Thales et al Posted October 31, 2016 Posted October 31, 2016 https://www.youtube.com/watch?v=tJ2AJJMcQXs I was watching this lecture by a MIT professor, and something didn’t seem quite right to me. Perhaps someone here can set me straight. If a light source and an observer are moving away from one another, the flash of light will be redshifted. If the light source is considered to be in motion and the observer is considered to be at rest, or if the observer is considered to be in motion and the light source is considered to be at rest, either way, the same redshift logically occurs. Conceptionally, it’s either drawn out as it leaves the light source or its drawn out as it’s being perceived. He then goes on to say that if the observer is considered to be in motion, and so the time dilation is with the observer from the perspective of the light source at rest, that this will also contribute to a redshift. That doesn’t seem right to me. From the perspective of the light source at rest and the observer in motion, with the observer’s time slowed down, that seems like it should contribute to a blue shift. Right? If the observer’s time slows down, then the oncoming wave, for the observer, should be oncoming faster, based on the moving observer’s slower time. If the observer is considered to be in motion, then it seem like the flash of light should tend to be redshifted based on the relative motion but also oppositely tend to be blue shifted based on time dilation. No?
J.C.MacSwell Posted October 31, 2016 Posted October 31, 2016 (edited) https://www.youtube.com/watch?v=tJ2AJJMcQXs I was watching this lecture by a MIT professor, and something didn’t seem quite right to me. Perhaps someone here can set me straight. If a light source and an observer are moving away from one another, the flash of light will be redshifted. If the light source is considered to be in motion and the observer is considered to be at rest, or if the observer is considered to be in motion and the light source is considered to be at rest, either way, the same redshift logically occurs. Conceptionally, it’s either drawn out as it leaves the light source or its drawn out as it’s being perceived. He then goes on to say that if the observer is considered to be in motion, and so the time dilation is with the observer from the perspective of the light source at rest, that this will also contribute to a redshift. That doesn’t seem right to me. From the perspective of the light source at rest and the observer in motion, with the observer’s time slowed down, that seems like it should contribute to a blue shift. Right? If the observer’s time slows down, then the oncoming wave, for the observer, should be oncoming faster, based on the moving observer’s slower time. If the observer is considered to be in motion, then it seem like the flash of light should tend to be redshifted based on the relative motion but also oppositely tend to be blue shifted based on time dilation. No? At what point in the hour + does he present this? From the perspective of the observer, the source is redshifted initially as it is sent due to time dilation, and is further redshifted by doppler effect as it is received. An observer at the source perspective of the same events would see extra doppler shifting of the unaffected signals (more than the combined redshifting above), but calculate that the receiver's time dilation would make them perceive it redshifted the same as above. Edited October 31, 2016 by J.C.MacSwell
Strange Posted October 31, 2016 Posted October 31, 2016 Remember time dilation is relative. The observer sees time dilation in the other persons clock (i.e. sees the other persons clock slowed and they light time shifted). This is symmetrical so if observer A sees B's clock run slow because of time dilation then B will also see A's clock run slow because of time dilation.
Tim88 Posted October 31, 2016 Posted October 31, 2016 J.C.MacSwell already gave the correct answer, but maybe a few more words can be helpful. The point to keep in mind is that pure Doppler is asymmetric, depending on who is supposedly moving. It's the main effect that in both perspectives results in a redshift. You can't change facts of observation due to a different interpretation! Combining the differing Doppler effects with the contrary time dilation effects, results in the same "relativistic Doppler" equation - as it should be. Most useful is if you try the derivation for yourself - it's quite easy to do.
Thales et al Posted October 31, 2016 Author Posted October 31, 2016 Thank you for the responses. The professor gets to the math of this question at about the 48-minute mark. I understand the math formulas work out to lead to a redshift, but what I’m having trouble with is the concept behind these formulas. From the perspective of the light source at rest and the observer in motion (moving away), I understand that the Doppler effect will cause the flash of light to be redshifted (due to this, from this perspective). However, also from the perspective of the light source at rest and the observer in motion, where the observer’s time slows down, this (or so it seems to me) would lead to an increase in the frequency of the light wave reaching the observer (again, also from this perspective). If, from the perspective of the light source at rest and the observer in motion, the velocity of the wave of light remains constant ©, while time for the observer slows down, then (from this perspective) this seems like it would lead to the observer seeing more of the wave of light per second; which would mean an increase in frequency; a blue shift not a redshift. I understand that overall there will be a redshift, based on the greater effects from the Doppler effect. What I can’t see is how a decrease in time for the moving observer from the perspective of the light source at rest adds to the overall redshift (as the professor explains it does) rather than subtracts from the overall redshift (given that slower time means an increase in frequency). (Tim88: you seem to agree with J.C.MacSwell (who seems to agree with the MIT professor) that the shift from the Doppler effect and the shift from time dilation will both be redshift and therefore additive. But then you also referenced "the contrary time dilation effects." This confused me.) Thank you all.
Strange Posted October 31, 2016 Posted October 31, 2016 If, from the perspective of the light source at rest and the observer in motion, the velocity of the wave of light remains constant ©, while time for the observer slows down The observer's time doesn't slow down. He see the the time of the (moving) source slow down.
Thales et al Posted October 31, 2016 Author Posted October 31, 2016 The observer's time doesn't slow down. He see the the time of the (moving) source slow down. Right. I’m believe I’m the same page with you when it comes to this. From the observer’s perspective at rest, his time remains the same and the moving light source’s time slows down. And, From the light source’s perceptive at rest, his time remains the same and the moving observer’s time slows down. Right? The problem I’m having is from the perspective of the light source at rest and the observer in motion. From this perspective, the time of the moving observer slows down. And, from this perspective, the velocity of the wave of light (of course) remains invariant (at c). And so, if the velocity of the wave of light is unchanged, but the observers time is slower, then … from the perspective of the light source at rest … the wave of light arriving at the moving observer is blue shifted (… due to time dilation. It is redshifted due to the Doppler effect.) If the velocity of light remains unchanged, but the time of the moving observer … from the perspective of the light source at rest … slows down, then frequency must go up. Or … so it seems to me. But, assuming that an MIT professor knows more than me, if he’s saying “no, from the perspective of the light source at rest, the time dilation of the moving observer will mean not a contrary blue shift but an additive redshift,” then I must be wrong. But I don’t see how we can get an additional redshift from the moving observer’s time slowing down … from the perspective of the light source at rest. Did I make my issue any clearer? I hope so.
Strange Posted October 31, 2016 Posted October 31, 2016 From the light source’s perceptive at rest, his time remains the same and the moving observer’s time slows down. Right? Yes. From this perspective, the time of the moving observer slows down. No. Any observer's time is always the same (unchanged). An observer with the light source would see the other observer's time slowed (and hence redshifted). The observer will always measure the frequency with their unchanged clock. You seem to be mixing frames of references, by trying to apply the time dilation seen by the light source to the observer.
Mordred Posted October 31, 2016 Posted October 31, 2016 An easy rule to remember. In an events own frame, his length and time is always undilated and the longest. Each observer observes other frames as dilated/contracted.
Thales et al Posted October 31, 2016 Author Posted October 31, 2016 (edited) In the two statements of mine you quoted, I tried to express the same thing. My apologies. No. Any observer's time is always the same (unchanged). An observer with the light source would see the other observer's time slowed (and hence redshifted). The observer will always measure the frequency with their unchanged clock. You seem to be mixing frames of references, by trying to apply the time dilation seen by the light source to the observer. I think I understand what you’re getting at. But, if so, I think we might be on two different pages. If I might, let me take a step back and try presenting this again, in a slightly different way. Say, we had a super sensitive clock that was able to measure the difference between each crest of the electric part of the electromagnetic wave. And say that clock is in relative motion with another body at rest. From the perspective of the other body at rest, if there was no time dilation, the time readings on that clock between each crest of the wave would be a certain amount. However, if there is time dilation then, from the perspective of the other body at rest, the time readings on that clock between each crest of the wave will be shorter. In other words, there is an increase in frequency as light wave crests encounter the clock. This increase in frequency occurs for the observer at rest relative to the moving clock. And these time recordings are physically real for him. No? Please let me know what I'm missing. ------ I wrote: “This increase in frequency occurs for the observer at rest relative to the moving clock.” But to be more accurate I should have written: “This increase in the frequency measurement occurs for the observer at rest on the clock in relative motion.” It’s not that the frequency of the wave of light increases for the observer at rest. Rather, it’s that from the perspective of the observer at rest the frequency of the wave of light increases for the clock in motion, and therefore for the observer in “motion”. That’s a distinction I think I failed to clearly make in my last post. Edited October 31, 2016 by Thales et al
Strange Posted October 31, 2016 Posted October 31, 2016 From the perspective of the other body at rest, if there was no time dilation, the time readings on that clock between each crest of the wave would be a certain amount. However, if there is time dilation then, from the perspective of the other body at rest, the time readings on that clock between each crest of the wave will be shorter. In other words, there is an increase in frequency as light wave crests encounter the clock. This increase in frequency occurs for the observer at rest relative to the moving clock. And these time recordings are physically real for him. No? Please let me know what I'm missing. The clock in motion with the light source will be time-dilated by exactly the same amount as the frequency of the light. So that clock will show the same period of the wave, regardless of the state of motion relative to the observer. Time dilation is a relative thing. So it is only a difference in the measurements made by the observer's clock and the moving clock. So there is only red-shift when measured by the other clock.
Tim88 Posted November 1, 2016 Posted November 1, 2016 [..] If I might, let me take a step back and try presenting this again, in a slightly different way. sf02.jpg Say, we had a super sensitive clock that was able to measure the difference between each crest of the electric part of the electromagnetic wave. And say that clock is in relative motion with another body at rest. From the perspective of the other body at rest, if there was no time dilation, the time readings on that clock between each crest of the wave would be a certain amount. However, if there is time dilation then, from the perspective of the other body at rest, the time readings on that clock between each crest of the wave will be shorter. In other words, there is an increase in frequency as light wave crests encounter the clock. This increase in frequency occurs for the observer at rest relative to the moving clock. And these time recordings are physically real for him. No? Please let me know what I'm missing. ------ [..] to be more accurate I should have written: “This increase in the frequency measurement occurs for the observer at rest on the clock in relative motion.” It’s not that the frequency of the wave of light increases for the observer at rest. Rather, it’s that from the perspective of the observer at rest the frequency of the wave of light increases for the clock in motion, and therefore for the observer in “motion”. That’s a distinction I think I failed to clearly make in my last post. If I correctly understand you, the light bulb on the right (the source) is supposed to be in rest, so that the receiver is moving. Correct? Then I agree with you, and I likely understood your issue from the start (see once more my post #4). Did you have a look at the Doppler formula? For that case you have, as correctly explained in Wikipedia, fr/f0 = (c+vr)/c Further, the time dilation effect makes that according to the observer in rest, the perceived frequency by the moving detector is increased by the time dilation factor - just as you say. It will thus appear less red-shifted than according to the Doppler equation here above. Combining Doppler with time dilation yields the so-called "relativistic Doppler" equation. Did you try to work it out? I just now verified it, it only takes a couple of minutes. If you don't know how, just ask, and I or someone else will show you.
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