What is the relationship between relativity and the Doppler effect?
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My sister just watched this video about space contraction (Spanish), and asked me if this is related to doppler effect.
In the clip they also introduce the idea that a bat would be affected by similar effects when measuring an object's length, due to the time it takes for sound to propagate.
I told her that:
Doppler effect is about alteration of the perceived frequency of a
signal produced by the relative movement between transmitter and
receiver. The quoted video is about relativity, which is a "deeper" effect.
Maybe doppler effect can be understood as the effect of relativity on
a wave phenomena.
Now I'm wondering about her intuition. If she's right, should I be able to take a sin function, apply a Lorentz transform to it, and arrive to same results as with the doppler formula? Unfortunately, the maths are beyond my skills.
Can someone shed some light about the relation between doppler and relativity, if any? Can be doppler effect explained by relativity/Lorentz alone?
special-relativity waves doppler-effect
New contributor
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add a comment |
$begingroup$
My sister just watched this video about space contraction (Spanish), and asked me if this is related to doppler effect.
In the clip they also introduce the idea that a bat would be affected by similar effects when measuring an object's length, due to the time it takes for sound to propagate.
I told her that:
Doppler effect is about alteration of the perceived frequency of a
signal produced by the relative movement between transmitter and
receiver. The quoted video is about relativity, which is a "deeper" effect.
Maybe doppler effect can be understood as the effect of relativity on
a wave phenomena.
Now I'm wondering about her intuition. If she's right, should I be able to take a sin function, apply a Lorentz transform to it, and arrive to same results as with the doppler formula? Unfortunately, the maths are beyond my skills.
Can someone shed some light about the relation between doppler and relativity, if any? Can be doppler effect explained by relativity/Lorentz alone?
special-relativity waves doppler-effect
New contributor
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Related: physics.stackexchange.com/q/61946
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– Chair
18 mins ago
add a comment |
$begingroup$
My sister just watched this video about space contraction (Spanish), and asked me if this is related to doppler effect.
In the clip they also introduce the idea that a bat would be affected by similar effects when measuring an object's length, due to the time it takes for sound to propagate.
I told her that:
Doppler effect is about alteration of the perceived frequency of a
signal produced by the relative movement between transmitter and
receiver. The quoted video is about relativity, which is a "deeper" effect.
Maybe doppler effect can be understood as the effect of relativity on
a wave phenomena.
Now I'm wondering about her intuition. If she's right, should I be able to take a sin function, apply a Lorentz transform to it, and arrive to same results as with the doppler formula? Unfortunately, the maths are beyond my skills.
Can someone shed some light about the relation between doppler and relativity, if any? Can be doppler effect explained by relativity/Lorentz alone?
special-relativity waves doppler-effect
New contributor
$endgroup$
My sister just watched this video about space contraction (Spanish), and asked me if this is related to doppler effect.
In the clip they also introduce the idea that a bat would be affected by similar effects when measuring an object's length, due to the time it takes for sound to propagate.
I told her that:
Doppler effect is about alteration of the perceived frequency of a
signal produced by the relative movement between transmitter and
receiver. The quoted video is about relativity, which is a "deeper" effect.
Maybe doppler effect can be understood as the effect of relativity on
a wave phenomena.
Now I'm wondering about her intuition. If she's right, should I be able to take a sin function, apply a Lorentz transform to it, and arrive to same results as with the doppler formula? Unfortunately, the maths are beyond my skills.
Can someone shed some light about the relation between doppler and relativity, if any? Can be doppler effect explained by relativity/Lorentz alone?
special-relativity waves doppler-effect
special-relativity waves doppler-effect
New contributor
New contributor
edited 19 mins ago
Chair
4,39072241
4,39072241
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asked 7 hours ago
jjmontesjjmontes
1214
1214
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Related: physics.stackexchange.com/q/61946
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– Chair
18 mins ago
add a comment |
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Related: physics.stackexchange.com/q/61946
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– Chair
18 mins ago
$begingroup$
Related: physics.stackexchange.com/q/61946
$endgroup$
– Chair
18 mins ago
$begingroup$
Related: physics.stackexchange.com/q/61946
$endgroup$
– Chair
18 mins ago
add a comment |
2 Answers
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The ordinary Doppler effect is independent of relativity; it's basically just a fact of kinematics. It's not even really a wave phenomenon; it also applies to particles. For example, the Doppler effect explains why your car windshield gets wetter faster when you're driving than when you're parked.
The formula for the Doppler effect is
$$f_o = frac{v - v_o}{v - v_s} f_s$$
where $f_o$ is the observed frequency, $f_s$ is the source's emitted frequency, and $v_0$ and $v_s$ are the velocities of the observer and source. These are absolute velocities; they have to be defined with respect to the medium, e.g. the air for a sound wave. Relativity adds a correction to this formula because both the source and the observer will experience time dilation, so we should really have
$$gamma_0 f_0 = frac{v - v_o}{v - v_s} gamma_s f_s.$$
This is a very small correction assuming the speeds are small.
When people talk about the relativistic Doppler effect, they usually mean the Doppler effect for light waves specifically, with full relativistic corrections. Light waves are exceptional because they have no medium, so we aren't tied to a specific frame. It's instead more convenient to go to the observer's frame, where we naively have
$$f_o = frac{c - v_r}{c} f_s$$
where $v_r$ is the relative velocity. Relativity corrects this formula in two ways. First, velocities don't quite add linearly, so $v_r neq v_o - v_s$ in general. Second, we have to remember the time dilation factor for the source,
$$f_o = frac{c - v_r}{c} gamma_s f_s = sqrt{frac{1 - v_r/c}{1 + v_r/c}} f_s.$$
There is no time dilation factor for the observer, because we're in the observer's frame, where they are at rest. This last formula is what people usually call "the relativistic Doppler effect", but again it's pretty close to the nonrelativistic result as long as $v_r ll c$.
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add a comment |
$begingroup$
There is a Doppler effect even without Special or General Relativity, just arising from Galilean relative motion. For example, neither of these theories is necessary to explain the fact that the pitch of an ambulance siren changes as it passes by.
However, relativity does have to be taken into account when calculating the Doppler effect for a fast-moving object or one in a strong gravitational field. In other words, there are relativistic corrections to the Doppler effect.
If you use a Lorentz transformation to derive the Doppler effect, you will get the right answer for any velocity, but you won’t get the Doppler effect for a gravitational field.
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add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
The ordinary Doppler effect is independent of relativity; it's basically just a fact of kinematics. It's not even really a wave phenomenon; it also applies to particles. For example, the Doppler effect explains why your car windshield gets wetter faster when you're driving than when you're parked.
The formula for the Doppler effect is
$$f_o = frac{v - v_o}{v - v_s} f_s$$
where $f_o$ is the observed frequency, $f_s$ is the source's emitted frequency, and $v_0$ and $v_s$ are the velocities of the observer and source. These are absolute velocities; they have to be defined with respect to the medium, e.g. the air for a sound wave. Relativity adds a correction to this formula because both the source and the observer will experience time dilation, so we should really have
$$gamma_0 f_0 = frac{v - v_o}{v - v_s} gamma_s f_s.$$
This is a very small correction assuming the speeds are small.
When people talk about the relativistic Doppler effect, they usually mean the Doppler effect for light waves specifically, with full relativistic corrections. Light waves are exceptional because they have no medium, so we aren't tied to a specific frame. It's instead more convenient to go to the observer's frame, where we naively have
$$f_o = frac{c - v_r}{c} f_s$$
where $v_r$ is the relative velocity. Relativity corrects this formula in two ways. First, velocities don't quite add linearly, so $v_r neq v_o - v_s$ in general. Second, we have to remember the time dilation factor for the source,
$$f_o = frac{c - v_r}{c} gamma_s f_s = sqrt{frac{1 - v_r/c}{1 + v_r/c}} f_s.$$
There is no time dilation factor for the observer, because we're in the observer's frame, where they are at rest. This last formula is what people usually call "the relativistic Doppler effect", but again it's pretty close to the nonrelativistic result as long as $v_r ll c$.
$endgroup$
add a comment |
$begingroup$
The ordinary Doppler effect is independent of relativity; it's basically just a fact of kinematics. It's not even really a wave phenomenon; it also applies to particles. For example, the Doppler effect explains why your car windshield gets wetter faster when you're driving than when you're parked.
The formula for the Doppler effect is
$$f_o = frac{v - v_o}{v - v_s} f_s$$
where $f_o$ is the observed frequency, $f_s$ is the source's emitted frequency, and $v_0$ and $v_s$ are the velocities of the observer and source. These are absolute velocities; they have to be defined with respect to the medium, e.g. the air for a sound wave. Relativity adds a correction to this formula because both the source and the observer will experience time dilation, so we should really have
$$gamma_0 f_0 = frac{v - v_o}{v - v_s} gamma_s f_s.$$
This is a very small correction assuming the speeds are small.
When people talk about the relativistic Doppler effect, they usually mean the Doppler effect for light waves specifically, with full relativistic corrections. Light waves are exceptional because they have no medium, so we aren't tied to a specific frame. It's instead more convenient to go to the observer's frame, where we naively have
$$f_o = frac{c - v_r}{c} f_s$$
where $v_r$ is the relative velocity. Relativity corrects this formula in two ways. First, velocities don't quite add linearly, so $v_r neq v_o - v_s$ in general. Second, we have to remember the time dilation factor for the source,
$$f_o = frac{c - v_r}{c} gamma_s f_s = sqrt{frac{1 - v_r/c}{1 + v_r/c}} f_s.$$
There is no time dilation factor for the observer, because we're in the observer's frame, where they are at rest. This last formula is what people usually call "the relativistic Doppler effect", but again it's pretty close to the nonrelativistic result as long as $v_r ll c$.
$endgroup$
add a comment |
$begingroup$
The ordinary Doppler effect is independent of relativity; it's basically just a fact of kinematics. It's not even really a wave phenomenon; it also applies to particles. For example, the Doppler effect explains why your car windshield gets wetter faster when you're driving than when you're parked.
The formula for the Doppler effect is
$$f_o = frac{v - v_o}{v - v_s} f_s$$
where $f_o$ is the observed frequency, $f_s$ is the source's emitted frequency, and $v_0$ and $v_s$ are the velocities of the observer and source. These are absolute velocities; they have to be defined with respect to the medium, e.g. the air for a sound wave. Relativity adds a correction to this formula because both the source and the observer will experience time dilation, so we should really have
$$gamma_0 f_0 = frac{v - v_o}{v - v_s} gamma_s f_s.$$
This is a very small correction assuming the speeds are small.
When people talk about the relativistic Doppler effect, they usually mean the Doppler effect for light waves specifically, with full relativistic corrections. Light waves are exceptional because they have no medium, so we aren't tied to a specific frame. It's instead more convenient to go to the observer's frame, where we naively have
$$f_o = frac{c - v_r}{c} f_s$$
where $v_r$ is the relative velocity. Relativity corrects this formula in two ways. First, velocities don't quite add linearly, so $v_r neq v_o - v_s$ in general. Second, we have to remember the time dilation factor for the source,
$$f_o = frac{c - v_r}{c} gamma_s f_s = sqrt{frac{1 - v_r/c}{1 + v_r/c}} f_s.$$
There is no time dilation factor for the observer, because we're in the observer's frame, where they are at rest. This last formula is what people usually call "the relativistic Doppler effect", but again it's pretty close to the nonrelativistic result as long as $v_r ll c$.
$endgroup$
The ordinary Doppler effect is independent of relativity; it's basically just a fact of kinematics. It's not even really a wave phenomenon; it also applies to particles. For example, the Doppler effect explains why your car windshield gets wetter faster when you're driving than when you're parked.
The formula for the Doppler effect is
$$f_o = frac{v - v_o}{v - v_s} f_s$$
where $f_o$ is the observed frequency, $f_s$ is the source's emitted frequency, and $v_0$ and $v_s$ are the velocities of the observer and source. These are absolute velocities; they have to be defined with respect to the medium, e.g. the air for a sound wave. Relativity adds a correction to this formula because both the source and the observer will experience time dilation, so we should really have
$$gamma_0 f_0 = frac{v - v_o}{v - v_s} gamma_s f_s.$$
This is a very small correction assuming the speeds are small.
When people talk about the relativistic Doppler effect, they usually mean the Doppler effect for light waves specifically, with full relativistic corrections. Light waves are exceptional because they have no medium, so we aren't tied to a specific frame. It's instead more convenient to go to the observer's frame, where we naively have
$$f_o = frac{c - v_r}{c} f_s$$
where $v_r$ is the relative velocity. Relativity corrects this formula in two ways. First, velocities don't quite add linearly, so $v_r neq v_o - v_s$ in general. Second, we have to remember the time dilation factor for the source,
$$f_o = frac{c - v_r}{c} gamma_s f_s = sqrt{frac{1 - v_r/c}{1 + v_r/c}} f_s.$$
There is no time dilation factor for the observer, because we're in the observer's frame, where they are at rest. This last formula is what people usually call "the relativistic Doppler effect", but again it's pretty close to the nonrelativistic result as long as $v_r ll c$.
edited 6 hours ago
answered 7 hours ago
knzhouknzhou
45k11122218
45k11122218
add a comment |
add a comment |
$begingroup$
There is a Doppler effect even without Special or General Relativity, just arising from Galilean relative motion. For example, neither of these theories is necessary to explain the fact that the pitch of an ambulance siren changes as it passes by.
However, relativity does have to be taken into account when calculating the Doppler effect for a fast-moving object or one in a strong gravitational field. In other words, there are relativistic corrections to the Doppler effect.
If you use a Lorentz transformation to derive the Doppler effect, you will get the right answer for any velocity, but you won’t get the Doppler effect for a gravitational field.
$endgroup$
add a comment |
$begingroup$
There is a Doppler effect even without Special or General Relativity, just arising from Galilean relative motion. For example, neither of these theories is necessary to explain the fact that the pitch of an ambulance siren changes as it passes by.
However, relativity does have to be taken into account when calculating the Doppler effect for a fast-moving object or one in a strong gravitational field. In other words, there are relativistic corrections to the Doppler effect.
If you use a Lorentz transformation to derive the Doppler effect, you will get the right answer for any velocity, but you won’t get the Doppler effect for a gravitational field.
$endgroup$
add a comment |
$begingroup$
There is a Doppler effect even without Special or General Relativity, just arising from Galilean relative motion. For example, neither of these theories is necessary to explain the fact that the pitch of an ambulance siren changes as it passes by.
However, relativity does have to be taken into account when calculating the Doppler effect for a fast-moving object or one in a strong gravitational field. In other words, there are relativistic corrections to the Doppler effect.
If you use a Lorentz transformation to derive the Doppler effect, you will get the right answer for any velocity, but you won’t get the Doppler effect for a gravitational field.
$endgroup$
There is a Doppler effect even without Special or General Relativity, just arising from Galilean relative motion. For example, neither of these theories is necessary to explain the fact that the pitch of an ambulance siren changes as it passes by.
However, relativity does have to be taken into account when calculating the Doppler effect for a fast-moving object or one in a strong gravitational field. In other words, there are relativistic corrections to the Doppler effect.
If you use a Lorentz transformation to derive the Doppler effect, you will get the right answer for any velocity, but you won’t get the Doppler effect for a gravitational field.
edited 7 hours ago
answered 7 hours ago
G. SmithG. Smith
8,98611427
8,98611427
add a comment |
add a comment |
jjmontes is a new contributor. Be nice, and check out our Code of Conduct.
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Related: physics.stackexchange.com/q/61946
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– Chair
18 mins ago