How do I use dimensional analysis to construct an energy for the system given the Newton constant $G$, the speed of light $c$ and the Planck constant $h$? I don't know of any energy formulas containing the constants $G$, $c$ and $h$.
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$\begingroup$ I don't know the answer, but you might want to check out proposals for defining the kilogram along with the Watt balance. $\endgroup$– John DuffieldAug 25, 2015 at 18:49
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2$\begingroup$ Hi and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$– John RennieAug 25, 2015 at 19:12
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$\begingroup$ See Planck energy. $\endgroup$– Qmechanic ♦Aug 31, 2015 at 12:40
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3 Answers
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"I don't know any equations..." is the point of dimensional analysis!
Let's make a table of the quantities you listed, and their dimensions:
M L T
G -1 3 -2 \
c 1 -1 +- given these inputs...
h 1 2 -1 /
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E 1 2 -2 - I need to get this output
If we assume there is an expression
$$E \propto G^A c^B h^C$$
then it follows that we need to solve for A, B, C such that
-A + C = 1 (balancing M)
3A + B + 2C = 2 (balancing L)
-2A - B - C = -2 (balancing T)
I will leave you the fun of solving for A, B and C - the coefficients will be non-integer.
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Say,
This implies m=-1/2 n=1/2 p=5/2
So this way you can formulate equations by dimensional analysis.
answered Aug 25, 2015 at 19:30
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Write down a few dimensionally correct equations. E.g.:
$$E = m c^2$$
$$E = m^2 G/r$$
$$E = h f = h c/r$$
Divide the last two:
$$1 = m^2 G/(h c)$$
So, we have that:
$$m = \sqrt{\frac{h c}{G}}$$
Therefore:
$$E = m c^2 = c^{\frac{5}{2}}\sqrt{\frac{h}{G}}$$
answered Aug 25, 2015 at 19:37
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$\begingroup$ This looks odd. $$E=mc^2$$ refers to the energy of a massive particle. $$E = hf$$ refers to the energy of a massless particle with a wavelength $r$. $$E=\frac{m^2G}{r}$$ refers to the energy of a system of two massive particles with a separation of $r$. Three different systems, the way I see it. $\endgroup$– AlecSep 4, 2015 at 13:12
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$\begingroup$ @Alec, as we're only interested in the dimensions, we can do the computations modulo the physics content. Two quantities are considered equivalent if they have the same SI dimensions. $\endgroup$ Sep 4, 2015 at 16:05