## Lorentz Transformations [block_0] The **Lorentz transformations **relate the coordinates of an event in one inertial frame to those in another moving at a constant velocity relative to the first. > They ensure that the **speed of light **remains constant in all inertial frames, a key postulate of special relativity. ### Deriving the Lorentz Transformations [block_3] Consider two inertial frames, $S$ and $S'$: 1. $S'$** **moves with velocity $v$ relative to $c$ along the x-axis. 2. The origins of $S$ and $S'$ coincide at $t = t' = 0$**.** [{"_key":"block_9","_type":"block","asset":null,"children":[{"_key":"65b41b017d2a","_type":"span","markDefs":[],"marks":[],"text":"The Lorentz transformations relate the coordinates $(x, t)$ in $S$ to the coordinates $(x', t')$ in $S'$."}],"markDefs":[],"style":"normal"}] #### Transformation for Position [block_10] 1. In Galilean relativity, the **position transformation** is: $$ x' = x - vt $$ 2. However, this **does not account** for the constancy of the speed of light. 3. The **Lorentz transformation** modifies this to: $$ x' = \gamma(x - vt) $$ where $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$ is the **Lorentz factor**. #### Transformation for Time [block_16] 1. In **Galilean **relativity, time is **absolute **($t' = t$). 2. However, in **special **relativity, time is **relative**. 3. The **Lorentz transformation** for **time **is: $$ t' = \gamma\left(t - \frac{vx}{c^2}\right) $$ [{"_key":"block_20","_type":"block","asset":null,"children":[{"_key":"486d7ca75952","_type":"span","markDefs":[],"marks":[],"text":"The term $\\frac{vx}{c^2}$ accounts for the "},{"_key":"e43a885a27bc","_type":"span","markDefs":[],"marks":["strong"],"text":"relativity of simultaneity"},{"_key":"54f30aee412f","_type":"span","markDefs":[],"marks":[],"text":", ensuring that events that are simultaneous in one frame may not be simultaneous in another."}],"markDefs":[],"style":"normal"}] #### Why the Lorentz Factor? [block_21] 1. The Lorentz factor $\gamma$ ensures that the speed of light is **constant **in **all inertial frames**. 2. It **approaches 1** at **low **speeds (recovering Galilean transformations) and **increases **significantly as $v$ **approaches **$c$. ![](https://cdn.sanity.io/images/w7j7fz60/production/bbd2cf8bafba9c4d6acc49117556c546799ffbba-1356x1050.png) Let’s apply the Lorentz transformations to an event at $x = 100$ m and $t = 2$ s in frame $S$. If $S'$ moves at $0.8c$ relative to $S$, what are the coordinates in $S'$? * Calculate $\gamma$:$$ \gamma = \frac{1}{\sqrt{1 - \frac{(0.8c)^2}{c^2}}}$$ $$ = \frac{1}{\sqrt{1 - 0.64}} = \frac{5}{3} $$ * Apply the transformations:$$ x' = \gamma(x - vt) = \frac{5}{3}(100 - 0.8 \cdot 3 \cdot 10^8 \cdot 2)$$ $$ = \frac{5}{3}(100 - 4.8 \times 10^8) \approx -8.0 \times 10^8 $$ $$t' = \gamma\left(t - \frac{vx}{c^2}\right) $$ $$= \frac{5}{3}\left(2 - \frac{0.8 \cdot 100}{(3 \cdot 10^8)^2}\right) \approx \frac{10}{3}$$ * The event occurs at $x' \approx -8.0 \times 10^8$ m and $t' \approx \frac{10}{3}$ s in $S'$. ## Time Dilation [block_27] **Time dilation **is the phenomenon where time passes more slowly for an observer in motion relative to a stationary observer. ### Deriving the Time Dilation Formula [block_29] 1. Consider a clock at **rest **in frame $S'$. 2. It ticks twice, separated by a proper **time interva**l $\Delta t_0$ (measured in the clock’s rest frame). 3. An observer in frame $S$ sees the clock moving with **velocity **$v$. 4. Using the Lorentz transformation for time: $$ t' = \gamma\left(t - \frac{vx}{c^2}\right) $$ 5. Since the clock is at rest in $S'$, $\Delta x' = 0$. 6. Therefore: $$ \Delta t = \gamma \Delta t_0 $$ [{"_key":"block_33","_type":"block","asset":null,"children":[{"_key":"d222df32f134","_type":"span","markDefs":[],"marks":[],"text":"The time interval measured by the observer in $S$ is $\\Delta t$."}],"markDefs":[],"style":"normal"}] [{"_key":"block_39","_type":"block","asset":null,"children":[{"_key":"48a2a38e11b9","_type":"span","markDefs":[],"marks":[],"text":"The proper time $\\Delta t_0$ is always the "},{"_key":"d32b9be4cad7","_type":"span","markDefs":[],"marks":["strong"],"text":"shortest time interval"},{"_key":"7a58d24097d9","_type":"span","markDefs":[],"marks":[],"text":". "}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"},{"_key":"903c8f6dcd0d","_type":"block","asset":null,"children":[{"_key":"decd718d7097","_type":"span","markDefs":[],"marks":[],"text":"Any observer measuring the interval in a frame where the clock is moving will measure a longer interval."}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"}] If a muon travels at $0.99c$, what is its lifetime as measured by an observer on Earth? * Calculate $\gamma$:$$ \gamma = \frac{1}{\sqrt{1 - \frac{(0.99c)^2}{c^2}}} \approx 7.09 $$ * Apply the time dilation formula:$$ \Delta t = \gamma \Delta t_0 $$ $$= 7.09 \times 2.2 \times 10^{-6} \approx 1.56 \times 10^{-5} \text{ s}$$ * The muon’s lifetime is extended to $1.56 \times 10^{-5}$ s in the Earth’s frame. ## Relativity of Simultaneity [e5025e37f4c6] > In special relativity, the concept of simultaneity is **not absolute,** events that are **simultaneous **in **one **inertial frame **may not be simultaneous** in **another **moving frame. [{"_key":"a8f1a1c9401d","_type":"block","asset":null,"children":[{"_key":"595266f79517","_type":"span","marks":[],"text":"This arises because the "},{"_key":"034c7e6dce07","_type":"span","marks":["strong"],"text":"measurement "},{"_key":"718342ca8ff4","_type":"span","marks":[],"text":"of time "},{"_key":"7e1ef27b9788","_type":"span","marks":["strong"],"text":"depends "},{"_key":"02e3d33d3934","_type":"span","marks":[],"text":"on the "},{"_key":"71fd09b3260e","_type":"span","marks":["strong"],"text":"relative motion "},{"_key":"92fd01256666","_type":"span","marks":[],"text":"between the observer and the events being measured."}],"markDefs":[],"style":"normal"}] 1. Consider two events occurring at different positions, $x_1$ and $x_2$, in frame $S$ but at the same time, $t_1 = t_2$. 2. An observer in **another **frame, $S'$, **moving **with velocity $v$ relative to $S$, will measure the **time difference **between the events as: $$\Delta t' = \gamma \left( \Delta t - \frac{v \Delta x}{c^2} \right)$$ 3. Since $\Delta t = 0$ in the **original **frame ($S$), the time interval observed in $S'$ becomes: $$\Delta t' = -\gamma \frac{v \Delta x}{c^2}$$ 4. That equation shows that the time difference between events depends on their spatial separation ($\Delta x$) and the relative velocity ($v$) of the frames. 5. Thus, simultaneity is **relative** and depends on the observer's **frame of reference**. [{"_key":"cfbe36565db6","_type":"block","asset":null,"children":[{"_key":"e3ec3e133eca0","_type":"span","marks":[],"text":"Two fireworks explode 2 km apart and are simultaneous in frame $S$. "}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"},{"_key":"6cbe57f0b630","_type":"block","asset":null,"children":[{"_key":"9e997af297e9","_type":"span","marks":[],"text":"An observer in frame $S'$, moving at $0.6c$ relative to $S$, perceives the explosions as occurring at different times due to the relativity of simultaneity. "}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"},{"_key":"02a154e3a739","_type":"block","asset":null,"children":[{"_key":"cb7518698b44","_type":"span","marks":[],"text":"The "},{"_key":"71a76d7ff9eb","_type":"span","marks":["strong"],"text":"time difference"},{"_key":"59fb0bc49a6f","_type":"span","marks":[],"text":" can be calculated using the "},{"_key":"d1a12c44499b","_type":"span","marks":["strong"],"text":"Lorentz transformations"},{"_key":"ee66d4fb4de1","_type":"span","marks":[],"text":"."}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"}] [{"_key":"072c2d9c002b","_type":"block","asset":null,"children":[{"_key":"6b99e37df5760","_type":"span","marks":[],"text":"Don’t assume simultaneity is "},{"_key":"a7eac730e9be","_type":"span","marks":["strong"],"text":"absolute"},{"_key":"93698478be00","_type":"span","marks":[],"text":". "}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"},{"_key":"bec50008fc0d","_type":"block","asset":null,"children":[{"_key":"18df62bf8c00","_type":"span","marks":[],"text":"In special relativity, two observers in relative motion can disagree on whether two events happened at the same time."}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"}] [{"_key":"0f8622233a45","_type":"block","asset":null,"children":[{"_key":"faeb543e92c10","_type":"span","marks":[],"text":"To "},{"_key":"4797df979da4","_type":"span","marks":["strong"],"text":"visualize "},{"_key":"d5c7d1432dba","_type":"span","marks":[],"text":"relativity of simultaneity, consider two "},{"_key":"f983b8061de7","_type":"span","marks":["strong"],"text":"lightning strikes "},{"_key":"5da3e54b06f4","_type":"span","marks":[],"text":"occurring "},{"_key":"bcbc997ecb72","_type":"span","marks":["strong"],"text":"simultaneously "},{"_key":"db4c32a20253","_type":"span","marks":[],"text":"as seen from a stationary observer. "}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"},{"_key":"c4c9f8f92642","_type":"block","asset":null,"children":[{"_key":"34fe7775ed34","_type":"span","marks":[],"text":"A "},{"_key":"faed3b232e7e","_type":"span","marks":["strong"],"text":"moving "},{"_key":"1ca8291a0e30","_type":"span","marks":[],"text":"observer may perceive one strike happening before the other due to their "},{"_key":"b646f8cd208e","_type":"span","marks":["strong"],"text":"motion relative "},{"_key":"a0bd34891b8d","_type":"span","marks":[],"text":"to the events."}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"}] ## Spacetime Interval: An Invariant Quantity [0c6515c8409a] 1. In special relativity, the **spacetime interval** ($\Delta s$) between two events remains **invariant** across all inertial reference frames. 2. This means that even though different observers may measure **different distances **and **time **intervals, the **spacetime interval **is the **same **for all observers. 3. It is mathematically expressed as: $$(\Delta s)^2 = (c\Delta t)^2 - (\Delta x)^2$$ where: 1. $\Delta s$ is the spacetime interval, 2. $c\Delta t$ represents the separation in time (converted into distance units by multiplying by $c$), 3. $\Delta x$ represents the spatial separation between the events. [{"_key":"4589b86d2f5b","_type":"block","asset":null,"children":[{"_key":"108aeef95abb","_type":"span","marks":[],"text":"This equation closely resembles the "},{"_key":"5b252c8544b3","_type":"span","marks":["strong"],"text":"Pythagorean theorem"},{"_key":"4d351b18646e","_type":"span","marks":[],"text":", but with a "},{"_key":"3e012f3274dd","_type":"span","marks":["strong"],"text":"minus sign"},{"_key":"7047765629f5","_type":"span","marks":[],"text":", highlighting the fundamental difference between space and time in relativity."}],"markDefs":[],"style":"normal"}] ### Why is the Spacetime Interval Invariant? [1addba851a4d] The **invariance** of the spacetime interval means that **all inertial** observers, regardless of their motion, **agree **on its value, even though they may **disagree **on $\Delta x$ and $\Delta t$. [{"_key":"76070404fd4d","_type":"block","asset":null,"children":[{"_key":"ce0770c6351c","_type":"span","marks":[],"text":"This property ensures that the laws of physics remain consistent in all reference frames and underpins the effects of "},{"_key":"e68c13812394","_type":"span","marks":["strong"],"text":"time dilation, length contraction, and the relativity of simultaneity"},{"_key":"56e05d90c07c","_type":"span","marks":[],"text":"."}],"markDefs":[],"style":"normal"}] ### Interpretation and Consequences [45fa6ab1cb9c] 1. If **$(\Delta s)^2 > 0$**, the interval is **timelike:** 1. $\implies$ The two events are causally **connected**, and a signal traveling at or **below **the speed of light could travel between them. 2. If **$(\Delta s)^2 = 0$**, the interval is **lightlike:** 1. $\implies$ The events are connected by a **signal traveling** exactly **at the speed of light**. 3. If **$(\Delta s)^2 < 0$**, the interval is **spacelike:** 1. $\implies$ **No signal**, not even light, could travel between the events, making them causally disconnected. [{"_key":"70476fa27b29","_type":"block","asset":null,"children":[{"_key":"b98a014ded0d","_type":"span","marks":[],"text":"The spacetime interval remains "},{"_key":"508962db460d","_type":"span","marks":["strong"],"text":"constant "},{"_key":"b98d496073b2","_type":"span","marks":[],"text":"across all inertial reference frames, even when distances and time intervals change for different observers."}],"markDefs":[],"style":"normal"}] [{"_key":"dce9d482ee42","_type":"block","asset":null,"children":[{"_key":"781a1a6cd489","_type":"span","marks":[],"text":"Don’t confuse the spacetime interval with the Euclidean distance formula. "}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"},{"_key":"b95371d541f7","_type":"block","asset":null,"children":[{"_key":"5fa733916251","_type":"span","marks":[],"text":"The "},{"_key":"84089a34e412","_type":"span","marks":["strong"],"text":"negative sign"},{"_key":"97a6e4f63e62","_type":"span","marks":[],"text":" in the equation reflects the fundamental difference between space and time in relativity."}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"}] [{"_key":"fb01132dd702","_type":"block","asset":null,"children":[{"_key":"30cafcc6984a","_type":"span","marks":[],"text":"If two events are separated by $\\Delta x = 3$ km and occur $\\Delta t = 2$ microseconds apart, what is the spacetime interval $\\Delta s$?"}],"markDefs":[],"style":"normal"}] [{"_key":"b7ea97cd9921","_type":"block","asset":null,"children":[{"_key":"b81908575aa2","_type":"span","marks":[],"text":"Two flashes of light occur 600 meters apart with a time interval of 2 microseconds in one frame. "}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"},{"_key":"c664a030657a","_type":"block","asset":null,"children":[{"_key":"86ae45390e19","_type":"span","marks":[],"text":"Using the spacetime interval formula: $$(\\Delta s)^2 = (3.0 \\times 10^8 \\times 2.0 \\times 10^{-6})^2 - (600)^2$$ $$\\Delta s^2 = (600)^2 - (600)^2 = 0$$"}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"},{"_key":"44ff8e3a51b6","_type":"block","asset":null,"children":[{"_key":"17e9191287fa","_type":"span","marks":[],"text":"Since $\\Delta s = 0$, these events are connected by a light signal."}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"}] ## Length Contraction [block_47] **Length contraction **is the phenomenon where an object in motion appears shorter along the direction of motion to a stationary observer. ### Deriving the Length Contraction Formula [block_49] 1. Consider a rod at **rest **in frame $S'$. Its **proper length** (measured in its rest frame) is $L_0$. 2. An observer in frame** $S$ **measures the **length of the rod** while it moves with velocity $v$. [{"_key":"block_53","_type":"block","asset":null,"children":[{"_key":"63fa1c38c160","_type":"span","markDefs":[],"marks":[],"text":"To measure the length in $S$, the observer must record the positions of the rod’s ends "},{"_key":"702017eb47e8","_type":"span","markDefs":[],"marks":["strong"],"text":"simultaneously"},{"_key":"a1c05fd602e9","_type":"span","markDefs":[],"marks":[],"text":"."}],"markDefs":[],"style":"normal"}] 1. Using the **Lorentz transformation for position**: $$ x' = \gamma(x - vt) $$ 2. The length measured in $S$ is: $$ L = x_2 - x_1 $$ 3. The **proper length** is: $$ L_0 = x_2' - x_1' = \gamma(x_2 - vt_2) - \gamma(x_1 - vt_1) $$ 4. Since the measurement is **simultaneous** in $S$ ($t_2 = t_1$), the equation simplifies to: $$ L_0 = \gamma L $$ 5. Solving for $L$ gives the **length contraction formula**: $$ L = \frac{L_0}{\gamma} $$ ![](https://cdn.sanity.io/images/w7j7fz60/production/5016bf96a126ed3e66e2c91c450a178dca8a476a-1971x873.png) [{"_key":"block_65","_type":"block","asset":null,"children":[{"_key":"d46b87057b86","_type":"span","markDefs":[],"marks":[],"text":"Length contraction only occurs"},{"_key":"cc903df7ed67","_type":"span","markDefs":[],"marks":["strong"],"text":" along the direction of motion"},{"_key":"fae6686cc2c5","_type":"span","markDefs":[],"marks":[],"text":". "}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"},{"_key":"e15ec51f57fe","_type":"block","asset":null,"children":[{"_key":"b60ba551426c","_type":"span","markDefs":[],"marks":[],"text":"Perpendicular dimensions remain unchanged."}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"}] A spaceship has a proper length of 100 m. What length does an observer measure if the spaceship travels at $0.8c$? * Calculate $\gamma$:$$ \gamma = \frac{1}{\sqrt{1 - \frac{(0.8c)^2}{c^2}}} = \frac{5}{3} $$ * Apply the length contraction formula:$$ L = \frac{L_0}{\gamma} $$ $$= \frac{100}{\frac{5}{3}} = 60 \text{ m} $$ * The observer measures the spaceship’s length as 60 m. ## Relativistic Velocity Addition [block_70] 1. At **relativistic **speeds, the classical formula for **adding velocities** ($u = u' + v$) **breaks down**. 2. Instead, the **relativistic velocity** **addition formula** is used: $$ u = \frac{u' + v}{1 + \frac{u'v}{c^2}} $$ [{"_key":"block_74","_type":"block","asset":null,"children":[{"_key":"8e7d2e089bb6","_type":"span","markDefs":[],"marks":[],"text":"This formula ensures that the resulting velocity "},{"_key":"b0f69517cbf6","_type":"span","markDefs":[],"marks":["strong"],"text":"never exceeds"},{"_key":"beef5f0a4756","_type":"span","markDefs":[],"marks":[],"text":" the speed of light."}],"markDefs":[],"style":"normal"}] A rocket moves at $0.8c$ relative to the ground. It fires a missile forward at $0.6c$ relative to the rocket. What is the missile’s speed relative to the ground? * Use the relativistic velocity addition formula:$$ u = \frac{u' + v}{1 + \frac{u'v}{c^2}}$$ $$= \frac{0.6c + 0.8c}{1 + \frac{0.6c \cdot 0.8c}{c^2}}$$ $$= \frac{1.4c}{1 + 0.48} = \frac{1.4c}{1.48} \approx 0.95c$$ * The missile’s speed relative to the ground is approximately $0.95c$. [{"_key":"block_79","_type":"block","asset":null,"children":[{"_key":"382834487eeb","_type":"span","markDefs":[],"marks":[],"text":"A common mistake is to assume that velocities simply add up. "}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"},{"_key":"28a41c630c2e","_type":"block","asset":null,"children":[{"_key":"25c06b667c92","_type":"span","markDefs":[],"marks":[],"text":"Always use the "},{"_key":"8d9eaada6d01","_type":"span","markDefs":[],"marks":["strong"],"text":"relativistic formula"},{"_key":"340235dda53c","_type":"span","markDefs":[],"marks":[],"text":" at"},{"_key":"e00f21e63005","_type":"span","markDefs":[],"marks":["strong"],"text":" high speeds"},{"_key":"9298fff5379a","_type":"span","markDefs":[],"marks":[],"text":"."}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"}] [{"_key":"block_83","_type":"block","asset":null,"children":[{"_key":"fad0bb53b665","_type":"span","marks":[],"text":""}],"markDefs":[],"style":"normal"},{"_key":"block_84","_type":"block","asset":null,"children":[{"_key":"ip9ZSUgFyzGIsQgTfxOB1m","_type":"span","markDefs":[],"marks":[],"text":"How do the Lorentz transformations differ from Galilean transformations?"}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"},{"_key":"block_85","_type":"block","asset":null,"children":[{"_key":"ip9ZSUgFyzGIsQgTfxOB3U","_type":"span","markDefs":[],"marks":[],"text":"What is the proper time, and why is it always the shortest time interval?"}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"},{"_key":"block_86","_type":"block","asset":null,"children":[{"_key":"ip9ZSUgFyzGIsQgTfxOB5C","_type":"span","markDefs":[],"marks":[],"text":"Why does length contraction only occur along the direction of motion?"}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"},{"_key":"block_87","_type":"block","asset":null,"children":[{"_key":"ip9ZSUgFyzGIsQgTfxOB6u","_type":"span","markDefs":[],"marks":[],"text":"How does the relativistic velocity addition formula ensure that no object exceeds the speed of light?"}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"}]