Question 1

Which pairs of events shown in the space time diagram for problem R7T. 1 could in principle be causally connected? Which could not? (For each of the ten pairs, answer $T$ if they could be causally connected, and $F$ if not.)

Accepted Answer

The answer of Which pairs of events shown in the...

Question 2

Suppose that event $A$ is the origin event in the Home Frame and that event $B$ occurs at $t=1 \mathrm{~ns}$ and $x=10 \mathrm{~ns}$ in that frame. What would be the minimum $x$-velocity $\beta$ that an Other Frame would have to have relative to the Home Frame if $B$ occurred first in that frame?
A) 10
B) 1
C) 0.60
D) 0.40
E) 0.10
F) B can't occur first.

Accepted Answer

The answer of Suppose that event $A$ is the origin...

Question 3

Two blinking warning lights are $3000 \mathrm{~m}$ apart along a railroad track. Suppose that in the ground frame the west light blinks (event $W$ ) $5 \mu$ s before the east light blinks (event $E$ ). Now imagine that both lights are observed by a passenger on the train who passes the west light just as it blinks. It is possible for the train to be moving fast enough that the two blinks are observed (not necessarily seen) by the passenger to be simultaneous.

Accepted Answer

According to the theory of relativity, the simultaneity of events is relative to the observer's frame of reference. If the train is moving fast enough towards the east light, the time it takes for the light from the east blink to reach the observer can be reduced, making it possible for the observer on the train to perceive both blinks as simultaneous. This is due to the relativistic effects of time dilation and length contraction.

Question 4

Two blinking warning lights are $1000 \mathrm{~ns}$ apart along a railroad track. A train moves from west to east at a speed of 0.5 . In the train frame, an observer registers flashes from each light, and after correcting for light travel time, concludes that the east light flashed 200 ns before the west light flashed. Which light flashed first in the ground frame? (Hint: Draw a space time diagram.)
A) The east light flashes before the west light.
B) The west light flashes before the east light.
C) The lights flash simultaneously.
D) One does not have enough information to say.

Accepted Answer

Given that in the train frame, after correcting for light travel time, the east light is observed to have flashed 200 ns before the west light, it implies that in the ground frame, the west light must have flashed first. This is because the train is moving towards the east light, making it seem like the east light flashed earlier due to the relative motion between the train and the lights.

Question 5

Suppose an explosion occurs at $x=0$ in the Home Frame (event $A$ ). Light from the explosion is detected by a detector at position $x=-100 \mathrm{~ns}$ (event $B$ ) and by a detector at position $x=+50 \mathrm{~ns}$ (event $C$ ). Events $B$ and $C$ are causally connected.

Accepted Answer

Events $B$ and $C$ are not causally connected because they are simultaneous detections of the same light signal from event $A$, and neither can influence the other.

Question 6

A laser beam is emitted on earth (event $A$ ), bounces off a mirror placed on the moon by Apollo astronauts (event $B$ ), and then returns to a detector on earth (event $C$ ). The detector is $12 \mathrm{~ns}$ of distance from the laser.
-(a) The space time interval between events $A$ and $B$ is:
A) Time like
B) Light like
C) Space like

Accepted Answer

The space-time interval between events A and B is light-like because the laser beam, which is light, travels from Earth to the Moon, meaning the separation in space and time fits the criteria for a light-like interval where the distance light travels in space equals the distance it travels in time, according to the speed of light.

Question 7

A laser beam is emitted on earth (event $A$ ), bounces off a mirror placed on the moon by Apollo astronauts (event $B$ ), and then returns to a detector on earth (event $C$ ). The detector is $12 \mathrm{~ns}$ of distance from the laser.
-(b) The space time interval between events $A$ and $C$ is:
A) Time like
B) Light like
C) Space like

Accepted Answer

The space time interval between events $A$ and $C$ is time like because the laser beam travels from the laser to the moon and back to the detector, which is a time like path.

Question 8

A laser beam is emitted on earth (event $A$ ), bounces off a mirror placed on the moon by Apollo astronauts (event $B$ ), and then returns to a detector on earth (event $C$ ). The detector is $12 \mathrm{~ns}$ of distance from the laser.
-(c) Events $A$ and $C$ are definitely causally connected.

Accepted Answer

Events $A$ (emission of the laser beam) and $C$ (detection of the laser beam) are causally connected because the detection of the laser beam (event $C$) is a direct consequence of its emission (event $A$). The fact that the laser beam was emitted is the cause of its detection upon return, establishing a clear cause-and-effect relationship between these two events.

Question 9

If the space time interval between two events is time like, then the temporal order of the two events is the same in every inertial reference frame.

Accepted Answer

In a time-like interval, the separation between events is such that all observers in inertial frames will agree on the temporal order (which event happened first), because the interval can be traversed at a speed less than or equal to the speed of light, ensuring causality is preserved across all frames.

Question 10

If the space time interval between two events $A$ and $B$ is spacelike and event $A$ occurs before event $B$ in some Home Frame, then it is always possible to find an Other Frame where the events occur in the other order.

Accepted Answer

In the case of a spacelike interval, the separation between events is such that it exceeds the distance light could travel in the time between them, meaning there exists a frame of reference (the Other Frame) where the order of these events can be reversed due to the relativity of simultaneity in special relativity.

Question 11

An object moves with speed $v_{x}=+0.9$ in the Home Frame. In an Other Frame moving at $\beta=0.60$ relative to the Home Frame, the object's $x$-velocity $v_{x}^{\prime}$ is
A) $v_{x}^{\prime}=1.5$
B) $1<v_{x}^{\prime}<1.5$
C) $0.9<v_{x}^{\prime}<1$
D) $0.3<v_{x}^{\prime}<0.9$
E) $v_{x}^{\prime}<0.3$
F) $v_{x}^{\prime}<0$

Accepted Answer

To find the object's velocity in the Other Frame, we use the formula for velocity transformation in special relativity: $v_{x}^{\prime} = \frac{v_{x} - v}{1 - \frac{v_{x}v}{c^2}}$, where $v$ is the velocity of the Other Frame relative to the Home Frame, $v_{x}$ is the object's velocity in the Home Frame, and $c$ is the speed of light. Given $v_{x} = +0.9c$ and $v = 0.6c$, the transformed velocity $v_{x}^{\prime}$ will be less than $v_{x}$ but greater than 0, as the relative motion reduces the object's speed in the Other Frame but does not reverse its direction. This places $v_{x}^{\prime}$ in the range $0.3 < v_{x}^{\prime} < 0.9$, making option D correct.

Question 12

Harry Potter points his wand at the sun and cries "Spotificus!" Within 8 minutes, observers all over the earth see a large string of sunspots form on the sun. Did Harry create the sunspots?
A) Yes. Harry is a very powerful wizard.
B) Yes, if magical effects travel at the speed of light.
C) No. This is not physically possible, even with magic.

Accepted Answer

The answer of Harry Potter points his wand at the...

Question 13

An unidentified spaceship cruises past Outpost 11 , a space station floating in deep space, at a speed of $\frac{3}{5}$, without acknowledging requests for identification. Ten seconds after the ship passes (according to Outpost 11 clocks), Outpost 11 sends a message to the departing ship, warning that it will commence firing if the ship does not respond. Fifteen seconds after that, Outpost 11 sensors indicate that the ship has raised its shields. This is clearly a response to the warning.

Accepted Answer

The scenario involves relativistic speeds, and due to the effects of time dilation, the elapsed time aboard the spaceship for events (such as receiving the message and raising shields) would not match the elapsed time as measured by Outpost 11. Therefore, it's not clear without further information that the ship's action is a direct response to the warning, as the timing of events would be perceived differently by each observer.

Question 14

A particle's Newtonian $x$-momentum $\chi_{x}$ is always either equal to or smaller than the $x$ component of the particle's four-momentum $p_{x}$.

Accepted Answer

Note that $p_{x}=\lambda_{\chi} / \sqrt{1-|\vec{v}|^{2}}$.

Question 15

A particle's mass is always either equal to or smaller than the time-component of its four-momentum.

Accepted Answer

The time-component of a particle's four-momentum is given by $E/c$, where $E$ is the energy of the particle and $c$ is the speed of light. Since the energy of a particle includes its rest mass energy ($mc^2$) and its kinetic energy (which is always positive), the time-component of the four-momentum will always be greater than or equal to the rest mass of the particle. 
Note that $p_{t}=m / \sqrt{1-\left|\vec{v}^{2}\right|}$.

Question 16

The absolute value of the $x$ component of a particle's four-momentum vector is always either equal to or smaller than its $t$ component.

Accepted Answer

The four-momentum of a particle is defined as $p^\mu = (E/c, \vec{p})$, where $E$ is the energy, $c$ is the speed of light, and $\vec{p}$ is the three-momentum. The magnitude of the $t$ component (energy term) is always greater than or equal to the magnitude of any spatial component due to the relationship $E \geq |\vec{p}c|$, which follows from the relativistic energy-momentum relation $E^2 = (pc)^2 + (mc^2)^2$, where $m$ is the rest mass of the particle. This ensures that the energy (or the $t$ component of the four-momentum) is always greater than or equal to the magnitude of the momentum multiplied by $c$ (the spatial components). 
Note that $m^{2}=p_{t}^{2}-p_{x}^{2}-p_{y}^{2}-p_{z}^{2} \Rightarrow p_{t}^{2}=p_{x}^{2}+\left[p_{y}^{2}+p_{z}^{2}+m^{2}\right]$.

Question 17

(a) The squared magnitude of a four-vector could be negative.

Accepted Answer

The squared magnitude of a four-vector in special relativity, defined as $s^2 = -(c\tau)^2 + \mathbf{x}^2$, where $c$ is the speed of light, $\tau$ is the proper time, and $\mathbf{x}$ is the spatial part of the four-vector, can be negative for spacelike intervals where the spatial separation is greater than the temporal separation when measured in the same units. 
(a) For example, the displacement four-vector $\Delta \boldsymbol{s}$ between space like events has a squared magnitude of $\Delta s^{2}<0$ by definition.

Question 18

(b) The squared magnitude of a particle's four-momentum could be negative.

Accepted Answer

The squared magnitude of a particle's four-momentum is equal to the square of its mass times the square of the speed of light, which cannot be negative. This quantity is invariant and always non-negative, reflecting the particle's rest mass energy. 
(b) The squared magnitude of a particle's four-momentum is its mass squared by definition. This squared mass must be positive.

Question 19

How does a particle's relativistic kinetic energy $K$ compare to its Newtonian kinetic energy when the particle's speed $|\vec{v}|=\frac{4}{5}$ ?
A) $K<\frac{1}{2} m|\vec{v}|^{2}$
B) $K>\frac{1}{2} m|\vec{v}|^{2}$
C) $K=\frac{1}{2} m|\vec{v}|^{2}$

Accepted Answer

The relativistic kinetic energy of a particle is given by $K = (\gamma - 1)mc^2$, where $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$ and $v$ is the particle's speed. For speeds $v < c$, $\gamma > 1$, making $K$ larger than the Newtonian kinetic energy $\frac{1}{2} m|\vec{v}|^{2}$, especially as $v$ approaches $c$. At $|\vec{v}|=\frac{4}{5}c$, $\gamma$ is greater than 1, thus $K>\frac{1}{2} m|\vec{v}|^{2}$. 
Note that $K=m / \sqrt{1-|\vec{v}|^{2}}-m=m / \sqrt{\frac{9}{25}}-m=\frac{5}{3} m-m=\frac{2}{3} m$, but $\frac{1}{2} m|\vec{v}|^{2}=\frac{1}{2} m \frac{16}{25}=\frac{8}{25} m$.

Question 20

Consider a particle with mass $m$. As its speed approaches the speed of light, its relativistic momentum magnitude $|\vec{p}|$ approaches:
A) infinity
B) $m c$
C) $m$
D) zero

Accepted Answer

As a particle's speed approaches the speed of light, its relativistic momentum increases without bound due to the Lorentz factor, which approaches infinity as velocity approaches the speed of light, making the momentum approach infinity as well. 
Note that $|\vec{p}|=m|\vec{v}| / \sqrt{1-|\vec{v}|^{2}}$.

Quiz 3: The Laws of Physics Are Frame-Independent Relativity

Which pairs of events shown in the space time diagram for problem R7T. 1 could in principle be causally connected? Which could not? (For each of the ten pairs, answer $T$ if they could be causally connected, and $F$ if not.)

Suppose that event $A$ is the origin event in the Home Frame and that event $B$ occurs at $t=1 \mathrm{~ns}$ and $x=10 \mathrm{~ns}$ in that frame. What would be the minimum $x$ -velocity $\beta$ that an Other Frame would have to have relative to the Home Frame if $B$ occurred first in that frame?

Suppose an explosion occurs at $x=0$ in the Home Frame (event $A$ ). Light from the explosion is detected by a detector at position $x=-100 \mathrm{~ns}$ (event $B$ ) and by a detector at position $x=+50 \mathrm{~ns}$ (event $C$ ). Events $B$ and $C$ are causally connected.

A laser beam is emitted on earth (event $A$ ) , bounces off a mirror placed on the moon by Apollo astronauts (event $B$ ) , and then returns to a detector on earth (event $C$ ) . The detector is $12 \mathrm{~ns}$ of distance from the laser. -(a) The space time interval between events $A$ and $B$ is:

A laser beam is emitted on earth (event $A$ ) , bounces off a mirror placed on the moon by Apollo astronauts (event $B$ ) , and then returns to a detector on earth (event $C$ ) . The detector is $12 \mathrm{~ns}$ of distance from the laser. -(b) The space time interval between events $A$ and $C$ is:

A laser beam is emitted on earth (event $A$ ), bounces off a mirror placed on the moon by Apollo astronauts (event $B$ ), and then returns to a detector on earth (event $C$ ). The detector is $12 \mathrm{~ns}$ of distance from the laser. -(c) Events $A$ and $C$ are definitely causally connected.

If the space time interval between two events is time like, then the temporal order of the two events is the same in every inertial reference frame.

If the space time interval between two events $A$ and $B$ is spacelike and event $A$ occurs before event $B$ in some Home Frame, then it is always possible to find an Other Frame where the events occur in the other order.

An object moves with speed $v_{x}=+0.9$ in the Home Frame. In an Other Frame moving at $\beta=0.60$ relative to the Home Frame, the object's $x$ -velocity $v_{x}^{\prime}$ is

Harry Potter points his wand at the sun and cries "Spotificus!" Within 8 minutes, observers all over the earth see a large string of sunspots form on the sun. Did Harry create the sunspots?

A particle's Newtonian $x$ -momentum $\chi_{x}$ is always either equal to or smaller than the $x$ component of the particle's four-momentum $p_{x}$ .

A particle's mass is always either equal to or smaller than the time-component of its four-momentum.

The absolute value of the $x$ component of a particle's four-momentum vector is always either equal to or smaller than its $t$ component.

(a) The squared magnitude of a four-vector could be negative.

(b) The squared magnitude of a particle's four-momentum could be negative.

How does a particle's relativistic kinetic energy $K$ compare to its Newtonian kinetic energy when the particle's speed $|\vec{v}|=\frac{4}{5}$ ?

Consider a particle with mass $m$ . As its speed approaches the speed of light, its relativistic momentum magnitude $|\vec{p}|$ approaches:

Quiz 3: The Laws of Physics Are Frame-Independent Relativity

Which pairs of events shown in the space time diagram for problem R7T. 1 could in principle be causally connected? Which could not? (For each of the ten pairs, answer TTT if they could be causally connected, and FFF if not.)

If the space time interval between two events is time like, then the temporal order of the two events is the same in every inertial reference frame.

If the space time interval between two events AAA and BBB is spacelike and event AAA occurs before event BBB in some Home Frame, then it is always possible to find an Other Frame where the events occur in the other order.

An object moves with speed vx=+0.9v_{x}=+0.9vx​=+0.9 in the Home Frame. In an Other Frame moving at β=0.60\beta=0.60β=0.60 relative to the Home Frame, the object's xxx -velocity vx′v_{x}^{\prime}vx′​ is

Harry Potter points his wand at the sun and cries "Spotificus!" Within 8 minutes, observers all over the earth see a large string of sunspots form on the sun. Did Harry create the sunspots?

A particle's Newtonian xxx -momentum χx\chi_{x}χx​ is always either equal to or smaller than the xxx component of the particle's four-momentum pxp_{x}px​ .

A particle's mass is always either equal to or smaller than the time-component of its four-momentum.

The absolute value of the xxx component of a particle's four-momentum vector is always either equal to or smaller than its ttt component.

(a) The squared magnitude of a four-vector could be negative.

(b) The squared magnitude of a particle's four-momentum could be negative.

How does a particle's relativistic kinetic energy KKK compare to its Newtonian kinetic energy when the particle's speed ∣v⃗∣=45|\vec{v}|=\frac{4}{5}∣v∣=54​ ?

Consider a particle with mass mmm . As its speed approaches the speed of light, its relativistic momentum magnitude ∣p⃗∣|\vec{p}|∣p​∣ approaches:

Which pairs of events shown in the space time diagram for problem R7T. 1 could in principle be causally connected? Which could not? (For each of the ten pairs, answer $T$ if they could be causally connected, and $F$ if not.)

If the space time interval between two events $A$ and $B$ is spacelike and event $A$ occurs before event $B$ in some Home Frame, then it is always possible to find an Other Frame where the events occur in the other order.

An object moves with speed $v_{x}=+0.9$ in the Home Frame. In an Other Frame moving at $\beta=0.60$ relative to the Home Frame, the object's $x$ -velocity $v_{x}^{\prime}$ is

A particle's Newtonian $x$ -momentum $\chi_{x}$ is always either equal to or smaller than the $x$ component of the particle's four-momentum $p_{x}$ .

The absolute value of the $x$ component of a particle's four-momentum vector is always either equal to or smaller than its $t$ component.

How does a particle's relativistic kinetic energy $K$ compare to its Newtonian kinetic energy when the particle's speed $|\vec{v}|=\frac{4}{5}$ ?

Consider a particle with mass $m$ . As its speed approaches the speed of light, its relativistic momentum magnitude $|\vec{p}|$ approaches: