New unit of mass in a different system of units....

[etotheipi]

Homework Statement

In this new system of units, the unit length is still 1 m, however the units of time and mass are chosen such that c and G are both of magnitude unity. What's the new unit of mass, in kg?

Relevant Equations

N/A

I'm finding it slightly tricky to just get a hold of where to start. I try $$\text{m}\text{s}^{-1} = \frac{c}{3\times 10^8}$$ If we then set $c = 1$, that would seem to imply $$3\times 10^8 = \text{s}\text{m}^{-1}$$For $G = 1$, I might also write $$\text{kg}\text{m}^{-3}\text{s}^{2} = 6.67\times 10^{-11}$$From here I'm slightly lost. I wondered whether I should try to isolate $\text{kg}$ somehow, like $$\text{s}^{2} = 9 \times 10^{16} \text{m}^{2} \implies 1 \text{kg} = 7.41 \times 10^{-28} \text{m}$$But this doesn't seem to help me much. I wondered whether someone could give me a little pointer in the right direction? Thank you!

 

[PeroK]

You could imagine converting units for $G$ from $m, s, kg$ to $m, T_e, M_e$, say.

If you want $G = 1 \ m^3 T_e^{-2} M_e^{-1}$, then that shoud give you the conversion from $kg$ to $M_e$ units.

 

[PeroK]

Spoiler

$T_e = 3.34 \times 10^{-9}s, \ \ M_e = 1.35 \times 10^{27}kg$

 

[etotheipi]

You could imagine converting units for $G$ from $m, s, kg$ to $m, T_e, M_e$, say.

If you want $G = 1 \ m^3 T_e^{-2} M_e^{-1}$, then that shoud give you the conversion from $kg$ to $M_e$ units.

So then $$G = 1m^3T_e^{-2}M_e^{-1} = 6.67 \times 10^{-11} m^{3} s^{-2} kg^{-1}$$ $$c = 1m T_e^{-1} = 3 \times 10^{8} m s^{-1}$$ After a bit of cancellation: $$T_e = 3.33 \times 10^{-9} s$$ $$M_e = 1.35 \times 10^{27} kg$$

I think I understand this method, but I thought that when we use some form of natural units we set the relevant quantities (in this case, $c$ and $G$) to be dimensionless, like $c=1$. However, in this method the quantities still have dimensions. I wonder if this is just because of how the question is formulated - i.e. it's not exactly a natural unit system in this case?

 

[etotheipi]

Spoiler

$T_e = 3.34 \times 10^{-9}s, \ \ M_e = 1.35 \times 10^{27}kg$

Pipped to the post

 

[PeroK]

However, in this method the quantities still have dimensions. I wonder if this is just because of how the question is formulated - i.e. it's not exactly a natural unit system in this case?

To consider dimensionless quantities is something further.

You can now measure time in metres and a speed becomes a dimensionless quantity. And, you can measure mass in metres. Your $1.35 \times10^{27} kg$ becomes $1m$ and that is, physically, half the Schwarzschild radius of something of that mass. And your gravitational constant becomes dimensionless if you measure mass, length and time all in metres.

 

[etotheipi]

To consider dimensionless quantities is something further.

You can now measure time in metres and a speed becomes a dimensionless quantity. And, you can measure mass in metres. Your $1.35 \times10^{27} kg$ becomes $1m$

Ah OK sure. So the next time someone asks me how tall I am, I'll just give it in kilograms...

and that is, physically, half the Schwarzschild radius of something of that mass.

That's quite a nice touch!