Re: Time Dilation (Basic silly bull)

In sci.physics.relativity, eratosthenes
<rehamkcirtap@xxxxxxxxx>
wrote
on Fri, 11 Jul 2008 14:43:21 -0700 (PDT)
<77c6cda3-c0df-4953-b9e8-25ecbbe3cb5a@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
My friend, who has no training beyond high school physics, told me
that he heard that heading out from the Earth for one hour and then
back at light speed would mean that fourteen years would pass on
earth.

Your friend is committing multiple logic errors.

[1] First, it is impossible to travel at lightspeed.
There is just not enough energy in rockets to accelerate
to lightspeed. If one takes v_e = 3000 m/s (which is
about what one would get from the Saturn V), and
c = 3 * 10^8 m/s, then

c = v_e * log(M_i/M_f)

M_i/M_f = exp(c/v_e) = exp(100000) = 2.806 * 10^43429

This is way more than the mass of the Universe, assuming
M_f is on the order of 100 kg. Even accelerating a proton
(1.67 * 10^-27 kg) in a tiny little rocket would have some
problems -- assuming such a rocket is possible anyway. ;-)

[2] Even with sufficient energy, we have yet to observe
any particle going faster than light. The Bevatron in
particular was operating in the early 1950's, and was so
named because it could accelerate electrons to 1 BeV of
energy. If one calculates m_e * c^2, one gets 511 kEv;
accelerating an electron to half this value
(KE = 1/2 m v^2 in Newtonian physics) should have
been enough.

[3] If one heads out near lightspeed for 1 light-hour --
about 7.2 AU, or about 65-75% of the way to Saturn --
and back, one might lose 2 hours at the most, if one
doesn't get squished into a pancake on the way. Newtonian
calculations suggest that
a = v/t = (3 * 10^8) / (3600) = 83000 m/s/s, or about
8300 'g's at a minimum -- hardly survivable for a mere human.

Now I am not an expert on the subject but I am not untrained. I'm one
semester from my B.S. (lol) in physics with a minor in math and am
currently applying to grad schools. I thought that it sounded a
little suspect.

So I investigated:

t = (t')/(1 - v^2/c^2)^(1/2) where t is the time as recorded at the
launch point, v is the ship velocity, c is obviously
light . speed, and t' is
the time recorded in the ship.

if v approaches c then v^2/c^2 approaches 1 and the earth time goes to
infinity. I don't like that either.

I used Newton's generalized binomial theorem and Mathematica and I get
either a directed infinity or the square root of two.

I believe I may be approaching it wrong, highly likely, or it may just
be that I don't have a great book on relativity and am being forced to
depend on the internet.

Any thoughts?

IMO, the proper way to approach this, assuming one doesn't
want to deal with the acceleration problem, might simply
be to ask about the clock ticks, if one sets up the Twin
Paradox problem so that they can observe each other via
ultrapowerful telescopes or sophisticated radio units.
It turns out:

stationary twin's age = 2L/v
accelerating twin's age = 2L/(vg)

where g = 1/sqrt(1-v^2/c^2). (The actual letter is the Greek
letter gamma, but ASCII being what it is, I use g.)

To derive this result, one first works out how light is affected.
If twin A sends out a light pulse t_A seconds outbound, then twin
S will see the result according to the Lorentz, which can
be expressed as:

(x_S,t_S)_S = (g*(x_A - v*t_A), g*(t_A - v*x_A/c^2))_S

(the ()_S indicates that I'm working in the stationary
twin's time/space coordinate system, and is my notation).

Since x_A is 0 (the accelerating twin's laser is mounted to his
spacecraft), we can simplify to:

(x_S,t_S)_S = (g*(- v*t_A), g*t_A)_S
= (-gvt_A, gt_A)_S

By necessity, x_S = 0, so this doesn't quite work, but we
know that light propagates at lightspeed *regardless of
coordinate space* (as long as the coordinates are consistently
measured). This means that

t_S = gvt_A/c + gt_A = (v/c+1)*g

is an acceptable solution (the light beam has to travel
a little farther, basically), and that eventually yields

t_S/t_A = sqrt(1+v/c)/sqrt(1-v/c).

Similarly, when coming back t_S/t_R = sqrt(1-v/c)/sqrt(1+v/c),
where I use 'R' for the return trip (the accelerating twin is
changing frames), though it turns out L_R = L_A anyway; space
is weird, but not *that* weird. :-)

If one wants to be quick and dirty about it, one can observe that
(1+v/c) and (1 - v/c) are the classical Dopplers.

And now we count ticks. The accelerating twin will
generate the same number of ticks outgoing and returning
(the flight is assumed symmetric, though one can work out
aging in an asymmetric flight with obvious modifications
to this logic). We assume L_A/v ticks are sent
(1 per subjective second); we cannot assume lengths are
unaffected so we leave L_A undefined for now, and in any
event the Lorentz Transformation suggests that the two
twins won't see eye to eye regarding the distance actually
traveled. [*]

It will take L_A/v*sqrt(1+v/c)/sqrt(1-v/c) for
the stationary twin S to observe these ticks; he
then sees the ship turn around, and it will take
L_A/v*sqrt(1-v/c)/sqrt(1+v/c) more time to see the ship
return back to Earth, and then he greets his now younger,
flatter [+] brother.

These two expressions must be equal, so:

2L_S/v = L_A/v*sqrt(1+v/c)/sqrt(1-v/c) + L_A/v*sqrt(1-v/c)/sqrt(1+v/c)
2L_S = L_A*((1-v/c) + (1+v/c))/((sqrt(1-v/c)*sqrt(1+v/c))
= L_A*(2)/sqrt(1-(v^2/c^2))
= 2*L_A*g

or L_A = L_S/g.

Since t_A = L_A/v and t_S = L_S/v,

t_A = t_S/g

and the accelerating twin is younger (since g > 1).

[*] They *will*, however, agree on the speed, though in opposite
directions.

[+] No, this isn't a relativistic effect, despite the
L_A/L_S ratio; the usual formulations of the Twin
Paradox squish the poor traveling twin to a meaty
pulp. A more realistic formulation would have the
twin traveling to Sirius, 8.6 lightyears distant,
with a v of about 0.1 c accelerating over the period
of about a month. Even to accomplish this, one has
to assume something along the lines of a futuristic
boron-hydrogen fusion rocket with rather large tanks --
and a very long-lived species such as terrapins.

--
#191, ewill3@xxxxxxxxxxxxx
Windows Vista. Now in nine exciting editions. Try them all!
** Posted from http://www.teranews.com **
.

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