# I Does continuous acceleration eventually create a black hole

1. Nov 23, 2015

### ddesaneis

2. Nov 23, 2015

### Orodruin

Staff Emeritus
In general relativity, mass is not the source of gravity, the energy-momentum-tensor is and it also contains information about mass.

What is relativistic mass and why it is not used much?
Physicists do not talk much about rest mass simply because it is an archaic concept which only tends to confuse the general public.

3. Nov 23, 2015

### ddesaneis

Consider the following equation:

https://en.wikipedia.org/wiki/Mass_in_special_relativity#Controversy

velocity (v) * the sum of m0 + mass associated with velocity.

By this logic, momentum (p) is also a function of gravitational constant. If a GPS satellite returns from orbit to launch site, the relativistic mass within the satellite will correspond to the gravity and velocity of the launch site.
https://www.uam.es/personal_pdi/ciencias/jcuevas/Teaching/GPS_relativity.pdf

4. Nov 23, 2015

### Staff: Mentor

That expression for momentum is correct, but it has nothing to do with the gravitational constant (neither G nor anything derived from it appears).

There's another way of thinking about the question in your original post: right now, even as we speak, you are moving at 99.9999% of the speed of light relative to something somewhere. Are you showing any signs of turning into a black hole?

5. Nov 23, 2015

### ddesaneis

The momentum equation does not include the influence of relativistic mass from gravity, but should.

If a GPS satellite returns from orbit to launch site, the relativistic energy within the satellite will correspond to the gravity and velocity (v) of the launch/return site.
https://www.uam.es/personal_pdi/ciencias/jcuevas/Teaching/GPS_relativity.pdf

6. Nov 23, 2015

### Orodruin

Staff Emeritus
No it should not. It is a special relativistic equation and there is no gravity in SR. In addition, when you go to GR it becomes a local statement which is still true. Furthermore, relativistic mass is an obsolete concept and you are trying to apply Newtonian gravity to SR rather than going into the actual GR description. This is doomed to fail and to create misunderstandings.

7. Nov 23, 2015

### Staff: Mentor

The answer to the OP question, as has been said several times, is "no". Enough said. Thread closed.