If two electron beams are travelling side-by-side, then, from the point of view of an observer relative to whom they are moving, the force between them has two components, one electrostatic and the other magnetic. But from the point of view of an observer who is travelling with them, the electrons in both beams are stationary so there is no current and the only force is electrostatic. The Lorentz transformation explains the difference in terms of a change of the distances between the charges in their direction of motion (and so a change of the charge densities in the beams); but regardless of that, what is seen as a sum of “electric” and “magnetic” forces in one frame is seen as just an “electric” field effect in the other, and so there is no inherent observer-independent distinction between the electric and magnetic components. The formulation in terms of a second order tensor allows us to express both the apparent electric and magnetic fields seen by an observer as aspects of a single Lorentz invariant tensor field; but while it describes how they are related as aspects of a single tensor field, I wouldn’t really say that it shows they are so related.
Category: All
Mass to Energy easier than Energy to Mass?
The reason we say mass and energy are the same thing is because how we distinguish them is just a matter of bookkeeping – which depends on our point of view. But if you think of heat as a form of energy rather than of mass, then it is indeed easier to convert mass to energy than vice versa.
When a chemical reaction reorganizes matter into a state of lower potential energy, such as for example whenever we burn a fuel, the energy released typically gets converted to kinetic energy and quickly distributed among many particles of a larger system in the form of what we call heat (which is impossible to fully recover in the form of usable work). An observer looking in detail at the system will see each molecule as having a tiny bit less mass than its component atoms (by an amount corresponding to the binding energy) with the total of all these differences also corresponding to the increased kinetic (heat) energy of the molecules.
On the other hand, when we break apart a chemical bond then we must provide some amount of energy and the resulting masses of the components will add up to that much more than the mass of the molecule. But if we want to do so in a consistent way (such as in electrolysis of H2O or carbon capture from CO2) then we need to provide the energy in a very specific way that is not so trivially easy to set up as just exposing fuel to oxygen.
However, if we measure the mass of the system as a whole (either gravitationally, or as inertia at starting from rest), then the result we get includes not just the masses of all its molecules but also their kinetic energies relative to its centre of mass. And that total (which includes all of the heat energy in the system) remains absolutely constant and never either increases or decreases.
Does mass increase with velocity?
The claim of “mass increase with velocity” is based on using the word “mass” for something that is not a property of the object itself but rather depends also on the relative velocity of the observer.
Although this usage is not “wrong” and was fairly common in the early days, most physicists soon chose to use the word “mass” just for the invariant “rest mass” and followed the title of Einstein’s initial paper on the effect by just using the word “inertia” for the velocity-dependent quantity that used to be called the “inertial (or relativistic) mass”.
In order to prove that the observed “inertia” depends on the relative speed (and direction) of motion of the observer it is necessary first to define it.
One way is to define it as the ratio of applied force to observed acceleration, but this in turn requires a clear definition of force. For example you could perhaps look at the dynamics of a collision with an object of known mass at rest in the observer’s frame.
Or perhaps you could just read Einstein’s original paper on the subject.
How do CO2 molecules heat their neighbours?
By hitting them!
A CO2 molecule in the atmosphere, after having absorbed an infrared photon, is vibrating or rotating more vigorously than it was before. When such a molecule collides with a neighbouring molecule (most likely an N2 or an O2) some of that vibrational energy contributes to the speed or vibration or rotation of the one it hits.
The atmosphere doesn’t continue heating up though, because in the equilibrium situation the molecules in the atmosphere collectively emit radiation at the same rate as they absorb it. But they do this equally in all directions, so what they are absorbing from below gets re-directed and only half of it ends up going outwards (with the other half going back down to warm the Earth’s surface).
If energy (E) is equal to mass (m), then why is E = mc^2?
It is not true that “energy (E) is equal to mass (m)”. What is true is that the mass (m) of a system is equal to the special case of the time component of its 4-momentum in the “rest” frame (where the net spatial component of momentum is zero), and that the units we typically use to measure the time component as “energy” (E) are chosen in such a way that the extra “kinetic” energy as seen by a moving observer is given approximately by (1/2)mv^2 rather than (1/2)m(v/c)^2 with units of space and time chosen in such a way that the limiting speed c is very large (and so that speeds we actually experience do not have to be associated with very tiny numbers).
Misconceptions about optics
One that just now surprised me is the idea that mirror reflection can be explained in terms of excitation and decay of atomic energy levels when in fact it depends on the existence of a “sea” of conduction band electrons that are not tied to any particular atom. (The excitation of atomic energy levels only happens at particular frequencies and because of the delay between absorption and re-emission does not lead to a coherent reflected wave.)
Photon Speed in Glass
A question on Quora asks about speed of photons in a refractive medium.
Most of the answers given so far are correct – even when they contradict one another. This is because the path of a photon is something that is not well defined. In quantum theories particles with definite properties don’t really exist between observations, and what we call the path is just something that we infer after observations have been completed and depends on how closely we look. If we set up an experiment to detect single photons after passing from a source through a uniform medium like glass or air, then we find that no photons arrive when the straight line path is obstructed (so they appear to have followed a straight line) and the time delay between emission and detection corresponds to a speed slightly less than that of light in a vacuum. But on the other hand it seems that experiments designed to estimate the speed of photons between adjacent atoms show them travelling at the same speed as they would in a vacuum. One way of explaining this discrepancy is to realize that what quantum theory predicts is that the probability of detecting a photon is the squared magnitude of a sum (actually a generalized integral) of many complex-valued contributions corresponding to paths that a point particle could be imagined to have taken by going either directly in a straight line or by going on a (longer) polygonal path bouncing off electrons and protons in the medium. If there are many charged particles between source and detector the contributions from all these paths may interfere with one another in such a way that they cancel out almost completely everywhere except near events on the straight line path but at times slightly later than for direct travel in a vacuum.
Apparent Earth Time Jump During Turn
Travellers on the spaceship don’t actually see (or in any other way directly experience) the sudden ageing of those on earth when they turn around, but rather they just infer it from what they do see. Just before the turn-around, what the travellers see of Earth is actually not what they think is happening in their version of “now” but rather what was occurring at an earlier time when the light signal was emitted (and what for them is actually occurring “now” hasn’t been seen yet). After a quick turn-around what they see of the Earth is essentially the same as just before, but now the correction for light travel time is different and the result is that it now seems to them that the events occurring “now” on Earth (which they haven’t seen yet) correspond to a later Earth time than they did just before the turn. So even though they don’t see the jump it seems to the travellers as if the Earth’s clocks quickly jumped ahead. This solves the mystery of the age difference because the amount of the apparent jump in Earth’s time is exactly twice the amount of progress lost during the two periods where the Earth seems to be ageing slowly so at the end of the trip the tavellers and homies all agree that the Earth has aged more.
Kronecker Tensor?
Kronecker delta is not a tensor of any kind. It is just a fixed numerical matrix.
Tensors of order $#n#$ are linear functions from sets of $#n#$ vectors to the field of numbers and can be described in terms of any particular basis by arrays of numbers called components of the tensor with respect to that basis.
If the vector space has an inner product, then tensors of second order (ie ones involving two vectors) can be identified with linear operators on the vector space by requiring the matrix of the operator in basis $#\{ b_{i}\}#$ to be given by $#T_{ij}=\langle b_{i}|Tb_{j}\rangle=T(b_{i},b_{j})#$ .
In general, these components depend on the choice of basis, and the transformation rules for tensors tell us how they vary with changes in the basis that is being used to determine them.
For the special case where the tensor function $#T#$ is just the inner product, then for any orthonormal basis we get $#T_{ij}=T(b_{i},b_{j})=\langle b_{i}|b_{j}\rangle=\delta_{ij}#$, (and the corresponding operator is just the entity operator $#Tv=v#$ for all $#v#$). These components will be the same for any orthonormal basis, so the components are invariant under orthogonal transformations; but the same tensor may have a different matrix with respect to a basis that is either skew or not normalized. So what is true is that for a tensor whose components are given by the Kronecker delta, those components are invariant with respect to orthonormal changes of basis. (But, while the components will be changed by changes of scale or angle so that the new ones will not be given by$# \delta_{ij}#$, the Kronecker delta will still be what it was – just corresponding now to a different tensor.)
Magnetic Work
The claim that “classical physics says magnetic fields shouldn’t do work” is false. What classical physics says is just that a magnetic field does no work on an isolated single-point charged particle (because a charge at a single point has no magnetic moment and the magnetic force on such a particle is always perpendicular to its direction of motion). But this does not preclude the doing of work on a wire loop carrying a current or on a rotating charge distribution. [To create such things though requires us to have some mechanism for confining or holding together the relevant charges, and although classical physics did not say these things were impossible it never got to the point of including a description of the required additional non-electromagnetic forces. Quantum mechanics, on the other hand, did provide a means (without even requiring non-electromagnetic forces) for describing situations in which charges could be confined in atoms or metallic solids (and even predicted that an isolated point charge would nonetheless have a nonzero magnetic moment).]