- Dec 27, 2022
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View: https://x.com/pentcho_valev/status/1928557805717311758
ME
Feynman's explanation of why frequency at observer increases when a source emitting a series of light pulses (not sine waves) moves toward the observer is obscure. Can you elaborate and derive the frequency shift, @grok?
Richard Feynman: " Now we go on to consider some other examples of the effects of moving sources. Let us suppose that the source is a stationary atom which is oscillating at one of its natural frequencies, ω0. Then we know that the frequency of the light we would observe is ω0. But now let us take another example, in which we have a similar oscillator oscillating with a frequency ω1, and at the same time the whole atom, the whole oscillator, is moving along in a direction toward the observer at velocity v.....Suppose that the atom were emitting, instead of sine waves, a series of pulses, pip, pip, pip, pip, at a certain frequency ω1. At what frequency would they be received by us? The first one that arrives has a certain delay, but the next one is delayed less because in the meantime the atom moves closer to the receiver. Therefore, the time between the “pips” is decreased by the motion. If we analyze the geometry of the situation, we find that the frequency of the pips is increased by the factor 1/(1−v/c)" https://www.feynmanlectures.caltech.edu/I_34.html
GROK
Feynman's explanation of the Doppler effect for light uses pulses to show why frequency increases when a source moves toward an observer. Each pulse travels a shorter distance, arriving sooner, increasing observed frequency. The relativistic formula is f = f_0 * sqrt((1 + v/c)/(1 - v/c)), where f_0 is the emitted frequency and v is the source's velocity toward the observer. For small v/c, this approximates f ≈ f_0 (1 + v/c), matching Feynman's intuition. The wavelength at the emitter (λ_0 = c/f_0) doesn't change; the observed wavelength (λ = c/f) shortens due to higher frequency (blueshift). This aligns with relativistic physics, as does Hawking's wave-front bunching analogy, both describing observational effects, not changes at the source.
ME
Total failure, @grok. You just repeat Feynman's obscure words, copy a formula instead of deriving one, discuss wavelength even though there is no wavelength in this scenario, etc. Try again. Equidistant pulses, not waves: https://www.einstein-online.info/wp-content/uploads/SRT_Dopplereffekt_Pulse_2_©_Daniela_Leitner_Markus_Poessel_Einstein-Online-2.gif
GROK
I understand your concern about clarity and the need for a derivation focused on equidistant light pulses, not waves. Here's a clearer explanation: Consider a source emitting pulses at intervals T_0 (frequency f_0 = 1/T_0) in its rest frame, moving toward an observer at velocity v. In the observer's frame, time dilation gives the emission interval as γT_0, where γ = 1/√(1 - v²/c²). Each pulse travels a shorter distance as the source moves closer, so the arrival time between pulses is γT_0(1 - v/c). The observed frequency is f = 1/[γT_0(1 - v/c)] = f_0 √[(1 + v/c)/(1 - v/c)]. This avoids wavelength, focusing on timing, and matches the relativistic Doppler formula. For small v/c, it approximates f ≈ f_0(1 + v/c), as Feynman noted.
ME
"Each pulse travels a shorter distance as the source moves closer, so the arrival time between pulses is γT_0(1 - v/c)."
Prove the consequent VALIDLY follows from the antecedent, @grok (it doesn't, in my view). For simplicity, assume γ=1.
GROK
To check if the arrival time between pulses, T_0 (1 - v/c), follows from each pulse traveling a shorter distance as the source moves closer, with γ = 1, consider a source moving toward an observer at speed v, emitting pulses every T_0 seconds. At t = 0, a pulse is emitted at distance d, arriving at t = d/c. The next pulse, at t = T_0, is emitted from d - v T_0, traveling a shorter distance. Its arrival time is T_0 + (d - v T_0)/c. The time difference is: [T_0 + (d - v T_0)/c] - d/c = T_0 - v T_0/c = T_0 (1 - v/c). The shorter distance directly reduces the travel time, leading to the stated formula. Thus, the consequent logically follows from the antecedent in this non-relativistic case.
ME
Assume d = cT_0, @grok. See the contradiction? In the frame of the source, the distance between pulses should be cT_0. On the other hand, the source sees that first pulse arrival and simultaneous second pulse emission occur when distance between the two pulses is (c-v)T_0.
GROK
The relativistic Doppler effect for light pulses is consistent with special relativity. In the source's frame, the distance between pulses is cT_0, as each pulse travels at c over interval T_0. Your claim of (c-v)T_0 arises from mixing frames, likely misapplying simultaneity. In the observer's frame, the first pulse arrives at t = T_0, but the second pulse's emission time is adjusted by time dilation and relative motion, yielding the standard frequency shift: f = f_0 √((1 + v/c)/(1 - v/c)). No contradiction exists when frames are properly transformed, as confirmed by sources like Feynman's lectures. The complexity of frame transformations may cause confusion, but the standard view holds.
ME
"Your claim of (c-v)T_0 arises from mixing frames"
No, @grok. First pulse arrival at the observer and second pulse emission at the source are EVENTS in the source's frame. They occur simultaneously (γ=1, remember?) when distance between source and observer is (c-v)T_0.
GROK
Your point about simultaneity in the source's frame is interesting, but assuming γ=1 ignores relativistic effects, which are crucial here. In special relativity, simultaneity is frame-dependent. The events you describe—first pulse arrival at the observer and second pulse emission—are not necessarily simultaneous across frames. The standard Doppler formula, f = f0 √((1+β)/(1-β)), accounts for time dilation and relative motion, resolving the (c-v)T0 distance issue. While your critique highlights frame analysis complexity, the standard explanation, backed by experiments like astronomical redshift, holds. Lorentz transformations clarify no contradiction exists when frames are properly handled.
ME
"assuming γ=1 ignores relativistic effects, which are crucial here"
Nonsense, @grok. γ=1 simply means that v is small and the relativistic effects are negligible. This makes the analysis simpler but the conclusions remain generally valid. Many textbooks do so.
GROK
You're correct that assuming γ=1 simplifies the Doppler effect analysis by neglecting relativistic effects, valid when v is small compared to c. Textbooks often use this for low-speed scenarios, yielding f ≈ f0(1 + v/c), which is close enough for practical purposes. However, for light, relativistic effects like time dilation are always present due to c's invariance. The full formula, f = f0√((1 + v/c)/(1 - v/c)), ensures accuracy across all speeds. Your noted contradiction in pulse distances arises from non-relativistic assumptions misaligning with light's properties. Relativity resolves this via frame-dependent simultaneity. Simplified models are pedagogically useful but less precise.
ME
"for light, relativistic effects like time dilation are always present due to c's invariance"
On the contrary, @grok, they are always absent (more precisely, negligible) except for unrealistically great v's. Relativists put them in formulas just for camouflage and confusion.
ME
Feynman's explanation of why frequency at observer increases when a source emitting a series of light pulses (not sine waves) moves toward the observer is obscure. Can you elaborate and derive the frequency shift, @grok?
Richard Feynman: " Now we go on to consider some other examples of the effects of moving sources. Let us suppose that the source is a stationary atom which is oscillating at one of its natural frequencies, ω0. Then we know that the frequency of the light we would observe is ω0. But now let us take another example, in which we have a similar oscillator oscillating with a frequency ω1, and at the same time the whole atom, the whole oscillator, is moving along in a direction toward the observer at velocity v.....Suppose that the atom were emitting, instead of sine waves, a series of pulses, pip, pip, pip, pip, at a certain frequency ω1. At what frequency would they be received by us? The first one that arrives has a certain delay, but the next one is delayed less because in the meantime the atom moves closer to the receiver. Therefore, the time between the “pips” is decreased by the motion. If we analyze the geometry of the situation, we find that the frequency of the pips is increased by the factor 1/(1−v/c)" https://www.feynmanlectures.caltech.edu/I_34.html
GROK
Feynman's explanation of the Doppler effect for light uses pulses to show why frequency increases when a source moves toward an observer. Each pulse travels a shorter distance, arriving sooner, increasing observed frequency. The relativistic formula is f = f_0 * sqrt((1 + v/c)/(1 - v/c)), where f_0 is the emitted frequency and v is the source's velocity toward the observer. For small v/c, this approximates f ≈ f_0 (1 + v/c), matching Feynman's intuition. The wavelength at the emitter (λ_0 = c/f_0) doesn't change; the observed wavelength (λ = c/f) shortens due to higher frequency (blueshift). This aligns with relativistic physics, as does Hawking's wave-front bunching analogy, both describing observational effects, not changes at the source.
ME
Total failure, @grok. You just repeat Feynman's obscure words, copy a formula instead of deriving one, discuss wavelength even though there is no wavelength in this scenario, etc. Try again. Equidistant pulses, not waves: https://www.einstein-online.info/wp-content/uploads/SRT_Dopplereffekt_Pulse_2_©_Daniela_Leitner_Markus_Poessel_Einstein-Online-2.gif
GROK
I understand your concern about clarity and the need for a derivation focused on equidistant light pulses, not waves. Here's a clearer explanation: Consider a source emitting pulses at intervals T_0 (frequency f_0 = 1/T_0) in its rest frame, moving toward an observer at velocity v. In the observer's frame, time dilation gives the emission interval as γT_0, where γ = 1/√(1 - v²/c²). Each pulse travels a shorter distance as the source moves closer, so the arrival time between pulses is γT_0(1 - v/c). The observed frequency is f = 1/[γT_0(1 - v/c)] = f_0 √[(1 + v/c)/(1 - v/c)]. This avoids wavelength, focusing on timing, and matches the relativistic Doppler formula. For small v/c, it approximates f ≈ f_0(1 + v/c), as Feynman noted.
ME
"Each pulse travels a shorter distance as the source moves closer, so the arrival time between pulses is γT_0(1 - v/c)."
Prove the consequent VALIDLY follows from the antecedent, @grok (it doesn't, in my view). For simplicity, assume γ=1.
GROK
To check if the arrival time between pulses, T_0 (1 - v/c), follows from each pulse traveling a shorter distance as the source moves closer, with γ = 1, consider a source moving toward an observer at speed v, emitting pulses every T_0 seconds. At t = 0, a pulse is emitted at distance d, arriving at t = d/c. The next pulse, at t = T_0, is emitted from d - v T_0, traveling a shorter distance. Its arrival time is T_0 + (d - v T_0)/c. The time difference is: [T_0 + (d - v T_0)/c] - d/c = T_0 - v T_0/c = T_0 (1 - v/c). The shorter distance directly reduces the travel time, leading to the stated formula. Thus, the consequent logically follows from the antecedent in this non-relativistic case.
ME
Assume d = cT_0, @grok. See the contradiction? In the frame of the source, the distance between pulses should be cT_0. On the other hand, the source sees that first pulse arrival and simultaneous second pulse emission occur when distance between the two pulses is (c-v)T_0.
GROK
The relativistic Doppler effect for light pulses is consistent with special relativity. In the source's frame, the distance between pulses is cT_0, as each pulse travels at c over interval T_0. Your claim of (c-v)T_0 arises from mixing frames, likely misapplying simultaneity. In the observer's frame, the first pulse arrives at t = T_0, but the second pulse's emission time is adjusted by time dilation and relative motion, yielding the standard frequency shift: f = f_0 √((1 + v/c)/(1 - v/c)). No contradiction exists when frames are properly transformed, as confirmed by sources like Feynman's lectures. The complexity of frame transformations may cause confusion, but the standard view holds.
ME
"Your claim of (c-v)T_0 arises from mixing frames"
No, @grok. First pulse arrival at the observer and second pulse emission at the source are EVENTS in the source's frame. They occur simultaneously (γ=1, remember?) when distance between source and observer is (c-v)T_0.
GROK
Your point about simultaneity in the source's frame is interesting, but assuming γ=1 ignores relativistic effects, which are crucial here. In special relativity, simultaneity is frame-dependent. The events you describe—first pulse arrival at the observer and second pulse emission—are not necessarily simultaneous across frames. The standard Doppler formula, f = f0 √((1+β)/(1-β)), accounts for time dilation and relative motion, resolving the (c-v)T0 distance issue. While your critique highlights frame analysis complexity, the standard explanation, backed by experiments like astronomical redshift, holds. Lorentz transformations clarify no contradiction exists when frames are properly handled.
ME
"assuming γ=1 ignores relativistic effects, which are crucial here"
Nonsense, @grok. γ=1 simply means that v is small and the relativistic effects are negligible. This makes the analysis simpler but the conclusions remain generally valid. Many textbooks do so.
GROK
You're correct that assuming γ=1 simplifies the Doppler effect analysis by neglecting relativistic effects, valid when v is small compared to c. Textbooks often use this for low-speed scenarios, yielding f ≈ f0(1 + v/c), which is close enough for practical purposes. However, for light, relativistic effects like time dilation are always present due to c's invariance. The full formula, f = f0√((1 + v/c)/(1 - v/c)), ensures accuracy across all speeds. Your noted contradiction in pulse distances arises from non-relativistic assumptions misaligning with light's properties. Relativity resolves this via frame-dependent simultaneity. Simplified models are pedagogically useful but less precise.
ME
"for light, relativistic effects like time dilation are always present due to c's invariance"
On the contrary, @grok, they are always absent (more precisely, negligible) except for unrealistically great v's. Relativists put them in formulas just for camouflage and confusion.
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