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Doppler Effect

When a wave is emitted, like a sound wave or a light wave, it has an associated wavelength and frequency, which are inversely related. The frequency (or the wavelength) is related to the energy a wave carries. The higher the frequency (the shorter the wavelength), the higher the energy. If the source of the wave or whoever is receiving the…

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Doppler Effect

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Jetzt kostenlos anmeldenWhen a wave is emitted, like a sound wave or a light wave, it has an associated **wavelength and frequency**, which are inversely related. The frequency (or the wavelength) is related to the energy a wave carries. The higher the frequency (the shorter the wavelength), the higher the energy. If the source of the wave or whoever is receiving the wave is moving, the measured frequency/wavelength will change. This is known as the **Doppler effect**.

The **Doppler effect** is the **measured change in the frequency of a wave** when the source that emits the wave moves relative to the observer.

The Doppler effect is only an **apparent shift**: the wave does not change its wavelength when measured in its own frame of reference.

- In
**sound**, the Doppler effect causes a**shift in the pitch of the sound**. - In
**light**, the Doppler effect causes a**shift in the perceived colour**.

We can perceive the Doppler effect with ambulance sirens. If you’re standing on a street and see an ambulance coming closer, you’ll notice an increase in the pitch of the ambulance siren.

Once the ambulance passes you, you’ll notice the pitch gets lower and lower as it moves further away from you.

This apparent change only occurs in **your frame of reference** with respect to the moving ambulance.

We can calculate the perceived frequency of waves due to the Doppler effect if we know the frequency of the wave in its frame of reference (f_{s }), the speed of the wave (v), the speed of the observer (v_{o}), and the speed of the object emitting the waves (v_{s}). The equation is the following:

\[f = \frac{v \pm v_0}{v \pm v_s} \cdot f_s\]

The frequency is measured in Hertz (Hz) and the speed in m/s. The signs depend on the relative orientation of the speeds of the observer and emitter with respect to the wave.

- If the
**observer**travels**towards the emitter**, its**speed**will be**positive**, and if the observer**goes away from the emitter**, its speed will be**negative**. - If the
**emitter**travels**towards the observer**, its**speed**will be**negative**, and if the emitter**travels away from the observer**, its speed will be**positive**.

An observer with a speed of 8000m/s is moving towards a wave emitter. He observes an object emitting a wave with a speed of 500m/s. The wave has a frequency in its own frame of reference of 900Hz, and the object moves closer to the observer with a velocity of 100m/s. What is the perceived frequency of the wave for the observer moving closer to the source?

All we have to do is apply the Doppler effect equation:

\[f = \frac{v \pm v_0}{v \pm v_s} \cdot f_s = \frac{500m/s + 8000m/s}{500 m/s - 100m/s} = 900 Hz = 19,125 Hz\]

When moving at speeds close to the **speed of light in a vacuum**, two important effects will occur:

- The lengths observed by the moving observer will contract.
- The time will dilate for the moving observer.

Any measured frequency is a measure of time, and any movement at speeds closer to the speed of light implies a change in position. As a result, the frequency of any wave observed when moving at relativistic speeds will be different from the ones calculated with the classical formula. In this case, we need a term that adds these effects to our equation. We can do this by using the **Lorentz transformations**.

The **Lorentz transformations** are a set of equations that map the geometrical coordinates (x,y,z) and time (t) from one space system (A) to another (B). In this case, B moves at a certain constant velocity observed from A.

The coordinates in system B observed from system A will be altered. If we make system B move in the x-direction away from A, the coordinates of z and y will not vary with respect to A.

Coordinate x, however, will change as it varies its position at a velocity of v as time t increases. In relativity, this happens at the speed of light, and time also dilates. Here are the equations to calculate the lengths and the time as seen from one frame when knowing the values as seen from the other.

\[L = L_0 \cdot \sqrt{1 -\frac{v^2}{c^2}}\]

\[\Delta t' = \frac{\Delta t}{\sqrt{1- \frac{v^2}{c^2}}}\]

Because of the change of length and time, any wave emitted while the observer or emitter moves at speeds close to relativistic speeds will present an extra deformation of its frequency. In these equations, L_{o} and Δt represent the original time and length, v is the observer’s velocity, and c is the speed of light.

When we take these effects into account, we get the **relativistic Doppler effect** equation:

\[f_{obs} = f_s \cdot \frac{\sqrt{1 - \frac{v^2}{c^2}}}{1-\frac{v}{c}}\]

Here, f_{obs} is the observed frequency, f_{s} is the source frequency, c is the speed of light in a vacuum, and v is the observed speed.

Objects emitting light with a certain frequency *f _{s}* will experience a frequency shift (the same as with sound). If the

Let’s compare the difference between the classical and the relativistic approach with the following example.

An observer moves at 80% of the speed of light in a vacuum in the direction of a star. The star is a giant blue in the constellation of Orion known as Rigel. The star has a radiation peak of 2.06 petaherz (PHz). The light velocity is approximated as 300,000,000m/s.

**Classical equation**

** **

Calculate the frequency observed by the observer moving closer to the star with the classical equation.

\[f = \frac{c+v_0}{c-v_s} \cdot f_s = \frac{3 \cdot 10^8m/s + 2.4 \cdot 10^8 m/s}{3 \cdot 10^8 m/s - 0 m/s}\cdot 2.06 \cdot 10^{15} Hz = 3.71 \cdot 10^{15} Hz\]

**Relatavistic equation**

Calculate the same using the relativistic formula:

\[f_{obs} = f_s \frac{\sqrt{1-\frac{v^2}{c^2}}}{1-\frac{v}{c}} = 2.06 \cdot 10^{15} Hz \cdot \frac{\sqrt{1 - \Big( \frac{2.4 \cdot 10^8 m/s}{3 \cdot 10^8 m/s} \Big)^2}}{1- \frac{2.4 \cdot 10^8m/s}{3 \cdot 10^8 m/s}} = 6.18 \cdot 10^{15} Hz\]

**The relativistic effects alter the perceived frequency by about 25%.**

When doing your calculations, don’t forget your conversions!

The American scientist Edwin Hubble observed that the light of faraway stars in other galaxies appears to be systematically **redshifted**. This redshift has two important characteristics:

- The red shift appears in all galaxies independent of the direction observed. This means galaxies are
**moving away from Earth**. - The
**greater the distance**to galaxies, the**more intense the redshift**(i.e. the faster they move away).

The observations made by Edwin Hubble help us to understand that **the universe is expanding**, which sheds some light on the history of the universe.

- The Doppler effect creates an apparent shift (Doppler shift) in the observed frequency of a wave.
- When the source and observer approach, there is a shift towards higher frequencies (shorter wavelengths), and when they move away, the opposite effect occurs.
- When objects move at relativistic speeds (close to the speed of light), the frequency shift still exists, but it acquires a different mathematical form.
- The differences between the classical Doppler effect and the relativistic Doppler effect are caused by the contraction of length and dilatation of time that appears when travelling at relativistic speeds.
- Edwin Hubble observed a red shift in faraway galaxies. This revealed that galaxies are moving and the universe is expanding.

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