Show that momentum and energy transform from one inertial frame to another as px^'=γ(px-v E / c^

Show that momentum and energy transform from one inertial...

Key Concepts

Gamma Factor

The gamma factor, defined as ? = 1/?(1 - v²/c²), plays a crucial role in relativistic transformations. It quantifies the amount of time dilation, length contraction, and the corresponding adjustments in energy and momentum observed when switching between inertial frames.

Relativistic Energy and Momentum

In relativity, energy and momentum are deeply interconnected. Their transformation properties, which involve the Lorentz factor, illustrate how measurements of energy and momentum depend on the relative motion between observers in different inertial frames.

Inertial Frames

Inertial frames are reference frames in which an object either remains at rest or moves at a constant velocity when no forces are acting upon it. They are fundamental to the formulation of the laws of physics in special relativity, ensuring that physical laws are the same for all observers in uniform motion.

Lorentz Transformation

Lorentz transformations are the mathematical formulas used in special relativity to convert space and time coordinates from one inertial frame to another. They guarantee that the laws of physics, including the invariant speed of light, hold true in all inertial reference frames.

Four-Momentum

Four-momentum is a four-dimensional vector that unifies energy and momentum into a single consistent framework in the context of special relativity. Its components transform under Lorentz transformations in a way that preserves the invariant mass of an object.

Transcript

00:01 So, it has been asked to show the momentum and energy transform from one reference frame to another, where the reference frames are moving relative to each other.

00:21 They are related as p x prime is equals to gamma times p x minus v e by c square, p y prime is p y, p z prime is p z and e z prime is gamma e minus v p x.

00:44 Now these expressions are for the reference frames which are moving along the p x direction.

00:50 So let us consider two momentum frames, reference frames s and s prime, such that the frame s prime is moving away from frame s with a velocity v along the p x direction.

01:15 So let this be the three coordinate axis, momentum coordinate axis p x, p y, p z of the frame s and these are for frame s prime, p x prime, p y prime and p z prime.

01:33 This is frame s and this is frame s prime and frame s prime is moving away from frame s with a velocity v along p x direction.

01:46 Let t be an event represented by a point, let us say this is the point, they are represented by the coordinates, momentum coordinates p x, p y, p z and e, energy e, whereas for the observer in frame s prime, the coordinates of the event t would be p x prime, p y prime, p z prime and the energy would be e prime.

02:12 So the momentum would be e prime by c.

02:14 The four momentum vectors for frame, for the observer in frame s would be e by c, p x, p y, p z and for frame s prime would be e prime by c, p x prime, p y prime, p z prime.

02:30 Now, how we have to find out the transformation relation between p x prime and p x, p y prime and p y, p z prime and p z and e prime and e.

02:43 For an observer having a rest mass m0 moving with a relativistic speed v, its energy e is given as mc square that is m0 c square by root over 1 minus v square by c square, which we can write it as gamma m0 c square where gamma is the lorentz factor, which is 1 by under root 1 minus v square by c square.

03:07 Similarly, the relativistic expression for momentum is m0 v by under root 1 minus v square by c square that is gamma times m0 v.

03:18 Now from equation 1, we can write gamma to be equal to e by m0 c square.

03:27 Putting this value of gamma in equation 2, we get p that is equal to m0 v divided, multiplied by e by m0 c square.

03:39 So, m0 m0 cancels, therefore p is ev by c square.

03:44 Let v be equation number 3.

03:48 Now, as per the special theory of relativity, the length of the vectors will be contracted when observed from different frames of reference, which are moving relative to each other, from different frame, reference frame, s and s prime moving relative to each other or moving with respect to each other.

04:25 So, let this be the x coordinate, o t prime is the x component of the vector, o t and o prime t.

04:35 So, o prime t as observed by the observer in frame s will appear shorter than o t as per special theory of relativity.

04:50 So, o prime t will appear shorter than o t to an observer in frame s by a factor root over 1 minus v square by c square or 1 by gamma.

05:03 The momentum of the event t, x component of momentum, the x component of momentum of the momentum of the event t with respect to the observer in frame s is then o t for the observer in frame s would be p x.

05:31 Similarly, the x component of the momentum of the event t with respect to the observer in frame s.

05:41 So, now o prime t with respect to the observer in frame s is p x with respect to the observer in frame s prime is if it is p x prime, then for the x component of the momentum of the event at the origin o prime with respect to the observer at o is e v by c square.

06:08 So, o prime t for an observer for the observer in frame s would be p x prime, it will be shortened by a factor 1 minus v square by c square root over.

06:20 So, it is p x prime root over 1 minus v square by c square.

06:25 So, for the observer in frame s, we can write this relation o t prime is o prime t plus o o prime.

06:32 O t is p x and o prime t is p x prime root over 1 minus v square by c square o o prime is e v by c square.

06:39 So, p x would be p x prime by gamma plus v e by c square.

06:45 So, p x prime by gamma is p x minus v e by c square.

06:51 So, we have to get an expression for p square p prime.

06:54 So, p prime is gamma whole multiplied by p x minus v e by c square...