What's the difference between Lorentz factor for frames and for particles?

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The Lorentz factor for moving inertial reference frames and for particles are both expressed similarly but serve different purposes. The Lorentz factor for frames, λ, relates to the relative velocity between two inertial frames, while the particle Lorentz factor, λp, pertains to the velocity of a particle itself. Time dilation for a particle is a measurable effect based on its proper time, while time dilation for coordinate transformations is a property of the coordinate systems involved. These concepts are interconnected, as the Lorentz transformation can be derived from the time dilation of particles and vice versa. Understanding these distinctions is crucial for grasping the principles of relativity.
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Lorentz factor for moving inertial reference frames is

λ = \frac{1}{\sqrt{1 - \frac{v<sup>2</sup>}{c<sup>2</sup>}}},

where v is the relative velocity between the frames. But in my textbook (I'm only just learning relativity), it says the Lorentz factor for a particle is

λp = \frac{1}{\sqrt{1 - \frac{u<sup>2</sup>}{c<sup>2</sup>}}},

where u is the velocity of the particle...but what's the difference between these two expressions?

The textbook introduces λp during the derivation of relativistic momentum, so I'm fine with where it came from, but then without explanation the book starts using it for time dilation Δt = λpΔ\tau. Little confused so would love it if someone could clear this up for me.

Thanks in advance!
 
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There are two different, but related, concepts:
  1. Time dilation for a particle.
  2. Time dilation for a coordinate transformation.

For particle time dilation: (For simplicity, let's just look at one spatial dimension)
If a particle travels along some trajectory described by an equation such as:

x = f(t)

where (x,t) are coordinates in an inertial Cartesian coordinate system, then the proper time \tau experienced by the particle as it travels along the trajectory is given by:

\tau = \int \sqrt{1-\dfrac{v^2}{c^2}} dt

where v = \dfrac{df}{dt}

For coordinate transformation time dilation: (Again, just using one spatial dimension): Suppose you have two inertial Cartesian coordinate systems (x,t) and (x&#039;,t&#039;), such that the spatial origin of the primed system moves at speed v relative to the unprimed system. Let e_1 and e_2 be two events (points in spacetime). Let (\delta(x),\delta(t)) be their separation according to the unprimed coordinate system, and (\delta(x&#039;),\delta(t&#039;)) be their separation according to the primed coordinate system.

If

\delta(x&#039;) = 0

(so the two events take place at the same location, according to the primed coordinate system), then

\delta(t&#039;) = \sqrt{1-\dfrac{v^2}{c^2}} \ \delta(t)

Time dilation for particles is a physical effect, which is actually measurable. Time dilation for coordinate systems is just a property of a particular coordinate transformation (the Lorentz transformation). However, they are related, in the sense that the Lorentz transformation is what you would expect to apply if you create a coordinate system based on time-dilated clocks (so you can derive the LT from particle time dilation), and the other way around, if you assume that the laws of physics governing clocks is Lorentz-invariant and insensitive to past history of accelerations, then clock time dilation is derivable.
 
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