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In the Lagrangian formalism of mechanics the momentum is defined as. p ≡ ∂L ∂v. Using the relativistic Lagrangian for a free massive particle. L = − mc2√1 − v2 c2. we obtain p = γmv from the definition. The conservation laws can be obtained by using the ordinary Lagrangian formalism.
Suppose that the momentum calculation turns out to be as follows: mAuA + mBuB = mCuC + mDuD. Now, lets move to a reference frame that is moving uniformly with some velocity v. The transformation of each velocity is given by: ui → ˉui + v 1 + (ˉuiv / c2) and hence our momentum calculation yields: mA ˉuA + v 1 + (ˉuAv / c2) + mB ˉuB + v 1 ...
Are relativistic momentum and relativistic mass conserved in special relativity? 1. Under which ...
Differentiating relativistic momentum. I am struggling to differentiate relativistic momentum formula. Considering special relativity, →F = d→P dt = d dt m→v √1 − v2 / c2. which I understand. The textbook proceeds, "when the net force and velocity are both along the x-axis," F = m (1 − v2 / c2)3 / 2a. This is where I am stuck.
To give an idea of the rigour I expect in an answer, in example, an answer I'd accept for the derivation of $ E = \frac{1}{2} m v^2$ in classical mechanics would have been as follows:
where m m is the relativistic mass here, hence. m = p/c m = p / c. In other words, a photon does have relativistic mass proportional to its momentum. De Broglie's relation, an early result of quantum theory (specifically wave-particle duality), states that. λ = h/p λ = h / p. where h h is simply Planck's constant.
Suppose the de Broglie wave-length is (non-relativistic) case: λ = h p = h mv In the case of RELATIVISTIC particle, the momentum is p = γmv. Therefore a way to recast the de Broglie wavelength is: λr = h√1 − v2 / c2 mv Suppose now that we focus on the kinetic energy. For a free particle, we get in the nonrelativistic case, KE = T = p2 ...
Using Relativistic Velocity Transformation, I find v′2 = v + v2 1 + v ⋅v2/c2. The momentum of the system shouldn't change in both frames of reference. As a result of conservation of momentum, equation (1) and (2) (shown below) should always be true. Just to make sure everyone is on the same page, I'm using γ(v) = 1 1 − v2 c2− −−− ...
1. In a recent discussion a friend of mine claimed that kinetic energy (K K) an momentum (p p) in relativity can me expressed by. K = p2 (1 + γ)m (1) (1) K = p 2 (1 + γ) m. This equation if holds, has some cool significance for me, because it may show a smoother connection between classical and relativistic mechanics, as it is extremely easy ...
Relativistic angular momentum in a related discussion suggests that the component of the relativistic angular momentum along the direction of motion between two inertial frames is the same for both frames. It can be inferred from this that the relativistic mass used for calculating the relativistic angular momentum in that direction is ...