37-2 The Relativity of Length#
Prompts
Why must you mark the positions of the front and back of a moving rod simultaneously (in your frame) to measure its length? How does the relativity of simultaneity imply that length is frame-dependent?
What is proper length \(L_0\)? If you measure a rod’s length when it moves past you at speed \(v\), will you get \(L_0\) or a smaller value?
Derive length contraction \(L = L_0/\gamma\) from time dilation. (Hint: consider Sally on a train and Sam on a platform; who measures proper time for the platform to pass?)
Length contraction occurs only along the direction of motion. Does a cube moving past you look like a flattened box? What about dimensions perpendicular to \(v\)?
A spaceship of proper length 230 m passes you in 3.57 \(\mu\)s. How can you find the relative speed \(v\)? Which frame gives proper length? Which gives proper time?
Lecture Notes#
Overview#
Length is relative: observers in different inertial frames measure different lengths for the same object or distance.
Proper length \(L_0\) is the length measured in the rest frame of the object. Any observer moving parallel to that length measures a contracted length \(L < L_0\).
Length contraction follows from time dilation and the relativity of simultaneity. Measuring length requires simultaneous marking of endpoints—and simultaneity is frame-dependent.
Contraction occurs only along the direction of relative motion; dimensions perpendicular to \(\vec{v}\) are unchanged.
Measuring length: why simultaneity matters#
To measure the length of a rod at rest, you note the positions of its endpoints on a stationary scale and subtract. For a moving rod, you must mark the positions of the front and back simultaneously (in your reference frame). If you mark the front at one time and the back later, the rod has moved—your “length” would mix space and time and would not be a length.
Because simultaneity is relative (section 37-1), length must also be relative. Different observers, using their own notion of simultaneity, will measure different lengths for the same rod.
Proper length and length contraction#
The proper length \(L_0\) (or rest length) of an object is the length measured in the inertial frame in which the object is at rest.
If you and the object are in relative motion at speed \(v\) along the length of the object, and you mark the endpoints simultaneously in your frame, you measure a contracted length:
where \(\beta = v/c\) and \(\gamma = 1/\sqrt{1 - \beta^2}\). Since \(\gamma > 1\) when there is relative motion, \(L < L_0\).
Length contraction occurs only along the direction of relative motion. Dimensions perpendicular to \(\vec{v}\) are unchanged.
The “length” can be the distance between two objects at rest relative to each other (e.g., two stars), not just an object like a rod.
Does the object really shrink?
What we measure is real. The object is measured to be shorter—motion affects the measurement. An observer moving with the rod would say you did not mark the endpoints simultaneously: you marked the front first, then the rear later, so you got a shorter length. Both are correct in their own frames.
Derivation from time dilation#
Length contraction is a direct consequence of time dilation. Consider Sally on a train and Sam on a platform. Sam measures the platform’s length as \(L_0\) (proper length, platform at rest). Sam sees Sally pass through this length in time \(\Delta t = L_0/v\) (using two clocks).
For Sally, the platform moves past her. The two events (Sally passes the back, Sally passes the front) occur at the same place in her frame—at her position. She measures proper time \(\Delta t_0\) with one clock. The platform length in her frame is \(L = v\,\Delta t_0\).
From time dilation, \(\Delta t = \gamma\,\Delta t_0\). Thus \(L_0/v = \gamma\,(L/v)\), so \(L = L_0/\gamma\).
Careful use of frames#
When solving problems, always use distance and time from the same frame in relations like \(v = L/\Delta t\). Do not mix measurements from different frames.
Observer |
Measures |
Proper length? |
Proper time? |
|---|---|---|---|
Sam (platform rest) |
Platform length \(L_0\), time \(\Delta t\) for Sally to pass |
Yes |
No |
Sally (train) |
Platform length \(L = L_0/\gamma\), time \(\Delta t_0\) for platform to pass |
No |
Yes |
Example: spaceship passing
A ship of proper length \(L_0 = 230\) m passes Sally in \(\Delta t = 3.57\,\mu\)s (Sally’s frame). In her frame the ship has contracted length \(L = L_0/\gamma\), so \(v = L/\Delta t = L_0/(\gamma\,\Delta t)\). Substituting \(\gamma = 1/\sqrt{1 - v^2/c^2}\) and solving for \(v\) gives \(v = L_0/\sqrt{(\Delta t)^2 + (L_0/c)^2} \approx 0.21c\).
Summary#
Proper length \(L_0\): length in rest frame of object
Length contraction \(L = L_0/\gamma\): moving observers measure shorter length along the direction of motion
Requires simultaneity in the measuring frame; relativity of simultaneity makes length frame-dependent
Contraction only along \(\vec{v}\); perpendicular dimensions unchanged
Length contraction derives from time dilation; keep distance and time in the same frame when computing \(v\)