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Old Jan 25th 2013, 05:19 AM   #1
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Lightbulb How fast can we arrive at a certain place, in theory?

Hi there,

In theory, we cannot travel faster than light.
Therefore, is the answer to this question "The speed of light" ?
Definitely NO.

According to Enstain's Special Theory of Relativity,
if someone travels in a inertial reference frame which moves at a speed that near the speed of light, then the time in the frame flows slower.
As the velocity approaches the speed of light, the ratio of time between the outside world and the inside world goes to infinity.
That is to say, if we travel in a speed that very close to light, then when we arrive at the destination, it may be hundreds of thousands of years later. It is not FAST at all.

So we should re-define "fast": Make the ETA(estimated time of arrival) as short as possible.
In this sense, we can formulate the question:

Assume a reference frame moves at constant speed V, and the speed of light is V_light. The distance is D.

T_i is the time that people in the inside world feel, which is
D/V .
T_o is the time that people in the outside world feel, which should be (D/V)*(V_light/sqrt(V_light^2 - V^2)) by applying Lorentz transformation.
Then, Find the MIN(T_o), which is a function of V.
By simple calculas, V should be V_light/sqrt(2).

What do you think about the deduction above?
Any reply would be appreciate.
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Old Jan 25th 2013, 10:04 AM   #2
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You need to be careful with your definitions. What is D? It's not the same value for people "outside" and those "inside" the moving reference frame.

Let D_o = distance between two end points which are at rest for the "outside" obeserver. Then the time the outside observer measures for the trip is T_o = (D_o)/v. Thisis minimimzed by makeing v as large as possible.

For an observer "inside," the distance D is a shorter length because the end point is rushing toward him at speed v; for him the distance of the trip is: D_i = D_o x sqrt(c^2-v^2)/c. So the time the insider measures to complete the trip is T_i = (D/v) x sqrt(c^2- v^2)/c. If you do the calculus the value of v that makes T_i a minimum is v = c. In other words to both observers to get to the end point in the least amount of time you want v to be as close to c as possible.

Last edited by ChipB; Jan 25th 2013 at 10:06 AM.
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Old Oct 9th 2013, 10:42 AM   #3
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I think it's not that time outside travels extremely fast, it's that time inside barely moves at all. So if something is one light-year away and I go there at 3/4 light speed, in the outside world it takes me 1.3 years, but to me it takes me around 8/9 of a year.
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