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## Quantum Gravity?

The mass of an object increases as a function of its velocity, I forget the exact equation, but it's elementary physics. Therefore, once even an elementary particle, such as an electron or baryon, nears the speed of light, there must a point where it falls inside its own event horizon, and it should become a tiny Black Hole. Those conditions surely existed at the time of the Big Bang. Why doesn't / didn't that happen? I presume that this is not a problem for massless particles such a photons.

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Originally Posted by samsara15
The mass of an object increases as a function of its velocity, I forget the exact equation, but it's elementary physics. Therefore, once even an elementary particle, such as an electron or baryon, nears the speed of light, there must a point where it falls inside its own event horizon, and it should become a tiny Black Hole. Those conditions surely existed at the time of the Big Bang. Why doesn't / didn't that happen? I presume that this is not a problem for massless particles such a photons.
While it can be thought of as the mass increasing, this requires two definitions of mass. An invariant (or rest mass) and a relativistic mass. This thought can also lead to confusion and the idea that a moving particle, if moving fast enough, can collapse into a black hole.

In current physics thought, it's not the mass that is increasing, its the total energy of the object that increases. (BTW, I think you are thinking of E = mc2 . While this is good for an object at rest, the equation for an object in motion is E2 = (mc2)2+(pc)2. That second term (the one with the "p" in it) the increase is for the kinetic energy (or momentum), which is dependent on the frame of reference. The rest mass (or stress-energy) is frame invariant, and this is what has to increase if an object is to collapse into a black hole.

3. Played with a few figures for a laugh. Turns out that even if rest mass wasn't invariant (and what a strange universe that would be!) you would need a gamma of about 10^20 to get even close to a proton collapsing on itself. Thats around .9999999999999999999999999999999999999999999999999 9999999999c or so as far as I can tell.

4. In Tensor's second equation what is P?

5. That second term (the one with the "p" in it) the increase is for the kinetic energy (or momentum),

6. I'm still lost. I understand the Energy, the speed of light (Now in furlongs/fortnight) and the mass. So p is the velocity or speed?

7. Sorry .. should have been more explicit. p is the standard symbol for momentum, where p=mv (mass x velocity).

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Originally Posted by loglo
Sorry .. should have been more explicit. p is the standard symbol for momentum, where p=mv (mass x velocity).
I should have been also. Thanks loglo.

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Thanks. I knew there had to be something awry with that idea!

10. Thanks, the equation for moving objects was completely new to me.

11. Originally Posted by loglo
Played with a few figures...Turns out that even if rest mass wasn't invariant (and what a strange universe that would be!) you would need a gamma of about 10^20 to get even close to a proton collapsing on itself. Thats around .9999999999999999999999999999999999999999999999999 9999999999c or so as far as I can tell.
Thanks...my calculator only goes to 64 decimal places, and with the rounding-errors, I got an answer of 1.0c, which I knew couldn't be right

12. Okay, a second question, pardon my ignorance.

In the equation E2 = (mc2)2+(pc)2, why doesn't all the exterior squaring cancel itself and become E = (mc2)+(pc)?

(Sorry the superscript notation didn't come out right from the cut and paste)

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Originally Posted by BigDon
Okay, a second question, pardon my ignorance.

In the equation E2 = (mc2)2+(pc)2, why doesn't all the exterior squaring cancel itself and become E = (mc2)+(pc)?

(Sorry the superscript notation didn't come out right from the cut and paste)
Because there are two terms on the right side. The squaring doesn't straight cancel, as you have done. You have to square both terms on the right side, then take the square root. When you take the square root of both sides to get E, the equation becomes : E= sqrt ((mc2)2+(pc)2).

14. Thank you for your patience Mr. Tensor.

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Originally Posted by BigDon
Thank you for your patience Mr. Tensor.
Heheheheheh. No problem. If you ask questions, and I can answer them, I will. If I can't, I'm sure there are several other here that can.

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