What About if It Is Shot Directly Away From the Earth Again Relative to C

Special Relativity

226 Relativistic Addition of Velocities

Learning Objectives

Calculate relativistic velocity addition.
Explain when relativistic velocity add-on should exist used instead of classical add-on of velocities.
Summate relativistic Doppler shift.

The total velocity of a kayak, like this one on the Deerfield River in Massachusetts, is its velocity relative to the h2o also as the h2o's velocity relative to the riverbank. (credit: abkfenris, Flickr)

A man with oar in his hand is kayaking downstream in a shallow fast-flowing river.

If you've ever seen a kayak move down a fast-moving river, you know that remaining in the aforementioned place would be hard. The river current pulls the kayak along. Pushing the oars back confronting the water can motion the kayak forward in the h2o, but that but accounts for office of the velocity. The kayak's motion is an case of classical improver of velocities. In classical physics, velocities add together equally vectors. The kayak'south velocity is the vector sum of its velocity relative to the water and the h2o'south velocity relative to the riverbank.

Classical Velocity Add-on

For simplicity, we restrict our consideration of velocity addition to one-dimensional motion. Classically, velocities add like regular numbers in one-dimensional motility. (See (Figure).) Suppose, for example, a girl is riding in a sled at a speed i.0 m/south relative to an observer. She throws a snowball showtime frontward, and so backward at a speed of 1.5 m/s relative to the sled. We announce management with plus and minus signs in ane dimension; in this example, forwards is positive. Let $v$ be the velocity of the sled relative to the Globe, $u$ the velocity of the snowball relative to the Globe-bound observer, and $u\prime$ the velocity of the snowball relative to the sled.

Classical Velocity Addition

$\text{u=v+u}\prime$

Thus, when the daughter throws the snowball forrad, $u=1.0 m/s+1.5 m/s=2.5 m/s$ . It makes good intuitive sense that the snowball will head towards the World-bound observer faster, considering information technology is thrown forward from a moving vehicle. When the daughter throws the snowball backward, $u=1.0 m/s+\left(-1.5 m/s\right)=-0.5 m/s$ . The minus sign means the snowball moves away from the Earth-jump observer.

Relativistic Velocity Add-on

The second postulate of relativity (verified past extensive experimental observation) says that classical velocity add-on does not apply to light. Imagine a car traveling at dark along a direct road, as in (Effigy). If classical velocity addition applied to light, then the light from the car's headlights would arroyo the observer on the sidewalk at a speed $\text{u=v+c}$ . But we know that light volition move away from the car at speed $c$ relative to the commuter of the car, and calorie-free will move towards the observer on the sidewalk at speed $c$ , too.

According to experiment and the 2nd postulate of relativity, light from the car's headlights moves away from the car at speed $c$ and towards the observer on the sidewalk at speed $c$ . Classical velocity addition is non valid.

A car is moving towards right with velocity v. A boy standing on the side-walk observes the car. The velocity of light u primed is shown to be c as observed by the girl in the car and the velocity of light u is also c as observed by the boy.

Showing that the Speed of Light towards an Observer is Constant (in a Vacuum): The Speed of Light is the Speed of Light

Suppose a spaceship heading direct towards the World at half the speed of calorie-free sends a signal to the states on a laser-produced beam of low-cal. Given that the light leaves the send at speed $c$ as observed from the transport, calculate the speed at which it approaches the Earth.

A spacecraft is heading towards earth v equals zero point five zero zero times c. A laser beam from the ship travels towards the Earth with velocity c as shown by a vector. A second spaceship traveling away from the Earth. The velocity of the second ship and second laser are the same as the first, but in the opposite direction.

Strategy

Because the low-cal and the spaceship are moving at relativistic speeds, we cannot use simple velocity addition. Instead, we can determine the speed at which the light approaches the Earth using relativistic velocity add-on.

Solution

Identify the knowns. $\text{v=}0\text{.}\text{500}c$ ; $u\prime =c$
Identify the unknown. $u$
Cull the appropriate equation. $u=\frac{\mathrm{v+u}\prime }{1+\frac{vu\prime }{{c}^{2}}}$
Plug the knowns into the equation.
$\begin{array}{lll}u& =& \frac{\mathrm{v+u}\prime }{1+\frac{vu\prime }{{c}^{2}}}\\ & =& \frac{\text{0.500}c+c}{1+\frac{\left(\text{0.500}c\right)\left(c\right)}{{c}^{2}}}\\ & =& \frac{\left(\text{0.500}+1\right)c}{1+\frac{\text{0.500}{c}^{2}}{{c}^{2}}}\\ & =& \frac{\text{1.500}c}{1+\text{0.500}}\\ & =& \frac{\text{1.500}c}{\text{1.500}}\\ & =& c\end{array}$

Discussion

Relativistic velocity addition gives the right outcome. Low-cal leaves the send at speed $c$ and approaches the Earth at speed $c$ . The speed of light is independent of the relative motion of source and observer, whether the observer is on the ship or Earth-bound.

Velocities cannot add together to greater than the speed of light, provided that $v$ is less than $c$ and $u\prime$ does not exceed $c$ . The following example illustrates that relativistic velocity addition is not as symmetric as classical velocity add-on.

Comparing the Speed of Light towards and away from an Observer: Relativistic Packet Delivery

Suppose the spaceship in the previous example is budgeted the Earth at half the speed of lite and shoots a canister at a speed of $0.750c$ . (a) At what velocity volition an Earth-bound observer see the canister if it is shot directly towards the Globe? (b) If information technology is shot directly away from the Earth? (See (Effigy).)

Strategy

Because the canister and the spaceship are moving at relativistic speeds, nosotros must make up one's mind the speed of the canister by an Earth-bound observer using relativistic velocity addition instead of simple velocity improver.

Solution for (a)

Identify the knowns. $\text{v=}0.500c$ ; $u\prime =0\text{.}\text{750}c$
Place the unknown. $u$
Choose the appropriate equation. $\text{u=}\frac{\mathrm{v+u}\prime }{1+\frac{vu\prime }{{c}^{2}}}$
Plug the knowns into the equation.
$\begin{array}{ll}u& =& \frac{\mathrm{v+u}\prime }{1+\frac{vu\prime }{{c}^{2}}}\\ & =& \frac{0.500\text{c +}0.750c}{1+\frac{\left(0.500c\right)\left(0.750c\right)}{{c}^{2}}}\\ & =& \frac{1.250c}{1+0.375}\\ & =& 0.909c\end{array}$

Solution for (b)

Place the knowns. $\text{v}=0.500c$ ; $u\prime =-0.750c$
Identify the unknown. $u$
Choose the appropriate equation. $\text{u}=\frac{\mathrm{v+u}\prime }{1+\frac{v\text{u}\prime }{{c}^{2}}}$
Plug the knowns into the equation.
$\begin{array}{ll}u& =& \frac{\mathrm{v+u}\prime }{1+\frac{vu\prime }{{c}^{2}}}\\ & =& \frac{0.500\mathrm{c +}\left(-0.750c\right)}{1+\frac{\left(0.500c\right)\left(-0.750c\right)}{{c}^{2}}}\\ & =& \frac{-0.250c}{1-0.375}\\ & =& -0.400c\end{array}$

Discussion

The minus sign indicates velocity abroad from the Earth (in the reverse direction from $v$ ), which ways the canister is heading towards the Globe in part (a) and away in office (b), as expected. But relativistic velocities practise not add as merely equally they exercise classically. In function (a), the canister does arroyo the Earth faster, merely non at the simple sum of $1.250c$ . The total velocity is less than y'all would get classically. And in part (b), the canister moves away from the World at a velocity of $-0.400c$ , which is faster than the $-0.250c$ you lot would expect classically. The velocities are not even symmetric. In part (a) the canister moves $0.409c$ faster than the ship relative to the World, whereas in part (b) it moves $0.900c$ slower than the ship.

Doppler Shift

Although the speed of lite does not change with relative velocity, the frequencies and wavelengths of calorie-free do. First discussed for sound waves, a Doppler shift occurs in any wave when there is relative motion between source and observer.

Relativistic Doppler Furnishings

The observed wavelength of electromagnetic radiation is longer (called a red shift) than that emitted by the source when the source moves away from the observer and shorter (chosen a bluish shift) when the source moves towards the observer.

${\text{=λ}}_{\text{obs}}{\text{=λ}}_{s}\sqrt{\frac{1+\frac{u}{c}}{1-\frac{u}{c}}}.$

In the Doppler equation, ${\lambda }_{\text{obs}}$ is the observed wavelength, ${\lambda }_{s}$ is the source wavelength, and $u$ is the relative velocity of the source to the observer. The velocity $u$ is positive for motility away from an observer and negative for movement toward an observer. In terms of source frequency and observed frequency, this equation can be written

${f}_{\text{obs}}{\text{=f}}_{s}\sqrt{\frac{1-\frac{u}{c}}{1+\frac{u}{c}}}.$

Notice that the – and + signs are different than in the wavelength equation.

Career Connection: Astronomer

If you are interested in a career that requires a noesis of special relativity, in that location's probably no better connection than astronomy. Astronomers must accept into account relativistic effects when they summate distances, times, and speeds of black holes, galaxies, quasars, and all other astronomical objects. To have a career in astronomy, y'all need at least an undergraduate degree in either physics or astronomy, simply a Master'south or doctoral degree is frequently required. You besides need a good background in high-level mathematics.

Computing a Doppler Shift: Radio Waves from a Receding Galaxy

Suppose a galaxy is moving away from the World at a speed $\text{0.825}c$ . It emits radio waves with a wavelength of $0\text{.}\text{525}\phantom{\rule{0.25em}{0ex}}\text{m}$ . What wavelength would we detect on the Earth?

Strategy

Because the galaxy is moving at a relativistic speed, we must determine the Doppler shift of the radio waves using the relativistic Doppler shift instead of the classical Doppler shift.

Solution

Identify the knowns. $\text{u=}0\text{.}\text{825}c$ ; ${\lambda }_{s}=0\text{.}\text{525}\phantom{\rule{0.25em}{0ex}}m$
Place the unknown. ${\lambda }_{\text{obs}}$
Cull the advisable equation. ${\lambda }_{\text{obs}}{\text{=λ}}_{s}\sqrt{\frac{1+\frac{u}{c}}{1-\frac{u}{c}}}$
Plug the knowns into the equation.
$\begin{array}{lll}{\lambda }_{\text{obs}}& =& {\text{λ}}_{s}\sqrt{\frac{1+\frac{u}{c}}{1-\frac{u}{c}}}\\ & =& \left(0.525 m\right)\sqrt{\frac{1+\frac{0\text{.}\text{825}\text{c}}{c}}{1-\frac{0\text{.}\text{825}\text{c}}{c}}}\\ & =& \text{1.70 m.}\end{array}$

Discussion

Considering the galaxy is moving abroad from the World, nosotros expect the wavelengths of radiation it emits to be redshifted. The wavelength nosotros calculated is i.70 m, which is redshifted from the original wavelength of 0.525 one thousand.

The relativistic Doppler shift is easy to observe. This equation has everyday applications ranging from Doppler-shifted radar velocity measurements of transportation to Doppler-radar tempest monitoring. In astronomical observations, the relativistic Doppler shift provides velocity information such every bit the motion and altitude of stars.

Bank check Your Understanding

Suppose a infinite probe moves away from the Globe at a speed $0\text{.}\text{350}c$ . It sends a radio wave message dorsum to the World at a frequency of 1.50 GHz. At what frequency is the message received on the World?

Answer

${f}_{\text{obs}}{\text{=f}}_{s}\sqrt{\frac{1-\frac{u}{c}}{1+\frac{u}{c}}}=\left(1\text{.}\text{50 GHz}\right)\sqrt{\frac{1-\frac{0\text{.}\text{350}\text{c}}{c}}{1+\frac{0\text{.}\text{350}\text{c}}{c}}}=1\text{.}\text{04 GHz}$

Department Summary

Conceptual Questions

Explain the meaning of the terms "red shift" and "blue shift" as they relate to the relativistic Doppler effect.

What happens to the relativistic Doppler outcome when relative velocity is zero? Is this the expected event?

Is the relativistic Doppler event consistent with the classical Doppler effect in the respect that ${\lambda }_{\text{obs}}$ is larger for motion away?

All galaxies farther away than almost $\text{50}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{ly}$ showroom a red shift in their emitted lite that is proportional to altitude, with those farther and farther away having progressively greater red shifts. What does this imply, assuming that the only source of red shift is relative motion? (Hint: At these large distances, information technology is space itself that is expanding, but the effect on light is the same.)

Issues & Exercises

Suppose a spaceship heading directly towards the Earth at $0\text{.}\text{750}c$ tin can shoot a canister at $0\text{.}\text{500}c$ relative to the ship. (a) What is the velocity of the canister relative to the Earth, if information technology is shot directly at the Earth? (b) If information technology is shot directly away from the Earth?

(a) $0\text{.}\text{909}c$

(b) $0\text{.}\text{400}c$

Repeat the previous problem with the transport heading directly away from the Earth.

If a spaceship is approaching the World at $0.100c$ and a bulletin capsule is sent toward it at $0.100c$ relative to the World, what is the speed of the sheathing relative to the ship?

$0\text{.}\text{198}c$

If a galaxy moving away from the Earth has a speed of $1000 km/s$ and emits $\text{656 nm}$ lite characteristic of hydrogen (the most common element in the universe). (a) What wavelength would nosotros observe on the Earth? (b) What blazon of electromagnetic radiation is this? (c) Why is the speed of the Earth in its orbit negligible here?

a) $\text{658 nm}$

b) red

c) $v/\text{c}=9\text{.}\text{92}×{\text{10}}^{-5}$ (negligible)

A infinite probe speeding towards the nearest star moves at $0\text{.}\text{250}c$ and sends radio information at a broadcast frequency of 1.00 GHz. What frequency is received on the Earth?

If two spaceships are heading straight towards each other at $0\text{.}\text{800}c$ , at what speed must a canister be shot from the first send to approach the other at $0\text{.}\text{999}c$ as seen by the 2d transport?

$0\text{.}\text{991}c$

Two planets are on a collision course, heading straight towards each other at $0\text{.}\text{250}c$ . A spaceship sent from one planet approaches the second at $0\text{.}\text{750}c$ equally seen by the second planet. What is the velocity of the ship relative to the offset planet?

When a missile is shot from one spaceship towards another, it leaves the first at $0\text{.}\text{950}c$ and approaches the other at $0\text{.}\text{750}c$ . What is the relative velocity of the two ships?

$-0\text{.}\text{696}c$

What is the relative velocity of two spaceships if 1 fires a missile at the other at $0.750c$ and the other observes it to approach at $0.950c$ ?

Near the center of our galaxy, hydrogen gas is moving directly away from us in its orbit virtually a blackness hole. We receive 1900 nm electromagnetic radiation and know that it was 1875 nm when emitted by the hydrogen gas. What is the speed of the gas?

$0\text{.}\text{01324}c$

A highway patrol officer uses a device that measures the speed of vehicles by bouncing radar off them and measuring the Doppler shift. The outgoing radar has a frequency of 100 GHz and the returning echo has a frequency fifteen.0 kHz higher. What is the velocity of the vehicle? Note that there are two Doppler shifts in echoes. Exist certain not to round off until the end of the trouble, because the effect is pocket-size.

Prove that for any relative velocity $v$ between two observers, a beam of light sent from ane to the other volition approach at speed $c$ (provided that $v$ is less than $c$ , of course).

$u\prime \phantom{\rule{0.25em}{0ex}}=c$ , so

$\begin{array}{ll}u& =& \frac{\text{v+u}\prime }{1+\left(\text{vu}\prime /{c}^{2}\right)}=\frac{\text{v+c}}{1+\left(\text{vc}/{c}^{2}\right)}=\frac{\text{v+c}}{1+\left(v/c\right)}\\ & =& \frac{c\left(\text{v+c}\right)}{\text{c+v}}=c\end{array}$

(a) All just the closest galaxies are receding from our own Milky way Galaxy. If a milky way $\text{12}\text{.}0×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{ly}$ ly away is receding from the states at 0. $0.900c$ , at what velocity relative to us must we ship an exploratory probe to approach the other galaxy at $0.990c$ , as measured from that galaxy? (b) How long will it accept the probe to accomplish the other milky way as measured from the Earth? You may presume that the velocity of the other galaxy remains constant. (c) How long will it then accept for a radio signal to be beamed back? (All of this is possible in principle, but non practical.)

a) $0\text{.}\text{99947}c$

b) $1\text{.}\text{2064}×{\text{10}}^{\text{11}}\phantom{\rule{0.25em}{0ex}}\text{y}$

c) $1\text{.}\text{2058}×{\text{10}}^{\text{11}}\phantom{\rule{0.25em}{0ex}}\text{y}$ (all to sufficient digits to show effects)

woodthicanat79.blogspot.com

Source: https://opentextbc.ca/openstaxcollegephysics/chapter/relativistic-addition-of-velocities/

What About if It Is Shot Directly Away From the Earth Again Relative to C

226 Relativistic Addition of Velocities

Learning Objectives

Classical Velocity Add-on

Relativistic Velocity Add-on

Doppler Shift

Department Summary

Conceptual Questions

Issues & Exercises

0 Response to "What About if It Is Shot Directly Away From the Earth Again Relative to C"

Post a Comment

Iklan Atas Artikel

Iklan Tengah Artikel 1

Iklan Tengah Artikel 2

Iklan Bawah Artikel