[Physics] Physics Digest, Vol 18, Issue 1

[carmam at tiscali.co.uk]

Carl. I think that you have missed the point. Einstein did not say or even infer that these relativistic effects were not real until 1921 when he admitted that they were not real (but nobody took any notice of this rather startling announcement). Take the case of mass increase. Page 44 "General Results Of The Theory" gives the details. Mass increase (kinetic energy in the book) stops any vehicle from reaching or surpassing the speed of light - you know the story, an increase of mass needs an increase of force to get the same reaction, so infinite mass needs infinite force...Quote "The velocity must therefore always remain less than c, however great may be the energies used to produce the acceleration."Einstein is here describing what he acknowledges to be a real effect. This itself is however wrong, as shown here : -This is an excerpt from my web page which unfortunately is unavailable until I find a new host web site.Imagine now a space rocket, which is propelled by ejecting a small amount of matter (the rocket exhaust) at high speed from the rear, so imparting a thrust in the opposite direction. We will assume that the exhaust velocity is 3,000 m/s and the mass of the rocket is 30,000 Kg (very similar to NASA's Mercury-Redstone rockets). Now we can use the Lorentz transformation to find the new mass. The velocity between exhaust and rocket is 3000 m/s, so :-

m = m0 / sqrt( 1 - ( v / c )^2)
m = mass of rocket at velocity v as measured by the essential observer
(Remember that Einstein's observer, properly called the essential observer, is always at rest relative to the motive force. In this example therefore, the essential observer is in the same frame as the rocket exhaust).
m0 = 30,000 Kg (proper mass of rocket or rest mass when v = 0)
v = 3,000 m/s - rocket's velocity relative to the exhaust
c = 300,000,000 m/s
m = 30000 / sqrt( 1 - (3000 / 3e8)^2) = 30000.0000015000000001125 Kg

The mass increase is therefore 0.0000015 Kg or 0.0015 gram which is simply not measurable compared to 30,000 kilograms. For all intents and purposes the mass increase is zero. A further point to note here is that the mass increase is measured against the exhaust which is providing the motive force, and no matter what the velocity of the rocket when measured against its starting point (or anything else for that matter), the velocity between rocket and exhaust never changes, so the rocket mass is always 30,000.0000015 Kg (disregarding the loss of mass due to fuel used). In other words, the mass is fixed at 30,000.0000015 Kg for the values used above between rocket and exhaust, and the extra 0.0000015 Kg is an insignificant amount. As there is no significant mass increase with velocity, and certainly no accumulative mass increase, there is no theoretical upper limit to the velocity of the rocket.
Regarding the muons.The velocity of the muons is (as far as I am aware) only measured in the iron block above the scintillator (if I am wrong on this please show me a reference to the technique used). This block slows down (or stops) the muons. As the muons move through the iron they experience a drag force. Assume that the force is of known fixed strength while the muon is in motion, and stops when the muon stops. So how do we determine how far it travels before stopping?  This can be done using energy conservation equations.The muon strikes the iron with a certain amount of kinetic energy.  As it passes through, this energy gets steadily converted to potential energy according to the equation : - E(p) = f * d (potential energy = force * distance).If the muon comes to a stop, this means the original kinetic energy of the muon has been fully converted. Kinetic energy is given by the equation : - E(k) = 0.5 m * V^2 .As the stopping distance is known, the initial velocity can be found.  IE it can be determined how fast the muon was going when it hit the iron. The 0.5m * v^2 equation however is for classical mechanics.  There is another equation that is used in relativity.  It is of the form :- E(k) = E(relativistic) - E(rest), or E - Eo = m*c^2 - mo * c^2 .Where mo is the rest mass and m is the relativistic mass.  m = mo / sqrt(1-(v^2/c^2)), i.e. the rest mass multiplied by the Lorentz Transform.For low velocities there is basically no difference in these two formulae, but for high velocities, the velocity calculated by the classical formula can allow v>c.  Whereas the relativistic formula ensures v<c.The above is simplified.  In reality the force experienced won’t be constant but a complicated function of velocity.  The actual function used is the Bethe Formula. 
Tom Hollings.


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