Is the Schwarzschild metric dimensionless?

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Homework Help Overview

The discussion revolves around the dimensionality of an expression derived from the Schwarzschild metric in general relativity, specifically whether the term on the right-hand side of the equation is dimensionless.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the dimensionality of the expression \(\frac{2GM}{c^2 r}\) and its implications for the validity of the Schwarzschild metric. There is an attempt to relate this to the concept of the Schwarzschild radius.

Discussion Status

Several participants express agreement on the dimensionless nature of the expression, noting that it must be dimensionless for the formula to be valid. The discussion reflects a shared understanding of the relationship between the metric and dimensional analysis.

Contextual Notes

Participants reference unit conventions and the necessity for the expression to be dimensionless in the context of the metric's validity.

help1please
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Homework Statement



The problem is I am wanting to know if the expression on the right hand side is dimensionless.

Homework Equations



[tex]ds^2 = (1 - \frac{2GM}{c^2 r})c^2 dt^2[/tex]

The Attempt at a Solution



Since the Schwarzschild radius is [tex]r = \frac{2GM}{c^2}[/tex] would I be right in saying that

[tex]\frac{2GM}{c^2 r}[/tex]

is dimensionless?
 
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help1please said:

The Attempt at a Solution



Since the Schwarzschild radius is [tex]r = \frac{2GM}{c^2}[/tex] would I be right in saying that

[tex]\frac{2GM}{c^2 r}[/tex]

is dimensionless?

Yes, it is.
 
Note that it would have to be if the formula is valid since you're subtracting that quantity from 1, which is dimensionless.
 
Of course it is. Note that sometime we write metric in this form:[tex]1-\frac{2GM}{r}[/tex]
just a matter of unit conventions.
 
thanks every1
 

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