[talanum1 (OP)]

I worked out the Planck charge and found that any real value of the Planck constant exponent is consistent. Thus one Planck charge may be equal to any real number. I set:

ε^a₀ħ^bc^cG^d = 1 Coulomb

to get this result. Leaving out "G" leads to an inconsistency.

Therefore something other than the particle should determine its charge.

[Origin]

Thus one Planck charge may be equal to any real number.

No, the Planck Charge is a constant,
  C

[Bored chemist]

I worked out the Planck charge and found that any real value of the Planck constant exponent is consistent. Thus one Planck charge may be equal to any real number. I set:

ε^a₀ħ^bc^cG^d = 1 Coulomb

to get this result. Leaving out "G" leads to an inconsistency.

Therefore something other than the particle should determine its charge.

It looks like you don't understand dimensional analysis.

[talanum1 (OP)]

No, the Planck Charge is a constant

What calculation did you use?

[evan_au]

Charge was not one of Planck's original units, but some people have tried to extend his model.

There are a few slightly different definitions shown for the Planck charge.

One is q_p = √e₀ħc

See: https://en.wikipedia.org/wiki/Planck_units#History_and_definition

[talanum1 (OP)]

Yes, I made an error for:

ε₀ħc,

this does not suppose G = 1,

but if you add G:

ε₀ħcG

you get infinitely many values.

[Bored chemist]

It looks like you don't understand dimensional analysis.

Still.

[alancalverd]

You get a more "natural" number if you multiply by A_p.

[talanum1 (OP)]

You get a more "natural" number if you multiply by Ap.

What is A_p?

[talanum1 (OP)]

It looks like you don't understand dimensional analysis.
Still.

I know how to calculate it: set:

ε₀^aħ^bc^dG^e = x Coulomb.

[Bored chemist]

It looks like you don't understand dimensional analysis.
Still.

I know how to calculate it: set:

ε₀^aħ^bc^dG^e = x Coulomb.

Thank for clarifying that you don't understand it.
Try doing the calculation for kinetic energy.

[Kryptid]

It looks like you don't understand dimensional analysis.
Still.

I know how to calculate it: set:

ε₀^aħ^bc^dG^e = x Coulomb.

Tell me what all of those symbols stand for (the exponents a, b, d, and e in particular). I would like to check your math.

[talanum1 (OP)]

Set the constants as shown, then calculate a, b, d, e  by equating the exponents of the dimensions in the constant raised to the power a, b, d, e to the corresponding exponent of the dimension of the left side. Then you get 4 equations in 4 variables. a = 1/2 you get from ε₀ which must have the exponent of it's C dimension = 2a = 1. Further you get:

(2) d+2b-3a+2e = 0
(3) -d-b+2a-e = 0
(4) b-a+e = 0

you get d = 1/2 and
b+e = 1/2. Just set b+e = 1/2 in (2) and (3). You get (2), (3) and (4) from a knowledge of what units the constants are made up of.

We have that b and e is left as b+e =1/2, so b and e can be chosen as any real numbers so long as their sum is 1/2.

[Bored chemist]

Set the constants as shown, then calculate a, b, d, e  by equating the exponents of the dimensions in the constant raised to the power a, b, d, e to the corresponding exponent of the dimension of the left side. Then you get 4 equations in 4 variables. a = 1/2 you get from ε₀ which must have the exponent of it's C dimension = 2a = 1. Further you get:

(2) d+2b-3a+2e = 0
(3) -d-b+2a-e = 0
(4) b-a+e = 0

you get d = 1/2 and
b+e = 1/2. Just set b+e = 1/2 in (2) and (3). You get (2), (3) and (4) from a knowledge of what units the constants are made up of.

We have that b and e is left as b+e =1/2, so b and e can be chosen as any real numbers so long as their sum is 1/2.

As I said; please try doing that with the calculation of kinetic energy.

[talanum1 (OP)]

What do you set kgm²s^-2 equal to?

[talanum1 (OP)]

Setting it equal to  h^ac^bG^c leads to a contradiction:

You get:

(1) 2a+b+2c = 2 (m)
(2) -a-b-c = -2 (s)
(3) a+c = 1 (kg)

from (3): c = 1-a...(4)
(4),(1): 2a+b+2-2a = -2 -> b = -4 ... (5)
(2),(4),(5): -a+4-1+a = -2 which is a contradiction.

Please check my mathematics.

[talanum1 (OP)]

That calculation is wrong. You get:

(1) 2a+b+3c = 2 (m)
(2) -a-b-2c = -2 (s)
(3) a-c = 1.

(3): a = 1+c ... (4)
(4),(1): 2+2c+b+3c = 2 -> 5c+b = 0 ... (5)
(4),(5),(2): -1-c+5c-2c = -2 -> c = -1/2
b = 5/2
a = 1/2

so E_kp = √(c⁵h¹G^-1) ~ 10^8.5 J

[Bored chemist]

What do you set kgm²s^-2 equal to?

How would you derive the formula for KE from mass and velocity?

[evan_au]

b and e can be chosen as any real numbers so long as their sum is 1/2

This is a familiar problem when solving simultaneous equations.

You started with 4 equations in 4 unknowns; this may possibly have a unique solution - but it's not guaranteed
- If some of the equations are linearly related to each other, they don't add any "extra" information
- So one of the equations is redundant
- And instead of a unique solution, you end up with a linear relationship between two variables (a line)
- Or even a linear relationship between 3 variables (a plane)

[Kryptid]

Set the constants as shown, then calculate a, b, d, e  by equating the exponents of the dimensions in the constant raised to the power a, b, d, e to the corresponding exponent of the dimension of the left side. Then you get 4 equations in 4 variables. a = 1/2 you get from ε₀ which must have the exponent of it's C dimension = 2a = 1. Further you get:

(2) d+2b-3a+2e = 0
(3) -d-b+2a-e = 0
(4) b-a+e = 0

you get d = 1/2 and
b+e = 1/2. Just set b+e = 1/2 in (2) and (3). You get (2), (3) and (4) from a knowledge of what units the constants are made up of.

We have that b and e is left as b+e =1/2, so b and e can be chosen as any real numbers so long as their sum is 1/2.

I have to admit that I don't understand what you're trying to say.