Why c=1 in relativistic equations

[bernhard.rothenstein]

As I see many Auhors present relativistic equations considering c=1. Has that certain advantages? Where from could I find out the power of c at the different points of the equation.
Thanks

 

[neutrino]

Where from could I find out the power of c at the different points of the equation.
Thanks

Dimensional analysis...

 

[MeJennifer]

As I see many Auhors present relativistic equations considering c=1. Has that certain advantages? Where from could I find out the power of c at the different points of the equation.
Thanks

The advantage is that it gives less clutter in the equations. But as you say you cannot read the implied powers of c when applicable.

I suppose for educational purposes it would be wise to include c.

 

[bernhard.rothenstein]

c=1

Dimensional analysis...

Which dimensional analysis if I do not know where is c located?

 

[malawi_glenn]

Which dimensional analysis if I do not know where is c located?

for example in particle physics we rewrite all units in terms of energy and c and h_bar. And it is very nice to get rid of the c's and h_bar's when you do forumulas and so on. If you for example have:

cross section = konstant*(energy^(-2))

we know cross section has dimensions: area = length^2

and the konstant is alaways dimensionless.

so from basic relationships we write length as E times/divided by some c's and/or h_bar. Then we add the proper c's and h_bar's so we have equal on both sides of the equation. Then we do the numerical stuff..

The same holds for each branchs of physics that uses some kind of "natural units".

 

[Meir Achuz]

If all components fo a vector have the same unit, then there is no
conversion constant needed. Since t,x,y,z are four components of a vector,
using the same unit for each means no conversion constant is needed.
This is done in astronomy. I good exercise for students
"for educational purposes", is to ask them to calculate c in light years per year, using SI units.

 

[rbj]

to me, this is just as issue of what Natural Units (the variant of which that i prefer are Planck Units) are about and why they are used. if Planck units are used exclusively, there simply is no c to worry about where it is or what power it is. same for G and $\hbar$. in fact, all equations of physical law become non-dimensionalized.

instead of

$$E = m c^2$$

we have

$$E = m$$

which is really a nondimensionalized way of saying

$$\frac{E}{E_P} = \frac{m}{m_P}$$.

we could normalize all expressions of physical quantity against their corresponding Planck unit and the have no dimensional analysis to worry about. all equations would be dimensionally correct. they might still be meaningless garbage, but dimensionally correct.

 

[ice109]

If all components fo a vector have the same unit, then there is no
conversion constant needed. Since t,x,y,z are four components of a vector,
using the same unit for each means no conversion constant is needed.
This is done in astronomy. I good exercise for students
"for educational purposes", is to ask them to calculate c in light years per year, using SI units.

what a silly exercise

$$c = \frac{1 light year}{60s \times 60s \times 24h \times 7d \times 52w} \times \frac{60s \times 60s \times 24h \times 7d \times 52w}{1 year} =1 \frac{light year}{year}$$

 

[Meir Achuz]

The correct answer is "why bother?".
The same answer holds in all of relativity and in high energy physics.