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What is the maximum ratio in the rate of change in time in reference to object $A$ which is standing still and object $B$ which is moving at the speed of light?

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Alex, I'll toss in a true nitpick: Your phrase "$A$ which is standing still" can be made more precise by saying "$A$ as viewed by someone who is motionless relative to $A$". That's because "standing still" is not an assertion that can be defined meaningfully in special relativity without specifying the frame of reference from which it is being asserted. (BTW, great answer by Mark Eichenlaub!) – Terry Bollinger Jan 26 '13 at 20:16
Thanks any corrections are allays appreciated. – Alex Voinescu Jan 26 '13 at 20:32
up vote 6 down vote accepted

Objects, defined as things with mass, don't move at the speed of light. The time dilation factor is

$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$

and it has no limit - it diverges at $v\to c$. For speeds very close to the speed of light, we could define $\epsilon = \frac{c - v}{c}$, then we'd have $\gamma \sim \frac{1}{\sqrt{2\epsilon}}$ This shows how much time slows down for speeds very near the speed of light. Here's a picture

enter image description here

As gamma shoots up to infinity, the time dilation factor becomes arbitrarily large. If you want the clock to go 1/100 as fast, or one millionth as fast, or one quadrillionth, that can be done by going very close to the speed of light (just solve for $\epsilon$ in the above).

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Thank you, this helps! – Alex Voinescu Jan 26 '13 at 20:03

There is no limit. The Lorentz factor

$$ \gamma = {1 \over {\sqrt{1 - {v^2 \over c^2}}}} $$

can become arbitrarily large if $v$ gets close enough to the speed of light.

Given any value of $\gamma$ (e.g., 1, 2, 1000, 1000000) use

$$ v = c \sqrt{1 - {1 \over \gamma^2}} $$

to compute the required velocity $v$ to achieve that $\gamma$.

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