Are these equation still valid in a relativistic context?

Question

It's been a while I took Physics and, being an engineering course, the emphasis was on Newtons's Physics.

In a relativistic context, how should we interpret classical equations like f = ma (f, force, m mass, acceleration). Is it still true that v = ds/dt (v the velocity vector, s the dsplacement vector and t the time). I think this has to do with Lorentz transform, right?

In electromagnetism, are results like Maxwell's equations and Biot Savat Law still valid?

Thank you

Answer 1

The Lorentz transform is L(v) = sqrt(1 - (v/c)^2) and this is the term that changes rest time (T), mass (M), and length (L) into relativistic time (t), mass (m), and length (l). As you can see L(v = c) = 0 and that so called singularity becomes an issue, when we look at t, m, and l.

t = T/L(v) is the relativistic relationship for time. When L(v = c), t approaches infinity. [Some may say "is" infinity, but that implies equals and there is no such thing as equals infinity because infinity is without bound.] This remarkable effect means that time stands still on the frame (e.g., the spaceship Enterprise) moving at v ~ c, light speed, when observed from outside that frame (e.g., from Earth).

But v is relative to the frame of the observer. This means, on the Enterprise, v = 0 because Captain Kirk is sitting in his Captain's chair and not moving relative to the bridge of his ship. Thus, t = TL(v = 0) = T for the Captain and his crew. That is, time passage has not changed for those on the moving frame of reference.

So what we have is, for example, a second ticks off on the Enterprise traveling near light speed, while, in that same time, depending on how fast the ship is traveling relative to Earth, thousands of years pass on Earth. When Captain Kirk finally returns after a year's star trek at near light speed, he could find Earth many millennia into its future. Theorectically, at least, time travel into the future is possible if not feasible.

By observing subatomic, unstable particles at near light speed in colliders, physicists have seen the slowing down of time. This shows up as a slowing down of the natural decay rates of the particles. Since f = 1/t in classical physics; where f is frequency (the rate of decay) and t is time, as t --> infinity, f ---> 0, which is predicted by f = 1/[T/L(v)] = L(v)/T

For mass, the relationship is m = M/L(v), which is similar to the time equation. Thus, on board the ship, m = M/L(v = 0) = M and the Enterprise's mass remains at rest mass for Captain Kirk and his crew.

But back at the Lick Observatory on Earth, gravitational warping of space (gravitational lensing) around the ship indicates the Enterprise's mass has become greater than the magnitude of the largest massive galaxies in the universe. That results because m = M/L(v ~ c) --> infinity as the Lick people see it from outside the ship.

Understand, mass doesn't get bigger in dimensions, but its inertia does get bigger. (In fact, as we see later, it will shrink in one of the dimensions.) That is, and here's Newton's first law, the mass of the Enterprise acts like it has gotten bigger because it takes more and more force to change its direction and/or to slow it down or speed it up.

In other words, its momentum p = mv = Mv/L(v) is the relativistic version of Newtonian physics for momentum. This effect has been observed on subatomic particles that have been sped up to near light speed in colliders. So it's a proved fact, not just a theory.

Finally, length....l = LL(v), which is a bit different from m and t because the Lorentz is in the numerator along with rest length L. Thus, l = LL(v ~ c) ~ 0; where L and l are measured in the direction of travel...in the direction of the pointy end for the Enterprise.

This remarkable relationship says that we at the Lick Observatory would see the magnificent Enterprise shrink to nothing in length if we saw it directly from the side. On the other hand, we'd see no change in its size if we saw it from end on. But, still, Captain Kirk et al would see no changes while on board the near light speed ship. That results, as with the other two, from l = LL(v = 0) = L because, relatively speaking, they are not moving wrt the Enterprise itself.

Maxwell's equations not only are affected by space-time curvature, a relativistic phenomenon, they were in fact in part used by Einstein to derive his relativity equations. [See source.] Relativity alters the frame within the electro-magnetic waves described by Maxwell's equations operate. So that alteration has to be accounted for through tensor operators when the frame's velocity exceeds around v = .1c where relativisitc effects start to become significant. [NB: Keep in mind, we are talking about effects as seen by an observer outside the moving frame. Inside, an observer would see no space-time warping because v ~ 0 so that L(~0) ~ 1.0]

I'm not familiar with Biot-Savart as a law. However, I did study the magnetic field equations as a physics student. We didn't call it the BS law back then. Anyhow, as vectors are involved and space-time gets warped at v ~ c speeds, I have to believe this law would require alteration by the Lorentz effects at near light speed.

Answer 2

I don't have anything especially new to add to what has been said already in total, but I thought it would be fun to offer my perspective:
- Instead of talking about 3-dimensional space and 1-dimensional time, we talk about 1+3 dimensional space-time. Position in space-time is represented as a 4-vector, with time being a bit separate in nature from the 3 spatial coordinates, but mixed in with them by Lorentz Transformations (LTs) which create the notorious "paradoxes" of relativity.
- In the same way, instead of talking about 3-dimensional momentum and 1-dimensional energy, we talk about 1+3 dimensional energy-momentum. This is also a 4-vector, which behaves in the same way under LT as does the space-time position 4-vector. It is proportional to the relativistic "velocity" 4-vector, which has components (c^2, v_x, v_y, v_z); the constant of proportionality is the mass (sometimes called "rest mass")
- Relativistic particle dynamics depends on conservation of 4-vector energy-momentum and on the 4-vector equivalent of Newton's laws. In the latter, instead of differentiating with respect to the time coordinate, one differentiates with respect to the "proper time", which is the quantity
sqrt(dt^2 - (dx^2 + dy^2 +dz^2)/c^2)

The 3-vector part of the 4-vector equation is then very similar to Newton's 2nd law, and attributes the change in 3-vector momentum to the imposed force; the 1-vector part of the 4-vector equation is very similar to the work/change-in-energy equation.

- Electromagnetic theory is unchanged by relativity, as Maxwell's equations are applicable perfectly in all frames; indeed, the LT was discovered a long time before Einstein (and before Lorentz, too!) as a transformation that leaves the MEs invariant; but no one quite knew what this meant at the time. However, under LT, the E-fields and B-fields are NOT 4-vectors. Instead, together they form the the electromagnetic field tensor, so they get mixed together when you change frames; and this actually explains the Lorentz force law,
F = qE + q(v/c) x B

The Biot-Savart law etc. are derived from the MEs and thus follow automatically.

- In the Lagrangian formulation of classical mechanics, the important function called the Lagrangian is
L = T - U = Kinetic energy - potential energy
= mv^2/2 - PE

But in relativity, the T becomes
T' = - m*c^2*sqrt(1 - (v/c)^2) = - m*c^2 + m*v^2/2 + ...

This seems odd: Why isn't it
T'' = m*c^2/sqrt(1-(v/c)^2) = m*c^2 + mv^2/2 + ...
instead? Because that doesn't give the right "Newton's Laws" equations! The relativistic lagrangian IS a generalization of the newtonian lagrangian - just not generalized in the way you might have guessed!

- Finally, in thermal physics, the heat dQ and the change in internal energy dU are energies, which transform like the 4th component of a 4-vector; but the entropy dS and temperature T are Lorentz invariant. (The equation dQ = T*dS only applies in the rest-frame.)

In general relativity, even stranger things happen to the temperature. But that is another story...

Answer 3

Mechanics, invariant with respect to the Lorentz transformation, is called relativistic mechanics. Relativistic mechanics is converted into the classical kind only when v/c much less than 1, where v is the velocity of the moving body or particle, and c is the velocity of light in vacuum.

The Lagrangian function for a free particle is

L = -m zero*c^2 sqrt(1-v^2/c^2)
Where m zero is the mass of the particle.

The Momentum Vector P is = P=m zero/sqrt(1-v^2/c^2)
The relativistic expression for the mass is
m = m zero/ (sqrt(1-v^2/c^2)
When v/c is much less than 1, then the equation for the momentum is immediately reduced to the classical Newtonian equation where p =mv

The equation of Force becomes F= dp/dt where p constitutes two variables, namely the velocity and the mass since m is no longer a constant in relativistic mechanics.
The equation of the second law of Newton becomes:
F = m zero/sqrt(1-v^2/c^2)*a
The total energy of a body or particle in the theory of relativity is E =(PV) = - L = m zero*c^2/sqrt (1-v^2/c^2) = mc^2
Time also becomes variable in relativistic mechanics and its equation is calculated as: t (Rel) = t-V/C^2*x/sqrt(1-v^2/c^2) where x is the displacement
Even x becomes variable and is no longer dealt with as a straight line; X in relativistic mechanics becomes:
X = (X prime + V*t prime)/sqrt(1-V^2/c^2)

Answer 4

You need to replace vectors with four-vectors and use the proper time in derivatives. So v=ds/dτ where s is the four-vector displacement and τ is the proper time. The idea of acceleration is a bit different in relativity as well. All objects move at the same speed in relativity, the speed of light. By speed I mean the magnitude of their four-vector velocity. Therefore, acceleration occurs only when an object changes directions in four dimensional space-time. Newton's 2nd law is best stated as the proper time derivative of the momentum four-vector. f=dp/dτ Since all objects move at the speed of light in relativity, the magnitude of their momentum four-vector is always mc. (where m is the rest mass of the object and c is the speed of light) It may interest you to know that the energy is proportional to the temporal component of the momentum four-vector. (the constant of proportionality is 1/c) So the energy is kind of like "time momentum". (You may find this interesting as well. For an object at rest with respect to you, the momentum four-vector is (E/c,0,0,0) and since the magnitude must be mc we have (mc)²=(E/c)² or mc=E/c or E=mc² Now you know where that comes from, notice this equation is only valid for an object at rest with respect to you) Maxwell's equations and the results of electromagnetism are still valid in relativity. They were relativistically correct before relativity and provided the major impetus for relativity's developement. Just note that not everyone will agree on whether or not a particular phenomena was electric in nature or magnetic in nature. One observer may attribute an event to an electric force acting on an object while an observer moving with respect to the first may say it was caused by a magnetic force. They both agree on the phenomena and that a force was acting on the object, they just call force by different names. Relativity reveals the true unity between electricity and magnetism.

Answer 5

No while you grow to be an Atheist you're not allowed to make use of that expression once more. It is facet of the Atheist code. Sort of like the 10 commandments is for Christians. You had greater stroll frivolously after pronouncing some thing like that. You are striking your self in risk of Hell hearth.

Answer 6

v=ds/dt is still the definition of speed, yes.

F=ma is not correct, but F=dp/dt is ok in special relativity, with a redefinition of p=(gamma)mv.