Transforming Y-axis table values into a separate formula?

  • #1
29
2

Homework Statement:

Hoping to get some help with the details of Special Relativity in Accelerating frames of reference.

Relevant Equations:

Integrals, Graphing, Velocity. Time dilation factor.
Hey everyone, I have generated a nice little velocity vs time graph that I would love if somebody could help me put to use.
I have marked data points on the x-axis for the Y-value for every second on the function.

Just to be clear: X-axis = time in seconds & Y-axis = velocity in meters/second. We have a constant acceleration.
phys.png


I have also made an integral of the function, calculating the total area under the curve to be exactly 4.999993*10^9

However, since the 100 data points on the table would be quite tedious to manually apply the time dilation to √1-v^2/c^2 for every seconds at the different velocities. I have wondered whether there is a way to insert the values in my table directly into a calculus such that I retrieve the accumulated amount of time dilation over the 100 second journey depicted on the graph. I think there should be a way integrate these values with the Time dilation factor, even if we have to assume that we travel in constant motion, that increase every second.
Thanks in advance:)
 

Answers and Replies

  • #2
anuttarasammyak
Gold Member
333
158
You draw the graph
[tex]v=at[/tex]
where a = 10^6 km/s^2, right?
and you get the value of
[tex]x=1/2 \ at^2[/tex].
These are Newtonian mechanics.

Then I do not understand your scenario to relate this with [tex]\sqrt{1-v^2/c^2}[/tex] of relativity? What situation you are thinking ?
 
Last edited:
  • #3
29
2
You draw the graph
[tex]v=at[/tex]
where a = 10^6 km/s^2, right?
and you get the value of
[tex]x=1/2 \ at^2[/tex].
These are Newtonian mechanics.

Then I do not understand your scenario to relate this with [tex]\sqrt{1-v^2/c^2}[/tex] of relativity? What situation you are thinking ?
I want to transform every velocity in the table, into the formula: [tex]\sqrt{1-v^2/c^2}[/tex]for 1 second each, then sum them all up to get a good approximation of the total amount of time dilation during this 100 second acceleration. You cannot use Newtonian mechanics, since the Lorentz transformation gets exponentially larger at higher velocities according to special relativity.

There must be somehow, in excel, word or otherwise. To take all these y(velocity) values and plug them all into the formula simultaneously. Or a calculus trick to sum the total amount of time dilation occurring over all these t-intervals.
 
  • #4
29
2
So for example, at second 1, my velocity is 1 000 000 m/s, at second 2 my velocity is 2 000 000 etc.

For every single time interval I want to plug in my y-values into the formula: [tex]\sqrt{1-v^2/c^2}[/tex] where we insert our 100 different y-values as the v^2 component.
 
  • #5
anuttarasammyak
Gold Member
333
158
So correspondence
time:speed:time dilation rate is
t : at : ##\sqrt{1-\frac{a^2t^2}{c^2}}##

The y-t graph y= ##\sqrt{1-\frac{a^2t^2}{c^2}}## would be what you want. Area under the graph

[tex]\int_0^T \sqrt{1-\frac{a^2t^2}{c^2}} dt < T [/tex]

is the time reading of the moving clock when time T in the Earth frame of reference.
 
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  • #6
29
2
So correspondence
time:speed:time dilation rate is
t : at : ##\sqrt{1-\frac{a^2t^2}{c^2}}##

The graph y= ##\sqrt{1-\frac{a^2t^2}{c^2}}## would be what you want. Area under the graph

[tex]\int_0^T \sqrt{1-\frac{a^2t^2}{c^2}} dt[/tex]

is the time reading of the moving clock when time T in the Earth frame of reference.
The graph shows t in stationary observer time, so that when we plug in the velocities into the formula we should get a lower value t' for the spaceship.
 
  • #7
29
2
So correspondence
time:speed:time dilation rate is
t : at : ##\sqrt{1-\frac{a^2t^2}{c^2}}##

The y-t graph y= ##\sqrt{1-\frac{a^2t^2}{c^2}}## would be what you want. Area under the graph

[tex]\int_0^T \sqrt{1-\frac{a^2t^2}{c^2}} dt < T [/tex]

is the time reading of the moving clock when time T in the Earth frame of reference.
Wow, seems like it worked perfectly. Thanks!! I'm getting two different values, depending on whether I square a & x separately. Not sure which one of these is correct and why - however it would be easy to estimate which approach is correct. That's awesome, been spending a bunch of hours on this and you were the saving grace.
calc1.png

calc2.png
 
  • #8
anuttarasammyak
Gold Member
333
158
Take a look at dimension of the form
[tex]\sqrt{1-X^2}[/tex]
1 is dimensionless. As well X = Z/c should be dimensionless.
 
  • #9
29
2
Take a look at dimension of the form
[tex]\sqrt{1-X^2}[/tex]
1 is dimensionless. As well X = Z/c should be dimensionless.
Just found something amazing with this approach, assuming a constant increase in velocity like depicted above, which goes directly to C over 100 seconds. (impossible scenario admittedly) This equation dictates that the minimum t value is equal to exactly t````````'= ((π/4)*100) seconds
 
  • #10
29
2
Take a look at dimension of the form
[tex]\sqrt{1-X^2}[/tex]
1 is dimensionless. As well X = Z/c should be dimensionless.
Am I correct in assuming that acceleration and kinetic energy values under the circumstances given in the graph above would transform to the inverse gamma(γ) formula ?
 
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