful measurable quantities in the real world. $\endgroup$ – Cort Ammon Jun 19 at 15:11
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    $\begingroup$ Thus there is merely "Frame B." But if you happen to choose the behavior of Frame B such that the position of the antenna does not change over time in Frame B, then the calculations done in Frame B translate into meaningful observed results when one looks at the signals received by the antenna. But the presence of an antenna did not actually change the math of the frame transform at all... it merely made the frame transform into a useful one for predicitng what we observe. $\endgroup$ – Cort Ammon Jun 19 at 15:13
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Relative speed exists only between source of light and observer. To compute doppler shift you need to know this relative speed. If you discard the source of light, then to what you assign a velocity?

No, the relevant speed for doppler shift is not the current speed between source and observer. The light emitted by the star 100 years ago exists independent of the star after being emitted. The doppler shifted light you see is the light that fills the space between you and the star. Only you and the light in the local space around you is important, thus the distance to the star or what happened to the star in those 100 years is irrelevant. The star might not even exist anymore. The relevant speed is the relative speed between your frame before the acceleration and your frame after acceleration. The doppler shift depends on how much you accelerated. Acceleration is absolute not relative.

After all, the light transmitter is far like 100 light years away, so when antenna moves, in antenna's frame we would expect some kind of delay to see a movement for the the far star (transmitter) and ofcourse the Doppler shift effect (antenna is at the rest in its frame, so the only reason for receiving doppler shifted frequency would be the movement of transmitter itself in this frame). But there is no such delay in formulas at least.

No we would only expect a delay if the star is the one accelerating. If we are accelerating with our antenna there is no delay, because we accelerate.

Maybe an analogy can help you:

Imagine a cloud at rest above your head. You are initially standing at rest at the ground. The cloud is emitting raindrops and each raindrop is falling for a time T until it hits the ground. Initially the raindrops will fall vertically on your head. However if you start moving the drops will hit you in the face at an angle. From the angle at which the drops hit you, you can calculate how fast the clouds are moving relative to you (assuming the clouds never accelerate). In this analogy you can only observe the clouds indirectly by measureing the rainsdrops hitting you. If you are the one who accelerates, the clouds will appear to move relative to you immediately after acceleration. The raindrops will also change their angle immediately. What you are asking ist basically: How can the raindrops change their angle immediately when you accelerate if there is a delay of T for the raindrops before they hit you. The answer is that the raindrops are already in the local space around you and when you accelerate the raindrops emitted before you accelerated will hit you at the new angle. The change in speed for both the clouds and the drops around you is the same, because you are the one who accelerated, therefore you can still use the angle to correctly calculate the clouds currect velocity relative to you. If instead the clouds accelerate and you stand still, only the clouds and all drops emitted after the acceleration will have a different speed. the drops already around you will fall on their old trajectory and you will only see the clouds moving after a delay T when the new drops hit you. So acceleration is not relative and it matters whether you or the cloud accelerates.

If transmitter starts its movement while it's 100 light years away, we will see transmitter movement and Doppler shift effect 100 years later. However, when antenna moves, there is no such delay which is very strange.

As you can see, this happens with raindrops too, so it isn't strange at all.

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  • $\begingroup$ Unlike raindrops which you can assign your velocity to them(and then to their clouds too),it's not possible to assign a relative velocity to the light.Assume that there is only a transmitter and receiver in the world,when receiver start its movement and become inertia,there is no way for him to understand that the source of light is moving now,because light speed is c,nevertheless,only its frequency is shifted.when antenna tries to explain why this frequency is shifted, he has to say it's the source who is moving,but with what evidence? He can't use the doppler shifted light it'd be a loophole $\endgroup$ – Paradoxy Jun 19 at 15:30
  • $\begingroup$ @Paradoxy : If there is only a transmitter and receiver in the world, there is not an electromagentic field, already filled with a long history of light emitted from the star. You aren't receiving light freshly emitted when you accelerate, you are receiving the light that was already in the space around you. Your receiver is a boat traveling over waves emitted long ago; their (apparent) frequency is a result of their wavelength (crest and trough spacing) and your motion through them. $\endgroup$ – Eric Towers Jun 19 at 15:47
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    $\begingroup$ @Paradoxy I think you mean the relative velocity to the light is always c and doesn't change? Thats true but instead the frequency changes. The point I was trying to make is, that there is no delay for angle or velocity change for the raindrops when the receiver accelerates. In the same way there is no delay for frequency change of light when the receiver accelerates. You said you find that strange, but the explanation in both cases is that the raindrops/light change their angle/frequency immediately because they are already at you location when you accelerate. $\endgroup$ – Azzinoth Jun 19 at 15:53
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    $\begingroup$ @Paradoxy There are electromagnetic wave peaks and valleys travelling towards you though space. When you accelerate towards them they appear to hit you faster. Calculate the frequency at which wave peaks hit your observer (considering time dilation and length contraction correctly) and you get your formula. The source where these waves originally came from doesn't matter anymore at this point. $\endgroup$ – Azzinoth Jun 19 at 15:53
  • $\begingroup$ @Azzinoth Ok let's drop the source. How are you going to define length contraction in empty space which only contains light, and time dilation (relative to what)? If there was indeed an ether everywhere, your method was absolutely correct, but there is nothing like that in SR, you need the source of light to assign a velocity to it, to assign a Lorentz contraction between antenna and star. $\endgroup$ – Paradoxy Jun 19 at 16:19
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The position and velocity of a source of light are completely irrelevant once the light exists. It does not have a frequency nor a wavelength, both of those are (inertial) frame dependent quantities that transform as a 4 vector:

$$ k_{\mu} = (\omega/c, \vec k) $$

Different frames see different $k_{\mu}$, which transform according to a Lorentz transformation:

$$ k'_{\mu} = \Lambda_{\mu}^{\nu} k_{\nu} $$.

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  • $\begingroup$ Relative speed exists only between source of light and observer. To compute doppler shift you need to know this relative speed. If you discard the source of light, then to what you assign a velocity? The light itself travels at a null path. relative speed is meaningless for it, and that lorentz boost matrix also need a relative speed between frames. Antenta itself consider itself at the rest, if v in matrix of lorentz does not belong to source of light, it is for what then? $\endgroup$ – Paradoxy Jun 18 at 22:27
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    $\begingroup$ @Paradoxy You're too fixated on the source. It doesn't matter. $\endgroup$ – JEB Jun 19 at 13:20

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