Traveling .999999 reapeating the speed of light .

AI Thread Summary
Traveling at .999999999 repeating the speed of light is theoretically impossible for massive objects, as they cannot reach the speed of light (c). The discussion clarifies that while .999999... is mathematically equivalent to 1, the distinction between finite and infinite series is crucial. Energy requirements increase dramatically as one approaches the speed of light, necessitating infinite energy for a massive object to reach c. The fastest a massive object could theoretically travel is just below the speed of light, with advancements in particle accelerators pushing this limit closer. Ultimately, while one can approach c, achieving it remains unattainable for anything with mass.
cragar
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Is it possible to travel .999999999 repeating the speed of light , because that would be a 1 in math .
 
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Sure. if you've got infinite enegy, or zero mass. If you are curious about the energy requirements the Kinetic energy required to reach a given velocity is given by:

\gamma m c^2 - m c^2 = \frac{m c^2}{\sqrt{1-\frac{v^2}{c^2}}} - m c^2 where m is the rest mass. Play around with it and you can get a sense of how the energy requirements increase to infinity as one approaches c.
 
Is this a math problem or a physics problem?


In any case, your question is really "Is it possible to travel at velocity c" and the answer is "yes, for light but not for any massive object".
 
cragar said:
Is it possible to travel .999999999 repeating the speed of light , because that would be a 1 in math .

And .999999999 doesn't equal 1, "in math". Those numbers differ by .00000001.
 
Vanadium 50 said:
And .999999999 doesn't equal 1, "in math". Those numbers differ by .00000001.

No, but ".999999 repeating" does. It seems clear that what he really means is the more conventional ".999999..."
 
negitron said:
No, but ".999999 repeating" does. It seems clear that what he really means is the more conventional ".999999..."
More precisely, the limit of the series 0.99, 0.99, 0.999, ... equals one.
 
And the limit of the series \frac{10}{\sqrt{19}},\frac{100}{\sqrt{199}},\frac{1000}{\sqrt{19<br /> 99}},... (the time dilation factor for v = .9 c, .99 c, .999 c, ...) is infinite.
 
cragar said:
Is it possible to travel .999999999 repeating the speed of light , because that would be a 1 in math .

You can't travel at 0.999999999... c, but you can travel at 0.99999999...9c, where in the first case the ... means an unending string of 9s, and the in the second it means a finite but undetermined number of 9s.
 
Hootenanny said:
More precisely, the limit of the series 0.99, 0.99, 0.999, ... equals one.
Of course, that's exactly what "0.999 repeating" means.
 
  • #10
Well, since 0.999... and 1 are different representations of the same number (like 2-1 or 4/4), since a massive body can't travel at c, it can't travel at (0.999...)c either.
 
  • #11
i guess i should have said can something with mass travel at .99999 reapeating c and i guess it would be no , cause it is 1
 
  • #12
Correct. And just fyi, you can write that like this: .999...

The thre periods at the end (an ellipsis) indicates they repeat forever.
 
  • #13
so then what could be that fastest you could travel if you had mass .
 
  • #14
cragar said:
so then what could be that fastest you could travel if you had mass .
You can get arbitrarily close to C. Every time a new, bigger particle accelerator is built, they add a couple of more 9's to that.
 
  • #15
interesting
 
  • #16
cragar said:
so then what could be that fastest you could travel if you had mass .


Theoretically? Well I guess if one could estimate all the energy in the universe and if all that went into the kinetic energy of a single things of mass m. Which I imagine woud be a great many 9's
 
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