http://www.ck12.org Chapter 22. The Special Theory of Relativity
Solution:
Let us first convert into 100, 000 kmh intokms.
A speed of 100, 000 kmh = 105 kmh =
(
105 kmh)( 1 h
3 , 600 s)
= 27. 8 kms.The speed of light isc= 300 , 000 kms. Therefore, the speed of the space ship in terms ofcis
27. 8
300 , 000 c=^9.^26 ×^10
− (^5) c.
Also, 1. 00 h= 3 , 600 s.
Thus,
t=√tp
1 −vc^22
=√^3 ,^600 s
1 −(^9.^26 ×^10 c 2 −^5 )^2 c^2
=√^3 ,^600 s
1 −(^9.^26 ×^101 −^5 )^2
= 3 , 600. 000015 s.
This represents a difference in time between inertial frames of 1. 5 × 10 −^5 s, which from an everyday point of view
is negligible. The fact thattdepends upon the ratiov
2
c^2 shows that whenv<<c,
√
1 −v
2
c^2
∼=1. Clearly, the commonspeeds that we experience are smaller than 100, 000 kmh, so there can be no question thattandtpcan be considered
essentially equal for day-to-day occurrences.
To date, the speed of the ship given in part B is about 40% greater than that achieved by any space ship or space
probe.
You may wonder if the effects of time dilation only apply to clocks. The answer is no! Clocks measure time
differently because time itself is different in moving frames of reference. Everything happens slower in such
circumstances. Biological systems, for instance, are affected. The process of aging, the beating of one’s heart,
the pace of breathing, are all subject to the effects of time dilation. The heart rate of a person traveling in a space
ship compared to what is measured on earth is slower. Recognition of this fact suggested what called today as the
twin paradox.
The twin paradox
Let’s say one of two twins decides to take a relativistic trip (a trip where the effects of time dilation cannot be
ignored). If the traveling twin leaves when she is 20 years old, traveling at about 0. 87 c, she will age half as slowly
as her twin who remains on earth. If, after 10 years, as measured by a clock aboard her ship, she returns, she’ll find
that 20 years will have passed on earth. When the two twins meet again, the traveling twin will be 30 years old and
the stay-at-home twin will be 40 years old.
Does this seem reasonable? The objection is sometimes raised that since all inertial frames are equivalent, the
stay-at-home twin could make the same argument and thus she should be younger than the traveling twin.
There are several arguments showing that special theory of relativity does not create a paradox and, indeed, it is the
twin on the ship who ends up younger. It’s important to recall that the special relativity equations above are valid
only for the inertial frames of reference, or frames that do not accelerate. In order for the traveling twin to return to
the Earth, she must turn around. Any change in direction indicates a change in velocity and, therefore, acceleration.
Even though the situation has an interval of acceleration, one can still use special relativity to correctly determine the
result. The inertial frame returning may not be the same inertial frame when leaving, but they are both still inertial
frames. And if the traveling twin never returns, thus introducing no acceleration, the paradox becomes moot, since
there is then no way for the twins to compare ages.
Of course, nothing prevents one from arguing that it was the Earth that had undergone the acceleration, and the space
ship that had remained motionless. But how would the Earth do this? The only way the Earth could move back and
forth is to suppose the entire universe must be moving back and forth. And even with this assumption it can be
shown using Einstein’s general theory of relativity, which deals with accelerating reference frames, that the results
of special relativity are still correct. The situation between the Earth and the rocket ship is not symmetrical.