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Author Topic: Does Redshift or Blueshift occur only when an object is moving?  (Read 3035 times)

Offline brainfreak97

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Does Redshift/ Blueshift occurs when only the observer is moving?
When only the observer moves, based on Doppler's effect, frequency for waves change due to the change in the velocity of the wave particles perceived by the user.
On that logic, if the wave is a light wave, for Red/Blueshift to occur, the relative velocity of the light wave has to change, which contradicts the constant speed of light, c.

Using the velocity addition in special relativity we get
v' = (u+v)/(1+uv/c2)
plug in u = c, we get v' = c, which the speed of light relative to observers remain constant, which (probably) means no red/blueshift.

Yet my highschool teacher(Yes I'm currently only a highschooler) claims that red/blueshift indeed occurs when only the observer is moving.

So, does Redshift/ Blueshift occurs when only the observer is moving?
Thanks in advance.
« Last Edit: 22/07/2015 14:31:54 by chris »


 

Offline PmbPhy

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Re: A Question on Redshift/Blueshift
« Reply #1 on: 21/07/2015 23:19:42 »
Quote from: brainfreak97
Does Redshift/ Blueshift occurs when only the observer is moving?
It's clear to me that you don't know what an observer is in special relativity (SR). It's a misconception to think of an observer as a person. An observer in SR is a coordinate system. See: https://en.wikipedia.org/wiki/Observer_(special_relativity)
Quote
In special relativity, an observer is a frame of reference from which a set of objects or events are being measured. Usually this is an inertial reference frame or "inertial observer".
So your questions are meaningless. If you study SR you'll learn that you can't really say that the observer is moving or the source is moving, etc. without it meaning that the coordinate system is moving. A Red/Blue shift occurs when there is a relative motion between the light source and the light detector.

Quote from: brainfreak97
When only the observer moves, based on Doppler's effect, frequency for waves change due to the change in the velocity of the wave particles perceived by the user.
What do you mean "only the mover moves"? Since you're using the term "observer" incorrectly let me ask you a question so that I might understand your question better.

Let S be an inertial frame. The coordinate system erected and at rest in S will serve as the observer. It should be noted that it makes no difference where an observer is since a coordinate system is not localized. Now consider the following two scenarios:

Scenario 1 Source moving, Detector at rest:

In frame S a light source is moving along the x-axis towards the origin from the positive side of the x-axis towards the origin. At rest at a negative value of x is a light detector which measures the frequency of the light that it absorbs. This is illustrated below:

-------------- <Light Detector> ------------------(0, 0)-------------------- <== <Light Source> --------------- X-axis
                                                               <Observer = Frame S>

Scenario 2 Source at rest, Detector moving:

In frame S a light source is at rest on the positive side of the x-axis. Moving towards the origin from the negative side is a light detector which measures the frequency of the light that it absorbs. This is illustrated below:

-------------- <Light Detector> ==> ------------(0, 0)-------------------------- <Light Source> --------------- X-axis
                                                               <Observer = Frame S>

Now please rephrase your question in terms of these two scenarios.

Until then I'm going to make some assumptions: You claim that When only the observer moves, based on Doppler's effect, frequency for waves change due to the change in the velocity of the wave particles perceived by the user.  I'm going to assume that by "observer" you really mean the light detector. As such your conclusion On that logic, if the wave is a light wave, for Red/Blueshift to occur, the relative velocity of the light wave has to change, which contradicts the constant speed of light, c. is quite wrong. The relative velocity of the wave doesn't change. What changed is the fact that as the light source moves towards the detector its emitting wave peaks (and wave crests) at shorter distances in space because the light source has moved during the time that the last peak was emitted.


Quote from: brainfreak97
Yet my highschool teacher(Yes I'm currently only a highschooler) claims that red/blueshift indeed occurs when only the observer is moving.
If that's what he said then he doesn't understand relativity that well. One of the very important things that came out of relativity is that there's no such thing as "when only the observer is moving". Please have him contact me and I'll explain it to him.

Quote from: brainfreak97
So, does Redshift/ Blueshift occurs when only the observer is moving?
Thanks in advance.
Please remember that the only thing that can be spoken of is relative motion. After all, that's why the theory is called relativity.
« Last Edit: 22/07/2015 03:51:06 by PmbPhy »
 

Offline evan_au

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Re: A Question on Redshift/Blueshift
« Reply #2 on: 22/07/2015 13:16:17 »
Quote from: brainfreak97
Does Redshift/ Blueshift occur when only the observer is moving?
Let's say that you measure everything from the light source (what is described above as "the frame of reference" of the light source). You have an astronaut who is moving relative to that light source, who will experience redshift or blueshift, depending on whether he is moving away or towards the light source.

So the answer is "Yes".

Quote from: paraphrased
Does Redshift/ Blueshift occur when only the light source is moving?
Equivalently, let's say that you measure everything from the astronaut (what is described above as "the frame of reference" of the astronaut). You have a light source which is moving relative to the astronaut, which will experience redshift or blueshift, depending on whether it is moving away or towards the astronaut.

So the answer is "Yes".

So far, we are just talking about Doppler shift, which works fine if the relative velocity is much less than the speed of light, so you can ignore any Lorentz corrections (unless you want to be really precise).

What the Michelson Morley experiment showed is that no matter who measures the speed of light, it is always constant. This was contrary to our experience with sound waves in air or water, and was one of the paradoxes which led Einstein to propose his theory of relativity.
 

Offline brainfreak97

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Re: A Question on Redshift/Blueshift
« Reply #3 on: 22/07/2015 13:51:57 »
Oh ho. You are right. I do really mean the light detector by "observer". Thanks for pointing it out.
As for my teacher, we study physics in Chinese language mostly, so it was my bad for translating it without looking into the correct phrase.

As for my question, it is of scenario 2. The light source is stationary relative to the "observer"(Am I using it right?), where the light detector is "moving".

Quote
What changed is the fact that as the light source moves towards the detector its emitting wave peaks (and wave crests) at shorter distances in space because the light source has moved during the time that the last peak was emitted.

Do correct me if I am wrong. I believe this is the explanation for when the light source is moving. Indeed it will cause red/blueshift to occur since wavelength has changed and therefore caused the frequency perceived by the detector to change.
 
Quote
Let's say that you measure everything from the light source (what is described above as "the frame of reference" of the light source). You have an astronaut who is moving relative to that light source, who will experience redshift or blueshift, depending on whether he is moving away or towards the light source.

So the answer is "Yes".

But if the light source is stationary, and the light detector moves relative to the source and observer, the wavelength of light should remain the same. And since the velocity of light perceived by the light detector remains constant at c, shouldn't the frequency perceived by the light detector remain unchanged since f = c/λ?

Please correct me if i have any misconceptions on these things. Again, thanks in advance.
 

Offline evan_au

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Quote from: brainfreak97
But if the light source is stationary, and the light detector moves relative to the source and observer, the wavelength of light should remain the same.
Have a look at: https://en.wikipedia.org/wiki/Doppler_effect

This tells you that if there is any relative velocity between the light source and the light detector, there will be a difference in the wavelength that is measured at the transmitter, and the wavelength that is measured by the receiver.

This effect is used every day, in police traffic radar, automatically-opening doors, and detecting exoplanets. It is quite real.

After coming to terms with the Doppler effect, then you can consider: https://en.wikipedia.org/wiki/Relativistic_Doppler_effect
 

Offline brainfreak97

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Well, after some research, I'm finally getting the idea although there was some part i find too complicated.
Guess I was wrong in my earlier assumptions.

Nevertheless, thanks again.
« Last Edit: 28/07/2015 15:00:21 by brainfreak97 »
 

Offline AndroidNeox

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Just remember that the speed of light is the same for all observers. Space and time (which are not constant, or even "things", since they are not conserved) adjust (dilate) to keep c a constant.

If you look at a spacetime diagram, in which time is just another dimension like the space dimensions, the speed of light, c, corresponds to the equal angle between space and time.
 

Offline PmbPhy

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Just remember that the speed of light is the same for all observers. Space and time (which are not constant, or even "things", since they are not conserved) adjust (dilate) to keep c a constant.

If you look at a spacetime diagram, in which time is just another dimension like the space dimensions, the speed of light, c, corresponds to the equal angle between space and time.
That depends on the units used for the scale of the axes. If the time axis is ct where t is in seconds then what you say is true.
 

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