- Explanation to Brownian Motion,
- Photo-Electric Effect, and
- The Special Theory of Relativity.
Let in a particular frame of reference, two events occur in a same point in space. The time interval between the two events as measured by an observer at rest in the same frame is 
. Then he will conclude that an another observer in second reference frame moving with constant speed v relative to the frame of reference, will measure time interval of 
.1
Now, let us study what the Twin Paradox is.
TWIN PARADOX is said to be, “A longer life, but it will not seem longer”2 .
This Paradox involves two identical clocks, one of them is kept on Earth, and the other one is brought to a journey with a speed very near to ‘c’, and then is brought back to Earth. We can replace them with two biological clocks, say twins George and Fred. It is assumed that we have made a spacecraft which can move with velocity very near to that of light in vacuum. George goes on a voyage with the spacecraft with a high velocity and Fred stays on Earth. Let us assume, George travels with v= 0.99995c.
Then, putting v=0.99995c in
give us,
[1- (v/c)2]1/2= 1 x 10-2= 1%.
This implies that Fred finds that, for every 100 beats of his heart, George’s heart beats only once. Similarly, for every 100 ticks of Fred’s wristwatch, George’s wristwatch ticks only once. That is, by the time Fred’s heart beats 100 times, George’s heart beats only one time. By the time George watches Interstellar once, Fred would have watched it 100 times consecutively.
So, by the time George travels for 1 year and then returns back to Earth, Fred will have spent 100 years of his life.
Now, why is this a “Paradox”?
- For more illustrations, see Young and Freedman, 2013.
- For more illustrations, see Beiser, 1997.
The solution states that, the length of the path travels by George contracts with respect to himself. If the total length of the path of travel, measured from Earth (or as measured by Fred) is L0, then, from George’s frame, it is (L0/100). From George’s frame, Fred is moving with V=0.99995c and the path traversed by Fred is (L0/100). So, the time required for Fred to travel, as measured by George, is L0/(100xV). However, from Fred’s frame, the length of the path is still L0, and from his frame, George is moving with V=0.99995c. So, the time required for George to travel, as measured by Fred, is (L0/V) which is 100 times higher than that measured by George.
- For more illustrations, see Beiser, 1997.
The above figure is when we have to find the relative time of Maria with respect to ours. Similar figure can be drawn for finding the Rest of Universe’s time with respect to Maria’s.
Now, if Maria travels with velocity v=0.99995c with respect to us, then what will happen? Will Maria travel in time?
From the Time-Dilation Equation, we know that Maria should travel in time with respect to us. We also have our practical proof in support of it. The Neutrinos4 in the Large Hadron Collider5 are time-travelers. When they are at rest with respect to observer, their lifetime is so small that human-made machine cannot detect them. However, when they are travelling with very high speed within LHC, according to Time-Dilation Equation, their lifetime increases with respect to us. And, then we can detect them.
Similar thing happens for the Muons6. Naturally, Muons are formed high above from the sea-level, because of the collisions between cosmic rays and atmosphere. But, Muons are found in the sea-level too. This can only happen if they travel to sea level. But, Muons have decay time of about 2.2×10-6 and they travel with approximately v=0.998c. Then, from simple calculations, they can travel a maximum distance of d=vt= (0.998x3x108x2.2×10-6x10-3) km
= 0.66 km, much low.
But, as they are travelling with v=0.998c with respect to us, so according to Time-Dilation Equation, their life-time becomes: t= 2.2×10-6/ (1-0.996)1/2= 34.8×10-6.
Now, d= vt= (0.998x3x108x34.8×10-6x10-3) km = 10.4 km.
Here, with respect to the observer, the Muons are time-travelers. So, if Maria travels with v=0.99995c with respect to us, she also travels in time with respect to us.
- Neutrino is an electrically neutral elementary particle with half-spin. Theoretically, it seems to be massless, but evidences are there to show that it has mass, no matter how small the mass can be. Neutrino is lepton, like charged electron, muon and tau. Neutrino is uncharged, so is unaffected to Electro-magnetic force. It is lepton, so is not affected by Strong-Force. It is affected only by Weak-Force and Gravity. However, weak-force is low-ranged and gravity is weak for subatomic particles. So, Neutrino typically can pass through our bodies, and is not detected. For this strange nature, it is also said “Ghost-Particle”. In LHC, it travels with very high speed and is said to travel in time.
- LHC, world’s largest and most powerful particle collider, is undoubtedly the most complex lab ever built. It was built from 1998 to 2008. It runs from CERN, near Geneva of Switzerland, to Sasso of Italy. The aim of it is to test different proposals of Particle Physics and High-Energy Physics, especially to find the existence of Higgs-Bosons. This is the particle which is said to be the reason behind matters having ‘mass’.
- Muon is elementary particle, with charge of -1e, like electron, and spin ½. Its mass is 105.7 MeV/c2. It is a lepton, like electron, tau and neutrino.
.Let us take an interval as ‘the interval when the second hand of a clock ticks for once, that is 1 second in a particular frame of reference.’
Then, putting v=0.99995c in [1- (v/c)2]1/2give us,
[1- (v/c)2]1/2= 1 x 10-2= 1%.
In this case, Maria is in motion with respect to us. So, for every 100 ticks of the second hand of our clocks, Maria’s clock’s second hand ticks only once. So, our 100 seconds is equivalent to her 1 second. Thus, her biological clock is also slowing down and she is aging less than us.
Now, from our reference frame, Maria is travelling for the same years that we spend (here it is100 years). So, while travelling, if her time goes forth with respect to our time, then at the end of voyage she would age more than us (100+x years; if v=0.99995c then x=99 years). And, if her time goes back with respect to our time, while travelling, then at the end of voyage she would end aging less than us (100-x years).
This implies that, with respect to our time, Maria’s time is going to ‘Past’ while she travels.
Now, let us study what is seen where Maria is going, when seen from Maria’s reference frame. From the theory of equivalence of reference frame, Maria should see that the Rest of Universe is travelling in time. With respect to her time, the time of Rest of Universe should go to ‘Past’. So, every artificial and biological clock of the Rest of Universe will go to ‘Past’ with respect to Maria. Now, the question is who will end up as older.
Now, let us study the following figure.
It appears from the above figure that before the start of the voyage, Maria’s time and the Rest of Universe’s time overlapse. However, when the voyage starts, Rest of Universe’s time is going to ‘Past’ with respect to Maria’s time. So, Maria’s time is going to ‘Future’ with respect to Rest of Universe’s time.
If we see from Maria’s frame, her time is ‘Present’ at every instant. So, Maria will conclude that, the Rest of Universe’s ‘Present’ is her ‘Present’ and their ‘Present’s would over lapse all the time if there were no voyage of the Rest of Universe. Their ‘Presents’ overlap until Rest of Universe starts its voyage. After the moment, Rest of Universe’s time is going to ‘Past’ with respect to Maria’s time- Present. So, Rest of Universe’s time is going to ‘Past’ of itself. Studying the above figure, we can say, so, Maria is going to Rest of Universe’s ‘Future’. And, as she will conclude that she is ‘Present’ now, so, she will conclude that she will spend 1 year (as seen from her reference frame) whereas, the Rest of Universe is aging as she is going to its ‘Future’.
We can take any two Frames having relative velocity between themselves, and apply this method.
In figure above, we have two Frames of references. At (a), both the Frames F and f are in the same position at time t=0.
At (b), Frame f is moving with velocity v relative to F.
At (c), Frame F is moving with velocity -v (as it is in the opposite direction to v) relative to f.
Now, each of the Frames travels in time with respect to other.
Let F concludes that f is going to f’s Past throughout f’s motion.
Then, f will conclude that f is going to F’s Future throughout F’s motion. This is equivalent to f’s conclusion of F is going to F’s Past throughout F’s motion.
The Observations:
If we see a body moving (i.e. the body is moving with respect to our frame of reference), we will conclude that its time is going to Past with respect to our time.
Anything, from a celestial body moving with respect to us, to the trees, houses which seem to move when we travel by train, is going to Past with respect to our time.
It deserves mention that the proposition revealed out from the above theoretical construction may support other conclusions which are as follows:
(i) As motion is relative, so going to Past with respect to the observer, is also relative. And, thus, there is no absolute Past, Present or Future, but actually they are relative.
(ii) From our conclusion we can say that a clock moving with respect to us, is going to Past with respect to us. Einstein had said that ‘moving clock runs slower’. So, there must be a link in between our conclusion and Einstein’s conclusion. Let us try to find it.
Suppose, t=0 is the initial time. We are looking for an incident to happen at t=t. Let us say, the incident is, the second-hand of a clock ticks for once. Now, the clock is moving with respect to us. So, it is going back in time every instant with respect to us. So, when we conclude, after watching a stationary watch, say our wrist watch, that one tick(or 1 second) has happened, i.e., t=t has come, the moving clock still does not reach to t=t. So, its second-hand still does not tick for once. So, the moving clock is going slow with respect to our wrist-watch. So, the moving clock is running slower.
(iii) Our conclusion supports the equivalent nature of Reference Frames.
(iv) An another Paradox can be solved by our method. It is the Rocket Warfare Paradox7. Consider, two identical rockets A and B of respective length l0 in their respective frames. When seen from B’s frame, length of A is l0/k, where k=1/(1-v2/c2). Similar is for A’s frame. Consider the situation when, as seen from B, a missile is fired to A from B when tip of B coincides with tail of B. In this case, the missile misses the target A.
The above figure is viewed from B. Now, when seen from A,
the missile should blast A up.
Solution to this paradox is Simultaneity. We used a concept of Simultaneity when we supposed that the missile was launched when tip of B met tail of A. But these two events of launching and the meeting of tip of B to tail of A are simultaneous from B’s frame. But they were not simultaneous from A’s frame.
Now, let us try to solve it using the conclusive observation of present study.
- For more illustrations, see Mani and Mehta, 2000.
Suppose, there are frame-f and frame-F. An explosion occurs as shown. Now, our concern is how the time of explosion changes with respect to f and F.
At any instant, we can imagine a frame f*, in between f and F, as shown: 
Here, mod(V)> mod(v).
So, from frame f*, velocity of frame F is –v.
And, from frame f*, velocity of frame f is (V-v).
From Lorentz Transform, we know, t=k(t0 – xv/c2), where k= (1 – v2/c2)-1/2.
So, from frame f*, for frame f:
tf = k[t0 – x(V-v)/c2]
And, from frame f*, for frame F:
tF =k(t0 + xv/c2)
And k>= 1.
So, tf< t0< tF.
So, from frame f*, frame F’s time is increasing, however, frame f’s time is decreasing.
So, with respect to a specific time t0, f’s time is going to Past, where as F’s time is going to Future.
Let us study the following figure:
In this case, x is negative. So,
tF= k(t0 – xv/c2)
And tf = k[t0 + x(V-v)/c2].
So, here tF< t0< tf.
So, with respect to a specific time t0, F’s time is going to Past, where as f’s time is going to Future.
It is amazing to see in former case, f’s time was going to Past, and F’s to Future whereas in second case the vice versa happens. But it is mostly like to phenomena when we run towards the finish line in a race. As we move, the finish line comes closer to us. But as we touch and cross it, and move further ahead, the finish line goes further away. Similarly, as we live, we go closer to event of man landing on Mars. But as we will land, and live more, we will be going further away from the event.
Thus, we see how the time of one frame goes to Past with respect to an event, where as the time of other frame goes to Future with respect to the event.
Now let us recall the paradox. George is moving with v=0.99995c by a space craft while Fred remains on earth. As both of them see the other moving, so both may to conclude that, after they meet again, the other one will be younger.
Now, let us use our conclusion of the thought-experiment to this Paradox. We can assume George with the space-craft as one frame and Rest of Universe with Fred as the other frame. Now, we can simply imagine George’s frame in place of Maria’s.
George travels with velocity v=0.99995c, as seen by Fred, i.e. George is in motion with respect to Fred. So, Fred will conclude that George has gone to Past, as we conclude for Maria. Using the Time-Dilation equation, or by using the conclusion of the present study, he will conclude that George’s time is lagging with respect to his time. So, George will end up as younger.
George will see everything besides himself is in motion, so will conclude, “Everything besides himself is going back in time with respect to him. So, he himself is going forth of the Rest of Universe’s time. So, he is going to ‘Future’ of Fred. So, at end of voyage he will see ‘Future’ Fred – that is, older Fred.
References
- Beiser, A. (1997): Concepts of Modern Physics, New Delhi: Tata McGraw-Hill Publication.
- Mani, H. S. and Mehta, G. K. (2000): Introduction to Modern Physics, New Delhi: East West Press.
- Young, H. D. and Freedman, R. (2013): University Physics, 12th Edition, New Delhi: Pearson Publication..
