Physics 62 Special Relativity (30 of 43) The Relativistic Triangle

In this video I will explain the relativistic triangle - an ingenious method of graphically representing the relationship between the momentum, the rest mass energy, the total energy, and the kinetic energy of a particle at relativistic speeds.

Next video in the Special Relativity series can be seen at:

[chavab]

This reminds me of Minkowski's contribution to Einstein; turning algebra into geometry & trig. Does that make sense to anyone?

[Lleanlleawrg]

I would in no way classify myself as especially smart. Maybe slightly above average at best - and I independently arrived at this over the span of a couple of weeks recently.
This just sort of happened because I knew the pythagorean theorem and I knew any increase in spatial velocity comes at the cost of time dilation, and so any increase in spatial velocity means a decrease in temporal velocity. That told me we should be able to represent time dilation in terms of a right triangle, and use the pythagorean theorem and trig to model the relationship and derive the ratio of spatial and temporal velocities, which at the time (like a couple months ago now I guess) I didn't know was the lorentz factor. I hadn't even heard of minkowski space, but I knew spacetime is a 4D geometry according to Einstein though..

Anyway, point is - you don't necessarily have to be some sort of genius to do this. I don't think I am, hell, I barely scraped through in school. But I'm curious about stuff, and I learn by associating new information to information I already know.

[kinjalpatel]

Why we draw that arc can you please explain?

[ammudhakshayani]

why we are not substituting K.E for total energy while calculating cos(theta)? But you did that for calculating sin(theta)... I couldn't get it can u explain?

[alistairfernandes]

I really appreciate how you explained it in the simplest way possible. Also when you expressed v in terms of gamma, I don't know if you noticed but if you substitute 1/gamma by cos(theta) and simplify, you get v = csin(theta) which is kind of fascinating. Please let me know if this is right.