Talk:Doppler effect

In the section "General", there is the below part (quoted) showing the wavelength ${\displaystyle \lambda }$. How to interpret it? I thought it it the wavelength observed by the receiver, but it is not when I consider a case that the source is stationary and the receiver is moving toward the source. In this case, the observed wavelength by the receiver should be shorter than the wavelength observed when the receiver is also stationary, but this formula shows the wavelength not shortened ${\displaystyle v_{ws}=c}$, so I'm confusing what this ${\displaystyle \lambda }$ is. It needs to be clarified.

"Equivalently, under the assumption that the source is either directly approaching or receding from the observer:

$${\displaystyle {\frac {f}{v_{wr}}}={\frac {f_{0}}{v_{ws}}}={\frac {1}{\lambda }}}$$ where

• ${\displaystyle v_{wr}}$ is the wave's speed relative to the receiver;
• ${\displaystyle v_{ws}}$ is the wave's speed relative to the source;
• ${\displaystyle \lambda }$ is the wavelength."

Goodphy (talk) 08:57, 21 September 2025 (UTC)[reply]

The paragraph had no sources. I deleted it. Johnjbarton (talk) 14:07, 21 September 2025 (UTC)[reply]

Hi.

In the Section "General", there is the below description (quoted with ""), that is mentioned as an approximation. However, if the frame of the reference is fixed to the wave Source, then I think it is not the approximation anymore; it is exact because ${\displaystyle v_{\text{s}}}$ in ${\displaystyle f=\left({\frac {c\pm v_{\text{r}}}{c\mp v_{\text{s}}}}\right)f_{0}}$ becomes 0, resulting in the formula of the below description.

Thus, I would like to confirm if the description is an approximation or or exact. I think it is exact as long as the special relativity is not concerned.

"If the speeds ${\displaystyle v_{\text{s}}}$ and ${\displaystyle v_{\text{r}}\,}$ are small compared to the speed of the wave, the relationship between observed frequency ${\displaystyle f}$ and emitted frequency ${\displaystyle f_{\text{0}}}$ is approximately^[1]

where

• ${\displaystyle \Delta f=f-f_{0}}$
• ${\displaystyle \Delta v=-(v_{\text{r}}-v_{\text{s}})}$ is positive when the source and the receiver are moving towards each other. The interpretation of it is easy when we consider ${\displaystyle v_{\text{r}}}$ and ${\displaystyle v_{\text{s}}}$ as vectors here; they are positive when they are travel along the right direction and opposite for the left direction travel, and the source is left to the receiver.
• This proof below is for a case where the receiver is moving toward the source and the source is moving away from it, but proofs for other cases can be similarly derived."

Goodphy (talk) 07:28, 29 September 2025 (UTC)[reply]

I agree and I deleted that content. The starting formula in that section already neglects relativity. The source does not mention this bit. Johnjbarton (talk) 18:16, 29 September 2025 (UTC)[reply]

Hi. Thank you to recognize my point of view and your effort of clarifying this article, but I think deleting the whole description in this topic seems too strong. I thought modifying the first sentence "If the speeds and are..." and simplifying the description of ${\displaystyle \Delta v}$ seems enough. Let me consider a possible good change, and implement it if needed. Let you know the change. Goodphy (talk) 06:35, 30 September 2025 (UTC)[reply]

The change has been made by me. @Johnjbarton Could you review my change? Goodphy (talk) 07:24, 30 September 2025 (UTC)[reply]

@Goodphy The problem I have with this paragraph is "why?" What is the point? What physical situation is modeled by these equations? The content is not sourced so we have no way to improve it without guessing. Johnjbarton (talk) 15:14, 30 September 2025 (UTC)[reply]

Hi. This is not about physics but these formulas as a simpler form is useful for calculation. In essence, we don't need it as it can be derived. I think it is good to have rather than what it must be.

If you remove it with your own reason, then I will not resist it. 39.7.15.211 (talk) 15:37, 30 September 2025 (UTC)[reply]

Calculation of what? In what case would the Doppler shift in the frame of the source be useful? Johnjbarton (talk) 16:26, 30 September 2025 (UTC)[reply]

Hi @Johnjbarton.

I have thought about the usefulness of the reference frame fixed to the wave source, as you mentioned, than today I decided to remove (removal done) the contents about this frame, because, in this frame, the medium of wave propagation moves, and that makes situation not simple. (Light, which speed is constant regardless of which reference frames are used in vacuum, seems more simple.)

Instead, in my vision, the approximation that you have removed on 2025-09-29 by this page TALK, seems useful for practical cases inc. engineering fields. I think there was a reference cited in the removed contents. I will try to find and review this reference to check its validity regarding the contents. Goodphy (talk) 06:12, 7 October 2025 (UTC)[reply]

Revived contents about small velocity approximation that has been removed on 2025-09-29, because similar contents can be found in different sources, even Google AI shows this contents based on different sources. This approximation formula seems widely accepted and would give an insight of how doppler shift frequency can be calculated. Goodphy (talk) 06:51, 8 October 2025 (UTC)[reply]

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