Physics 1600 11. Special relativity II

• 4 vector (t,x,y,z)

  • Is the length of this vector an extension of the invariant length?
  • It seems that the length remains the same even when rotated or boosted.

• $\vec{x}\cdot \vec{x}=$ invariant length

  • Why is this? Is it not ignoring the definition of dot product?
  • Oh, it’s in a non-Euclidean space.

• Introducing operators

  • Let $\Lambda ({\beta }_B)$ be the matrix of Lorentz transformations.

• Covariant

  • Lorentz transformations are called covariant transformations because the length remains the same.
  • Are 4-vector length, 4-vector velocity, and 4-vector momentum covariant?

• 4-vector

  • Length

  • Velocity

    • What does it mean to “transform covariantly between inertial frames”?
    • It’s saying that if we just do dx/dt normally, the denominator dt would change depending on the frame, which is not right.
    • So we need a Lorentz-invariant definition of time.
    • Let γt = τ and make τ the proper time.
    • Proper time: measured in the rest frame.
    • Oh, is this what was mentioned in time dilation?
    • Since x, y, z are all 0, τ = γt, right?
    • I understand u.

    • One demonstration of the utility of the four-velocity is the fact that its components perpendicular to a boost – uy and uz for boosts in the x direction are invariant under the LT.

    • I don’t understand any of this.
    • So, $u=(\gamma ,\gamma \dot{\beta })$
    • Of course, this is different from Newtonian.

  • Momentum

    • In special relativity, Newtonian momentum is not conserved, so we want to consider 4-vector momentum.
    • We can consider the momentum in the directions not boosted as invariant, such as $p_y=m\gamma {\beta }_y,p_z=m\gamma {\beta }_z$.
    • [+ However, $m{\beta }_y,m{\beta }_z$ are not invariant like 4-vector momentum.]
    • That’s because if we consider the definition of βy = Δy/Δt, when Δt changes due to a boost, βy and βz also change.
    • If 4-vector momentum is conserved in one frame, it is conserved in all frames.
    • It is trivial because if 4-vector momentum is conserved, Δp = 0.
    • It is obvious from the fact that if all components are not 0 when performing LT, the result after the LT will not be 0.
    • Details on p. 408.

  • Energy

    • The t-component of the 4-vector momentum becomes energy.
    • That is, $m\gamma$.
    • Let’s just think of it as that for now.
    • Relationship with non-relativistic energy:
    • Adding m to non-relativistic energy gives 4-vector energy (?).

  • Force

    • $F=\frac{dp}{dt}$ (Newton’s 2nd law) holds if we consider the proper SR momentum (3-vector momentum $=m\beta \gamma$).
    • (This momentum is called 3-vector momentum $=m\beta \gamma$), right?
    • dK/dt = Power = Fv.

  • ==p = (E, 3vec SR momentum) = m(γ, βγ)==

    • This relationship is important.
    • Using this, we can transform $p\cdot p=m^2$.
    • It’s a non-Euclidean invariant length, so be careful.
    • ==If we transform it, we can obtain E^2 = p^2 + m^2.==
    • I don’t really understand, but the units seem to match…?
    • Is it because of natural units?
    • If you understand this, you’re mostly good.
    • 
    • It summarizes what we have done so far.

  • Massless

    • As mentioned above, mass is the invariant length of momentum.
    • So, mass is invariant.
    • E=m is ultimately wrong.
    • If it is velocity-invariant, E=m is incorrect.
    • The correct one is E=mγ.
    • Some people call “mγ” velocity-variant mass, but that is a misunderstanding.
    • I’m not really grasping the discussion about wrong mass and real mass.
    • 
    • In the case of massless particles, the energy is equal to the momentum, E=|p|.
    • So there’s no problem.
    • At that time, β becomes 1, which matches with photons.

  • 

    • If I stare at this, I might understand.
    • If it’s in natural units, all the c’s cancel out.

  • Although 4-vectors are all covariant, they are not invariant.

    • They change when LT is applied.- (Since we are using proper time, the component perpendicular to the boost remains unchanged)

• By saying covariant, does it mean that if we boost in the x-direction, the y and z components remain invariant?

• And can we transform them using the same calculations as Lorentz transformations? (blu3mo)(blu3mo)

• Very important: the results for the velocity of the center of mass (COM) and velocities of particles relative to the COM that we obtained previously no longer apply in special relativity (SR).