Speed of the light and dilation of time

[ghwellsjr]

I want to find out if the approach I used is valid or not and if it is equivalent to the dot product method. I am now thinking it is equivalent but want to make sure.d
It appears to me that in the case at 90 deg. that I used,, the cos (45deg)=.707 times the direct distance (70.7 )does equal the radius. SO in that particular instance there is agreement.
But I am not sure if this is true all around the circumference
This a drawing of a further point: is my approach of orthogonal projection equivalent to the dot product?

Yes, I believe so.

Looking at your graph it looks like my result is very close to yours for 90 deg. estimating values for a quarter of the circuit from your chart.
How do you get attached images to show up full size??

I use the "Insert Image" button in the Advance Editing mode (after uploading the image with the "Manage Attachments" button down below).

 

[Austin0]

Here's what I get for that problem (r = 1, gamma = 2.0, CADO_H(0) = 0):

For any given angular position 0 <= theta <= 2pi of the traveler, the traveler's age is t = r * theta / v.

CADO_H(t) is of course equal to gamma * t.

CADO_T(t) = gamma * t - r * v * sin(v * t / r).

could you explain what you are doing here?


d{CADO_T(t)}/dt = gamma - v * v * cos(v * t / r).

The above derivative gives the slope of the CADO_T vs t graph (the "age-correspondence" graph). It is the time dilation of the home twin, according to the traveler. Since it is greater than one, the traveler says that the home twin is aging more quickly than he himself is, so it would more reasonably be called "time contraction", rather than "time dilation", in this case.

Here are a few values from the above equations:

theta t CADO_H CADO_T slope

0 0 0 0 1.25
pi/2 1.81 3.63 2.76 2.0
pi 3.63 7.26 7.26 2.75
3pi/2 5.44 10.88 11.75 2.0
2pi 7.26 14.51 14.51 1.25

If the home twin were at the center of the circle, the traveler would always conclude that the home twin was aging twice as fast as he himself was aging ... i.e., he would always exactly agree with the home twin about their respective ages.

But if the home twin is located on the circle (as in the above scenario), the traveler would conclude that near theta = 0, the home twin is aging faster than he himself is aging, but less than twice as fast. The home twin would be aging exactly twice as fast at theta = pi/2 and at theta = 3pi/2. And the home twin would be aging more than twice as fast at theta = pi.

The fact that the home twin is aging less than twice as fast as the traveler (according to the traveler) at theta = 0 was a surprise for me ... I had expected that the home twin's "time contraction" would be very nearly equal to gamma whenever the twins were very close together, but that's not what happens.

Thanks

 

[GrammawSally]

[...]

In arriving at the final results

CADO_T(t) = gamma * t - r * v * sin(v * t / r)

d{CADO_T(t)}/dt = gamma - v * v * cos(v * t / r),

you need some intermediate results:

The argument of the sin and cosine functions above is just the angular position theta, in radians, of the traveler relative to the home twin.

The triangle formed by the center of the circle and the positions of the home twin and the traveler has the angle theta between the two position vectors, and the other two angles are equal.

The length of the vector L is

|L| = 2 * r * cos{ (pi - theta) / 2}.

The angle alpha between L and v is theta / 2.

To get the final result for CADO_T(t), you need the two trig identities

cos(pi/2 - alpha) = sin(alpha)

sin(alpha) * cos(alpha) = (1/2) * sin (2 * alpha).

 

[Simplyh]

Unfortunately at the present moment of my work I don't have time to come here except on Sundays. This pace, when everybody writes daily, turns my participation impossible. I'll come some other time.
Meanwhile ...
Divirtam-se (have fun)!

 

[GrammawSally]

[...]
The fact that the home twin is aging less than twice as fast as the traveler (according to the traveler) at theta = 0 was a surprise for me ... I had expected that the home twin's "time contraction" would be very nearly equal to gamma whenever the twins were very close together, but that's not what happens.

I've realized that my above expectation (that the traveler would conclude that the home twin is aging twice as fast as she (the traveler) is, whenever they are close together (theta near zero)) wasn't reasonable. A reasonable (although incorrect, as it turns out) expectation would be that the traveler would conclude that the home twin is aging half as fast as she (the traveler) is, whenever they are close together.

When they are very close together, the traveler would be very nearly either directly approaching or directly receding away from the home twin, just like in the one-dimensional case where both the traveler and the home twin are inertial. So one might reasonably expect that the standard reciprocal time dilation result should apply. In other words, one might reasonably think that d{CADO_T(t)}/dt should be very close to 1/gamma when theta is very close to zero. But the previously quoted result, for the derivative of CADO_T(t) at theta = 0, was gamma - v*v = 1.25, not 0.5, so the revised expectation is incorrect.

The expectation is wrong because the traveler is undergoing a centripetal acceleration. The traveler's acceleration changes the time dilation of the one-dimensional unaccelerated case into a "time contraction" for the centripetal acceleration case, according to the traveler. On the other hand, according to the home twin, the traveler is aging half as fast as he (the home twin) is, regardless of whether the traveler is accelerating or not. The home twin doesn't ever conclude that the traveler's age is "time contracted".

The fact that acceleration has an effect on the traveler's conclusions about the relative rate of aging of the home twin, occurs for all situations where the traveler accelerates. For example, it also occurs when the home twin is located at the center of the traveler's circle (the "time contraction" is constant, and equal to gamma, in that case). And it also occurs in the one-dimensional twin paradox scenarios (during the turnaround).

 

[GrammawSally]

Here's a pic of the age-correspondence graph (CADO_T vs t) for the case where the inertial person is located on the circle:

http://img221.imageshack.us/img221/5648/xeq1.jpg

CADO_T is the age (in years) of the inertial person, according to the accelerating person, when the accelerating person is age t (in years). The radius r of the circle is 1 ly.

The dashed line shows CADO_H vs t, where CADO_H is the age of the inertial person, according to the inertial person, when the accelerating person is age t. And for the alternative scenario where the inertial person is located at the center of the circle, the CADO_T vs t curve is also the dashed line (because the two people always agree about their corresponding ages in that scenario).

The endpoints of the curve corresponds to the instants where the accelerating person is whizzing by the inertial person. The midpoint of the curve corresponds to the instant where the accelerating person is at the opposite side of the circle from the inertial person.

 

[GrammawSally]

I finally did the analysis to determine what the age-correspondence graph looks like when the inertial person is located anywhere outside the circle. As before, the analysis is purely a matter of geometry and trig (since the CADO equation has already taken care of the relativity part), but the geometry and trig are a bit more complicated than for the case where the inertial person is located on the circle (and MUCH more complicated than for the trivial case where the inertial person is located at the center of the circle).

Here's the graph for the case where the inertial person is located 20 ly away from the center of the circle, and when the radius of the circle is 1 ly. As before, the speed is 0.866c, giving a gamma factor of 2.0.

http://img846.imageshack.us/img846/259/xeq20.jpg

I started the plot at one of the instants where the traveler is whizzing past the inertial person, and ended the plot one full cycle later. At the midpoint of the plot, the traveler is on the opposite side of the circle from the inertial person.

Comparing the previous age-correspondence graph with this new one shows that the amplitude of the oscillations of the CADO_T curve about the straight CADO_H line gets larger as the distance of the inertial person from the center of the circle is increased (at least for distances on or outside the circle ... I haven't yet done the analysis for non-zero distances inside the circle).

This case is similar to the example given by Brian Greene in the NOVA series on "The Fabric of the Cosmos", in which someone located in an extremely far-away galaxy is riding a bicycle around in a small circle. The rider concludes that the current date and time on our Earth is fluctuating over centuries each time he completes a circle.

Brian's choice of the magnitudes of x (the distance of the inertial person from the center of the circle) and v (the speed of the traveler around the circle) is different from my choices in the above graph: he picks an extremely large x and an extremely small v, whereas I picked a much smaller x and a much larger v. But in both cases, the quantity (L dot v) is large enough compared to CADO_H to cause large fluctuations of CADO_T for each trip around the circle.

 

[GrammawSally]

Here's the last piece of the puzzle (of the circular motion problem), for the case where the inertial person is located somewhere inside the circle, but not at the center. The particular age-correspondence graph below is for x = 0.5 ly. And as before, the speed is 0.866c, giving a gamma factor of 2.0.

http://img833.imageshack.us/img833/7961/xeq0point5.jpg

The results were what I had suspected I'd see: the farther the inertial person is from the center of the circle (whether inside or outside the circle), the larger the amplitude of the oscillation about the CADO_H line.

 

[GrammawSally]

Here is a brief description of the analytical results for the case where the inertial person is outside the circle:

All of the following angles and lengths are as measured in the inertial person's reference frame.

Theta is the angle which specifies the position of the traveler on the circle, when the traveler's age is t years old (and I have arbitrarily set theta = pi/2 when t = 0). Since the traveler's speed |v| is constant, the angle alpha increases as a linear function of t. The line from the center of the circle to the inertial person is arbitrarily taken to be at pi/2 (90 degrees) from the theta = 0 line (so the traveler and the inertial person are momentarily co-located when theta = pi/2).

(Beware that the above choice of position for the inertial person is different from the choice I used in the "inertial person located on the circle" case, where that position was taken to be at theta = 0. So the two definitions of theta aren't exactly the same: theta_new = theta_old + pi/2.)

The quantity x is the distance of the inertial person from the center of the circle, in ly. The quantity r is the length of the radius of the circle, in ly.

The angle beta gives the direction of the L vector (the vector from the inertial person to the traveler), measured from the theta = 0 line. beta can be determined from the equation

beta = arctan{ -( x/r - sin(theta) ) / cos(theta) } .

The ambiguity of the arctan function is resolved by requiring that beta lie in the range (-pi, 0), or equivalently, in the range (pi, 2pi).

The angle alpha is the angle between the vector L and the vector v. We need that angle in order to determine (L dot v). Alpha can be calculated from

alpha = theta + pi/2 - beta .

To determine the quantity (L dot v), which we need in the CADO equation, we also need the length of the vector L, denoted |L|. |L| can be calculated from the equation

|L| = sqrt{ ( x - r * sin(theta) )^2 + ( r * cos(theta) )^2 } .

The age-correspondence graph that I gave, produced from the above equations, is for the case where x = 20 ly, r = 1 ly, and |v| = 0.866c (giving gamma = 2.0). I arbitrarily chose the ages of both of the two people to be zero at theta = pi/2, and I started drawing the graph at theta = pi/2 (when the two people are momentarily co-located), and stopped drawing at theta = 2pi + pi/2 (which shows one complete cycle of the circular motion).

 

[GrammawSally]

Here is a brief description of the analytical results for the case where the inertial person is inside the circle, but not at the center:

The following definitions are unchanged from the previous case (where the inertial person was located outside the circle):

All of the following angles and lengths are as measured in the inertial person's reference frame.

Theta is the angle which specifies the position of the traveler on the circle, when the traveler's age is t years old (and I have arbitrarily set theta = pi/2 when t = 0). Since the traveler's speed |v| is constant, the angle alpha increases as a linear function of t. The line from the center of the circle to the inertial person is arbitrarily taken to be at pi/2 (90 degrees) from the theta = 0 line (so the traveler and the inertial person are momentarily co-located when theta = pi/2).

(Beware that the above choice of position for the inertial person is different from the choice I used in the "inertial person located on the circle" case, where that position was taken to be at theta = 0. So the two definitions of theta aren't exactly the same: theta_new = theta_old + pi/2.)

The quantity x is the distance of the inertial person from the center of the circle, in ly. The quantity r is the length of the radius of the circle, in ly.

The angle beta gives the direction of the L vector (the vector from the inertial person to the traveler), measured from the theta = 0 line.

__________________________________________

Except as noted, the results below are not the same as in the previous (the outside-the-circle) case:

The angle beta can be determined from the equation

beta = arcsin{ -( r/L) * sin(theta) - (x/L) } .

The ambiguity of the arcsin function is more complicated to resolve than the arctan ambiguity in the previous case. First of all, the two possible angles that have the required sine don't differ by pi degrees as they did for the arctan: they are now reflections of each other about the theta = pi/2 line. So either of the two alternatives for beta can be obtained from the other via the equation

beta2 = pi - beta1 .

Before, the ambiguity was easily resolved by noting that the vector L always pointed in the general direction of the circle, never away from it. In the current case, the vector L has no such simple restriction. There IS a resolution, but it turns out to be more complicated to implement. The resolution of the ambiguity is fairly simple to state geometrically: pick the angle beta so that the difference in the directions of the vectors L and r is minimized. (This can be directly seen by sketching two extreme values of theta in each of the four quadrants). But implementing that rule analytically was harder than I expected it to be. The cause of the difficulty has to do with the discontinuities that angles inherently have either at their zero positions (when expressed as positive angles between zero and 2pi), or at their pi positions when they are expressed in principle-value form (positive between 0 and pi, and negative between -pi and zero). After many incorrect attempts, I finally got it right by first expressing beta and theta in principle-value form (PV-form), and then calculating the absolute value of (beta - theta) for each of the two alternatives of beta. Then I expressed those two results in PV-form, and finally I took the absolute value of those two results. The correct choice for beta is then the beta alternative which minimizes that final quantity.

(The above process seems harder than it should be, but I haven't been able to see an easier way to do it. If anyone sees a simpler way, please PM me.)

The angle alpha is the angle between the vector L and the vector v. We need that angle in order to determine (L dot v). Alpha can be calculated from

alpha = theta + pi/2 - beta .

(The above equation for alpha is the same as before).

To determine the quantity (L dot v), which we need in the CADO equation, we also need the length of the vector L, denoted |L|. |L| can be calculated from the equation

|L| = sqrt{ r^2 + x^2 - 2 * r * x * sin(theta) } .

The age-correspondence graph that I gave, produced from the above equations, is for the case where x = 0.5 ly, r = 1 ly, and |v| = 0.866c (giving gamma = 2.0). I arbitrarily chose the ages of both of the two people to be zero at theta = pi/2, and I started drawing the graph at theta = pi/2 (when the two people are momentarily co-located), and stopped drawing at theta = 2pi + pi/2 (which shows one complete cycle of the circular motion).