ryuunoseika said:
Since light speed requires an infinite amount of time to achieve it's effectively an acceleration right? Does anyone know what I'm trying to say? And better yet, if it's right?
I found this quite baffling. After all, speed is measured in m/s and acceleration in m/s
2, so how can a speed be an acceleration?
But then tiny-tim said this:
tiny-tim said:
it doesn't move away as you're trying to catch it!
Then I realized what you were thinking.
If you are accelerating, trying to catch up with a photon, you find that the speed of the photon is still
c. From your point of view, it
appears that the photon is accelerating as you accelerate, so that the difference in speeds remains constant.
But this is an illusion, based on the assumption that it is valid to subtract speeds.
You actually need to use the formula
\frac{u - v}{1 - uv/c^2}
There is another way to measure motion in relativity, and that is "rapidity".
The relationship between speed
v and rapidity
ϕ (measured in the same units as speed) is given by the equation
\frac {v}{c} = \tanh \frac{\phi}{c} = \frac {e^{\phi/c} - e^{\phi/c}} {e^{\phi/c} + e^{\phi/c}}
That may look complicated, but rapidity has lots of nice properties compared with speed.
- For objects all moving in the same straight line, rapidities can be added and subtracted.
- The rapidity of light is infinite.
- At low speeds, rapidity and speed are approximately the same.
- For an object undergoing constant proper acceleration α (as measured by its own accelerometer), rapidity increases linearly with proper time τ (as measured by its own clock), ϕ = ατ
Looked at from this point of view, the rapidity of light isn't an acceleration, it's just infinite so no wonder you can't reach it.
Technical note: I've deliberately chosen to define rapidity with the dimensions of speed, but many authors define it dimensionlessly via v = c \tanh \phi
