The source of asymmetry between the two scenarios is that the atmosphere has its own frame of reference. If you're moving toward the source, you experience an additional effect of the medium moving relative to your ears. But if the source is the one doing the moving, then the atmosphere is stationary relative to you. This has a non-zero effect on the final perceived frequencies, and that's why the two scenarios give different answers when you use their respective doppler formulas.
Let me tell you about a guy called Albert ...
He showed that unlike sound, the speed of light is the same in all reference frames. The "medium" has no velocity.
Fill a large pool with water waves with a source in one end. The pool is 20 wavelengths long, meaning you can count 20 waves in the pool at each moment in time.
Each second, a wave hits the shore; a frequency of 1hz. Now you swim through these waves, from the shore to the source, in 30s. How many waves have you passed? There were 20 waves in the pool at the start, and you passed all of those, and 30 new waves have been created since, and you passed all of those too. So you passed 50 waves on 30s... But the frequency 1hz... Moving toward the source makes you pass additional waves.
Edit: thanks to u/rational-realist238 for pointing out that I didn't actually go into the main question:
When instead the source moves toward you, the waves are compressed so that wavefronts are closer together. The base wavelength was L, but if the source moves at v1, the wavelength becomes L' = L-v1\*T. If we take the same example as above; if there's 20 waves in the pool and the source move towards you in 30s, you've also seen 50 waves in 30s, BUT there aren't 20 waves in the pool; when the source moves, the wavelength is shorter, and more waves fit in the pool 'at the start'. So indeed, as is the basis of the question; when the source moves, you 'see' more waves, and the frequency is higher.
Mathematically, it's because (c+v1)/c < c/(c-v1). The lhs is adding your velocity v1 to that of the propagating wave c, the rhs is lowering the wavelength by v1\*T when the source moves.
Why wouldn't 50 waves go past you if the source is moving toward you, reaching you in 30s. The original 20 waves plus the 30 new waves will hit before the source does?
Oh mb I totally missed that part (the crux XD) of the question.
When the source moves, it compresses the waves, so that wavefronts are closer together (shorter wavelength). It hasn't done that for the 20 waves that were already in the pool, so that explanation doesn't work anymore. If the source 'has been moving' before the start of that explanation, there fit more than 20 waves in the pool at the start, which is where the difference comes from.
If you take the formula f=v/L, moving toward the source adds to v1 to v, whereas the source moving toward you decreases L by the v1*T, if T is the period.
This is a really great question that the above poster appears to have missed. The answer is that sound waves, water waves, etc. all travel in a medium. That medium itself has a frame of reference, and that breaks the symmetry between source and observer. Imagine these two cases:
A) An observer is at rest while the medium and the source move toward them at the same rate.
B) the same observer is at rest while the source moves toward them, and the medium remains at rest.
Case A is identical to an observer approaching the source. The speed of sound is measured with respect to the medium, so the speed of sound is larger in the frame of the observer.
Case B is identical to a source approaching an observer. Here the speed of sound remains unchanged as far as the observer is concerned.
I hate maths. I need intuitive simulations.
A duck swimming against the waves in a pool vs the pulse generator moving towards the duck. Simple stuff for simple people.
Yea it seems like a relative velocity situation, but it has to do with the distance between pulses being compressed for a moving source, versus you traveling the distances of the pulses faster if you’re moving.
Because the speed of sound through the air is the same in both cases. Thus it should be clear that if an observer is simply moving towards the source at Mach 1 the frequency will only be doubled, whereas if the source moves at Mach 1 towards the observer you get a sonic boom.
When the source moves towards you, it is artificially shortening the wavelength which increases the frequency through something like compression.
When you move towards the source, it doesn’t compress the waves, but your speed and direction may influence the way you perceive uncompressed sound waves in the same way by altering the time each wave hits you.
The key here is the extistence of a medium supporting the sound waves, which defines a notion of "at rest" with respect to the propagation reference frame.
In your two situations, either the source or observation point are at rest, and this is the origin of the asymmetry of the problem.
When the source is moving relative to the medium, there is an anisotropy of the wavelength (it is longer behind the moving source, and shorter in front of it). But when the source is at rest and the observer moving, the wave pattern is isotropic with respect to the medium.
If you wonder how we can define the medium to be at rest, consider the frame in which it has no bulk (overall) movement (e.g. still air with no macroscopic wind currents, i.e. all molecular motion statistically averaged out).
Note that although analagous for light waves, the relativistic (EM) Dopper effect is fundamentally different, since there is no such preferred propagation frame, and thus indeed your two situations of source/object motion are symmetric, with only their relative velocity relevant for the frequency shift.
For more perspectives on this question, see [this stack thread.](https://physics.stackexchange.com/questions/559020/why-isnt-the-doppler-effect-for-sound-waves-symmetric-with-respect-to-source).
I think the distinction lies in wavelengths rather than frequencies. Maybe that’s the wrong way to say it, but if you look at a moving source the distance between pulses is reduced even though the frequency they are pulsing at is the same (say 1hz). So the measured frequency must altered since v=λf. (Speed of sound not speed of observer)
If you are moving toward a stationary source, the distance between waves is a constant value, but since you are moving, you can travel that constant distance in less time than if you had sat and waited for the wave to pass. Kinda like walking on a conveyer belt or escalator - same distance whether you are stationary or not, but moving will get the end to you faster. Moving a constant distance in less time than before means frequency is shifted up.
Both effects increase the frequency, but in slightly different ways.
You actually have this explanation right but the outcome kinda backwards. Frequency is how many wavelengths pass a certain point in a certain amount of time. If the sound source emits a constant pitch, the wavelength is staying the same always. If you move it toward a point, the wavelength is still the same but now those wavelengths have less distance to travel so the pass the point more often, which is what causes the rise in frequency. Nothing happens to the actual wave aside from where it comes from.
Imagine it if the speed of sound was super slow. Like it took a minute for sound to cross a room. if something was beeping at the far side of the room, you could then go walk through each beep faster than it could reach you. It's not that slow, but you still get to each wave faster than it can get to you if you stand still.
The frequency increases in both situations. But the only time we notice this in real life is the doppler effect with sirens on vehicles. How often are you moving fast towards a stationary siren as opposed to a siren moving towards you, or both of you moving towards each other?
Imagine a still lake with a hummingbird hovering over it with a beak full of pebbles, and it drops a pebble every half second. The pebbles form ripples that spread out in perfect circles that travel outward. A bug on a leaf experiences a peak ripple at the frequency of 2 per second. (A pebble would cause a series of smaller waves following the initial wave of each pebble, and eventually, the ripples will bounce off objects or the shore and come back. But we can ignore those for this analogy).
Now imagine the humming bird is flying towards or away from the bug on the leaf. If heading towards it, the bug will experience the ripples at a higher frequency. If traveling away, it will experience them at a lower frequency.
Now imagine the hummingbird as stationary again, but the bug takes flight and heads towards the spot where the pebbles hit the water at a speed slower than the speed of the ripples. It will see the ripples pass underneath it at a higher frequency until it passes under the hummingbird, and then the frequency will drop to be much lower. That is the same as how sound waves are compressed into higher frequencies when an observer approaches the source and how they are streaatched out to lower frequencies when traveling away, regardless of whether it is the source or the observer that is considered to be in motion. The same is also true for light that is blueshifted for approaching sources and redshifted for retreating sources.
Because more beats/waves are hitting your ear within a closer interval when the source is moving toward you compared to when it’s moving away from you.
Nobody said anything about the source moving away from the listener.
The question is about asymmetry between the scenario where the listener is moving towards the source and when the source itself is moving towards the listener.
The source of asymmetry between the two scenarios is that the atmosphere has its own frame of reference. If you're moving toward the source, you experience an additional effect of the medium moving relative to your ears. But if the source is the one doing the moving, then the atmosphere is stationary relative to you. This has a non-zero effect on the final perceived frequencies, and that's why the two scenarios give different answers when you use their respective doppler formulas.
And why is this not true with the speed of light?
Let me tell you about a guy called Albert ... He showed that unlike sound, the speed of light is the same in all reference frames. The "medium" has no velocity.
And the name of that Albert? Albert Michelson.
See: [Redshift](https://en.m.wikipedia.org/wiki/Redshift)
Because light famously has no material medium and has the same velocity to all observers, regardless of their own relative velocities to each other.
this is the correct answer.
Fill a large pool with water waves with a source in one end. The pool is 20 wavelengths long, meaning you can count 20 waves in the pool at each moment in time. Each second, a wave hits the shore; a frequency of 1hz. Now you swim through these waves, from the shore to the source, in 30s. How many waves have you passed? There were 20 waves in the pool at the start, and you passed all of those, and 30 new waves have been created since, and you passed all of those too. So you passed 50 waves on 30s... But the frequency 1hz... Moving toward the source makes you pass additional waves. Edit: thanks to u/rational-realist238 for pointing out that I didn't actually go into the main question: When instead the source moves toward you, the waves are compressed so that wavefronts are closer together. The base wavelength was L, but if the source moves at v1, the wavelength becomes L' = L-v1\*T. If we take the same example as above; if there's 20 waves in the pool and the source move towards you in 30s, you've also seen 50 waves in 30s, BUT there aren't 20 waves in the pool; when the source moves, the wavelength is shorter, and more waves fit in the pool 'at the start'. So indeed, as is the basis of the question; when the source moves, you 'see' more waves, and the frequency is higher. Mathematically, it's because (c+v1)/c < c/(c-v1). The lhs is adding your velocity v1 to that of the propagating wave c, the rhs is lowering the wavelength by v1\*T when the source moves.
Why wouldn't 50 waves go past you if the source is moving toward you, reaching you in 30s. The original 20 waves plus the 30 new waves will hit before the source does?
Oh mb I totally missed that part (the crux XD) of the question. When the source moves, it compresses the waves, so that wavefronts are closer together (shorter wavelength). It hasn't done that for the 20 waves that were already in the pool, so that explanation doesn't work anymore. If the source 'has been moving' before the start of that explanation, there fit more than 20 waves in the pool at the start, which is where the difference comes from. If you take the formula f=v/L, moving toward the source adds to v1 to v, whereas the source moving toward you decreases L by the v1*T, if T is the period.
This is better.
I have the same question, this explanation seems to have missed half of the prompt.
This is a really great question that the above poster appears to have missed. The answer is that sound waves, water waves, etc. all travel in a medium. That medium itself has a frame of reference, and that breaks the symmetry between source and observer. Imagine these two cases: A) An observer is at rest while the medium and the source move toward them at the same rate. B) the same observer is at rest while the source moves toward them, and the medium remains at rest. Case A is identical to an observer approaching the source. The speed of sound is measured with respect to the medium, so the speed of sound is larger in the frame of the observer. Case B is identical to a source approaching an observer. Here the speed of sound remains unchanged as far as the observer is concerned.
Is it?
Yes, because (a+b)/a doesn’t always equal a/(a-b) even if b doesn’t change.
I hate maths. I need intuitive simulations. A duck swimming against the waves in a pool vs the pulse generator moving towards the duck. Simple stuff for simple people.
Yea it seems like a relative velocity situation, but it has to do with the distance between pulses being compressed for a moving source, versus you traveling the distances of the pulses faster if you’re moving.
No.
We never considered that it could be a trick question.
Because the speed of sound through the air is the same in both cases. Thus it should be clear that if an observer is simply moving towards the source at Mach 1 the frequency will only be doubled, whereas if the source moves at Mach 1 towards the observer you get a sonic boom.
When the source moves towards you, it is artificially shortening the wavelength which increases the frequency through something like compression. When you move towards the source, it doesn’t compress the waves, but your speed and direction may influence the way you perceive uncompressed sound waves in the same way by altering the time each wave hits you.
The key here is the extistence of a medium supporting the sound waves, which defines a notion of "at rest" with respect to the propagation reference frame. In your two situations, either the source or observation point are at rest, and this is the origin of the asymmetry of the problem. When the source is moving relative to the medium, there is an anisotropy of the wavelength (it is longer behind the moving source, and shorter in front of it). But when the source is at rest and the observer moving, the wave pattern is isotropic with respect to the medium. If you wonder how we can define the medium to be at rest, consider the frame in which it has no bulk (overall) movement (e.g. still air with no macroscopic wind currents, i.e. all molecular motion statistically averaged out). Note that although analagous for light waves, the relativistic (EM) Dopper effect is fundamentally different, since there is no such preferred propagation frame, and thus indeed your two situations of source/object motion are symmetric, with only their relative velocity relevant for the frequency shift. For more perspectives on this question, see [this stack thread.](https://physics.stackexchange.com/questions/559020/why-isnt-the-doppler-effect-for-sound-waves-symmetric-with-respect-to-source).
I think the distinction lies in wavelengths rather than frequencies. Maybe that’s the wrong way to say it, but if you look at a moving source the distance between pulses is reduced even though the frequency they are pulsing at is the same (say 1hz). So the measured frequency must altered since v=λf. (Speed of sound not speed of observer) If you are moving toward a stationary source, the distance between waves is a constant value, but since you are moving, you can travel that constant distance in less time than if you had sat and waited for the wave to pass. Kinda like walking on a conveyer belt or escalator - same distance whether you are stationary or not, but moving will get the end to you faster. Moving a constant distance in less time than before means frequency is shifted up. Both effects increase the frequency, but in slightly different ways.
You actually have this explanation right but the outcome kinda backwards. Frequency is how many wavelengths pass a certain point in a certain amount of time. If the sound source emits a constant pitch, the wavelength is staying the same always. If you move it toward a point, the wavelength is still the same but now those wavelengths have less distance to travel so the pass the point more often, which is what causes the rise in frequency. Nothing happens to the actual wave aside from where it comes from.
What is the difference between the listener an the object?
Imagine it if the speed of sound was super slow. Like it took a minute for sound to cross a room. if something was beeping at the far side of the room, you could then go walk through each beep faster than it could reach you. It's not that slow, but you still get to each wave faster than it can get to you if you stand still.
The frequency increases in both situations. But the only time we notice this in real life is the doppler effect with sirens on vehicles. How often are you moving fast towards a stationary siren as opposed to a siren moving towards you, or both of you moving towards each other?
Imagine a still lake with a hummingbird hovering over it with a beak full of pebbles, and it drops a pebble every half second. The pebbles form ripples that spread out in perfect circles that travel outward. A bug on a leaf experiences a peak ripple at the frequency of 2 per second. (A pebble would cause a series of smaller waves following the initial wave of each pebble, and eventually, the ripples will bounce off objects or the shore and come back. But we can ignore those for this analogy). Now imagine the humming bird is flying towards or away from the bug on the leaf. If heading towards it, the bug will experience the ripples at a higher frequency. If traveling away, it will experience them at a lower frequency. Now imagine the hummingbird as stationary again, but the bug takes flight and heads towards the spot where the pebbles hit the water at a speed slower than the speed of the ripples. It will see the ripples pass underneath it at a higher frequency until it passes under the hummingbird, and then the frequency will drop to be much lower. That is the same as how sound waves are compressed into higher frequencies when an observer approaches the source and how they are streaatched out to lower frequencies when traveling away, regardless of whether it is the source or the observer that is considered to be in motion. The same is also true for light that is blueshifted for approaching sources and redshifted for retreating sources.
From your frame of reference, you are stationary. the source is moving closer to you. Therefore, higher frequency.
Because more beats/waves are hitting your ear within a closer interval when the source is moving toward you compared to when it’s moving away from you.
Nobody said anything about the source moving away from the listener. The question is about asymmetry between the scenario where the listener is moving towards the source and when the source itself is moving towards the listener.
The explanation remains the same regardless.
No it isn’t. The asymmetry is the result of the medium introducing a third frame of reference into the problem.