# Setting c=1, how does that work?

1. Feb 25, 2007

### quasar987

1. The problem statement, all variables and given/known data
In the usual units where c=3x10^8 m/s and g=10m/s², what is g in the units where c=1?

2. Relevant equations

3. The attempt at a solution

I said, let's redefine the meter like so: $\tilde{m}:=3\times 10^8 m$ so that now $c=1\tilde{m}/s$. And then I calculated g by substituting m for $(3\times 10^8)^{-1}\tilde{m}$... but is this what the question is asking? Because I might as well have chosen to derefine the seconds and since it is s² that pops in in the units of g, the answer would have been different. And the question seems to talk like there is only one answer: "what is g in the units where c=1?"

2. Feb 25, 2007

### arildno

The question is, indeed, ill-posed.
Perhaps you are meant to use a standard meter as your new length scale as well, but who knows?
Perhaps you are meant to use the second as your new time scale..

3. Feb 25, 2007

### George Jones

Staff Emeritus
In the context of interstellar travel with g = 1, it is often convenient to use years and lightyears, and g then also has a fairly nice value.

4. Feb 25, 2007

### arildno

Eeh, you meant c=1, George?
Of course using both c=1 and g=1 make those two values very nice, indeed..

5. Feb 25, 2007

### George Jones

Staff Emeritus
Yikes!

Instead of "with g = 1", I meant "with proper acceleration a = g".

Thanks arildno.

6. Feb 25, 2007

### Meir Achuz

c=1 LY/Y, as any reader of the New York Times knows.
Just calculate g in LY/Y^2. That proves that God exists,
and God is 1.

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