What Happens to Time Dilation at a Black Hole's Event Horizon?

JohnGano
Messages
4
Reaction score
0
I'm looking at this equation for gravitational time dilation:

<br /> T = \frac{T_0}{\sqrt{1 - (2GM / rc^2)}} <br />

I understand the relation of time dilation and velocity, and how v must be less than c, but I don't understand what exactly is implied here. At a certain point, M could be great enough such that the square root becomes negative or 0, or r could become small enough that the same thing happens. So what exactly does that mean? Is it possible that M or r could be a size such that you get an imaginary or undefined answer?
 
Physics news on Phys.org
JohnGano said:
I'm looking at this equation for gravitational time dilation:

<br /> T = \frac{T_0}{\sqrt{1 - (2GM / rc^2)}} <br />

I understand the relation of time dilation and velocity, and how v must be less than c, but I don't understand what exactly is implied here. At a certain point, M could be great enough such that the square root becomes negative or 0, or r could become small enough that the same thing happens. So what exactly does that mean? Is it possible that M or r could be a size such that you get an imaginary or undefined answer?
Sure. If M were infinite or r were zero, your answer could be 0. But how useful a solution is that in describing anything in the universe?

But no, it could never be negative.
 
JohnGano said:
I'm looking at this equation for gravitational time dilation:

<br /> T = \frac{T_0}{\sqrt{1 - (2GM / rc^2)}} <br />

I understand the relation of time dilation and velocity, and how v must be less than c, but I don't understand what exactly is implied here. At a certain point, M could be great enough such that the square root becomes negative or 0, or r could become small enough that the same thing happens. So what exactly does that mean? Is it possible that M or r could be a size such that you get an imaginary or undefined answer?

The square root becomes zero just at the event horizon of a black hole, where r=2GM/c^2. This is an indication of the fact that one cannot have a stationary observer exactly at the event horizon (one could have a stationary light beam, but a light beam isn't an observer).

One also cannot have a stationary observer inside the event horizon, i.e r < 2GM/c^2.

r here is the schwarzschild r cooridnate, btw.
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Does the speed of light change in a gravitational field depending on whether the direction of travel is parallel to the field, or perpendicular to the field? And is it the same in both directions at each orientation? This question could be answered experimentally to some degree of accuracy. Experiment design: Place two identical clocks A and B on the circumference of a wheel at opposite ends of the diameter of length L. The wheel is positioned upright, i.e., perpendicular to the ground...
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Back
Top