PhanthomJay said:
Thanks for responses but I don't get it, because I guess I asked the wrong question.
There are no "wrong" questions ...
PhanthomJay said:
Also if c = 1, then t must be in length units?
Yes, and vice versa... lenght in "time" units
to Alpha Centauri = 4 light years
a foot = one light nano second
to the Moon = approximately one light second
to the Sun = 8 light minutes
to the planet Neptune = 4.2 light hours
PhanthomJay said:
I need some numbers.. What happens when ds is imaginary?.
The distance between Earth "now" and Alpha Centauri "now" ##\Delta t=0## is 4 light years.
If you leave a washing machine :-) in the same place for 4 years ##\Delta x=0## , it will pass through space-time the same interval of 4 years.
However, if the light starts from Earth now and reaches Alpha Centauri in 4 years, it will cross the zero (null) distance. Means ##0^2=4^2-4^2##
If an event on Alpha Centauri happens in, for example, 6 years, then it is theoretically possible to reach it. (We say that it is inside the
light cone)
If some event on Alpha Centauri happens in, for example, 2 years, then it is impossible to reach it.
There are two conventions:
either ( the same thing has several names space-time interval ##\Delta s##, proper length ##\Delta L##, proper distance ##\Delta \sigma##)
##\Delta s^2=\Delta x^2 - \Delta t^2##
or ( the same thing has two names space-time interval ##\Delta s##, proper time ##\Delta \tau##)
##\Delta s^2=\Delta t^2 - \Delta x^2##
It's a matter of personal choice. (Only the plus and minus signs will swap places in formulas and results)
If you choose the first one, then you will get negative (imaginary) numbers for events inside the light cone and positive (real) numbers outside.
If you choose second convention, then the spatial distances will be negative(imaginary) and the events inside the light cone, in the future or the past, will be positive (real) numbers.
PhanthomJay said:
In the Minkowski space time equation in one dimensional space , ds^2 = dx^2 - (ct)^2, what is the value to use for x and t, and what does the space time interval ds represent?
Imagine that you want to make it to an event that takes place on Alpha Centauri in 5 years.
(important question)
At what time should you set your clock to wake you up from hibernation?
You enter your spaceship :-) ... and set the speed to 4/5 = 0.8 c and the ship's computer prints the following data:
destination: Alpha Centauri
speed: 0.8 c
gamma contraction: 0.6
distance: 2.4 light years
travel time: 3 years
How?
Lorentz factor
## \gamma=\sqrt{1-0.8^2}=\sqrt{1-0.64}=\sqrt{0.36}=0.6##
Length contraction ...
##L=0.6 \cdot 4 ly=2.4 ly ##
Time dilation
##t'=0.6 \cdot 5 yrs =3 yrs##
If we multiply the right side of the equation ##\Delta s^2=\Delta x^2 - \Delta t^2 ## by ##1=\frac{\Delta t^2}{\Delta t^2}## we get:
##\Delta s^2=\Delta x^2\frac{\Delta t^2}{\Delta t^2} - \Delta t^2\frac{\Delta t^2}{\Delta t^2 } ##
##\Delta s^2=\left(\frac{\Delta x^2}{\Delta t^2} - \frac{\Delta t^2}{\Delta t^2} \right) \Delta t^ 2##
##\Delta s^2=\left(v^2 - 1 \right) \Delta t^2##
##\Delta s=\sqrt{\left(v^2 - 1 \right) \Delta t^2}##
##\Delta s=\gamma i \cdot \Delta t##
The space-time interval ##\Delta s## is in this case the same as the travel time ##\Delta t'=3 yrs##
##\Delta s## is the time at what traveler should set his clock to wake him up from hibernation?
PhanthomJay said:
For example, if Alpha Centauri is 4 light years away, what values are. used for x and t, based on speed I guess, and what is the meaning of ds?
For each event, the spatial position (for example, Alpha Centauri = 4 ly) and time (for example, 10 years ago, now, in 5 years, ...) should be determined.