Revisiting classical Physics

1. Revisiting classical Physics So we have to start reconsidering old Physics in order to make it compatible with the 4-dimensional spacetime and special relativity. For the main part, that means turning vectors in 4_vectors. Which means at least to find the “time component” of the 4-vector. Starting, as usual, from the simplest case. And the simplest is: the “position vector” (x,y,z).In this case, the answer is obvious, since this is where it all started: The “position” 4-vector is what we called the “event” 4-vector, [t,x,y,z]. We know how we found it: it was the Physics that forced it upon us. In this case, the 4th component is the time. The next case follows directly from this one: it is the “displacement”, [δt,δx,δy,δz]. It not used by itself, but as a step in the calculation of the derivatives, like velocity which is a bit more complicated; but at least we have a 4-vector to start from. And, even more complicated, the acceleration. Another fundamental vector in classical physics is the Force. In classical physics it is a 3-vector, of experimental origin, and we do not have any idea on what the 4th term should be. Although for defining force we may start from the potential. It is –or so it seems- a scalar. And, as such, it is not dependent on the reference frame. But, on the other hand, we have one case (electromagnetism!!) in which we have a scalar potential and a vector one… It could as well happen that they are components of a sole 4-vector! So we may invert the argument and consider that from a scalar potential we obtain the gradient and therefore the force. For doing this, we need to check the transformation properties of that hybrid thing that the gradient is. Advanced EM - Master in Physics 2011-2012

2. The 4-vectorization of nabla Or, the transformation properties of gradients, curls, divergencies and laplacians in 4-dimensional spacetime. We have a vector field Awhose contravariant components we know. Since we are dealing with a vector fieldthe components are functions of the space point and possibly also of time. Anyway, since we are discussing 4-vectors, we have 4 components, which having a dependence on t,x,y,z will also have derivatives wrt them, and all others. If we want to change to a new reference system, we have to change all the components of our vectors… For every change of IRF exists a matrix of transformation of coordinates, [Aij] which has an inverse [A’lk], and which has coefficients ; an example of transformation is: For the covariant component a similar equation holds. The difference with the contravariant one is that in this case the coefficients are , i.e. they are those of the inverse matrix. And… what about the gradient? For the initial convention (x=x1, y=x2, z=x3), the 4-gradient extension of the 3-gradient will have the derivatives wrt contravariant coordinates. Advanced EM - Master in Physics 2011-2012

3. Then, well. It turns out that such definition of gradient makes it defined with covariant components. Because it turns out that, if we make a change of reference axes from IRF to IRF’ the coefficients of the transformation matrix will be determined by the equation The are the coefficients of the covariant components! For further reference then, the components of nabla are: Covariant components Contravariant components The same convention is valid for gradient, divergence and curl. For the Laplacian, in the first place, it has 4 addends and becomes the D’Alambertian; in the second place, it being a sort of norm of the vector, we do not have to choose: Where this new symbol, , is called the D’Alambertian. Note that the D’Alambertian can also be written as 2 2 2 Advanced EM - Master in Physics 2011-2012

4. Let us now go back to the infinitesimal change of position: the “displacement”, [δt,δx,δy,δz]. . It is obviously a 4-vector too, in the Minkowski space. That means that it is an invariant 4-vector, i.e. that its position with respect to other objects is unchanged when changing reference frame. Of course, the displacement we are talking about now is an infinitesimal part of a movement, in space and time, of an object. It is therefore only timelike, unless we are talking about light, in which case the LI will be null. Let us now deal with velocity v. In classical mechanics it is a relative quantity, i.e. a quantity whose value does depend on the IRF in which we are studying the system. In ordinary space its components are (dx/dt, dy/dt, dz/dt) and depend, as well as its absolute value, on the reference frame chosen. Now, if we translate literally the 3-vector components in spacetime language we get (in 3 dimensions): At first one would think that the 0-component could be a term like dt/dt. It does not work, because such 4 terms are NOT a 4-vector. For being a 4-vector the 4 terms should vary (for change of reference frame) as a 4-vector. And in fact the 4 differentials in the numerator do that. What does not do it is the denominator, i.e. ”dt”. That should have been a scalar, for the 4 terms to behave as a 4-vector. But it is not a scalar, because under a Lorentz transformation it varies as the 0-term of a 4-vector. Of course, dt is a time, and from that point of view it is suitable. But we need a time which is actually a scalar in spacetime. There are only two “times” in the market; and the other one, which is the proper timedτ of that displacement, would do. Advanced EM - Master in Physics 2011-2012

5. If we divide the 4-vector δx by dτ we obtain a 4-vector. This 4-vector is called Four-Velocity Its norm (i.e. its invariant) is obviously equal to 1, owing to the invariant of the infinitesimal displacement being equal to (dτ)2. This is of course a significant change wrt classical physics. Remark that the norm of the Four-Velocity is 1 only in a system in which time and space have the same units. Otherwise it norm is “c”. Now there is still another step we must make to be done with four-velocity: make its components intelligible. Well, we know from the dilatation of time that dt=γdτ. Replacing this value in the last equation we give the components of the four-velocity in 3-vector, understandable language. • In classical mechanics what is conserved in isolated systems is not the velocity of a system of interacting bodies, but its total linear momentum. We can extend to 4-vector space the definition of linear momentum: Obviously, Advanced EM - Master in Physics 2011-2012

6. At the limit for β0 we should find back the values of classical, newtonian physics. The three space components actually do tend to mv. What is still to be understood is what does the 0-term (mγ) mean in terms of physics. In order to understand it, we have to make a connection with the Classical physics formulae, i.e. the limit for β0. A hint we already have: The space part of this 4-vector is the linear momentum, a three component vector which is always conserved in isolated systems, just the same as a scalar quantity, the energy. Now, the quantity under scrutiny is: For small values of β2 we approximate the function (1-x)-1/2 with the Taylor formula Limx0(1-x)-1/2=1+x/2 And therefore: which is the classical formula for the knetic energy, which is added to the energy at rest of the particle. Advanced EM - Master in Physics 2011-2012

7. Four-acceleration, four-force We have already seen the Velocity 4-vector. To calculate the four-acceleration we repeat the procedure that allowed us to calculate without major trouble the four-velocity, i.e. we derive the 4-velocity wrt dτ. and writing We obtain: In the case of uniform motion(a = 0) the 4-acceleration is null, as it should be. In the case instead of a particle at rest (v=0 in a certain frame): where the time component is null and the space component is the classical acceleration: the four-acceleration is therefore spacelike. Moreover, the module (norm) of the four-acceleration (in any IRF, of course) is the square of the three-dimensional acceleration in the IRF in which the particle is at rest. Advanced EM - Master in Physics 2011-2012