special case: proper time

Special Case: Proper Time

LTE’s: x1 = (x2 + v*t2) / [1-(v/c)2]1/2

t1 = (t2 + v*x2/c2) / [1-(v/c)2]1/2 IF a time interval is measured using ONE

CLOCK, then there is no problem with simultaneity - but the beginning and ending have to happen at the same place:

x2 i = x2 f , so we get:

tnon-proper = tproper / [1-(v/c)2]1/2, tNP > tP

Special Case: Proper Length

LTE’s: x1 = (x2 + v*t2) / [1-(v/c)2]1/2

t1 = (t2 + v*x2/c2) / [1-(v/c)2]1/2 IF a length is measured using NO CLOCKS,

then there is no problem with simultaneity - but the beginning and ending have to happen at the same time in the other system:

t2 i = t2 f , so we get:

xproper= xnon-proper /[1-(v/c)2]1/2, xP > xNP

Review of Equations:

LTE’s: good for one EVENT (x,t)

x1 = (x2 + v*t2) / [1-(v/c)2]1/2

t1 = (t2 + v*x2/c2) / [1-(v/c)2]1/2 .

Time Dilation: good for a time interval:

tnon-proper = tproper / [1-(v/c)2]1/2, tNP > tP

Length Contraction: good for space interval

xproper= xnon-proper /[1-(v/c)2]1/2, xP > xNP

LTE Velocity Transformationx1 = (x2 + v*t2) / [1-(v/c)2]1/2

t1 = (t2 + v*x2/c2) / [1-(v/c)2]1/2 .

v1 = x1 / t1 = (x1 f - x1 i) / (t1 f - t1 i) =

{[(x2 f + v*t2 f) - (x2 i + v*t2 i) ] / [1-(v/c)2]1/2 } /

{[(t2 f+v*x2 f/c2) -(t2 i +v*x2 i/c2)] / [1-(v/c)2]1/2}

= [(x2 f-x2 i) + v*(t2 f-t2 i)] /

[(t2 f-t2 i) + v*(x2 f-x2 i)/c2] now divide numerator and denominator by (t2f-t2i)

= (v2+v)/(1+v*v2/c2) = v1

LTE Velocity Transformation

v1 = (v2+v) / (1+v*v2/c2)

Note: If v and/or v2 << c, then get Galilean result

NOTE: IF v2 = c, THEN

v1 = (c+v) / (1+vc/c2) = (c+v) / (1+v/c)

= (c+v) / (1/c)*(c+v) = c .

This result DOES AGREE with the Michelson-Morley experiment!

Relativity Example #1

Question:

Can there be a moving system that sees the explosion from a cannon shell happen before the cannon fires the shell?

NOTE: If this is possible, we have a problem with cause and effect!

Relativity Example #1Answer:This situation entails two events: the cannon

fires at x1e=0 m at t1e=0 s, and the cannon shell explodes at x2e at t2e, where x2e=vcst2e .

[BOTH x2e AND t2e are greater than zero.]

From the moving frame, we are suggesting t2s< t1s (shell explodes before cannon fires).

Since we are dealing with two events, we need the LTE:

Relativity Example #1

LTE: ts = (te +/- vxe/c2)/[1-(v/c)2]1/2

for the first event: ts1= (0 +/- 0)/[1-(v/c)2]1/2 = 0s.

for the second event: ts2 < ts1 = 0 s, so we need: 0 s > ts2 = (te2 +/- vxe2/c2)/[1-(v/c)2]1/2 or te2 < vxe2 /c2, but xe2 = vcste2 , so

te2 < v*vcste2/c2 , or v > c2/vcs

The shell exploding before the cannon fires

(the effect precedes the cause) is possible only if either the cannon shell, vcs, or the ship, v, can go faster than light!

Speeds greater than cBecause of the square root in the

denominator, if we have a speed, v, greater than c we will get imaginary results!

In the previous example, if we have a speed, v, greater than c we will mess up cause and effect.

Our theory will have to eventually explain why we can’t have speeds of objects go faster than light if we are to have a consistent theory.

Relativity Example #2

Question:

A spaceship is 300 m long as measured by the spaceship captain. The spaceship is moving at v=0.8*c relative to the space station. How big is the spaceship as measured by the space station personnel?

Relativity Example #2

Answer:Since the spaceship is at rest with respect to

the captain of the spaceship, this length of 300 m is a proper length. The length as determined by the space station personnel is NOT a proper length. Therefore, we can use: Lnp < Lp , or

Lnp = Lp* [1-(.8)2]1/2 = 300 m * 0.6 = 180 m

Relativity Example #3

Question:

A muon has a half life of 2.2 s in the lab.

How fast are a beam of muons moving if they travel 1200 m on average before decaying?

Relativity Example #3

Answer:

Since the muons were at rest in the lab, the 2.2s is the proper time interval.

Since the distance of travel is xearth = v*tearth the non-proper time interval, tearth, is (1200 m) / v.

Since tnp > tp ,

(1200 m)/v = tearth = 2.2s / [1-(v/c)2]1/2 , or

V = (1200m) * [1-(v/c)2]1/2 / 2.2s , or squaring both sides: [1-(v/c)2]*(1200 m/2.2s)2 = v2

Relativity Example #3

[1-(v/c)2]*(1200 m/2.2s)2 = v2

the quantity (1200 m / 2.2 s ) can be simplified to:

(1200 m / 2.2 s) = 5.45 x 108 m/s = 1.82 * c

[1-(v/c)2]*(1.82*c)2 = v2

Now bring all the terms with v to the left:

(1.82*c)2 = [1+(1.82)2]*v2 So we finally get:

v = (1.82*c)/[1+(1.82)2]1/2 = .877*c

Relativity Example #4

Question: A rocket moving toward the earth at a speed of 0.8*c fires a missile going at a speed of 0.95*c (relative to the rocket) directed toward the earth. How fast does the earth see the missile approaching?

Relativity Example #4

Answer:

v1 = (v2+v) / (1+v*v2/c2)

vearth = (vmissile/rocket +/- vrocket) / [1 +/- vmr*vr/c2]

use + sign in both since both rocket and missile are approaching the earth

vearth = (0.95*c + 0.80*c) / [1 + 0.95*0.80] =

1.750*c / 1.760 = 0.994*c .

Relativity Example #4

If the missile were fired away from the earth, we would have instead:

vearth = (vmissile/rocket +/- vrocket) / [1 +/- vmr*vr/c2]

vearth = (-0.95*c + 0.80*c) / [1 - 0.95*0.80] =

-0.150*c / 0.240 = -0.625*c .

The minus sign means the missile is headed away from the earth.

Relativity Example #4

• If the missile approaching the earth (first case) then fires a laser pulse at the earth, how fast does the earth measure for the speed of the laser pulse?

• This problem is a giveaway, since everybody always measures the speed of light (that is what a laser pulse is) as c = 3 x 108 m/s !

Relativity

• Note: for all these equations to work, the speed of one system with respect to another must be less than c (v < c). The equations will break down if v>c, since there is the square root of [1 - (v/c)2] in the denominator!

• To be a good theory, then, the theory must explain WHY v<c !

• If x and t change between systems, does m? See the introduction to the Relativity 2 computer homework to see whether m changes as x and t do.

Determining mass in different systems

As was demonstrated in the Relativity 2 computer homework introduction, the determination of mass of an object depends on the system. The “best” mass is the mass determined in a frame in which the mass is at rest (call this mass mo).

m = mo / [1-(v/c)2]1/2 , m(v) > mo

tnon-proper = tproper / [1-(v/c)2]1/2, tNP > tP

xproper= xnon-proper /[1-(v/c)2]1/2, xP > xNP

Kinetic Energy

KE = Workdone = Force thru distance

and if mass is constant, then dp = m*dv, so

KE = F ds = (dp / dt)ds = (ds / dt)dppi

pf

xi

xf

xi

xf

KE = v m dv = 1

2mv -

1

2mv

vi

vf

f2

i 2

Relativistic Kinetic Energy

However, if mass is not constant, then we need to do something different:

Note that dp = d(mv) = vdm + mdv, so

KE = F ds = (dp / dt)ds = (ds / dt)dppi

pf

xi

xf

xi

xf

KE = [v + ] where m = m(v).2 dm mvdv

Relativistic Energy

Since the integral on the previous slide looked hard, we will turn “tricky”: consider

m = mo/[1-(v/c)2]1/2 , or m2[1-v2/c2] = mo2, or

m2c2 - m2v2 = mo2c2 ;

now take d( ) of each side:

2mc2dm - 2mv2dm - 2m2vdv = 0 , or

c2dm = v2dm + mvdv !

Relativistic Energy

c2dm = v2dm + mvdv

so we can now write:

KE = [v + ] where m = m(v).2 dm mvdv

KE = c = m c - m c2

mi

mf

f2

i2 dm

Relativistic Energy

If we start from rest, then we can have for the kinetic energy: KE = mc2 - moc2 .

Recall the m = m(v), so we can also write:

KE(v) = moc2 ({1/[1-(v/c)2]1/2 } - 1) . This does not look at all like KE = (1/2)mv2, but if v/c is small, then using the approximation:

1/[1-(v/c)2]1/2 = 1 + (1/2)v2/c2 + ...

we do get: KE = (1/2)mov2 + negligible terms

Relativistic Energy

kinetic energy: KE(m) = mc2 - moc2 , and

KE(v) = moc2 ({1/[1-(v/c)2]1/2 } - 1) .

Note that as v approaches c, that KE approaches infinity! This means that we would need infinite energy to reach the speed of light - an impossibility! Thus we cannot go faster than light!

Experimental Evidence• Michelson-Morley experiment• Pair Production and Pair Annihilation

Note: E = mc2 and pair annihilation says that if we had 1 lb of anti-matter, we could combine it with 1 lb of regular matter and have about 1 kg for m. The the energy released would be: 1 kg x (3x108

m/s)2 = 9x1016 J, or 1000 MW*9x107 sec =

1,000 MW for 3 years !

Experimental EvidenceIn the cyclotron, accelerate a charge particle

with voltage difference, but make particle go in a circle with magnetic field:

F = qv x B and F = ma, so qvB=m(v2/r), orqB = mv/r (but v = r), so = qB/m .Normally, is a constant, independent of r

and v. But as v approaches c, we need to slow of Voltage source down to keep giving particle more energy: SYNCHROTRON

Experimental Evidence

In radioactive decay, the mass of the radioactive atom is larger than the masses of the resulting particles.

In addition, the mass of carbon-12 (which consists of 6 protons and 6 neutrons) is less than the mass of 6 protons and 6 neutrons!

How can these be?Where is the “missing mass” in each case?

Experimental Evidence

In the labs, when muons are relatively slow moving, they have a half life of about 2.2s

But when they are created in the upper atmosphere and have a very high speed, they seem to go further than the x=vt = 3x108 m/s * 2.2x10-6 s = 660 m .

This can be explained by time dilation and length contraction.

Consequences

If we can not go faster than c, and if we cannot live longer than 100 years, are we confined to a radius of 100 lt-yrs in our universe?

Consider which time (proper or non-proper) the 100 years is.

Consider which distance (proper or non-proper) the 100 lt-yrs is.

Relativity and Light

• Does light have mass?

But m = mo /[1-(v/c)2]1/2 , but with v=c, m=mo/0, or mo = m * 0 ! This means that mo=0 ! But light never stops (if it remains light), so can’t directly measure mo!

• Does light have energy? E = hf, E = mc2

therefore, light does have mass: m = hf/c2 , just no rest mass

• If light has mass and is moving, does light have momentum? p = mv = mc = (E/c2)c = E/c = hf/c, but f=c, or =c/f, so p = h/(The DeBroglie relation!)

Relativity and LightIf light has mass, does gravity act on it (bend it)?

Since light moves so fast, gravity will not have much time to deflect it. If we have starlight go very close to the sun, we do indeed see a slight deflection!

We even have BLACK HOLES in which the gravity attracts the light so strongly that it cannot escape!

Relativity Example #5

How much energy would it take to accelerate a 1 kg object from rest to 0.5*c

a) classically?

b) relativistically?

a) KE = (1/2)mv2 - 0 =

(1/2)*(1 kg)* (1.5x108 m/s)2 - 0 = 1.125x1016 J

Relativity Example #5

b) KE = moc2(1/[1-(v/c)2]1/2) - moc2

= (1 kg)*(3x108 m/s)2 *{1/ [1-(0.5)2]1/2 - 1} = 9x1016 J *{1.155 - 1} = 1.392x1016 J .(Recall from part a, that classically, KE =

1.125x1016 J .)The relativistic amount is larger than the

classical amount, as we should have expected.

Relativity Example #6

How fast would an electron be going and what would be the mass of this electron if we gave it 1.2 MeV of kinetic energy?

[The rest mass energy of an electron is:

moc2 = (9.1x10-31 kg)*(3x108 m/s)2 = 8.19x10-14 J x (1

eV / 1.6x10-19 J) = 511,000 eV = .511 MeV.

KE > moc2 makes the problem relativistic.]

Relativity Example #6

KE = moc2 ({1/[1-(v/c)2]1/2 } - 1) , so

1.2 MeV = 0.511 MeV * ({1/[1-(v/c)2]1/2 } - 1)

(1.2 + 0.511)MeV = 0.511 MeV/[1-(v/c)2]1/2 }

or 1 - (v/c)2 = (0.511 / 1.711)2 ,

or (v/c) = [1 - (0.511/1.711)2]1/2 ,

or v = .954*c ; and m = mo / [1-(v/c)2]1/2

m = mo/2.99 = 3.348*mo = 3.05x10-30 kg .

Relativity Example #7

Let’s contrast this with a photon with the same 1.2 MeV of energy. What is the mass of an x-ray of KE=1.2 MeV?

KE = E (since mo for photon=0) = mc2 = hf = hc/ = 1.2 MeV*1.6x10-13J/MeV = 1.92x10-13 J.

m = E/c2 = 1.92x10-13 J / (3x108 m/s)2 = 2.13x10-30

kg . (Note that this is a little less than the mass of an electron of KE=1.2 MeV.)

= hc/E = (6.63x10-34Js)*(3x108m/s)/1.92x10-13J

= 1.036x10-12 m (in the x-ray range).