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 Special and General Relativity Special and General Relativity Physics Help Forum Dec 20th 2016, 10:31 AM #1 Junior Member   Join Date: Dec 2016 Posts: 2 Question about Time Im writing a philosophy paper and need a little help with some maths on Time. To use the example within the documentary The Illusion of Time (Example Link: https://www.youtube.com/watch?v=YRwZ55zjzxc - Example illustrated From: 24:15min  26:50min in the vid). The presenter explains how if we are standing at Point A (lets say us on earth), and an Alien is standing at Point B 10 billion light years away from us, that if the alien gets on a bike and starts riding away or towards us at a speed then the Aliens perception of NOW time would change possibly as drastic as being 200 years in the future or past, depending on the direction the Alien was to ride. My question is, while Im sure there isnt a simple equation is there possibly a simple equation that can help explain this effect? Possibly an equation that shows, if the distance between point A and point B is X distance, and the Alien at point B moves directly further away (being a positive speed), or closer towards (being a negative speed), over a total distance of Y then when the Alien stops moving how much would its now time have change (in terms of seconds) compared to the person who stayed still at Point A. I'm no physicist but I'd really like to find a simple way of explaining this phenomenon and showing it's working through numbers to use within my paper about Time and our perceptions of it. If anyone out there can help that would be amazing!!!! Many thanks, - Anthony   Dec 20th 2016, 10:46 AM #2 Senior Member   Join Date: Aug 2010 Posts: 434 I think you are referring to the "Lorentz transformation": If A is moving at speed v relative to B then A measures time elapsing at a rate equal to $\displaystyle \frac{1}{\sqrt{1- \frac{v^2}{c^2}}}$$\displaystyle \left(t- \frac{vx}{c^2}\right)$ where t is the rate at which B measures time elapsing. benit13 likes this.   Dec 20th 2016, 12:12 PM #3 Junior Member   Join Date: Dec 2016 Posts: 2 Thanks for that - I think we're on the right track ... with the Lorentz transformation, for my limited understanding in its basic formula it more so shows what B's perception of time would be when in a similar space or when visualizing a similar space as A, but while B is traveling at a different speed to A. Any ideas how i might account for a different distance between them? In the documentary they show how over a large distance a very small speed can change the perception of time quite greatly. I guess I'm still having trouble figuring out how to calculate that side of things. Thanks for the help so far    Dec 21st 2016, 09:04 AM #4 Senior Member   Join Date: Jun 2010 Location: NC Posts: 418 Non-relativistic Time Hi, There are some other aspects of time. 1.21 Events and Time | THERMO Spoken Here! Perhaps this will be of value. JP   Sep 1st 2018, 10:43 AM   #5
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 Originally Posted by Yappa888 [...] The presenter explains how if we are standing at Point A (lets say us on earth), and an Alien is standing at Point B 10 billion light years away from us, that if the alien gets on a bike and starts riding away or towards us at a speed then the Aliens perception of NOW time would change possibly as drastic as being 200 years in the future or past, depending on the direction the Alien was to ride. My question is, while Im sure there isnt a simple equation is there possibly a simple equation that can help explain this effect? [...]
There IS such an equation. The details are given here:

https://sites.google.com/site/cadoeq...eference-frame

Here is a very brief excerpt from that webpage:

__________________________

First, it is important to understand that, for any given instant t in the traveler's life, the home-twin and the traveler will generally disagree with one another about how old the home-twin is at that instant of the traveler's life. There are two quantities in the CADO equation which represent each of the twins' conclusions about the home-twin's age when the traveler's age is t. The quantity CADO_T denotes the traveler's conclusion about the home-twin's age, when the traveler's age is t, whereas the quantity CADO_H denotes the home-twin's conclusion about the home-twin's age, when the traveler's age is t.

The CADO equation can be written (most simply) as

CADO_T = CADO_H - v * L

where

v is their current relative speed, according to the home-twin, at the given instant t in
the traveler's life, with v taken as positive when the twins are moving apart,

L is the distance from the home-twin to the traveler, at the given instant t in the
traveler's life, according to the home twin,

and

the asterisk denotes multiplication.

[...]

The above equation gives the relationship between those four quantities (CADO_T, CADO_H, v, and L), at the given instant t of the traveler's life. I.e., although it is not shown explicitly, each of the four quantities in the equation are functions of t.

[...]

For example, suppose that immediately after the twins are born, the traveling twin moves away from the home-twin at a constant relative velocity of 0.866 lightyears/year for 20 years of his life. That complicated-looking value of the velocity was chosen for this example because it produces the very nice value of 2 for the gamma factor (the time-dilation factor):

gamma = 1 / sqrt( 1 - v * v) ,

where "sqrt( )" denotes the square-root operation, and where, again for simplicity, the factor c*c that should actually be dividing the v*v term has been omitted.

The traveler then instantaneously reverses course, and spends the next 20 years of his life returning to his home-twin. The magnitude of his velocity is still 0.866 ly/y, but since he is now moving toward his twin, by convention his velocity is now negative, -0.866 ly/y. Since gamma depends only on the magnitude of the velocity, gamma is still equal to 2.

So, the traveler is 20 years old at his turnaround, and 40 years old when he is reunited with his twin. Since the home-twin is perpetually inertial, she is entitled to use the time dilation result for his entire trip. Since gamma = 2 for the entire trip, she concludes that the traveler ages half as fast as she herself does, so she concludes that she is 40 years old when he turns around, and 80 years old when they are reunited. (Of course, when they are reunited, they will each know both of their ages). So, just from the time-dilation result, we've been able to quickly determine that

CADO_H(20) = 40 years old.

Now, from the definition of the CADO frame, the MSIRF(t) for all t from 0 years up to, but not including, 20 years, is the same inertial frame ... it's the one which is moving at a velocity relative to the home-twin of 0.866 ly/y, and in which the traveler is located at the spatial origin. During that entire segment, 0 <= t < 20, the traveler (by definition) agrees with that single MSIRF about the age of any distant inertial object or person, and thus he also agrees with that MSIRF about how fast or how slowly any distant person is ageing, compared to his own ageing. So, during that outbound leg (but not including the instant at t = 20), the traveler is entitled to use the time-dilation result, and he concludes that the home-twin is ageing half as fast as he himself is. So he concludes that, right at the end of his constant-velocity outbound leg (but before he does his instantaneous turnaround), that the home-twin is 10 years old. Therefore we've been able to determine that

CADO_T(immediately before turnaround) = 10 years old.

The fact that the traveler is entitled to use the time-dilation result, during his entire unaccelerated outbound segment, is also true of any unaccelerated segment, of finite duration, in his life. During any unaccelerated finite segment of his life, he is a full-fledged inertial observer during that entire segment, and he is entitled to use the Lorentz equations to determine simultaneity at a distance, and he is entitled to use the time-dilation and length-contraction results that follow from the Lorentz equations.

So, for the entire outbound leg, we didn't need to use the CADO equation at all ... the time-dilation result was all that we needed. But we do need the CADO equation in order to determine what happens during the turnaround, right at the instant t = 20 years. How do we do that?

To make use of the CADO equation during the turnaround, we need to know the values of the three quantities on the right-hand-side of the CADO equation (CADO_H, v, and L), immediately before and immediately after the instantaneous turnaround. CADO_H and L are quantities that are computed in the home-twin's inertial frame, and they are always continuous ... they never change discontinuously, even when v changes discontinuously. So CADO_H and L don't change during the turnaround, but v does change.

We can denote the instant in the traveler's life, immediately before the turnaround, as t = 20-, and the instant immediately after the turnaround as t = 20+. So, we have

v(20-) = 0.866 ly/y,

and

v(20+) = -0.866 ly/y.

We also already know that

CADO_H(20-) = CADO_H(20+) = CADO_H(20) = 40 years.

So all we still need to determine is L(20). How do we do that? We know that, in the home-twin's frame, the velocity of the traveler is 0.866 ly/y during the outbound frame, and we know that that outbound leg lasts for 40 years of the home-twin's life, so she will conclude that the traveler's distance from her at the turnaround is

L = 0.866 * 40 = 34.64 ly.

Since, in the CADO equation, all of the quantities need to be specified as functions of the variable t (the traveler's age), we therefore have

L(20-) = L(20+) = L(20) = 34.64 ly.

So, we've got all the quantities we need, to evaluate CADO_T(20-) and CADO_T(20+) using the CADO equation. We actually were already able to determine CADO_T(20-) using only the time-dilation result for the outbound leg ... we got the value 10 years. But it is instructive to use the CADO equation for the instants immediately before and immediately after the instantaneous turnaround, just to understand why the CADO frame concludes that the home-twin's age abruptly changes during the instantaneous turnaround. Immediately before the turnaround, we get

CADO_T(20-) = CADO_H(20-) - v(20-) * L(20-) = 40 - 0.866 * 34.64,

so

CADO_T(20-) = 40 - 30 = 10 years.

And, immediately after the turnaround, we get

CADO_T(20+) = CADO_H(20+) - v(20+) * L(20+) = 40 + 0.866 * 34.64,

so

CADO_T(20+) = 40 + 30 = 70 years.

So, the CADO equation says that, according to the traveler, the home-twin instantaneously get 60 years older during his instantaneous turnaround.
__________________
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