A.5 Galilean and special relativity

Describe the same event in two frames

For an event, record its position $x$ and time $t$ in frame S, then $x′$ and $t′$ in frame S′. If S′ moves at constant velocity relative to S, both frames are inertial. The event is the same physical occurrence even though its coordinates may differ.

Check the frame type

An inertial reference frame is non-accelerating. Newton’s laws can be used in their usual form in such a frame. If the frame accelerates or rotates, extra apparent forces may be needed and it is not an inertial frame for this syllabus treatment.

Relate the observations

The same event occurs in both frames, but the observers can assign different positions. In the classical model, time is absolute: both observers use the same $t$, while the moving frame changes the position coordinate according to $x′=x-vt$.

Check the boundary

Use Galilean relativity for inertial frames in the non-relativistic limit. It is not the correct model for measurements involving speeds comparable with $c$, where the assumptions of absolute time and unchanged light speed fail.

Use the sign convention

Start with the event coordinates $(x,t)$ in S. Substitute the frame velocity with its signed value, calculate $x′$, and carry the unchanged time $t′=t$ into S′. A positive $x′$ means the event is on the positive side of S′’s origin.

Check the assumptions

These equations describe the classical model: inertial frames, common synchronized time, and an origin coincidence at $t=0$. They are not the Lorentz transformations and do not preserve the speed of light between frames.

Symbolic example from local Question Bank row 31622

An event has $x'=L$ and $t'=L/c$ in S′. Galilean time is absolute, so $t=t'=L/c$. Using the inverse position transformation,

$$x=x'+vt=L+v\frac{L}{c}$$

The extra $vL/c$ is the distance travelled by the S′ origin during the shared time interval.

Calculate in two steps

Choose the positive direction, write the velocity of the object and the relative frame velocity with signs, then subtract. If the object and S′ move in the same direction, the relative speed is reduced; if they move in opposite directions, the signed relative velocity has greater magnitude.

Check the model

Galilean addition assumes inertial frames, common time, and non-relativistic speeds. It can predict a resultant speed greater than $c$, which signals that special relativity—not the arithmetic—is required for a high-speed light problem.

Worked example from local Question Bank row 31297

In the classical model, two spacecraft moving in opposite directions at $0.80c$ have signed velocities $u=-0.80c$ and $v=+0.80c$.

$$u'=u-v=-0.80c-0.80c=-1.60c$$

The magnitude $1.60c$ shows why Galilean addition cannot describe relativistic relative velocity; the sign only gives direction in S′.

Postulate 2: invariant light speed

Every inertial observer measures the same speed of light in vacuum, $c$, independent of the motion of the source or observer. This replaces the Galilean expectation that measured velocities simply add.

Use the consequence

Together, the postulates require space and time coordinates to transform differently from the Galilean model. They lead to Lorentz transformations, time dilation, length contraction and relativity of simultaneity. The syllabus does not require deriving those equations.

Transform one event

For an event with coordinates $(x,t)$ in S, the syllabus equations are $x′=\gamma(x-vt)$ and $t′=\gamma(t-vx/c^2)$ in S′. Use the same signed direction for $x$ and $v$, and calculate both coordinates from the same event.

Show the coordinate method

Write $\gamma$ first, substitute the given $x,t,v$, then report $x′$ and $t′$ with units. If the question provides a space–time diagram, reading the coordinates is an alternative only when the axes and scale are correctly interpreted.

Worked example from local Question Bank rows 31630–31631

For $v=0.745c$, $\gamma=1/\sqrt{1-0.745^2}=1.499$. An event at $x=1.0\,\mathrm{m}$ and $t=0$ transforms to

$$x'=\gamma(x-vt)=1.499(1.0)=1.50\,\mathrm{m}$$
$$ct'=\gamma(ct-\tfrac{v}{c}x)=1.499(0-0.745\times1.0)=-1.12\,\mathrm{m}$$

Thus $t'=-1.12/c\,\mathrm{s}$; the negative coordinate is valid and reflects the chosen origins.

Substitute signed velocities

Choose one positive direction, write signed values for $u$ and $v$, calculate $uv/c^2$, and evaluate the full fraction. Report the magnitude if the question asks for speed; retain the sign if it asks for velocity or direction.

Check the limiting cases

When $u\ll c$ and $v\ll c$, the denominator is close to 1 and the result approaches $u-v$. If $u=c$, the equation gives $u′=c$ for any sub-light $v$, so light does not gain or lose speed between inertial frames.

Worked example from local Question Bank row 31056

A train has $u=-0.70c$ in the ground frame and observer P moves at $v=+0.60c$.

$$u'=\frac{-0.70c-0.60c}{1-(-0.70)(0.60)}=\frac{-1.30c}{1.42}=-0.915c\approx-0.92c$$

The negative sign means the train moves opposite P's positive direction; its speed remains below $c$.

Calculate carefully

Read the time and position differences between the same two events. Convert $\Delta t$ into the distance $c\Delta t$, square both terms, and subtract the spatial term: $(c\Delta t)^2-(\Delta x)^2$. Keep the sign; a negative result is physically meaningful.

Compare frames

Calculate the interval from either frame’s coordinates. Matching values demonstrate invariance and provide a check on transformed coordinates. For a light signal, $(\Delta s)^2=0$; this null interval is consistent with $\Delta x=c\Delta t$.

Worked example from local Question Bank row 32720

For two events with $c\Delta t=100\,\mathrm{ly}$ and $\Delta x=20\,\mathrm{ly}$,

$$(\Delta s)^2=(100)^2-(20)^2=10{,}000-400=9{,}600\,\mathrm{ly^2}$$

Any inertial frame must calculate the same $9{,}600\,\mathrm{ly^2}$ from its own coordinate differences.

Identify proper length

The proper length $L_0$ is the length measured in the rest frame of the object. Its endpoints are measured simultaneously in that frame. It is the maximum length assigned to the object by inertial observers.

Choose from the event conditions

For time, ask: do the two events happen at the same place in this frame? For length, ask: is the object at rest in this frame, and are both endpoints measured at the same time? These conditions, not the observer’s label, determine whether the measurement is proper.

Identify the proper interval first

Find the frame in which the two events occur at the same place; that frame measures $\Delta t_0$. Calculate $\gamma$ using the relative speed, then multiply by $\Delta t_0$ to obtain the longer interval measured in the other inertial frame.

Check the direction of the effect

Because $\gamma\ge1$, the non-proper observer measures a time interval at least as large as the proper interval. At $v=0$, $\gamma=1$ and the two measurements agree.

Worked example from local Question Bank rows 30639–30640

A spacecraft crosses $1.80\times10^{11}\,\mathrm{m}$ at $0.750c$. The station-frame interval is

$$\Delta t=\frac{1.80\times10^{11}}{0.750(3.00\times10^8)}=800\,\mathrm{s}$$

With $\gamma=1/\sqrt{1-0.750^2}=1.51$, the spacecraft clock measures the proper time

$$\Delta t_0=\frac{800}{1.51}=530\,\mathrm{s}$$

The events occur at one place on the spacecraft, so its interval is proper.

Select the proper length

Find the frame in which the object is at rest; that frame measures $L_0$. Calculate $\gamma$, then divide the proper length by $\gamma$. The endpoints must be measured simultaneously in the observer’s frame.

Check the result

Since $\gamma\ge1$, a moving observer measures $L\le L_0$. At low speed the contraction is negligible; as $v$ approaches $c$, the measured length along the direction of motion becomes substantially smaller.

Worked example from local Question Bank row 30450

A rocket's proper length is $L_0=450\,\mathrm{m}$ and $\gamma=5/3$. An observer who sees it moving measures

$$L=\frac{L_0}{\gamma}=\frac{450}{5/3}=270\,\mathrm{m}$$

Only the dimension parallel to the relative motion is contracted.

Use the transformed time

For two events, $\Delta t'=\gamma(\Delta t-v\Delta x/c^2)$. If $\Delta t=0$ in S but $\Delta x\neq0$, then $\Delta t'\neq0$ in a relatively moving frame.

Worked example from local Question Bank row 30453

Two lamps are simultaneous in S and separated by $9.00\times10^3\,\mathrm{m}$. For $v=0.80c$ and $\gamma=5/3$,

$$\Delta t'=\frac53\left(0-\frac{(0.80c)(9.00\times10^3)}{c^2}\right)=-4.0\times10^{-5}\,\mathrm{s}$$

The negative sign fixes the event order in S′; it is not an error.

Keep the order test local

To decide which event occurs first in a frame, calculate or read the sign of $\Delta t′$ using the same pair of events. A negative time difference means the event assigned as the second reference event occurs earlier in that frame.

Interpret a world line

A vertical world line represents an object at rest in that frame. A straight tilted line represents constant velocity. A light ray has $v=c$ and lies on the 45° light line when the axes use equal scales; no physical world line may be steeper toward the x-axis than the light line.

Read simultaneity and coordinates

To find an event’s time, project horizontally to the $ct$ axis; to find position, project vertically to the $x$ axis. Lines parallel to an observer’s $x′$ axis represent equal $t′$, while lines parallel to $ct′$ represent equal $x′$.

Read the line

A vertical line has $\theta=0$ and represents rest. As the line tilts toward the x-axis, $\theta$ and the speed increase. The light line has $\theta=45^\circ$ on equal scales and represents $v=c$.

Calculate from a diagram

Measure or read the angle from the time axis, evaluate $\tan\theta$, and multiply by $c$. If the diagram gives a rise/run ratio, use that ratio as $\tan\theta$ only after confirming the axes and angle definition.

Worked example from local Question Bank row 31477

For a world line representing $v=0.80c$ on equal-scale axes,

$$\theta=\tan^{-1}(v/c)=\tan^{-1}(0.80)=38.7^\circ\approx39^\circ$$

The angle is measured from the $ct$ axis, not the $x$ axis.

Explain it in the Earth frame

In the Earth frame the moving muon’s lifetime is dilated: $\Delta t=\gamma\Delta t_0$. The increased lifetime allows more muons to travel the atmospheric distance before decaying. This is experimental evidence for time dilation.

Explain it in the muon frame

In the muon’s frame, the atmosphere is moving and its thickness is length-contracted: $L=L_0/\gamma$. The shorter distance can be crossed within the muon’s proper lifetime. Both frames predict the same detection rate; together they support time dilation and length contraction.

Evidence calculation from local Question Bank row 30447

Muons are produced $2.0\,\mathrm{km}$ above ground, move at $0.98c$, have proper lifetime $2.2\,\mu\mathrm{s}$ and $\gamma=5.0$.

$$t_{flight}=\frac{2000}{0.98(3.0\times10^8)}=6.8\,\mu\mathrm{s}$$
$$t_{Earth}=\gamma t_0=(5.0)(2.2)=11\,\mu\mathrm{s}$$

The dilated Earth-frame lifetime exceeds the flight time, explaining why many more muons reach the ground than the non-relativistic model predicts.