Derivation of E=mc^2 and Lorentz force from relativistic Lagrangian

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Күн бұрын

Classical Mechanics and Relativity: Lecture 4
0:00 Introduction
2:22 Invariants in 3d space and 4d spacetime
13:50 The Action as a relativistic invariant
15:15 Relativistic Action
22:14 Relativistic Lagrangian
31:23 Derivation of E=mc^2
35:31 Example: Electromagnetism
45:14 Electromagnetic Action
47:05 Derivation of Lorentz force
Theoretical physicist Dr Andrew Mitchell presents an undergraduate lecture course on Classical Mechanics and Relativity at University College Dublin. This is a complete and self-contained course in which everything is derived from scratch.
In this lecture we generalize the Lagrangian formalism to include Einstein's special theory of relativity, starting from the requirement that the action be the same in all reference frames, to all observers. From this we can quickly derive E=mc^2. We then apply the relativistic Lagrangian formulation to the classical theory of electromagnetism, as an example, and derive the Lorentz force for magnetism.
A more in-depth discussion of relativity in electromagnetism can be found here:
• Relativistic Theory of...
Full lecture course playlist: • Classical Mechanics an...
Course textbooks:
"Classical Mechanics" by Goldstein, Safko, and Poole
"Classical Mechanics" by Morin
"Relativity" by Rindler

Пікірлер: 10

@strippins 15 күн бұрын

Phenomenal lecture, if there was just one change I would have loved you to have explained where the minus sign came from in the four vector

@amiralivanaki1150 3 жыл бұрын

Very instructive and comprehensive. Thanks a lot.

@haniefsofi 8 ай бұрын

You missed 1/2. Your lectures are nice. Thanks

@aname1362 2 жыл бұрын

At 26.09 is it supposed to be kc - 1/2k/cv^2; so there's a 1/2 missing. Then I can get the k = - mc. Also, this is fantastic series so far. And is appreciated.

@oni8337 10 ай бұрын

How do you arrive at k=-mc? Did you gauge fix with V=mc^2 when equating that expression with k on the LHS with T-V on the right?

@JSHD2010 Жыл бұрын

Dr Mitchell, since the Euler Legrange equations can also be presented or morphed into the geodesic metric could this mean that for non relativistic particles the total rest mass energy E=mc2 can be made equivalent to some geodesic distance (e.g. Work Done = Force x Distance (joules))? NB/ Where,the geodesic distance would be Lorentz invariant and the mass from Force = mass × acceleration would be relativistic (and a rest mass).

@jolez_4869 3 жыл бұрын

At around 30:00 you asserted that the Hamiltonian equals the total energy. I have watched the proof of this for nonreletavistic Hamiltonian, but I have not seen the proof for this specific Hamiltonian you are using in the video. Do you know where I can find one?

@oded2304 Жыл бұрын

On min.26:12 there is a half factor missing.

@abcdef2069 2 жыл бұрын

at 21:56 ds^2

@drmitchellsphysicschannel2955 2 жыл бұрын

The spacetime interval ds^2 can be defined with either sign -- this is just a convention depending on the signature of the metric. Since the action has to be real, we want |ds^2|, which for our signing convention means -ds^2. This is obviously the right choice because it reduces to the classical kinetic energy in the non-relativistic limit. One can incorporate the relative minus sign between time- and space-like four vector components in a Lorentz scalar by letting t -> i*t. This transformation to imaginary time is called a Wick rotation. However, the formalism involving four vectors is much more powerful than that and has much deeper geometrical meaning. The full structure of the theory is needed and exploited in general relativity where the metric is not so simple, and in fact changes from point to point in space depending on the mass-energy distribution. We didn't get on to that in this course.