Derive Planck Length in terms of c, G, and h_bar

In summary, the homework statement says to derive a formula for Planck length in terms of c, G, and h_bar. The first part of the problem was to derive the escape speed (Vesc) for a star of mass M and radius R. The second part is where I am having trouble. It says to equate the Vesc calculated above and derive a formula for Planck length in terms of c, G, and h_bar. I tried a couple of different methods and I can't get this to simplify down to the form I need. According to Wikipedia the Planck length is defined as sqrt(hG/c^3) and I think it is just a way to get units of meters out
  • #1
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Homework Statement



a. The first part of this problem was to derive the escape speed (Vesc) for a star of mass M and radius R.

b. The second part is where I am having trouble. It says to equate the Vesc calculated above and derive a formula for Planck length in terms of c, G, and h_bar.

Homework Equations



This is from the Heisenberg uncertainty section for position and momentum so:

[tex]\Delta[/tex]X * [tex]\Delta[/tex]P = h_bar/2

E = mc2

The Attempt at a Solution



The escape velocity I calculated from part a was Vesc = [tex]\sqrt{2GM/R}[/tex]

Equating Vesc to C gives:

C = [tex]\sqrt{2GM/R}[/tex]

Am I supposed to assume that [tex]\Delta[/tex]p = Mc and solve for [tex]\Delta[/tex]x using the Heisenberg uncertainty principle?

I have tried a couple of different methods and I can't get this to simplify down to the form I need. I am pretty sure this is an easy problem, I just am not seeing the trick.
 
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  • #2
According to Wikipedia the Planck length is defined as sqrt(hG/c^3)
I think it is just a way to get units of meters out of those fundamental constants.
I don't think it has a physical meaning, except in quantum mechanics which doesn't seem to apply to the problem you have.
 
  • #3
Ok I think I am getting close to this one:

c=[tex]\sqrt{\frac{2GM}{R}}[/tex]

M=[tex]\frac{c^2 R}{2G}[/tex]

Using the Heisenberg Principal:
[tex]\Delta[/tex]X [tex]\Delta[/tex]P = [tex]\frac{h_bar}{2}[/tex]

[tex]\Delta[/tex]P = c M

Solving this for [tex]\Delta[/tex]X
[tex]\Delta[/tex]X = [tex]\frac{G h_bar}{c^3 R}[/tex]

How do I get the square root and get rid of the R? :frown:
 
  • #4
Gravity level
c =under root of (2GM/R)
where R=L (Planck's Length) on gravity level
so c =under root of (2GM/L)

Uncertainty principle
uncertainty in momentum x uncertainty in position = reduced Planck constant/2
mv x L = reduced Planck constant/2
here on quantum level, Uncertainty in position is Planck length and velocity is c (velocity of light)
mcL= reduced h/2
m=reduced h/(2cL)

put the value of m in value of c in gravity level
c square = 2 G reduced h/(2cL*R)
c square = 2 G reduced h/(2cL*L)
where R=L (Planck length on gravity level)
Derive L (Planck length) from above equation.
 
  • #5


To derive the Planck length in terms of c, G, and h_bar, we can start by equating the escape velocity (Vesc) to the speed of light (c):

Vesc = c

Next, we can substitute the expression for the escape velocity from part a:

\sqrt{2GM/R} = c

Squaring both sides and rearranging, we get:

2GM = c^2R

Next, we can use the definition of gravitational constant (G) and the mass-energy equivalence equation (E=mc^2) to rewrite this expression:

2G(Mc^2) = c^2R

Using the Heisenberg uncertainty principle, we can also write:

\Deltax * \Deltap = h_bar/2

Substituting \Deltap = Mc, we get:

\Deltax * Mc = h_bar/2

Solving for \Deltax, we get:

\Deltax = h_bar/2Mc

Substituting this value for \Deltax into our previous equation, we get:

2G(Mc^2) = c^2R

h_bar/2Mc * Mc = c^2R

Simplifying, we get:

h_bar/2 = c^2R

Finally, we can solve for the Planck length (l_p) by rearranging this equation:

l_p = \sqrt{h_bar*c^2/2}

Substituting the value of c (speed of light in a vacuum) and h_bar (reduced Planck constant) in terms of their numerical values and units, we get:

l_p = \sqrt{(1.0545718*10^-34 m^2 kg/s)*(2.99792458*10^8 m/s)^2/2}

Simplifying, we get the final expression for Planck length in terms of c, G, and h_bar:

l_p = \sqrt{1.616199*10^-35 m^2 kg}
 

1. What is the Planck length?

The Planck length is the smallest unit of length that has physical significance in the universe. It is approximately 1.616 x 10^-35 meters.

2. How is the Planck length derived?

The Planck length is derived using the fundamental constants of the universe: the speed of light (c), the gravitational constant (G), and the reduced Planck constant (h_bar). It is expressed as the square root of (h_bar * G / c^3).

3. Why is the Planck length significant?

The Planck length is significant because it is the scale at which the effects of quantum gravity become important. It is also the smallest length that can be measured with any certainty according to the Heisenberg uncertainty principle.

4. Can the Planck length be measured?

No, the Planck length is far too small to be measured with current technology. It is also beyond the limits of our current understanding of physics.

5. How does the Planck length relate to other Planck units?

The Planck length is one of the Planck units, which are derived using the same fundamental constants. It is the smallest of the Planck units and is used to define other fundamental units such as Planck time and Planck mass.

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