sorular ilk konu

========================================================================================

Adjacent dots on the screen may light up as if the dot had moved from one position to the next at a speed greater than the speed of light, but in fact, no dot moved. Each lit dot is produced by a different electron.

The image of a moving object created by the successive impacts of the electrons on the screen is not an object. The information carried by the electrons moves from the electron gun to the screen at a speed below that of light.

It is possible for the electron beam in a television picture tube to move across the screen at a speed faster than the speed of light. Why does this not contradict special relativity?

Problem 1.2Physics 107

========================================================================================

When G(v) is "near" 1, relativistic effects are not obvious. When G(v) deviates significantly from 1, relativistic effects are more obvious.

Notice the rapid dropoff in GG(v)Notice how G(v) falls rather slowly.

0 1 .108 2 .108 3 .1080

0.5

1

GG v( )

v0 1 .108 2 .108 3 .108

0

0.5

1

G v( )

v

Plot:Plot:

GG v( ) 1v

2

cc2

−:=G v( ) 1v

2

c2

−:=

v 0 c⋅ 0.01 c⋅, c..:=

cc 3 107

⋅:=c 3 108

⋅:=

Set the range of v:Set the range of v:

This column is for c=3x107.This column is for c=3x108.

If c were smaller, then G(v) would differ from unity at much lower velocities v thus making relativistic effects more conspicuous than they are now. Let's plot G(v) as a function of v/c to see the effect.

The square box at the end of this equation indicates it is just text; i.e., not "live."G v( ) 1v

2

c2

−≡

All of the phenomena of special relativity depend upon the factor G(v):

If the speed of light were smaller than it is, would relativistic phenomena be more or less conspicuous than they are now?


1

========================================================================================

Something to think about: what time would an observer who remained stationary on earth measure?

t0 28.566=

t0 t 1v

c

2−⋅:=

t 40:=v 0.7 c⋅:=c 1:=

Assign values and solve:

t0 t 1v

c

2−⋅:=

tt0

1v

c

2−

:=

Solution: we apply the time dilation equation. An observer on the spacecraft measures a dilated time t = 40 min (why is it the dilated time?). We need to calculate the proper time, as measured by the driver of the car.

An observer on a spacecraft moving at 0.7c relative to the earth finds that a car takes 40 minutes to make a trip. How long does the trip take to the driver of the car?


========================================================================================

This answer ignores the complication of length contraction, which we will get into later in this chapter. Attempting to account for length contraction would introduce the problem of simultaneity (see the pole-barn paradox in the class notes). The time measured by the spacecraft observer would not be useable by an earth observer.

If the observer in the spacecraft times the run by watching a clock in the spacecraft, it appears as if the clock on the earth ran slow, so that, in fact, MORE time elapsed during the run. The spacecraft observer would actually measure a longer time.

If the observer in the spacecraft times the run by watching a clock on earth, nothing is gained because the clock and athlete are in the same reference frame (the athlete's speed is so small compared to c that we can ignore his motion relative to the clock).

Is it a good idea for an athlete trying to set a world record 100-m dash tie to have his time taken by an observer on a moving spacecraft?


2

========================================================================================

Solution (b): According to B, A is in motion relative to B. Moving watches (relative to the observer) always run slow. Therefore, B claims A's watch runs slow.

secondst 3.927=

t1

1 Γ−:=Calculated t:

Γ 1v

c

2−:=

v 200000000:=

Or I could say c=1 and v=2c/3.

c 300000000:=

Assign values to parameters:

t1

1 Γ−:=Result:

t Γ⋅ t− 1−:=t Γ⋅ t 1−:=tt 1−

Γ:=

Solve the above two equations for t (Let Γ be the square root):

t t0 1+:=andt

t0

1v

c

2−

:=

The two equations we need to solve are the time dilation equation and the relationship t=t0+1.

Solution (a): by A's reckoning, B's watch runs slow. Suppose A has two identical watches. Watch 1 is used as the timer and watch 2 is used to provide a time interval. If both watches are in A's frame of reference, watch 1 will measure the proper time t0 for the time interval. If A gives watch 2 to B, and B moves relative to A, A can use watch 1 to measure the time t it takes for watch 2 to tick off the time interval. Since the moving watch 2 ticks more slowly (according to A, who is doing the measuring), A's watch 1 must record a longer time for watch 2 to indicate the same time interval that it indicated when it was in A's reference frame.Thus, when the watches differ by 1 s, A's watch 1, which is being used to measure the time t, has ticked 1 more second. Thus, t=t 0+1; t is always greater than t0.

Two observes, A on earth and B in a spacecraft whose speed is 2x108 m/s, both set their watches to the same time when the ship is abreast of the earth. (a) How much time must elapse by A's reckoning before the watches differ by 1 s? (b) To A, B's watch seems to run slow. To bB, does A's watch seem to run fast, slow, or keep the same time as his own watch?


3

========================================================================================

Or v=0.866 c, because I let c=1.v 0.866=

v c 1t0

t

2

−⋅:=

t 2:=t0 1:=c 1:=

Define the values and plug them in:

v c 1t0

t

2

−⋅:=1

v

c

2−

t0

t

2

:=t

t0

1v

c

2−

:=

Solution: an observer on the spacecraft measures a proper time t0=1 d, and a dilated time t=2 d for the same event as it takes place on the "moving" earth. We solve the time dilation equation for v.

How fast must a spacecraft travel relative to the earth for each day on the spacecraft to correspond to 2 d on the earth?


========================================================================================

You can use c=2.998x108 if you want and get a slightly different answer.

secondst 1.9998 1012

×=

t1

1 Γ−:=Calculate t:

Γ 1v

c

2−:=

v 300:=

c 300000000:=

Assign values to parameters:

t1

1 Γ−:=

As in problem 1.5,

Solution: this is just problem 1.5 with an airplane replacing a rocket ship. Just plug in the new numbers.

An airplane is flying at 300 m/s (672 mi/h). How much time must elapse before a clock in the airplane and one on the ground differ by 1 s?


4

seconds.

========================================================================================

Physics 107 Problem 1.9

A certain particle has a lifetime of 1x10-7 s when measured at rest. How far does it go before decaying if its speed is 0.99c when it is created?

Solution: our first job is to figure out that the problem is really asking us to calculate how far a human observer would observe this particle to travel.

The observer sees the particle moving at a speed of 0.99c, and sees the particles "clock" dilated according to equation 1.3, where t0 is the time the observer sees the particle in motion.

We need to calculate d=vt, where v=0.99c and t is given by eq. 1.3. I'm going to include units in this solution. You could append the file "units.mcd" to for use with this problem. Instead, I will do the units here for you to see.

Define units: m 1L≡ kg 1M≡ s 1T≡

Define parameters: c 3 108

⋅m

s⋅:= t0 10

7−s⋅:= v 0.99 c⋅:=

Pertinent equations:t

t0

1v

c

2−

:=d v t⋅:= d 210.5m=

========================================================================================


The Apollo 11 spacecraft that landed on the moon in 1969 traveled there at a speed relative to the earth of 1.08x104 m/s. To an observer on the earth, how much longer than his own day was a day on the spacecraft?

This is another time dilation problem. We are given a relative velocity. We are given a time interval t0 of one day on the spacecraft moving relative to an observer. We want to find the dilated time t measured by the observer.Let's define our variables first, so I don't have to make the time dilation equation into text and then later on re-enter it as an equation.

We don't need to worry about units here because velocities have same units.c 3 10

8⋅:= v 1.08 10

4⋅:=

t0 1:= The time here is in days, so my answer t will be in days. t

t0

1v

2

c2

−

:= This is the equation for t. Below I will type "t=" to see the answer.

There are not enough digits to show any effect. Below I will type "t=" again and then type "f" to allow me to show more digits.t 1=

t 1.000000000648= Click on the number and select Math, Numerical Format to see that I picked a precision of 12 to display t.

The problem asks how much longer t is than t, so I'd better calculate t-t0.

t t0− 6.48 1010−

×= days

or t t0−( ) 24⋅ 3600⋅ 5.599 105−

×=

5

L0L

1v

c

2−

:=L0 1.625= meters

========================================================================================


How much time does a meter stick moving at 0.1c relative to an observer take to pass the observer? The meter stick is parallel to its direction of motion.

The relative speed is v: c 3 108

⋅:= m/s v 0.1 c⋅:= L0 1:=

We need to solve for the contracted length L, and then determine how long it takes to travel this length at a speed of 0.1*c.

L L0 1v

c

2−⋅:= L 0.995=

The equation d=vt still works in relativity, so we can solve it for t:

tL

v:= t 3.317 10

8−×= seconds

========================================================================================


An astronaut whose height on the earth is exactly 6 ft is lying parallel to the axis of a spacecraft moving at 0.9c relative to the earth. What is his height as measured by an observer in the same spacecraft? By an observer on earth?

An observer in the same spacecraft, at rest relative to the astronaut, measures the proper length, 6 ft, of the astronaut.

If the astronaut were lying perpendicular to the velocity vector of the spacecraft, the observer on earth would also measure his proper length. But with the astronaut parallel to the direction of relative motion, the observer on earth measures a contracted length.

Define parameters, then do calculation.

L0 6:= c 1:= v 0.9 c⋅:=

L L0 1v

c

2−⋅:= L 2.615= feet

========================================================================================


An astronaut is standing in a spacecraft parallel to its direction of motion. An observer on the earth finds that the spacecraft speed is 0.6c and the astronaut is 1.3 m tall. What is the astronaut's height as measured in the spacecraft?Solution: this is just problem 1.17 but solving for L0 instead of L.

Define parameters:

L 1.3:= c 1:= v 0.6 c⋅:=

6

That last equation looks "circular" (we already know tan=sin/cos) except that we can use our original angle to calculate the numerator, and we can apply the length contraction equation to the denominator, because it represents the component along the direction of relative motion.

L sin θ( )⋅

L cos θ( )⋅:=tan θ( ) y

x:=

Let's let y be the projection of the antenna length perpendicular to the motion. When the spacecraft is in motion, the antenna has a length L, a projection x=L cos(θ) along the direction of motion, and a projection y=L sin(θ) perpendicular to the direction of motion. The angle θ is given by

except that you can't apply the length contraction formula to L0 in order to calculate L, because the antenna has a component of length perpendicular to the direction of motion which has not been contracted.

x L cos θ( )⋅:=

When the spacecraft is in motion, both L0 and x0 appear contracted to an observer on earth, and the angle can be found from

x0 L0 cos θ0( )⋅:=

Let L0 be the length of the antenna when the spacecraft is at rest, and x0 be the projection of the antenna parallel to the direction of spacecraft travel when the spacecraft is at rest. Then when the spacecraft is at rest, the angle of the antenna is found from

This problem is tricky because when the spacecraft moves, the projection of the antenna along the spacecraft contracts, which means that the apparent length of the antenna also contracts. You have to be careful to take both contractions into account. It also helps to have a drawing.

A spacecraft antenna is at an angle of 10 degrees relative to the axis of the spacecraft. If the spacecraft moves away from the earth at a speed of 0.7c, what is the angle as seen from the earth? This problem is not assigned or "testable" and the solution is included only for "interest."


========================================================================================

st0 1.92 10

9−×=


A meter stick moving with respect to an observer appears only 500 mm long to her. What is its relative speed? How long does it take to pass her? The meter stick is parallel to its direction of motion.

L0 1:= L 0.5:= c 3 108

⋅:= m/s

First we need to solve L L0 1v

c

2−⋅:= for relative speed v.

L

L0

21

v

c

2−:=

v

c

1L

L0

2−:= v c 1

L

L0

2−⋅:=

v 2.6 108

×= m/s

Time to pass observer: (I call it t0 because the event is timed in the observer's reference frame.

t0L

v:=

7

Let M be the initial mass of the ice, and let L be the latent heat of fusion, as above. The energy added to melt the ice is ML. The mass equivalent of this energy is given in the statement of the problem as m=1 kg.

It takes 80 calories to melt a gram of ice, and a calorie is equivalent to 4.19 joules, so in SI units, it takes L=335.2 joules to melt a gram of ice, or 3.35x105 joules to melt a kilogram of ice.

A certain quantity of ice at 0 degrees C melts into water at 0 degrees C and in so doing gains 1 kg of mass. What was its initial mass?


========================================================================================

f 6 1011−

×=f5.4 10

6⋅

c2

:=Then

Here's how the units work out: if you choose the mass to be 1 kg, the units on top are joules, and the m=1kg times c2 also gives joules in the denominator, so the result is a pure number, or fraction.

(SI units).c 3 108

⋅:=We need to usef5.4 10

6⋅

c2

:=

The total energy content of this kg of dynamite is mc2. The fraction is simply

Dynamite liberates about 5.4x106 J/kg when it explodes. What fraction of its total energy content is this?

tan θ( ) L0 sin θ0( )⋅

L0 cos θ0( )⋅( ) 1v

2

c2

−⋅

:= tan θ( ) L0 sin 10( )⋅

L0 cos 10( )⋅( ) 1v

2

c2

−⋅

:=

tan θ( ) tan 10( )

1v

2

c2

−

:= θ atantan 10( )

1v

2

c2

−

:=

To solve, plug in values, and remember to convert angles to radians when you are using Mathcad.

c 1:= v 0.7 c⋅:=θ0 10

2 π⋅

360⋅:=

θ atantan θ0( )

1v

2

c2

−

:= θ 0.242= radians

θ θ360

2 π⋅⋅:= θ 13.869= degrees

========================================================================================


8

divide by 2 and square both sides 0.25 1v

2

c2

−:= v2

0.75 c2

⋅:= v 0.866 c⋅:=

========================================================================================


How many joules of energy per kilogram of rest mass are needed to bring a spacecraft from rest to a speed of 0.9c?

Again, we use γ m⋅ c2

⋅ m c2

⋅ K+:=

At a speed of 0.9c, mm

1v

2

c2

−

:= where v/c=0.9. c 3 108

⋅:= m 1:= v 0.9 c⋅:=

Solving for K gives Km c

2⋅

1v

2

c2

−

m c2

⋅−:= K 1.165 1017

×= joules per kilogram.

========================================================================================

m c2

⋅ M L⋅:= L 3.352 105

⋅:=M m( )

m c2

⋅( )L

:= M 1( ) 2.685 1011

×= kg

Using the density of ice, you can calculate that this mass would require a block of ice 16.5 km long, 16.5 km wide, and 1 km high. (Or about 10 miles by 10 miles by 0.6 miles high, if I remember my conversion factors correctly).

Yes, ice really does gain mass when it melts. Or rather, the energy that "goes into" the ice is manifested as mass. Similarly, water would "lose mass" when it freezes. No, that's not the main reason why ice floats on water.

========================================================================================


At what speed does the kinetic energy of a particle equal its rest energy?

rest_energy m c2

⋅:= KE γ m⋅ c2

⋅ m c2

⋅−:=

When KE=rest_energy,

γ m⋅ c2

⋅ m c2

⋅− m c2

⋅:= γ m⋅ c2

⋅ 2 m⋅ c2

⋅:= γ m⋅ 2 m⋅:= I cancelled out the c2's.

m

1v

2

c2

−

2 m⋅:=Cancel the m's and re-arrange: 1 2 1

v2

c2

−⋅:=

9

c 3 108

⋅:=

If you start with γmc2=mc2+K and solve for v/c, you get an "unofficial" but extremely useful equation that is sure easier to use than solving "by hand" every time. Make sure this equation is on your 3x5 card.

vrel c 11

1Krel

m c2

⋅+

2−⋅:= vrel 1.644 10

8×= m/s

Let's put the two answers side-by-side for comparison: vcl 1.875 108

×= vrel 1.644 108

×=

Classically-calculated speeds are always too large, although the error is not significant for low speeds and low energies.

========================================================================================


A particle has a kinetic energy 20 times its rest energy. Find the speed of the particle in terms of c.

Be lazy and use our "unofficial" equation (it's unofficial because it is derived, not fundamental).

v K m,( ) c 11

1K

m c2

⋅+

2−⋅:=

v 20 m⋅ c2

⋅ m,( ) 2.997 108

×=

Expressed "in terms of c:" v 20 m⋅ c2

⋅ m,( )c

0.999= or v=0.999c

========================================================================================

========================================================================================


An electron has a kinetic energy of 0.1 MeV. Find its speed according to classical and relativistic mechanics.

Let's begin by converting MeV to mks units.

eV 1.602 1019−

⋅:= This converts eV to Joules MeV eV 106

⋅:= This converts eV to MeV

Classical calculation.

We use our familiar equations of classical mechanics. Kcl mvcl

2

2⋅:=

m 9.109 1031−

⋅:= kg Kcl 0.1 MeV⋅:=

vcl 2Kcl

m⋅:= vcl 1.875 10

8×= m/s

Relativistic calculation. Krel 0.1 MeV⋅:=

10

========================================================================================

Setting MeV=1 above let me stick in my expression for me so that it looked like a unit.

W 0.294 MeV⋅:=

W me c2

⋅1

12.4 10

8⋅( )2

c( )2

−

1

11.2 10

8⋅( )2

c( )2

−

−

⋅:=

me0.511 MeV⋅

c2

:=To get the work in units of MeV, I could calculate the numbers out and convert joules to eV to MeV. Or I could be clever and use the electron mass in energy units.

MeV 1:=Remember Ef-Ei=[Wother]i-->f from Phys. 23?W E2 E1−:=

The work done on the electron is just the difference in the energies E at the two different velocities:

Em c

2⋅

1v

2

c2

−

:=E γ m⋅ c2

⋅:=

Solution: a body in motion has a total energy

How much work (in MeV) must be done to increase the speed of an electron from 1.2x108 m/s to 2.4x108 m/s?


========================================================================================

Mathcad made this solution very easy. If you're doing this "by hand," it will take several lines of algebra and computations.

Doubling the speed does not increase KE by the classically-expected factor of 4, and the discrepancy is larger as the speeds get greater.

K mproton 0.8 c⋅,( )K mproton 0.4 c⋅,( ) 7.319=

K mproton 0.4 c⋅,( )K mproton 0.2 c⋅,( ) 4.417=

mproton 1.67 1027−

⋅:=

We could do this algebraically, without using the proton mass, but here it is for those who like numbers:

K m v,( ) m c2

⋅1

1v

2

c2

−

1−

⋅:=K m v,( )m c

2⋅

1v

2

c2

−

m c2

⋅−:=

Here's another handy "unofficial" variant of our KE equation:

The speed of a proton is increased from 0.2c to 0.4c. (a) By what factor does its kinetic energy increase? (b) The speed of the proton is again doubled, this time to 0.8c. By what factor does its kinetic energy increase now?


11

The E in here is total energy; you are given proton KE.Ephoton2

Eproton2

mproton c2

⋅

2−:=

this is p2c2 for photon, which is equal to p2c2 for photonp2

c2

⋅ Eproton2

mproton c2

⋅

2−:=

both proton and photon have same momentum pEproton2

p2

c2

⋅ mproton c2

⋅

2+:=

Ephoton2

p2

c2

⋅:=E2

p2

c2

⋅ m c2

⋅( )2+:=

What is the energy of a photon whose momentum is the same as that of a proton whose kinetic energy is 10 MeV?


========================================================================================

If the proton is moving with the speed of light, and the galaxy is 105 light years across, the observer will say it takes 105 years (not 5 minutes) for the proton to cross the galaxy. The observer say "proton, your time was way too short." The proton will say "observer, your distance across the galaxy was way too long.

Part (b). The proton's energy is about 1019 eV, or 1013 MeV. That's about 1010 times its rest energy of roughly 1000 MeV. This is one fast proton. We might as well use vproton=c. I'm guessing our answer might be off in the 10th decimal place or so. Double-check by calculating the proton's speed, if you don't believe me.

You could be still more clever and express the proton mass in energy units (938 MeV/c2) and save multiplying by c.

1.58

1.6 1019−

⋅9.875 10

18×=To get the energy in eV, use the conversion factor 1 eV=1.6x10-19 joules:

E=1.58 joulesE 1.67 1027−

⋅ 5 60⋅, 105

365⋅ 24⋅ 3600⋅,( ) 1.58=

E m L, L0,( )L0 m⋅ c

2⋅

L:=

E

m c2

⋅

L0

L:=

E

mc2

γ:=L0

Lγ:=

so that

Or we could be clever and note that

This equation lets us solve for v, and we could plug v into E=γmc2.L L0 1v

c

2−⋅:=

Part (a). The only way for the proton to "think" it crosses the galaxy in 5 minutes is for the proton to see the galaxy's length contracted:

In its own frame of reference, a proton takes 5 min to cross the Milky Way galaxy, which is about 105 light-years across. (a) What is the approximate energy of the proton in eV? (b) About how long would the proton take to cross the galaxy as measured by an observer in the galaxy's frame of reference?


========================================================================================

12

========================================================================================

See lecture notes lect03.ppt for better "pencil-and-paper" version.pelectron 0.383MeV

c⋅:=

This answer has the factor of 1/c numerically embedded in the result. To see the answer expressed in units of MeV/c, I have to multiply through by c. If you do this using pen and paper (see the student solutions manual, available on reserve at the library), you will get the textbook answer "automatically."

pelectron 1.278 109−

×=

pelectron1

1velectron

c

2

−

melectron⋅ velectron⋅:=

I have c defined above numerically as 3x108. When I combine symbolic and numerical work, like I do here, Mathcad actually divides by a c2 to calculate the mass of the electron. When I multiply by v=0.6c, Mathcad "puts back" one of those "two" c's I divided out. To get the answer in "energy units" I have to put the other factor of c back in "by hand." You won't encounter this confusion on an exam.

velectron 0.6 c⋅:=melectron 0.511MeV

c2

⋅:=

If I use mass in "energy units" of MeV/c2 and v in terms of c, the answer will come out in units of MeV/c, because γ is unitless. Or you can work the problem in SI units and convert to "energy units."

p γ m⋅ v⋅:=

Find the momentum (in MeV/c) of an electron whose speed is 0.6c.


========================================================================================

Ephoton 137 MeV⋅:=

Ephoton Kproton mproton c2

⋅+

2mproton c

2⋅

2−:=

mproton 938MeV

c2

⋅:=Kproton 10 MeV⋅:=

here are the specific values for this problem:

Ephoton2

Kproton mproton c2

⋅+

2mproton c

2⋅

2−:=

Eproton Kproton mproton c2

⋅+:=Kproton Eproton mproton c2

⋅−:=

13


Find the momentum of an electron whose kinetic energy equals its rest energy of 511 keV.

First I'll do the algebra.

E2

p2

c2

⋅ m c2

⋅( )2+:= p

2c2

⋅ E2

m c2

⋅( )2−:= K E m c

2⋅−:=

E K m c2

⋅+:= p2

c2

⋅ K m c2

⋅+( )2m c

2⋅( )2

−:= pK m c

2⋅+( )2

m c2

⋅( )2−

c:=

Now I'll plug in the numbers.

keV 1:= melectron 511keV

c2

⋅:= Kelectron 511 keV⋅:=

pelectronKelectron melectron c

2⋅+

2melectron c

2⋅

2−

c:=

pelectron 885keV

c⋅:=

========================================================================================


Find the total energy and kinetic energy (in GeV) and the momentum (in GeV/c) of a proton whose speed is 0.9c. The mass of the proton is 0.938 GeV/c2.

Let's do the momentum first, because it is "like" the previous calculation.

GeV 1:= This definition just lets me include "GeV" in a problem as if I were writing the units by hand.

mproton 0.938GeV

c2

⋅:= vproton 0.9 c⋅:=

pproton1

1vproton

c

2

−

mproton⋅ vproton⋅:=pproton 1.937

GeV

c⋅:=

Eproton2

pproton2

c2

⋅ mproton c2

⋅

2+:= Eproton pproton

2c2

⋅ mproton c2

⋅

2+:=

Eproton 2.152= GeV

I could have also used E=γmc2, but I already had the Eproton equation available for cut and paste from problem 42.

Kproton Eproton mproton c2

⋅−:= Kproton 1.214 GeV⋅:=

========================================================================================

14

p2

c2

⋅ E2

E02

−:= p2 E

2

c2

E02

c2

−:= pE

2

c2

E02

c2

−:=

p3.5 GeV⋅( )

2

c2

0.938 GeV⋅( )2

c2

−:= p 3.52

0.9382

−( ) GeV

c⋅:= p 3.37

GeV

c⋅:=

========================================================================================


Find the total energy of a neutron (m=0.940 GeV/c2) whose momentum is 1.2 GeV/c.

E2

1.2GeV

c⋅

2c2

⋅ 0.940GeV

c2

⋅ c2

⋅

2+:=

E2

p2

c2

⋅ m c2

⋅( )2+:=

E2

1.2 Gev⋅( )2

0.940 GeV⋅( )2

+:= E2

1.2( )2

0.940( )2

+ GeV2

⋅:=

E 1.2( )2

0.940( )2

+ GeV⋅:= E 1.52 GeV⋅:=

========================================================================================

========================================================================================


Find the speed and momentum (in GeV/c) of a proton whose total energy is 3.5 GeV.

First think: this proton's total energy is about 4 times its rest energy of 0.938 GeV, so the speed had better be close to c. Let's do the algebra first, working across the line and then down to save paper.

E γ m⋅ c2

⋅:= E γ E0⋅:= γE

E0:=

1

1v

2

c2

−

E

E0:=

1v

2

c2

−E0

E:=

1v

2

c2

−E0

E

2

:=v

2

c2

1E0

E

2

−:= v2

c2

1E0

E

2

−

⋅:=

v c 1E0

E

2

−⋅:= v c 10.938 GeV⋅

3.5 GeV⋅

2−⋅:= v 0.963 c⋅:=

To find the momentum, we can use this equation from the previous problem (no sense re-deriving here):

p2

c2

⋅ E2

m c2

⋅( )2−:=

15

========================================================================================

See the student solution manual for a really handy algebra trick that eliminates the potential extraneous root problem.

v 0.358 c⋅:=v c 11

162

874+

2−⋅:=

v c 11

162 MeV⋅

874MeV

c2

⋅ c2

⋅

+

2−⋅:=

Those of you who observant will notice I quit making my equations "live." This eliminates the glitch mentioned in problem 43. I'm just using Mathcad as a symbolic text processor now.

v c 11

1K

m c2

⋅+

2−⋅:=

Now that I have the mass (a bit less than that of a proton or neutron) I can use our handy unofficial equation introduced in problem 31. There are other ways to get v--see the student solution manual.

m 874MeV

c2

⋅:=m335

262

2−( ) MeV⋅

2 62⋅ c2

⋅:=m

3352

622

−( ) MeV2

⋅

2 62⋅ MeV⋅ c2

⋅:=

m

3352 MeV

2

c2

⋅ c2

⋅ 622

MeV2

⋅−

2 62⋅ MeV⋅ c2

⋅:=m

p2

c2

⋅ K2

−( )2 K⋅ c

2⋅

:=2 K⋅ m⋅ c2

⋅ p2

c2

⋅ K2

−:=

K2

2 K⋅ m⋅ c2

⋅+ p2

c2

⋅:=K2

2 K⋅ m⋅ c2

⋅+ m c2

⋅( )2+ p

2c2

⋅ m c2

⋅( )2+:=

K m c2

⋅+( )2p

2c2

⋅ m c2

⋅( )2+:=

Makes me a bit nervous; any time you square an expression, you run the risk of introducing an extraneous root. Let's see where this takes us anyway.

Put E from the first equation into the second equation.E2

p2

c2

⋅ m c2

⋅( )2+:=E K m c

2⋅+:=

Here are two equations that look like candidates to me:

You have two knowns and two unknowns. None of our equations contain one of the unknowns expressed only in terms of the knowns. Looks like we are going to have to solve a system of equations.

A particle has a kinetic energy of 62 MeV and a momentum of 335 MeV/c. Find its mass (in MeV/c2) and speed (as a fraction of c).


16

E 4.23 GeV⋅:=E 22

3.732

+( ) GeV⋅:=E 22

GeV2

⋅ 3.732

GeV2

⋅+:=

E 22 GeV

2

c2

⋅ c2

⋅ 3.73GeV

c2

⋅ c2

⋅

2+:=E p

2c2

⋅ m c2

⋅( )2+:=E

2p

2c2

⋅ m c2

⋅( )2+:=

Part (b): mass is relativistically invariant, so we can use the mass from part (a) along with the new momentum.

m 3.73GeV

c2

⋅:=

m 42

1.452

−GeV

c2

⋅:=m42

GeV2

⋅

c4

1.452

GeV2

⋅

c4

−:=m42

GeV2

⋅

c4

1.452 GeV

2

c2

⋅

c2

−:=

m2 E

2

c4

p2

c2

−:=m c2

⋅( )2E

2p

2c2

⋅−:=E2

p2

c2

⋅ m c2

⋅( )2+:=

mE

2

c4

p2

c2

−:=

Part (a):

(a) Find the mass (in GeV/c2) of a particle whose total energy is 4 GeV and whose momentum is 1.45 GeV/c. (b) Find the total energy of this particle in a reference frame in which its momentum is 2 GeV/c.


17

sorular ilk konu

Documents