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    The special theory of relativity

    Flashcards for topic The special theory of relativity

    Intermediate41 cardsphysics

    Preview Cards

    Card 1

    Front

    Explain time dilation in special relativity with its mathematical formulation and physical interpretation.

    Back

    Time Dilation: Moving clocks run slower than stationary clocks.

    Mathematical Formula: Δt′=Δt1−v2/c2=γΔt\Delta t' = \frac{\Delta t}{\sqrt{1-v^2/c^2}} = \gamma \Delta tΔt′=1−v2/c2​Δt​=γΔt

    Where:

    • Δt\Delta tΔt = proper time (time measured in frame where events occur at same location)
    • Δt′\Delta t'Δt′ = improper time (time measured in frame where events occur at different locations)
    • γ=11−v2/c2\gamma = \frac{1}{\sqrt{1-v^2/c^2}}γ=1−v2/c2​1​ (Lorentz factor, always > 1)

    Physical Interpretation:

    • The proper time interval between two events is always the smallest possible measured time
    • For a moving observer, time passes more slowly by a factor of γ
    • This is a real physical effect, not an illusion or measurement error
    • Time dilation is reciprocal: each observer sees the other's clock running slower
    • This effect becomes significant only at speeds approaching c
    • Verified experimentally with muon decay rates and atomic clocks on aircraft

    Example: A spaceship traveling at 0.8c relative to Earth will experience only 6 years of elapsed time during what Earth observers measure as 10 years.

    Card 2

    Front

    How does length contraction work in special relativity, and what is the crucial constraint on when it applies?

    Back

    Length Contraction: Objects moving relative to an observer appear shortened along their direction of motion.

    Formula: L′=L1−v2/c2=LγL' = L\sqrt{1-v^2/c^2} = \frac{L}{\gamma}L′=L1−v2/c2​=γL​

    Where:

    • LLL = proper length (length measured in rest frame of object)
    • L′L'L′ = contracted length (length measured in frame where object is moving)
    • γ=11−v2/c2\gamma = \frac{1}{\sqrt{1-v^2/c^2}}γ=1−v2/c2​1​ (Lorentz factor)

    Crucial Constraint: Length contraction occurs ONLY in the direction parallel to motion. Dimensions perpendicular to motion remain unchanged.

    Key aspects:

    • Only affects the dimension along the direction of relative motion
    • The proper (rest) length is always the maximum possible measured length
    • Length contraction is reciprocal between reference frames
    • This is a real physical effect, not an optical illusion
    • No "stress" is involved—the object appears perfectly normal in its own rest frame

    Example: A 10-meter rod moving at 0.87c (where γ = 2) would measure only 5 meters in length to a stationary observer.

    Card 3

    Front

    How does special relativity connect energy and mass? Explain the famous equation, its complete form, and its profound implications.

    Back

    Mass-Energy Equivalence: Basic form: E=mc2E = mc^2E=mc2

    Complete relativistic energy equation: E2=(m0c2)2+(pc)2E^2 = (m_0c^2)^2 + (pc)^2E2=(m0​c2)2+(pc)2

    Where:

    • EEE = total energy
    • m0m_0m0​ = rest mass
    • ppp = relativistic momentum
    • ccc = speed of light

    For an object with velocity v: E=m0c21−v2/c2=γm0c2E = \frac{m_0c^2}{\sqrt{1-v^2/c^2}} = \gamma m_0c^2E=1−v2/c2​m0​c2​=γm0​c2

    Profound implications:

    1. Mass and energy are fundamentally interchangeable forms of the same entity
    2. Even objects at rest contain enormous energy (E0=m0c2E_0 = m_0c^2E0​=m0​c2)
    3. Energy contributes to inertia and gravitational effects
    4. Mass can be converted to other forms of energy (nuclear fission/fusion)
    5. Pure energy can create particles with mass (pair production)
    6. Conservation laws must include both mass and energy together
    7. A small mass defect accounts for the binding energy in atomic nuclei

    This equivalence revolutionized physics, leading to nuclear energy, explaining stellar processes, and forming a foundation for particle physics and cosmology.

    Card 4

    Front

    How does a light-beam clock demonstrate time dilation in special relativity, and what is the mathematical relationship between proper time (Δt) and improper time (Δt')?

    Back

    A light-beam clock consists of two mirrors separated by distance L with light bouncing between them:

    • In a stationary reference frame S, light travels distance 2L at speed c, with time interval Δt = 2L/c
    • In a moving reference frame S' (velocity v), light follows a longer diagonal path while mirrors move
    • The light must travel further but still moves at speed c, requiring more time

    Mathematical relationship: Δt′=Δt1−v2/c2=γΔt\Delta t' = \frac{\Delta t}{\sqrt{1-v^2/c^2}} = \gamma \Delta tΔt′=1−v2/c2​Δt​=γΔt

    Where:

    • Δt = proper time (measured in frame where events occur at same location)
    • Δt' = improper time (measured in frame where events occur at different locations)
    • γ = Lorentz factor = 1/√(1-v²/c²)

    This demonstrates the central principle: moving clocks run slower than stationary clocks by a factor of γ.

    Card image
    Card 5

    Front

    What happens to the simultaneity of events when observed from different reference frames in special relativity?

    Back

    • Events that are simultaneous in one reference frame are generally NOT simultaneous in another reference frame that moves relative to the first.

    • For two events occurring at different spatial locations:

      • If simultaneous in reference frame S
      • Will NOT be simultaneous in reference frame S' moving at velocity v relative to S
    • The time difference between events in the moving frame is: Δt′=Lvc2γ\Delta t' = \frac{Lv}{c^2}\gammaΔt′=c2Lv​γ Where:

      • L = spatial separation between events in the stationary frame
      • v = relative velocity between frames
      • γ = 1/√(1-v²/c²)
    • Example: Light emitted from the center of a box reaches both ends simultaneously for a stationary observer, but for a moving observer, the light reaches the rear end before the front end.

    Note: This directly contradicts our intuitive notion that simultaneity is absolute and demonstrates that time ordering depends on the observer's reference frame.

    Card image
    Card 6

    Front

    In a relativistic thought experiment with trains T₁ and T₂ moving in opposite directions, how would the opening sequence of doors D₁ and D₂ differ when viewed from different reference frames?

    Back

    • Analysis from different reference frames reveals contradictory but equally valid door-opening sequences:

    • Ground frame:

      • Box of length L with doors D₁ and D₂ at ends
      • Light source at center C emits signals simultaneously
      • Both doors open simultaneously (t = L/2c)
    • Train T₁ frame (moving left at velocity v):

      • Box appears length-contracted
      • Door D₁ opens first
      • Door D₂ opens later by time interval Δt = Lv/c²γ
      • Observer concludes: "D₁ definitely opened first"
    • Train T₂ frame (moving right at velocity v):

      • Box appears length-contracted
      • Door D₂ opens first
      • Door D₁ opens later by time interval Δt = Lv/c²γ
      • Observer concludes: "D₂ definitely opened first"
    • This demonstrates that the temporal ordering of spatially separated events is not absolute but depends on the observer's state of motion.

    Example: A referee standing midfield sees two football players touch boundary lines simultaneously, while a moving observer sees one player touch first—both observations are correct in their respective frames.

    Card image
    Card 7

    Front

    Explain how clock synchronization varies between different inertial reference frames in the context of relativity.

    Back

    In relativity, clock synchronization is frame-dependent: • Events that are simultaneous in one frame are not simultaneous in another frame • In a frame where objects are moving, clocks at different positions appear desynchronized • For objects moving at velocity vvv, distant clocks appear offset by γvx/c2\gamma vx/c^2γvx/c2 where xxx is their separation • Front clocks (in direction of motion) appear to lag behind rear clocks when viewed from a stationary frame • When changing reference frames, this synchronization difference must be accounted for • This effect explains why the "leading" clock in one frame becomes the "lagging" clock in another frame moving in the opposite direction

    Card 8

    Front

    Given a relativistic scenario with two objects moving at 0.8c relative to each other, calculate the time dilation factor and explain its significance.

    Back

    For objects moving at velocity v=0.8cv = 0.8cv=0.8c relative to each other:

    Time dilation factor: γ=11−v2/c2=11−(0.8c)2/c2=11−0.64=10.36=10.6=1.67\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = \frac{1}{\sqrt{1-(0.8c)^2/c^2}} = \frac{1}{\sqrt{1-0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} = 1.67γ=1−v2/c2​1​=1−(0.8c)2/c2​1​=1−0.64​1​=0.36​1​=0.61​=1.67

    Significance: • Moving clocks run slower by this factor (appear to tick at 0.6× normal rate) • 1 second in the moving frame corresponds to 1.67 seconds in the stationary frame • 12 years on a stationary clock corresponds to 12 × 0.6 = 7.2 years on the moving clock • This factor is reciprocal - each observer sees the other's clock running slow by the same factor • This creates the apparent paradox until reference frame changes are properly accounted for

    Card 9

    Front

    How does the ordering of events differ between reference frames, and why is this critical to resolving the Twin Paradox?

    Back

    The ordering of events differs between reference frames due to: • Events that are simultaneous in one frame are not simultaneous in another • The relativity of simultaneity means that temporal ordering can change between frames • Spatially separated events with timelike separation maintain their order in all frames • Events with spacelike separation can have different ordering in different frames

    This is critical to resolving the Twin Paradox because: • When Ram changes frames, he must account for different event orderings • In Earth's frame, the Planet and Earth clocks read zero simultaneously • In S₁-frame, "Planet's clock reading zero" occurred years before "Earth's clock reading zero" • When switching to S₂, the Earth's clock is leading rather than lagging • These ordering differences account for the final age discrepancy, resolving the paradox

    Card 10

    Front

    Explain why two twins moving relative to each other don't experience a truly symmetrical situation despite the principle of relativity.

    Back

    The situation is asymmetrical because:

    • One twin (Ram) changes reference frames while the other (Balram) remains in a single inertial frame • The twin who changes frames experiences proper acceleration, which is absolute and frame-independent • This acceleration breaks the symmetry of special relativity's inertial reference frames • The twin who stays in one frame follows a geodesic (straight worldline) in spacetime • The traveling twin follows a non-geodesic path with direction changes • Direction changes require accounting for different synchronization conventions between frames • The twin who changes frames must account for the discontinuity in how distant clocks are synchronized

    This fundamental asymmetry is why the paradox is only apparent, not real.

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