 In the last video, we showed that Newtonian momentum, p equals mv, is not conserved in special relativity. Instead, we need to consider p equals gamma mv. So does this mean Newtonian momentum is wrong? Well, let's see what happens when v is much smaller than c. Well, in that case, gamma is approximately equal to 1, and our relativistic momentum reduces to p equals mv, the Newtonian case. So it's not that Newtonian momentum is wrong, per se. It's just that it's an approximation valid in a particular regime when our speed is much less than the speed of light. Relativistic momentum is also only valid in particular regimes. It works well in a lot of cases, but if you were to try and apply it to a black hole or to something on the quantum level where special relativity breaks down, then you'd find problems, and you'd have to consider a different expression. So what about if your velocity approaches c? Well, in that case, gamma is equal to 1 on 1 minus v squared on c squared, and approaches infinity. So in other words, if you're trying to move a massive object at the speed of light, you need to give it infinite momentum, which would require an infinite force. And this is why you can't accelerate anything to the speed of light. So now we know about relativistic momentum, let's derive an expression for relativistic energy. Force is defined as the rate of change of an object's momentum. Or writing that in terms of calculus, f equals dp by dt. So if you have an object with some initial momentum, and I apply a force to it for time t, its momentum at time t will be the initial momentum, p0, plus the force times time. Substituting in our expression for momentum, that's the gamma at time t times n times the velocity at time t. Or expanding gamma. So now let's solve this for velocity. Now this would be very tedious and difficult to do by hand, so we just shove that all into a computer, and it spits out this. So what does this mean? It means if we have an object of mass m, and an initial momentum p0, and we apply a force f for a time t, we plug those values into the right hand side, and then this tells us what velocity we end up with. Okay, so another expression you might remember is work is equal to force times distance. So now the work I do on an object is equal to the change in energy of that object, and the distance an object will travel is the integral of its velocity v over the time period. So we have delta e equals f times the integral of v. Or substituting in our expression for v, we get this. So let's assume we start off at time zero, and we finish at time capital T. Now to evaluate this integral, we just shove it into a computer, and it gives us this. Now this term over here is the momentum at time capital T, the momentum the object ends up with. So change in energy of an object is its final energy minus its initial energy. So maybe the first term here represents the final energy of the object, and the second term represents its initial energy. This would give us the formula that the energy at time t is the square root of m squared c to the 4 plus p squared c squared. Or in other words, squaring both sides and leaving the time dependence of energy and momentum implicit. E squared equals m squared c to the power 4 plus p squared c squared. So now we can't just directly believe this because we derived it using all sorts of things that we know from Newtonian mechanics. Work equals force times distance and f equals dp by dt. And at the moment we don't know if these hold up in special relativity or not. So what we do is we take this as a candidate and we go to experiment. We try and see does this equation seem to be valid? And what thousands and thousands of experiments have shown us is that yes, this is the equation for the energy of an object.