Special Relativity Revisited (5): The Inertia Of Energy

We now conclude our introduction to Einstein's theory of special relativity with what many regard as its most fundamental conclusion: the inertia of energy, or light. This is embodied in Einstein's famous equation:

E = m c²

which is more accurately posed as:

E = (D m) c²

where D m is the "mass defect" or difference, say in a nuclear reaction, and c is the velocity of light.

    Before looking at examples, it's useful to consider the relativistic mass of a particle, in terms of its rest mass m_o. The rest mass, as the term implies is the mass of the object at rest or:

m_o = m [(1 - v²/c²)^½]

Thus, if v = 0 (particle at rest) then we have:

m_o = m(1)^½ = m

so the mass and rest mass are identical.

Now, the relativistic mass is then:

m = m_o / [(1 - v²/c²)^½]


and again, if v = 0 then m_o = m

But what if v = c? (Object moving at the speed of light?)

Then:

m = m_o / [(1 - c²/c²)^½] = m_o / [1 -1]^½ = m_o /0 = ¥

     In other words, m would be infinite! This is another way of saying that to try to achieve the velocity of light one would have to overcome infinite inertia! In other words, it can't be done...not for a material object.

From this, we can also see the relativistic momentum must be:

p = mu = m= m_o u / [(1 - c²/c²)^½ ]

This approaches the classical value (p = mu) as u ® 0

Newton's 2nd law in the relativistic format is simply:

F = ma = m (du/dt) = d/dt[m_o u / [(1 - v²/c²)^½ ]


The relativistic energy is found by taking the integral of : (dp/du) u du

® ò _o ^u  (dp/ du) u du

from 0 to u and obtaining:

W = mc²/ [(1 - u²/c²)^½ ] - mc²



And by the work -energy theorem:

W = K(f) - K(i)

where K(i) is just the initial rest energy, or m_o c²

Then W = m_o  c²/ [(1 - u²/c²)^½ ] - m_o c² =

 (total energy - rest energy) 

A variation entails finding the work done (W') between two points x1 and x2 with velocity v2 at x2 and time t2, and velocity v1 at x1 and time t1.

Then we may write:

W'  =  ò _x1 ^x2  (dp/ dt) dx  =    ò _t1 ^t2  (dp/ dt) (dx/dt) dt = ò _t1 ^t2  v (dp/ dt) dt


=   ò _v1 ^v2  v  dp   =    ò _v1 ^v2  v  (dp/dv)  dv


Note in the above we used the following facts:

v = dx/ dt   and dp = (dp/dv) dv

Here p is given as a function of v such that:


W'=   ò _v1 ^v2  v  (d/dv) [ m_o  v / [(1 - v²/c²)^½ ]   dv

On integrating the preceding equation by parts we find:

W'  =   m_o  c²/ [(1 - v2 ²/c²)^½ ] - m_o c² / [(1 - v1 ²/c²)^½ ]

This equation immediately shows that the effect of the work done is to produce a change in the quantity:

E' =   m_o  c²/ [(1 - v²/c²)^½ ] 

Note how this is different from the classical kinetic energy equation: 

 E = 1/2  ( m v² )


In particular E' does not become 0 when v = 0 (Instead it reduces to: 

E' = m_o c²

Hence, if we desire a quantity which correspond as closely as possible to classical KE we need to define:

E_k   =   m_o c² / [(1 - v ²/c²)^½ ]    -    m_o c²


If indeed this is a correct relativistic  generalization of kinetic energy it must reduce to approximately  1/2  m v²   when  v << c.

This can easily be shown by expanding the binomial   (1 - v ²/c²)^½   in the last eqn. using the binomial theorem, i.e.


(1 - v ²/c²) ^-1/2    =    _{1  +}    v ²/2 c²   +   (3 v ⁴/ 8c⁴  ) +    ......


Example Problem:

Apply the basic mass-energy equation, E = (D m) c², to the case of nuclear fusion.



Consider:

₁H² + ₁H² ®  ₂He³ + ₂He³ + _o n ¹

which actually occurs in the Sun.

We now compile the masses (in atomic mass units) on each side:

2.015 u + 2.015 u ® 3.017 u + 1.009 u

or:

4.030 u ® 4.026 u


Now, the right side is less than the left by an amount equal to the mass defect or:

D m = 4.030 u - 4. 026 u = 0.004 u

To get the energy E:

E = (0.004 u)(931 MeV/u) = 3.7 MeV

where 931 MeV/u is the conversion factor incorporating c²

To transfer to Joules:

3.7 MeV = 3.7 MeV x (1.6 x 10^-13 J/MeV)= 6.0 x 10^-13 J 


Example Problem (2):

Determine the energy required to accelerate an electron from 0.50c to 0.90c.

By the work -energy theorem:

W = K(f) - K(i)

K(i) = m_o c²/ [(1 - u²/c²)^½ ]

u1 = 0.50 c

K(f) = m_o c²/ [(1 - u²/c²)^½ ]

u2 = 0.90c   (where: m_o = 9.1 x 10^-31 kg )

K(f) - K(i) = m_o c²/ [(1 - (0.90c)²/c²)^½ ]

- m_o c²/ [(1 - (0.50c)²/c²)^½ ]

K(f) - K(i) = m_o c²/[(1 - 0.81]^½ - m_o c²/[(1 - 0.25]^½

K(f) - K(i) = 2.294 m_o c² - 1.155 m_o c² = 1.134 m_o c²

Or:

K(f) - K(i) = 9.32 x 10^-14 J = 0.583 MeV 



Problems for Budding Physicists:

1) A student proposes to compute the kinetic energy of a particle relativistically by using the expression  1/2  m v²  with the 'relativistic mass of the particle.   Would this be correct? Explain why or why not.

2) Determine the energy required to accelerate a proton from 0.25c to 0.50c. 

3) Protons emerge from a particle accelerator with a kinetic energy equal to 0.49 mc².   What is the speed of these particles? Compare the result to that obtained from the non-relativistic relation between mass and energy.

4) What is the speed of a particle whose kinetic energy is equal to its rest energy? What percentage error is made if the non-relativistic kinetic energy expression is used?

5) Show that the relativistic  kinetic energy equation:

E_k   =   m_o c² / [(1 - v ²/c²)^½ ]    -    m_o c²

^{Reduces to approximately:}

1/2  m v²   when  v << c.

Solutions To Special Relativity Problems (4)

1)      Let t = 10⁶ yrs be the quasar's lifetime in its own rest frame. Then the total Earth-based time that its radiation will be received is:

t' = t [1 - v²/c²] ^½

where v = 0.8c

so:

t' = (10⁶ yrs.) [1 - (0.8c)²/c²]^½

t' = (10⁶ yrs) [1 - 0.64]^½ = 10⁶ yrs (0.36) ^½

t' = 0.6 (10⁶ yrs) = 6 x 10⁵ yrs. 

(2) Let D t' = (t2' - t1')  = 9  s

t2' = (t2 - 3c/4)(1 - 9/16))^-½

t1' = (t1 - 3c/4)(1 - 9/16)^-½

or:

t2' = (t2 - 3c/4) (7/16)^-½

t1' = (t1 - 3c/4) (7/16)^-½

whence:

(t2' - t1') = [(t2 - 3c/4) - (t1 - 3c/4)]/0.661

or: 9 sec = (t2 - t1)/0.661 and

D t = 5.95 sec 


(3) The speed of the astronaut is given by:

v = (2 gr)^½

and r = 7 x 10⁶ m, g = 10 ms^-2

v = 11, 832 m/s = 3.94 x 10^-5 (c)

In the astronaut's frame, 1 week - 86,400 s x (7) = 604, 800 s

For the twin on Earth:

t = t'/ [1 - t²/c²] ^½

t = 604, 800.0002 s

and so the time difference = 0.0002s, or the twin in orbit will be (t' - t) or 0.0002 s younger on his return. 


(4) The speed of A observed in B = 0.7c, exactly equal to the speed of B observed in A, by principle of relative velocities.

To find the speed of C observed in B, we use relativistic addition of velocities or:

u = (u' + v)/ 1 + u'v/c²

u = (0.7c + 0.7c)/ [1 + (0.7c)(0.7c)/c²]

u = (1.4c)/ 1 + 0.49 = 1.4c/1.49 = 0.94c

Special Relativity Revisited (2)

Let’s now consider the implications of the Lorentz-Einstein transformation and refer again to the original diagram (Fig. 1). Recall:

(11)  x = x' + vt'/(1 - v²/c²)^½

and

x' = x - vt/(1 - v²/c²)^½

(12) y = y' and y' = y

(13) z = z' and z' = z

(14) t = t' + x'v/c²/(1 - v²/c²)^½

and

t' = t - xv/c²/(1 - v²/c²)^½

    Suppose there’s a clock located in system S’ and an observer in system S sees this clock moving with velocity v. At any time t, the position of the clock with reference to the S –observer’s system is given by x = vt. If the length of time between two ticks is of this clock is T’ in the S’ system, the transformation to the S system (assuming x’ = 0) makes the time interval T appear to be:

T = T’/ (1 - v²/c²)^½

    Since T > T’ then it seems to Observer S his clock is running slower than it does to S’. This applies not only to the clock but to all physical processes that depend on time, e.g. the vibrations of electrons in atoms, rates of chemical reactions, biological processes (heart beat, respiration) etc. In effect it appears to the observer in S that his counterpart in S’ is living at a slower rate than he is. However, to the observer in S’ it is the observer in S who seems to be living at a slower rate.

    This is a paradoxical result but one which can’t be escaped if we carry special relativity to its logical conclusion as Einstein did. So long as the two systems have a constant relative motion, we cannot say that one is moving and the other standing still, or that one’s clock is moving slowly and the other quickly.

    A number of subtle consequences arise out of this. For one thing, the notion of simultaneous events, i.e. for observers in two different reference frames, is no longer tenable. Einstein, in fact, seems to have been the first human being to recognize that simultaneity between two events is a provincial illusion. In his landmark (1905) paper on special relativity (“Does the Inertia of a Body Depend on Its Energy Content?’) he remarked:

“We have to take into account that all our judgments in which time plays a part are always judgments of simultaneous events. If, for instance, I say ‘That train arrives here at 7 o’clock’ I mean something like this: ‘The pointing of the small hand of my watch to seven o’clock and the arrival of the train are simultaneous events.”

    Einstein agreed that observing such simultaneity was a reasonably accurate way for a person holding a watch to tell the time of an event happening next to the watch, but insisted that on principle the method couldn’t be relied upon for timing an event far away from the watch, especially by someone moving in relation to the other things involved.

    As an illustrative example, consider Fig.2 which features an enormously long train (defined as system S’) with an observer ‘Bob’ at its very center. (E.g. if the total length of the train is L, he is at L/2).

Two bolts of lightning then strike the extremities of the train, and do so simultaneously according to Bob’s watch. Bill, however, is situated on a stationary platform as shown – defined in system S- at the exact midpoint between the strikes and swears that he recorded the rear flash a sizeable fraction of a second earlier than the forward flash.

    Who is correct? A non-relativistic thinker would undoubtedly incline towards the view of Bill, the stationary observer. The argument here is that Bob’s simultaneity is only apparent, not actual since the train would have carried him closer to the forward flash by the time he received their simultaneous light from points that were, by then, no longer equidistant from him.

    For a relativist, however, it is just as true to say the train was standing still and the ground sped by it, so the ground observer (Bill) could have been carried beyond the equidistant point as easily as Bob.

    In Einstein’s cosmic viewpoint, the lightning flashes (and their midpoint) are free to be considered an integral part of the train OR Earth, OR any other frame of reference. Each and every observer may select his own viewpoint. Any viewpoint is true and right, none is wrong. This example embodies the essence of relativity, that is, that time is truly relative and not absolute as Newton had believed.

    As an illustration of a type of experiment that attempts to reckon simultaneity, consider Fig. 3.

Here a light pulse is directed at a moving mirror M traveling with velocity v to the right. By the time the pulse reaches the mirror it has moved a distance D =  v (D t)/2 horizontally.

    According to an outside observer, if light is to hit the mirror it must depart at an angle to the vertical as shown.

    By Pythagoras’ theorem (square of the hypoteneuse equals the sum of the squares of the other 2 sides):

(c (Dt)/2)² = (v (D t)/2)² + d²

This implies:

D t = 2d/(c² – v²)^½ = 2d/ c[(1 - v²/c²)^½ ]
But, for the stationary observer:

D t’ = 2d/ c

which implies:

D t = D t’/ (1 - v²/c²)^½

We call delta t’ the “proper time” or that time interval between two events as measured by an observer who sees the events at the same place. It’s always the time measured by a single clock at rest in the frame.


Example Problem:

    The period T of a pendulum is measured to be T= 3.0 s in the inertial frame of reference of the pendulum. What is the period of the pendulum when measured by an observer moving at a sped of 0.95c with respect to the pendulum?

    We note that the time measurement taken in the inertial frame is the proper time, and this is t' = 3.0 s.

Then we need to obtain t, for which:

t = t’/ [1 - v²/c²] ^½


where v = 0.95c, so:

t = 3.0 s/ [1 - (0.95c)²/c²] ^½ = (3.0s) 1/ [0.0975] ^½

and t = (3.0 s)(3.2) = 9.6s 

Other Problems:

 1) With what speed would a clock have to be moving to run at a rate that is one half the rate of a clock at rest? 

2) An atomic clock is placed on a Jumbo Jet. The clock measures a time interval of 3600 s when the jet is moving at v = 300 m/s. What corresponding time would an identical clock left on the ground measure? (Hint: whenever v << c (e.g. v/c << 1), note that we have 1 + v²/2c²  and not [1 - v²/c²]^1/2)

3) A muon formed high in the Earth's atmosphere travels at v = 0.99c for a distance of 4.6 km before it decays into an electron, a neutrino and an anti-neutrino.

a) How long does the muon survive as measured in its rest frame?

b) How far does the muon travel as measured in its frame?

4) The average lifetime of a pi meson in its own frame of reference is 2.6 x 10^-8 s. If the meson moves with v = 0.95c, what is its mean lifetime as measured by an observer on Earth?

Time to Ditch Those Faster-than-Light Ideations!

Albert Einstein: Proven right yet again!


Readers will recall in a previous blog to do with the claimed CERN detection of superluminal or faster-than the speed of light neutrinos, I showed that if this were so then it would require the enhanced mass (i.e. over the rest mass) to be imaginary, as well as the associated kinetic energy. See, e.g. the very end of:http://brane-space.blogspot.com/2011/09/faster-than-light-neutrinos-not-so-fast.html


I also noted in a later blog:

http://brane-space.blogspot.com/2011/09/more-questions-concerning-those.html

that much more plausible than that the CERN team actually created FTL- imaginary mass neutrinos, is that they're having made assorted systematic or other errors they didn't detect.

I then noted the sort of corrections that would need to be made, or factored in, for these CERN FTL results to hold up.

Now it appears I was correct, and that a number of measurement "balls" were dropped that were never recognized at the time. This was reported in today's WSJ ('NO, You still can't travel faster than light', p. A13). Among the errors documented since last month:

1) A "carefully calibrated oscillator" used to synchronize the two clocks that assessed the neutrino flight duration was actually itself miscalibrated. Thus the two endpoint clocks, one at CERN and the other in Italy were out of step.

2) The fiber optic cable that linked the GPS satellite used to measure the distance from CERN (in Switzerland) to Italy was evidently loose, which would have introduced errors into the distance measurement, d, which needed to be incredibly precise. (I.e. accuracy of the measurement far exceeding the magnitude of the uncertainty in v.)

The errors (1) and (2) alone would easily have increased the velocity of the neutrinos by 20 parts per million, thereby accounting for the discrepancy reported of 20 pts. per 10^6.

But that isn't all! Apparently an independent group last week "redid the entire experiment and found that Einstein was correct all along: the neutrinos traveled precisely at the speed of light, and not faster or slower".

The moral of the story is clearly to take with a grain of salt any unusual or highly unorthodox results, until they are independently confirmed. In the case of the FTL neutrinos the results didn't make sense from the get-go since as I showed (first link above) they'd have required an imaginary mass for the FTL objects. Since no one has ever measured or determined such before, it made sense that this premature result was bollocks and that a more prosaic miscalculation or miscalibration of the instruments explained them.

And so it has evidently been found!

More Questions Concerning Those Superluminal Neutrinos

Readers will recall in my previous blog to do with the claimed CERN detection of superluminal or faster-than the speed of light neutrinos, I showed that if this were so then it would require the enhanced mass (i.e. over the rest mass) to be imaginary, as well as the associated kinetic energy. See, e.g. the very end of:

http://brane-space.blogspot.com/2011/09/faster-than-light-neutrinos-not-so-fast.html

But much more plausible than that the CERN team actually created imaginary mass neutrinos, is that they made assorted systematic or other errors they didn't detect. One can almost call this a variant of David Hume's "miracle" principle:

"No testimony is sufficient to establish a miracle, unless the testimony be of such a kind that its falsehood would be more miraculous than the fact which it endeavors to establish."

In this case we need only substitute the words "fantastic claim" for "miracle". Then by the Hume test one would ask whether the prosaic claim of an experimental error is more or less fantastic than the discovery of superluminal neutrinos violating a centuries old postulate that holds up most of modern physics? The answer, of course, is that the experimental error would be less fantastic and hence the more probable explanation.

Now, to be sure, the researchers insist they left no stone uncovered in their desire to eliminate any errors: using the GPS or global positioning system they measured the distance accurately to within 8 inches, they factored in the rotation of the Earth, and even stopped traffic in a tunnel through the Gran Sasso mountain in order to calibrate their instruments.

They made it certainly sound like they dealt with all conceivable errors, but have they? Consider first that small perturbative glitches could have transpired due to anomalous differences in the Earth's crust over the time the experiment was conducted. Thus, a tiny deviation in the actual shape of the Earth itself (based on a science called Geodesy) could have incepted the 60 ns difference in speeds.

In geodetic refinements of shape (which also have a bearing on the value of g, the acceleration of gravity) a first correction is made relative to the initiating station. (E.g. see Spherical And Practical Astronomy Applied to Geodesy, 1968, p.
471). In other words, a first correction will be made such that (ibid.):

delta φ = φ - φ(T) = -H/g' (@g'/@x) cosec (1")

where H is the orthometric height of the observation station, and g' and @g'/@x (a partial derivative) are the gravity and the north component of its horizontal gradient, respectively at the initializing station.

A secondary correction is then made to the eccentricity of the station. The correction is to be applied when the latitude is observed off the reference station, and thus reduced the observed latitude to that of the reference station. For example, in the lower diagram, let P be the reference station and P' the eccentric station where the latitude φ(T) is observed. If the distance PP' is d, and the azimuth of the line is A, the identity for the length of the applicable meridian arc is:

R (delta φ) sin (1") = d cos (A)

and hence the correction:

delta φ = φ - φ(T) = d/ (R sin (1") x [cos (A)]

where R is the mean radius of the Earth.

If these corrections aren't factored in, then the timing error could amount to as much as d{e(t)} ~ 6 x 10^-8 s or a factor of ten in excess of the 60 nanoseconds of difference in the speed of the neutrinos. In that case, the CERN team's results would be found insignificant.

Stay tuned!

Faster than light Neutrinos? Not so fast!

Yes, I have read the news that a team associated with CERN and based in Switzerland, is claiming the detection of neutrinos travelling faster than light. If correct, it would indeed overturn one of the foundational precepts of modern physics: that no material object can exceed the speed of light. But the $64 question is whether it's true or a belief based on premature data. (In any case, to be accepted, the results will have to be confirmed by independent groups!)

Let's also not forget we've had these 'false starts' before. In 2007, a team at Chicago's Fermilab believed they had similar faster-than -light results only to learn subsequently their findings couldn't be taken very seriously because the margins of error were so great that they extended to the insignificant!

Earlier, in a paper appearing in the journal Nature, in July 2000, physicists at the NEC Research Institute in Princeton, New Jersey claimed to have "broken the limit set by the speed of light". They achieved this by firing a laser pulse into a glass chamber filled with a cesium (Cs) atom vapor. NEC’s Lijun Wang, in fact, insisted he’d created an experiment in which light speed was not merely exceeded by an added increment, but by a factor of three hundred! This meant that an almost identical light pulse exited the chamber and traveled about sixty feet before the main part of the laser pulse finished entering the chamber, Wang said.

Needless to say, this flouts well-known accepted axioms of causality. (Wherein causes are assumed to precede effects). The problem also was that the experiment has never been confirmed, so we take it with more or less a grain of salt.

Basically, in this new experiment, CERN fired a neutrino beam from a particle accelerator near Geneva to a lab 454 miles away in Italy. In so doing it travelled 60 nanoseconds faster than the speed of light, defined to be at 186,282 mi./sec. The CERN team calculated the margin of error at 10 nanoseconds, which made the result statistically significant. Evidenlty also, appreciating the enormity of the find, they spend months checking and re-checking their results. According to one University of Bern physicist, Antonio Ereditato:

"We have not found any instrumental effect that could explain the result of the measurement"

Maybe so, but that doesn't mean we instantly accept it. Recall the hallowed halls of science are littered with experiments that appeared to prove or show some fantastic phenomenon or violation of normal physical laws, but could never be confirmed. One of them was the claim for cold fusion, by Pons and Fleischmann in the late 90s, which would have revolutionized the energy industry. But no one could replicate it and it was later found that they had not taken into account all the experimental errors that migh have led to heat being evolved.

Recall (from the last instalment of 'Introducing Special Relativity), that by the work -energy theorem:

W = K(f) - K(i)

K(i) = 0 (u = 0)

K(f) = m(o)c^2/ [(1 - u^2/c^2)^½ ]

Where in this case, K(f) would be the kinetic energy acquired by the neutrinos en route to the lab in Italy. We assume the neutrinos start from rest, so K(i) = 0 and u = 0 at that point.

Here we would need to find the velocity, u. If one knows in advance the rest mass m(o) of the particular species of neutrino (there are three types: electron, mu and tau) than one can obtain u by using:


K(f) = (m'/ m(o))

where m' denotes the enhanced mass over m(o)

For example:

The normal rest mass of an electron = 0.511 MeV (Millions of electron volts)

Then, if the rest mass is found to be 3.25 MeV while undergoing K(f) one can find u. We have:

K(f) = (3.25/ 0.511) = 6.36x the rest mass

So: E = 6.36mc^2 = mc^2/ [(1 - u^2/c^2)^½ ]

and:

6.36 = 1/ [(1 - u^2/c^2)^½ ] or:

(1 - u^2/c^2) = 1/(6.36)^2 = 1/40.44

or: u^2/c^2 = 1 - 0.0024 = 0.9976

u = [0.9976c^2]^½ = 0.9987c

Note that even such a large increase doesn't translate into superluminal velocity for the electron. Now we know the three species of the neutrino have rest masses as follows:

Electron neutrino: less than 8 eV

Mu neutrino: less than 250 keV

Tau neutrino: less than 35 MeV

Let's just deal with the last and assume for the sake of argument m(o) = 30 MeV. Then we ask what it would have to be to even be at the speed of light.

In performing the computations one would find (and I leave this as an exercise for ambitious readers) that no matter how high the enhancement of the rest mass you cannot obtain a superluminal velocity.

Even if the tau neutrino mass had increased to 10 million MeV this would not disclose a superluminal speed. One would have:

K(f) = (10^6 MeV/ 30 MeV) = 3.33 x 10^4 the rest mass

whence:

3.33 x 10^4 = 1/ [(1 - u^2/c^2)^½ ]

or:

(1 - u^2/c^2) = 1/(3.33 x10^4)^2 = (9 x 10^-10)

u^2/c^2 = 1 - (9 x 10^-10) = 0.9999

u = 0.9999c

What would yield a superluminal speed? Well, an imaginary mass! Let's take the example of the earlier electron again, and allow an imaginary rest mass of 3.25i MeV (E.g. employing the imaginary number, i). Then:

K(f) = (3.25i/ 0.511) = -6.36i over the rest mass of 0.511 MeV

So: E = -6.36i (mc^2) = mc^2/ [(1 - u^2/c^2)^½ ]

6.36 i= 1/ [(1 - u^2/c^2)^½ ] or:

(1 - u^2/c^2) = 1/(6.36i)^2 = 1/(- 40.44)

(Since: (i)^2 = -1 )

or: u^2/c^2 = 1 + 0.0024 =

u = [1.0024 c^2]^½ = 1.0011 c

In other words, in order to have a superluminal speed, it would imply the increase in mass has to be imaginary! I.e. an imaginary value would have to be assigned to the mass, and hence to K(f).

Now, of course, the counter-relativists will simply insist that these equations derived from special relativity are all wrong.

Fine, then we await their new supra-relativistic physics to deliver the correct equations!

In the meantime we take their results as well......imaginary!

Of course, I'm not alone with this view. Evidently, University of Maryland's Chairman of the Physics Dept., Drew Baden, called it a "flying carpet" or something that was too fantastic to be believable. Until these guys at CERN show they don't require imaginary masses for the neutrino beam's trip to Italia, we have to concur!

Solutions to Special Relativity Problems (6)

We now examine the solutions from the previous introduction to special relativity instalment;

(1) A spaceship of mass 10^8 kg is to be accelerated to 0.6c using a matter-antimatter mix engine.

(a) How much energy does this require?

(b) How many kilograms of matter and antimatter will it take to provide this much energy?

Solutions

(a) by the work -energy theorem:

W = K(f) - K(i)

or:

W = m(o)c^2/ [(1 - u^2/c^2)^½ ] - m(o) c^2

where m(o) = 10^8 kg and u = 0.6c

Then:

W = (10^8 kg))c^2/ [(1 - (0.6c)^2/c^2)^½ ]

W = (10^8 kg)c^2/ ([1 - 0.36]^½) = (10^8 kg)c^2/ ([0.64]^½)

W = (10^8 kg)c^2/ (0.8) = 1.25 x 10^8 kg(c^2)= 1.12 x 10^25 J

(b) According to theory, equal amounts (masses) of matter and antimatter are required for complete annihilation and total conversion of the initial masses into energy. The energy needed as shown in (a) is 1.12 x 10^25 J. Thus: m(a) = m(m) = 6.25 x 10^7 kg.


(2) Consider the decay:

24 Cr 55 -> 25 Mn55 + e

The Cr 55 nucleus has a mass of 54.9279 u and the Mn 55 nucleus has a mass of 54.9244u.

(a) Calculate the mass difference between the two nuclei.

Solution:

delta m = 54.9279 u - 54.9244u = 0.0035 u


(b) What is the maximum kinetic energy of the emitted electrons?

The mass defect is thus 0.0035u and:

E = (delta m) c^2 =(0.0035u)(931 MeV/u) = 3.25 MeV

By the work -energy theorem:

W = K(f) - K(i)

K(i) = 0 (u1 = 0)

K(f) = m(o)c^2/ [(1 - u^2/c^2)^½ ] = 3.25 MeV

where we need to find u.

Using a table one finds the rest energy of the electron = 0.511 MeV

and K(f) = (3.25/ 0.511) = 6.36x the rest mass

So:

E = 6.36mc^2 = mc^2/ [(1 - u^2/c^2)^½ ]

6.36 = 1/ [(1 - u^2/c^2)^½ ] or:

(1 - u^2/c^2) = 1/(6.36)^2 = 1/40.44

or:

u^2/c^2 = 1 - 0.0024 = 0.9976

u = [0.9976c^2]^½ = 0.9987c



(3) Find the energy required to remove a single proton from 19 K 41.


In removing a single proton the atom remaining is: 18A 40

whose measured mass is: 39.96238 u

The mass of the final system becomes:

39.96238 u + 1.007825 u = 40. 970205 u

The effective mass increase of the system:

delta m = 40.970205 u - 40.96184 u = 0.008365 u

Then the energy needed to remove a single proton, from Einstein's eqn.

E = (delta m) c^2 = (0.008365 u) (931 MeV/u) = 7.79 MeV

(Which can also be converted to MJ using: 1 MJ = 1.6 x 10^-13 MJ/MeV)


(4)Find the speed and mass of an electron whose kinetic energy is 50 MeV.

Solution:

As per Problem #2, we know the electron rest mass = 0.511 MeV

Then, the mass of this electron is (50 MeV/0.511 MeV) = 97.84

or 97.84 times the normal electron rest mass.

The speed, following the same procedure applied in #2:

E = 97.84 mc^2 = mc^2/ [(1 - u^2/c^2)^½ ]

and we need u:

(1 - u^2/c^2) = 1/(97.84)^2 = 1/9572

or:

u^2/c^2 = 1 - 0.0001 = 0.9999

u = [0.9999c^2]^½ = 0.9999c (approx.)


(5) A rocket ship is to be accelerated to a speed of 0.5c. If propulsion is to be by using nuclear fuel, what fraction of the intial rest mass of the ship would have to be converted into kinetic energy to attain the desired speed?

What time dilation results if the speed is v = 0.5c?

Would this be sufficient to allow one generation of humans to reach the star Proxima Centauri (4.2 light years distant)?

Solution

by the work -energy theorem:

W = K(f) - K(i)

where K(i) is just the initial rest energy, or m(o)c^2

Then W = mc^2/ [(1 - u^2/c^2)^½ ] - m(o) c^2 = (total energy - rest energy)

Then total energy E:

= mc^2/ [(1 - u^2/c^2)^½ ] = mc^2/ [(1 - 0.25)^½ ]= 1.16mc^2

Then the excess (E - E' = E_k) is that required from the fuel, or:

E_k = (1.16 mc^2 - mc^2 = 0.16 mc^2)


Mas of nuclear fuel - call it m'- is then related to E_k by:

E_k = m'c^2 = 0.16m' c^2 or

(m'/m) = 0.16 or m' = 0.16m

In other words, the nuclear fuel mass(m') needs to be at least 16% of the total initial mass of the rocket. So, if the rocket's mass is 100,000 kg then the nuclear fuel must be at least (0.16)(100,000 kg) = 16, 000 kg.

For time dilation:

t' = t [1 - v^2/c^2]^½

again, v = 0.5 so:

t' = t [1 - 0.25]^½ = t [0.75]^½ = 0.866t

Now, there are 9.5 x 10^15 meters per light year

So the distance to Proxima Centauri is:

D = (4.2 Ly) x (9.5 x 10^15 m/Ly)= 4 x 10^16 m

Or: D = 4 x 10^13 km

The time required is then:

(0.866/0.500) x (4 x 10^13 km)/ (300,000 km/s]

= 1.73 x (4.22 yrs) = 7.3 years

Since a generation is generally figured as 40 years, the time factor will not be an issue.

Solutions to Special Relativity Problems (3)

Recall the problems set at the end of (3):

1) With what speed would a clock have to be moving to run at a rate that is one half the rate of a clock at rest?

Solution

We have: t = t’/ [1 - v^2/c^2]^ ½

But the proper time is defined such that:

t' = t/2 or t'/ t = ½

Then:

[1 - v^2/c^2] = (t'/ t)^2

and:

v^2/c^2 = 1 - (t'/ t)^2

v^2 = c^2[1 - (t'/ t)^2]

so:

v = c[1 - (t'/ t)^2]^½ = c[1 - 0.5^2]^½ = c[0.75]^½ = 0.866c


2) An atomic clock is placed on a Jumbo Jet. The clock measures a time interval of 3600 s when the jet is oving at v = 300 m/s. What corresponding time would an identical clock left on the ground measure? (Hint: whenever v << c (e.g. v/c << 1), we have 1 + v^2/2c^2 and not [1 - v^2/c^2]^1/2)

Solution

The proper time t' = 3600 s

Since v = 300 m/s = (10^-6) c and hence v/c << 1 we need the form:

t = t’/ [1 + v^2/2c^2]


t = 3600s/ [1 + (10^-12)c^2/2c^2]

Since the numerator is only slightly larger than 1, then the time t will be:

3600 s/(1.000000000001)= 3600.0000000018

3600 + 1.8 x 10^-9 s

or slightly longer than one hour.


3) A muon formed high in the Earth's atmosphere travels at v = 0.99c for a distance of 4.6 km before it decays into an electron, a neutrino and an anti-neutrino.

a) How long does the muon survive as measured in its rest frame?

Solution

The proper time t' applies to the muon's reference frame.

So:

t = t’/ [1 - v^2/c^2]^ ½


and t' = t [1 - v^2/c^2]^ ½

v = 0.99 c and v^2 = (0.99c)^2 = 0.98c^2

Then: t' = t [1 - 0.98c^2/c^2]^ ½ = t [0.02]^ ½ = t(0.14)

recall distance travelled = 4.6 km = 4600 m

To get t' we need to find t first, e.g.

t = 4600m/ (2.97 x 10^8 m/s) = 1.55 x 10^-5 s

Then: t' = (1.55 x 10^-5 s) (0.14) = 2.1 x 10^-6 s


b) How far does the muon travel as measured in its frame?

Solution

The distance traveled in its frame is just the proper length, L'

so:

L' = 4600 m [[1 - v^2/c^2]^ ½ = 4600m (0.02)^ ½

L' = 4600 m (0.14) = 644 m


4) The average lifetime of a pi meson in its own frame of reference is 2.6 x 10^-8 s. If the meson moves with v = 0.95c, what is its mean lifetime as measured by an observer on Earth?

Solution

The proper time t' = 2.6 x 10^-8 s

t = t’/ [1 - v^2/c^2]^ ½

and v = 0.95c


so:

t = (2.6 x 10^-8 s)/ [1 - (0.95c)^2/c^2]^ ½

t = (2.6 x 10^-8 s)/ [0.0975]^ ½ = (2.6 x 10^-8 s)/ 0.312

t = 8.3 x 10^-8 s

An Introduction to Special Relativity (2)

In a last ditch efort to satisfactorily explain the riddle of the null result (from the Michelson-Morley experiment), a fantastic idea was put forward by George F. Fitzgerald in 1890. Using the analogy of a rubber ball which is deformed upon striking a wall, Fitzgerald conceived that the ether would distort matter.

This distortion would take the form of a contraction of length in the direction of the motion through the ether. Such a contraction would explain the null result of the Michelson -Morley experiment. That is, the arm L, of the apparatus, moving against the ether would be shortened by "ether pressure" just enough to compensate for the slowing down of light by the ether wind.

A similar but more mathematical theory was worked out by the physicist Hendrik A. Lorentz, who expressed the length contraction in the direction of motion by:

L' = L[1 - v^2/c^2]^½

This famous equation became known as the "Lorentz-Fitzgerald contraction".

Understandably, the hypothesis of Lorentz and Fitzgerald gradually gained general acceptance, except for a Swiss Patent clerk, who refused to be deluded by such a contrived idea (based as it was on absolute motion). The clerk's name was Albert Einstein, born in Ulm, Bavaria, in 1879.

Einstein had graduated as a physics major, but he managed to produce such a poor impression as a teacher that he was dismissed from three teaching jobs in succession. Having been reduced to a 'hand-to-mouth' existence, he was lucky to find a job processing patent applications in Bern, Switzerland. Fortunately, he also has a leisurely schedule that permitted him to while away a good many hours contemplating space, time and energy. (A good thing his time was nearly a century ago, otherwise - today- he'd be accused of "gold bricking"!)

After much thought, Einstein was forced to conclude that motion is never observable as motion with respect to spact and that there is no basis for the introduction of "absolute motion". In Einstein's mind, the only kind of motion was relative rest perceived from different viewpoints (e.g. "reference frames"). Einstein called this insight the Principle of Relativity. According to Einstein's Principle of Relativity: "All the laws of physics are the same in all inertial reference frames."

It's instructive at this point to commence a quantitative approach to see exactly how Einstein reasoned. This necessitates we first get accustomed to the idea of a coordinate system. Basically, the purpose of any coordinate system is to enable us to identify a particular point in space. The standard procedure (for 3-dimensional space) is to take three mutually perpendicular axes with coordinates in x, y and z, representing distanced to where the axes meet at the origin, or 0. In special relativity such coordinate systems are typically designated: S, S', S" etc. or alternatively, S1, S2, S3.

In Fig. 1 we have two coordinate systems, S and S', with S having coordinate x, y and S' having coordinates x', y'. (We confine the systems to 2 dimensions here for initial simplicity.) Thus, S' has its origin at 0' and S at 0. In the figure, S' is shown moving with constant velocity v in the x - x' direction (which obviously coincide for S and S').

Keep in mind though we arbitrarily assigned S' as moving this depends on which coordinate system is taken to be at rest. An observer attached to system S may very well consider himself moving and S' stationary. This would resemble the well-known example of a passenger in a stationary train observing a train parallel to his through his window and deducing he is moving, though it is actually the train parallel to his. The key point is that the systems S and S' have a constant relative velocity. Such coordinate systems occupy a special place in relativity and are called "inertial reference frames" or "inertial coordinate systems". Their primary feature is that they lack and acceleration of one to the other.

Now, think about this carefully: if the origins O and O' coincide at time t = 0, and we are observers in S', then we will see 0' move along OX with velocity v. Then the two sets of coordinates, representing the same point (in S and S') are related by:

x' = x - vt and y' = y

If we chose we could append a 3rd axes (z) i.e. coming out of the page, and have also:

z' = z

The preceding equations describe the "Galilean transformation". From these it is very easy to obtain velocities, and we need only differentiate with respect to t:

E.g.

dx'/dt = dx/dt = v

dy'/dt = dy/dt

dz'/dt = dz/dt

(Note here that never once did we write t' = t since the notion of an absolute, universal time was the very cornerstone of Newtonian theory!)

We now start with the preceding transformations and see how Einstein reasoned. Reference may be made to Fig. 2 which displays a three-dimensional perspective of the relative motion and hence is a bit more complicated. Our aim in using it is to find an alternative to the Galilean transformations - which obviously can't be correct for all situations.

For example, according to the Galilean transformations, a pulse o flight sent out from S' would move with the velocity c, the speed of light, as measured from S', but with the velocity (c + v) measured from S. But this directly contradicts the demopnstrated fact (Michelson-Morley) that the speed of light is always constant when computed from one reference frame relative to another.

Einstein thus began afresh by using not only the Principle of Relativity (just stated, i.e. the laws of physics are the same in all inertial reference frames)) but also:

The speed of light is always found to have the same value no matter what the motion for the source or the observer.

From these two postulates, Einstein deduced a number of surprising results which would have been totally unacceptable to a more conservative mind.

Start then with the two systems depicted in Fig. 2 which are coincident in space at at the instant a flash bulb (say) is set off when the origins coincide. The observer S sees S' moving in the x-direction with the velocity v and observer S' sees S moving with the same velocity in the opposite (-x) direction.

Both observers see the flash of light travel away from the origin with the same velocity c. The distance the light travels in the S system during time t, is ct. This must be true in any direction. If the light spreads out equally in all directions, then by the end of time t it has extended to fill a sphere whose radius is r, and we can write:

(1) r^2 = x^2 + y^2 + z^2 = c^2 t^2

The exact same equation holds in the S' system since the flash was set off when O coincided with O. Thus, while the 2 systems are separating from each other, the observer in S' also sees the light fill a spherical shell in his own system, so writes (for r'):

(2) r'^2 = x'^2 + y'^2 + z'^2 = c^2 t'^2

Note the velocity of light is the only thing that's the same in both systems. The two preceding equations thus describe a point lying on the spherical shell. Even though they describe the same point in space, the observer in S sees the point at position (x, y,z, t) abd the observer S' at position (x', y', z', t').


Subtracting equation (2) from equation (1) and transposing terms:

x^2 + y^2 + z^2 - c^2 t^2 = x'^2 + y'^2 + z'^2 - c^2 t'^2


We now look for a transformation similar to the Galilean transformation, but which will allow c to be the same in both S and S'. Since the y and z coordinates of the position are not affected by the motion in the x-direction we can say y' = y and z' = z. For the x-coordinate, we try a transformation of the form: x = a(x' + vt') and x = a(x - vt), where a is an invariant determined by the two fundamental postulates (i.e. the same quantity is used in going from x to x' as from x' to x).

Further, we expect a to depend on the velocity v in such a way that it becomes equal to 1 when v becomes very small compared with the speed of light. When this happens, the x and x' transformations become the same as the ordinary Galilean transformations.

So we begin by using x = a(x' + vt') to solve for t' and obtain:

(3) t' = 1/v (x/a - x')

For x' above, we now insert the value for x' (e.g. x' = a(x- vt)):

(4) t' = 1/v(x/a - ax - avt) = at - x^2(a^2 -1)/ va

Similarly, we find for t:

(5) t = -at' + x'(a^2 - 1)/ va


If we now substitute x' = a(x - vt) and equation (4) into the right hand side of equation (2), we obtain:

(6) x^2 + y^2 + z^2 -c^2t^2 = a^2(x - vt)^2 + y^2 + z^2

= c^2[at - x/v (a^2 - 1)^2/a]

Now, re-arrange terms and cancel the z and y terms, which are the same on both sides of the equation, to get:

(7) x^2 - c^t^2 = [a^2 - (a^2 - 1)c^2/ a^2v^2]x^2

+ 2[(a^2 - 1)c^2/v^2 - a^2]xvt - (c^2 - v^2)a^2t^2

If the preceding is to hold true for any value of x and t, each term on the left side must equal each term on the right. Since there are no terms with the combination xt on the left, the xt term on the right must be zero. This means:

(8) (a^2 - 1)c^2/v^2 - a^2 = 0

Solving for a:

(9) a^2 = 1/ (1 - v^2/c^2) and a = 1/ (1 - v^2/c^2)^½

Finally:

(10) (a^2 - 1)/a = v^2/c^2/ (1 - v^2/c^2)^½

Substituting the preceding into our x, x' transformation equations and equation (3), we arrive at the following transformations to replace the Galilean:

(11)

x = x' + vt'/(1 - v^2/c^2)^½

and

x' = x - vt/ (1 - v^2/c^2)^½


(12) y = y' and y' = y

(13) z = z' and z' = z

(14) t = t' + x'v/c^2/ (1 - v^2/c^2)^½

and

t' = t - xv/c^2/(1 - v^2/c^2)^½


Equations (11)- (14) are known as the Lorentz transformation or the Lorentz-Einstein transformation.

Note the important feature is that the time must be given as well as the position, because the respective clocks in S and S' will cease to read identical times after they have parted from one another. This is the significance of (14) in the above set. Also, the fact that time is given the same importance as space (i.e. as another dimension) shows there's nothing special or mystical about "the fourth dimension".

Problems:

1- Given that x' = 1/a (x - vt) and t' = 1/a (t - vx/c^2), derive similar equations for x and t in terms of x' and t'.
(Recall: 1/a = (1 - v^2/c^2)^½)

2- An event in space-time occurs at x' = 60 m, t = 8 x 10^-8 s, in a frame S' (y' = 0, z' = 0). The frame S' has a velocity of 0.6c along the x-direction with respect to a frame S. The origins O and O' coincide at time t = t' = 0. Find the space-time coordinates of the event in S.

3- Suppose an astronaut is traveling at 0.9c in a space ship with respect to the Earth. How long a time interval will his clock indicate when the Earth has revolved once around the Sun? (Take the duration of one standard revolution of Earth around the Sun to be 365 ¼ days.)

An Introduction to Special Relativity (1)

As promised, we'll now commence a six part blog on a basic introduction to Einstein's theory of special relativity. This will also include some basic problems at the end of each segment and subsequently the solutions. Not much more than basic algebra will be needed to follow most of the development.

In the popular mind, at least, the word "relativity" usually conjures up visions of space travelers returning youthful to Earth after journeys of many decades (measured on Earth). Of course, there is much more to relativity than this. As a matter of fact many aspects of relativity as in the case of motions, aren't new at all. Basically, it merely entails the assertion that the laws of physics appear to be the same in given reference frames or coordinate systems. This is made more accurate in Einstein's special relativity by referring to inertial reference frames.

Though Henri Poincare came close to discovering the principles of relativity, it was Albert Einstein first and foremost who ruthlessly and relentlessly pursued the basic principles to their utterly logical conclusions, including that lengths shrink as velocities tend toward the speed of light, c, and time slows or "dilates".

Like Ernst Mach before him, Einstein adopted the view that in considering two objects in relative motion, it is futile and meaningless to attempt to decide which object is "really" in motion and which is at rest. If you are in a Jumbo jet traveling at a speed of 900 km/hr relative to the Earth, it makes no difference whether someone says you are moving at that speed, or the Earth is moving at that speed. In either case, the operation of the laws of physics in your jet and on the Earth will be the same. Balls will still drop, and you will still fall off your chair if not careful. There exist no absolute frame of reference to contradict you.

The Special Theory of Relativity, interestingly, may be said to have had its origins in the null result of an experiment the primary aim of which was to detect relative motion. This experiment was first carried out by the American phuysicist Albert A. Michelson in 1881, and subsequently repeated in 1887, with the help of Edward W. Morley. The experiment has thus come to be known as "the Michelson-Morley experiment".

The basic idea was to time the transits of light in two distinct directions: perpendicular to the Earth's orbital motion, and parellel to the orbital motion. This was to be accomplished by using an arrangement of mirrors and light beams such as depicted in Fig. 1. A difference in light velocities (transit distance/ time) would reveal itself by a delicate interference pattern formed by two separate beams after rejoining each other.

The implication of a difference in velocities would mean the confirmation of a remarkable entity called "the Ether". In effect, if light represents waves propagating through the Ether, the velocity of light as recorded by instruments on Earth's surface must be distorted by the motion of the Earth through space. An analogous principle is that a swift river must retard a swimmer's combined upstream and downstream speeds more than his cross-current speed. Similarly, a large difference in light velocities (along the two different paths) should show Earth is moving rapidly through the Ether, while a small difference would show it's moving slowly.

Astonishingly, on each occasion the experiment was conducted the result was virtually negative. There was no evidence of any Ether flowing past the Earth in any direction. Any minor deviations fell within the purview of the experimental errors.

Now, imagine the apparatus in Fig. 1 to be moving with velocity v toward the right, then only in the event of a null result should there should there also be a relative velocity of the Ether of magnitude v to the left. To get a positive result the apparatus needs to move a speed v relative to the Ether. The round trip time for a light beam following path X-M2-X is:

t1 = 2Lc/ (c^2 - v^2)

Meanwhile, the light beam traveling from X to M1 must have a component of velocity v along X-M2, relative to the hypothesized Ether or it will not strike the mirror at M1. Since the velocity of the light relative to Ether is c, subtracting the preceding component leaves a velocity of (c^2 - v^2) E.g. see Fig. 2. The same is true for the return journey to the total time t2 for the path X-M1-X is:


t2= 2L/ (c^2 - v^2)^½

Now, the recomibination of two light beams (originally split in two) will produce interference fringes as depicted in Fig. 3. Any difference in the times taken to traverse the paths will show up as a shift in the position of the bright and dark fringes since it will indicate different path lengths.

Hence, a difference in time delta t is equivalent to a path difference:

c (delta t)/ lambda (d)

where d = the width of one fringe and lambda is the wavelength of the light (taken to be 6 x 10^-7 m).

From this, the time difference between the two beams can be computed from:

delta t = t1 - t2

= 2L/c{ 1/ (1 - v^2/c^2) - 1/ (1 - v^2/c^2)^½}

This equation can be approxiamtely expressed as:

delta t ~ L/c (v^2/c^2)

Since v << c, this corresponds to a fringe shift of:

delta d - c (delta t)/ lambda = Lc^2/ lambda (c^2)

The measurement of the fringe shift is accomplished by rotating the whole apparatus through 90 degrees. The effect of this is to interchange the arms, X-M1 and X-M2, thereby reversing the sign of the fringe shift so that an overall shift of 2 (delta d) should be observed.

For the Michelson -Morley experiment of 1887, L = 11 m, v = 0.0001c.

Using the path difference equation for delta d, we arrive at:

delta d = 0.183

or, approximately 0.2 fringe.

The overall shift expected was:

2 x (delta d) = 2 x (0.2) = 0.4 fringe

But the largest shift actually detected was only 0.01 fringe within the experimental error.

This null result flabbergasted physicists of the time. They were simply unable to conceive that light required no medium within which to propagate. Hence, it's not difficult to see why so many clung to the Ether mcguffin for so long, even after it was disproven. Thus, fanciful and elaborate schemes were thought up to explain the null result, much like today's intelligent design proponents have confected fanciful explanations to try and disavow Darwinian evolution.

For his own part, Michelson naturally assumed that the local ether had to be adhering to the Earth, travelling with it through space. All other scientists were incredulous that an orbital velocity of 30 km/s in relation to the Sun could not generate the tiniest ether "breeze". To many, the situation was not unlike a ship maintaining constant speed and direction in the sea, irrespective of current changes.

Problems

1. In the Michelson-Morley experiment, the length L of each arm of the interferometer was 11 meters. Sodium light of wavelength 5.9 x 10^-7 m (590 nm) was used. The experiment would have revealed any fringe shift > 0.005 fringe.

What upper limit does this place on the Earth's velocity through the supposed Ether?

2. Using Fig.1, say the time of travel to the right is: t(r) = L/(c - v) and the time of travel to the left is t(L) = L/(c + v).

a) Find the "total time of travel" by adding both left and right contributions.

b) find the time consumed for "a half-trip".

c) Find the time consumed for a round trip.

d) Add the two "half trips" and what do you obtain?

e) Why does this not agree with the value obtained for (a)?


Next: How Fitzgerald attempts to rescue the situation.