Ep-14HC Verma- 4-Vectors for Velocity and momentum by HC Verma || in hindi || theory of relativity

Most famous book of Prof. HC Verma sir
concept of physics volume1 & 2
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quantum physics book by HC Verma sir
hc verma part 1 amzn.to/3GvdDR8
foundation science physics for class 9(cbse)
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foundation science physics for class 10 (cbse)
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In special relativity, four-momentum is the generalization of the classical three-dimensional momentum to four-dimensional spacetime. Momentum is a vector in three dimensions; similarly four-momentum is a four-vector in spacetime. The contravariant four-momentum of a particle with relativistic energy E and three-momentum p = (px, py, pz) = γmv, where v is the particle's three-velocity and γ the Lorentz factor, is
{\displaystyle p=(p^{0},p^{1},p^{2},p^{3})=\left({E \over c},p_{x},p_{y},p_{z}
ight).}{\displaystyle p=(p^{0},p^{1},p^{2},p^{3})=\left({E \over c},p_{x},p_{y},p_{z}
ight).}
The quantity mv of above is ordinary non-relativistic momentum of the particle and m its rest mass. The four-momentum is useful in relativistic calculations because it is a Lorentz covariant vector. This means that it is easy to keep track of how it transforms under Lorentz transformations.
The above definition applies under the coordinate convention that x0 = ct. Some authors use the convention x0 = t, which yields a modified definition with p0 = E/c2. It is also possible to define covariant four-momentum pμ where the sign of the energy is reversed.

[@hcvermacourse8310]

Most famous book of Prof. HC Verma sir concept of physics volume1 & 2 amzn.to/3GvdDR8 quantum physics book by HC Verma sir hc verma part 1 amzn.to/3GvdDR8 foundation science physics for class 9(cbse) amzn.to/31cjESO foundation science physics for class 10 (cbse) amzn.to/3jK1v5i

[@prithamallik3454]

Thank you, Sir !! You're really great !!

[@MaheshSharma-fo9lj]

respected sir, you said that variable mass concept can work with momentum but will not work with force and kinetic energy . but if i talk about force , according to newton's second law of motion F ⃗=(ⅆP ⃗)/ⅆt and mass is not constant it will not come out of this integration so force is never equal to ma. so if i derive m(v)v with time i get (1/(1-v^2/C^2 ))^(3/2) as force an kinetic energy is 1/2 mv^2 only when mass is constant if mass is not constant we write K.E= ∫F ⃗⋅ⅆS . from here we get K.E= m_0 (c^2-v^2 ) . so, even with variable mass concept we are able to do it so why we need this proper velocity concept ? if there has been any misunderstanding in understanding this from my side , please please explain, sir

[@rajashreelipi783]

Sir ek question thi? Plz

[@sahilnegi4326]

But the velocity vector doenst transform the way we dervied without the 4vectors

[@rahulgupta2youtub100]

Shouldn't d(tou)=dt×(lorentz constant) and not divided by it?

[@nishitkrsingh]

In S frame of reference: η=u/√(1-u²/c²) where, u=dx/dt so, η=dx/dt√(1-u²/c²) where dτ=dt√(1-u²/c²) because in special relativity, the faster you travel, slower time becomes (Time Dilation) so, √(1-u²/c²) gives the desired result... As you'll increase u, dτ decreases (τ- Time which is constant for all frame of references). If you considered 1/√(1-u²/c²), you can check, time will increase with the increase in u. Hope it Helped!