Given eqns

• Also P4€

γ= 11−v2c2

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f=f0 1−v/c1+v/cΔs= c2t2−x2

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vo/a=vo/b+vb/a1+vo/bvb/ac2

A spaceship passes earth at t=t’=0. The earth frame is unprimed, the spaceship frame is primed.

What line is the worldline of the spaceship?

A spaceship passes earth at t=t’=0. The earth frame is unprimed, the spaceship frame is primed.

SS worldline = c

What line is the x’ axis for the spaceship?

A spaceship passes earth at t=t’=0. The earth frame is unprimed, the spaceship frame is primed.

X’ axis for ss = 3

What line is the ct’ axis for the spaceship?

Ct’ axis = worldline = c

A supernova explosion occurs at x = 1 light year, t = 2 years.

We will find the x’,ct’ coords of this event, by reading off x, ct axes and employing γ

Ct’ axis = worldline = c

A supernova explosion occurs at x = 1 light year, t = 2 years.

What line represents all events that happen at this same place, according to (or in the frame of) the spaceman?

Ans. b

A supernova explosion occurs at x = 1 light year, ct = 2 years.

To find the coordinates of the explosion in the spaceship frame, we need to find the point where the explosion occurred in the ss frame, but at the time when earth clocks read 0.

What point is that?

Ans. a

A supernova explosion occurs at x = 1 light year, ct = 2 years.

To find the coordinates of the explosion in the spaceship frame, we need to find the point where the explosion occurred, but at the time when earth clocks read 0.

What is the x-coordinate

of this point (in light years)?

The ship is moving at 0.6c.

a) -2.2

b) -1.2

c) -0.4

d) -0.2

e) -0.1

Ans. 1 - 1.2 = -0.2

A supernova explosion occurs at x = 1 light year, ct = 2 years.

Finally, what is the x’ coordinate of this point? For v/c = 0.6, gamma = 5/4.

a) +4/25

b) -4/25

c) +1/4

d) -1/4

e) -1/3

Answer: multiply by γ

to get - 1/4.

A supernova explosion occurs at x = 1 light year, ct = 2 years.

What line shows all the events

that, in the spaceship frame,

occurred at the same time

as the supernova?

Ans. c, parallel to x’

A supernova explosion occurs at x = 1 light year, ct = 2 years.

To find the time when the supernova occurs (in the spaceship frame) we need to find the point on the graph which is at the same time as the supernova according to the spaceman), but at the same place as Earth according to Earth. Where is this point?

Ans. b

A supernova explosion occurs at x = 1 light year, ct = 2 years. To find the time when the supernova occurs (in the spaceship frame) we need to find the point on the graph which is at the same time as the supernova according to the spaceman), but at the same place as Earth according to Earth.

What is the earth time

at this point (in years)?

a) 1.6

b) 1.5

c) 1.4

d) 1.3

e) 1.2

2 - 0.6x1 = 1.4

A supernova explosion occurs at x = 1 light year, ct = 2 years. To find the time when the supernova occurs (in the spaceship frame) we need to find the point on the graph which is at the same time as the supernova according to the spaceman), but at the same place as Earth according to Earth.

What is the spaceship time at this point?

Gamma = 5/4

a) 1.12

b) 1.4

c) 1.75

d) 2.8

Ans. c.

Two events happen in the earth frame - event A at x=t=0 and event B at x= 4 light seconds and t = 5 second.

Could event A have been the cause of event B?

a) Yes

b) No

c) No way to tell

Two events happen in the earth frame - event A at x=t=0 and event B at x= 4 light seconds and t = 5 second.

Could event A have been the cause of event B? Yes. These are timelike separated.

In what frame would these events be simultaneous?

a) V = +0.8 c

b) V = -0.8 c

c) V = 1.25 c

d) V = 4/√5 c

e) In no frame

Ans. e. How could they be simultaneous if A causes B??

Two events happen in the earth frame - event A at x=t=0 and event B at x= 4 light seconds and t = 5 second.

Could event A have been the cause of event B? Yes. These are timelike separated.

Smallest time interval = proper time (slowest clock) = 3 s = spacetime “length”

What is the speed of a frame in which this time interval is found (x c)?

a) 3/5

b) 3/4

c) 4/5

d) -3/5

e) -3/4

Two events happen in the earth frame - event A at x=t=0 and event B at x= 4 light seconds and t = 5 second.

Could event A have been the cause of event B? Yes. These are timelike separated.

Smallest time interval = proper time (slowest clock) = 3 s = spacetime “length”

Frame speed = 4/5 c. That puts the events in the same place.

Is there any frame where the time interval between these events is 4,567 seconds?

a) Yes

b) No

c) No way to determine

Two events happen in the earth frame - event A at x=t=0 and event B at x= 4 light seconds and t = 5 second.

Could event A have been the cause of event B? Yes. These are timelike separated.

Smallest time interval = proper time (slowest clock) = 3 s = spacetime “length”

Frame speed = 4/5 c. That puts the events in the same place.

Is there any frame where the time interval between these events is 4,567 seconds? Yes.

What is the motion of such a frame?

a) Very fast (nearly c) toward +x

b) Very fast toward -x

c) Very fast toward either +x or -x

d) Very fast in y or z.

Two events happen in the earth frame - event A at x=t=0 and event B at x= 4 light seconds and t = 5 second.

Smallest time interval = proper time (slowest clock) = 3 s = spacetime “length”

Frame speed = 4/5 c. That puts the events in the same place.

Is there any frame where the time interval between these events is 4,567 seconds? Yes.

Any frame moving nearly c in ±x will record a very long time between these events. Approximately how far apart are these events in such a frame?

a) 4 light seconds

b) 4562 light seconds

c) 4563 light seconds

d) 4564 light seconds

e) 4567 light seconds

Two events happen in the earth frame - event A at x=t=0 and event B at x= 4 light seconds and t = 5 second.

Smallest time interval = proper time (slowest clock) = 3 s = spacetime “length”

Frame speed = 4/5 c. That puts the events in the same place.

Is there any frame where the time interval between these events is 4,567 seconds? Yes.

Any frame moving nearly c in ±x will record a very long time between these events. Approximately how far apart are these events in such a frame?

Answer e) 4567 light seconds (Exact answer, from invariant spacetime length = 4566.999)

A particle has momentum 4 MeV/c and total energy 5 MeV.

This particle is:

a) A photon

b) Not a photon

c) Could be either a photon or an ordinary particle

A particle has momentum 4 MeV/c and total energy 5 MeV.

This particle is NOT a PHOTON.

A second particle has momentum - 4 MeV/c and total energy 5 MeV.

The two particles collide and annihilate, producing two new particles.

The momentum of new particle A is 5 MeV/c in the +y direction.

What is the momentum of new particle B?

a) 5 MeV/c in the +y direction

b) 5 MeV/c in the -y direction

c) 3 MeV/c in the +x direction

d) 3 MeV/c in the -y direction

e) Not enough info is given

A particle has momentum 4 MeV/c and total energy 5 MeV.

This particle is NOT a PHOTON.

A second particle has momentum - 4 MeV/c and total energy 5 MeV.

The two particles collide and annihilate, producing two new particles.

The momentum of new particle A is 5 MeV/c in the +y direction.

What is the momentum of new particle B? Answer = 5 MeV/c in -y.

What ARE particles A & B?

a) Photons

b) Electrons

c) Protons

d) Not enough information

A particle has momentum 4 MeV/c and total energy 5 MeV.

This particle is NOT a PHOTON.

A second particle has momentum - 4 MeV/c and total energy 5 MeV.

The two particles collide and annihilate, producing two new particles.

The momentum of new particle A is 5 MeV/c in the +y direction.

What is the momentum of new particle B? Answer = 5 MeV/c in -y.

What ARE particles A & B?

They have energy 5 MeV and momentum 5 MeV/c, so they have rest mass = 0. They are either photons or gravitons. (It happens that we have no idea how to make a 5 MeV graviton, though.)

Practice with velocity addition.

A rocket moving at 0.7c fires a bullet at 0.7 c in the forward direction. About how fast does the bullet move with respect to the earth?

a) 0.7 c

b) The average of 0.7 c and c = 0.85 c

c) (0.7+0.7)/(1 + 0.49) c = 1.4/1.5 c

d) √0.7 = 0.8366 c

e) 0.7 x 0.7 c = 0.49 c

Ans. C. Velocity addition (along x)

Practice with velocity addition.

Here’s a pretty difficult problem (simply because we haven’t spent much time on this.)

A spaceship is moving at 0.707 c in the +y direction (wrt Earth).

It is moving away from the earth.

You are moving away from the earth in the +x direction at 0.707c.

What is the speed of the spaceship wrt YOU?

a) C; lengths (and so speeds) in y are unaffected by relativity

b) Earth clocks run slow (according to YOU), so the y component of the spaceships velocity is 0.707 / γ = 0.5. So I can just add the x-velocity of -0.707 and the y-velocity of 0.5 in quadrature.

c) Velocity components don’t add in quadrature in relativity (Pythagorean theorem doesn’t apply normally.) There is no easy way to do this problem.

Practice with velocity addition.

Here’s a pretty difficult problem (simply because we haven’t spent much time on this.)

A spaceship is moving at 0.707 c in the +y direction (wrt Earth).

It is moving away from the earth.

You are moving away from the earth in the +x direction at 0.707c.

What is the speed of the spaceship wrt YOU?

Answer B! Earth clocks run slow (according to YOU), so the y component of the spaceships velocity is 0.707 / γ = 0.5. So I can just add the x-velocity of -0.707 and the y-velocity of 0.5 in quadrature.

Pythagoras still works for x & y, NOT for x & ct on a Minkowski diagram.