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From my pervious Question:What are the units of the quantities in the Einstein field equation?

i noticed that the unit of this constant $\frac {G}{c^4}$ is the unit of tenstion

$$\frac {m^3}{kg.s^2}\frac {s^4}{m^4}=\frac {s^2}{kg.m}={N^{-1}}$$

How could the unit of a constant be unit of tension $N^{-1}$!?

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3 Answers 3

First of all, the inverse newton is the unit of inverse tension ($1/F$, one over force), not tension itself.

Second, it follows from dimensional analysis. It's the constant that has to multiply the energy density (energy per cubic meter) to get the curvature (one per squared meter). Clearly, the ratio taken in one way is "energy per meter" i.e. force or tension, and the coefficient is the inverse to it, so it has to be inverse tension.

What you seem to confuse are the words "constant" and "dimensionless". These are different things. A "constant" is something that isn't allowed to change in time (or depend on space or other things). A "dimensionless" quantity is a quantity with no units. These are two totally different things.

There can be dimensionful constants, like your $G/c^4$ or the mass of the electron $m_0$, among infinitely many other examples. They don't depend on time but they do require units (e.g. the electron mass requires a unit of mass, e.g. a kilogram). And there can also be dimensionless quantities that are non-constant, like the albedo of the Earth's surface or the inflation rate. They don't need units because they're "ratios" of various things of the same type, but they do depend on time because the numerators and denominators do.

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Using G,$\hbar$, and c together you can make any unit you want, so it isn't surprising that you get inverse Newtons from G,c and $\hbar$. What is somewhat interesting is that you get it without using $\hbar$. This means that there is a fundamental force unit in General Relativity, and you should understand what it means.

The proper way to understand it is using some standard unit intuition. The combinations $G\over c^2$ and $G\over c^3$ are the conversion factors from kg to m and s respectively. They tells you the mass of a black hole when you multiply by (half) the Schwarzschild radius, or by the time it takes light to cross the Schwarzschild radius (time and space are interchangable in relativity).

$$c^3/G$ has units kg/s, and it tells you how many seconds it takes to cross the event horizon of a black hole, per kg of mass. You noticed that this quantity, times c, is a force.

A force is a flow of momentum from one place to another. One way of making a force is to have a certain amount of water smack into a wall with velocity v. In this case, the force is the amount of water hitting the wall (in Kg/s) times the velocity (in m/s).

The Kg/s unit c^3/G is the maximum amount of mass you can fit into a region of size 1 light-second. Multiplying by c, you get the maximum momentum you can transmit into a wall per second, since this is the maximum mass times the maximum velocity. So this constant is like a rough measure of the maximum force you can exert.

The value is large: $1.24 \times 10^{44} N$.

This is the intuition for the quantity, but when you boost an object, it becomes skinnier, and there is no limit to how skinny you can make a black hole. If you make a black pancake moving at ultra-relativistic speeds, you can exert a larger force than this by as much as you want, so this "maximum force" business is a total fake.

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What is somewhat interesting and also funny is that you get planck force without even using planck constant.

a. the Planck mass, denoted by $m_P$, is the unit of mass in the system of natural units known as Planck units. It is defined so that:

$$m_p=\sqrt {\frac{\hbar G}{c}}$$

b. planck time:

$$t_p=\sqrt {\frac{\hbar G}{c^5}}$$

c. planck force:

$$F_p=m_p.c/t_p=c^4/G$$

in other words:

$$R_{\mu\nu}-\frac {1}{2}g_{\mu\nu}R+g_{\mu\nu}\Lambda=\frac {8\pi }{F_p}T_{\mu\nu}$$

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@Ron Maimon how is it possible!? –  funny Aug 27 '12 at 11:17

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