What would happened to the mass of your body if you instantly accelerated to 99.9999% the speed of light?
Now give that one some deep thinking my friends. Thank you for your detailed answer. Tom
- OldPilotLv 71 decade agoFavorite Answer
The Rest Mass does not change
Many contemporary authors such as Taylor and Wheeler avoid using the concept of relativistic mass altogether:
"The concept of "relativistic mass" is subject to misunderstanding. That's why we don't use it. First, it applies the name mass - belonging to the magnitude of a 4-vector - to a very different concept, the time component of a 4-vector. Second, it makes increase of energy of an object with velocity or momentum appear to be connected with some change in internal structure of the object. In reality, the increase of energy with velocity originates not in the object but in the geometric properties of spacetime itself.
Scroll Down to “Relativistic Mass”
When we get into Relativity, “mass” has 2 different meanings. There is “Rest Mass” and “Relativistic Mass” and they are NOT the same thing.
A better approach is Relativiastic Momentum which is:
Rel Mo = gamma * velocity * rest mass
gamma = 1 / (sq rt (1 - (v^2/c^2)))
For your question:
gamma = 1 /( sq rt 1 - 0.99998)
gamma = 1 / sq rt 19.9999 * 10^-6) = 1 / 4.4721 * 10^-3)
===> Multiply normal momentum by 223.61
- Anonymous1 decade ago
To continue from Old Pilot's correct response.....
First, in Relativity, you need to consider that your speed, relative to YOU, is zero. You may need to think about this, but is is really an important consideration. There is no "real" or absolute speed anywhere in the universe! Speed is always ALWAYS relative to some other object.
So to you, your mass and dimensions and passage of time are always just as you have always known them. It is just as Newton had predicted. You are in a "Newtonian reference frame." Nothing relativistic at all.
But you are moving at some near-to-light speed - relative to what? To the Earth? OK. From Earth, we might measure your "apparent" mass (see Old Pilot's comments on this) as being very large. But you would measure the "apparent" mass on earth to be just as large!
Then, imagine that there is a third observer, travelling at .5c relative to Earth, and nearly .5c relative to you. This observer would measure the "apparent" mass to be very nearly the same on both Earth and your spaceship, but very different from what you were observing on Earth.
This is complicated stuff, once you begin to look more closely.