I know how to travel faster than light :\

[eNathan]

I travel at .9c for 1 second. From a stationary frame of reference, only 0.316 seconds has passed for me. So from there frame of reference, I traveled 269813212 meters in only 0.316 seconds or 2.8c or 280% the speed of light. I know that you cannot literally travel faster than light itself, because these same effects would apply to light itself or anything else traveling near c, I just wanted to point this out. I typed this quickly so if there are any mathematical errors in it please tell me and I will edit it.

 

[NewImage]

yeah i always wondered the same thing if you were traveling to a star 10 light years away, if you were almost going the speed of light only a few months or days might pass to you, but to the rest of us it took you 10 years to get there.

Guess you woudlnt need cryogenics like in the alian movies, you just need to go really realy fast :)

 

[JesseM]

yeah i always wondered the same thing if you were traveling to a star 10 light years away, if you were almost going the speed of light only a few months or days might pass to you, but to the rest of us it took you 10 years to get there.

That's true that the time to get there would be shorter from your point of view, but you have to remember that the distance from the Earth to the star would also be shrunk down to a lot less than 10 light years from your point of view, so in your frame the star doesn't approach you faster than light.

 

[kleinwolf]

Nice...but I think you should maybe use Lorentz transformations : let v=.9c, then we have to events in the moving frame (for simplicity, you stay at the origin of the frame) the second event is just 1 second (in that frame), after the first one :

A) (x',t')=(0,0)
B) (x',t')=(0,1),

This gives in the rest frame :

A) (x,t)=(0,0)
B) (x,t)=(gamma*v,gamma)

where gamma=1/Sqrt(1-v^2/c^2)

hence, in the rest frame, the measured speed is x/t=v=.9c ??

(I always get mixed when it concerns relativity, sorry)

 

[Doc Al]

I travel at .9c for 1 second.

I assume you mean that according to some observer's frame of reference, you traveled at 0.9c for 1 second. In other words, they measure you as moving from point A to point B at .9c for 1 second of their time, thus traveling a distance of .9 light-seconds according to them.

From a stationary frame of reference, only 0.316 seconds has passed for me.

I assume you mean from a frame in which you are stationary (your rest frame). According to your clock only .44 seconds have passed. But according to your measurements, the distance from A to B is only about 0.39 light-seconds.

So from there frame of reference, I traveled 269813212 meters in only 0.316 seconds or 2.8c or 280% the speed of light.

Huh? Where do you get this? In the other frame, you are measured to travel at 0.9c. In your frame, your speed is zero of course; but you measure the speed at which point B approaches you to be 0.9c.

 

[Doc Al]

So from there frame of reference, I traveled 269813212 meters in only 0.316 seconds or 2.8c or 280% the speed of light.

I finally see what you are saying (after JesseM pointed it out to me in another thread). Say another planet is X light years away from Earth (as measured in our Earth frame). You jump into a rocket traveling at .9c relative to earth. Yes, you can reach that planet in only .44 X years (according to your clock). So, yes, effectively you can travel greater than the speed of light (in a sense). But as others have pointed out, your actual speed is still just .9c relative to earth. But your point is valid: If you can somehow get moving fast enough, you can traverse any distance (measured from the original "stationary" frame) in a reasonable amount of (your) time.

(Sorry if I was being overly pedantic.)

 

[Enos]

Receding machines maybe?

shuttle}==[]=={}==[]=={}==[]=={}==[]=={}==[]=={}==[]=={earth

Each machine is }==[]=={ which } and { are pushing against each other causing the receding.

When the signal to each machine is sent it is sent when the machines are close to each other and each machine recedes to the set amount of time to fit the set amount of total distance. Each machine recedes slower then c but when pushing away from each other after the signal was sent the total receding with respect to the Earth is faster than c. But the receding from each machine to the next is less than the speed of c. But of course the resources of such a machine is the complicated part.