This discussion centers on the implications of traveling at relativistic speeds, specifically at 0.999999999999999 times the speed of light (c). Participants clarify that within the spaceship's reference frame, the observer experiences normal conditions, and their mass does not increase to infinity. The relativistic velocity addition formula, (u+v)/(1+uv/c²), ensures that speeds never exceed c. The conversation emphasizes that all relativistic effects remain valid regardless of the observer's frame, and no frame of reference is objectively preferred.
PREREQUISITESPhysicists, students of theoretical physics, and anyone interested in the principles of special relativity and the behavior of objects at high velocities.
No. Unless I'm mistaken though, an external observer looking at the craft as it whizzes by would be measuring its relativistic mass.lightarrow said:Are you talking about invariant mass?
Integral said:You arrive at the back of the ship, turn around and walk to the front. Meanwhile you note that since the food is so terrible, not only is your mass not nearing infinite, you have actually lost weight.
DaveC426913 said:Stop right there.
There is no such thing is an objective frame of reference. There is no "reality versus "less than reality" in the sense you mean it.
The whole point of relativity - the very core - is exactly this: no frame of reference is preferred over any other. i.e. the pilot's frame of reference is equally valid. As far as he's concerned, he is stationary and the rest of the universe is moving. If he looked out his window, he'd see the Earth whizzing past him at .99c. If he measured the mass of a guy walking around on Earth, he'd measure the guy's mass as very high and would wonder how the guy can walk around at all.
Neither of them is wrong. Neither of the is seeing an illusion. Neither of their experiences are any less real.
For which reason should it suggest ftl travels? It doesn't suggest it at all. They've given you the formula:Nickelodeon said:Doesn't this suggest that faster than light travel, relative to something, lightarrow's beam of neutrinos, for instance (see prev thread). is possible. This would contravene 'c' being tops.
I think you are mistaken. How would you measure the mass of a moving object? I think the only thing you could measure is the momentum and then you would have to calculate the mass. At that point you can choose if you want to calculate the rest mass or the relativistic mass.DaveC426913 said:No. Unless I'm mistaken though, an external observer looking at the craft as it whizzes by would be measuring its relativistic mass.
You can measure momentum, energy, kinetic energy, what you want. The point is that what they call "relativistic mass" has already a name: "energy".DaleSpam said:I think you are mistaken. How would you measure the mass of a moving object? I think the only thing you could measure is the momentum and then you would have to calculate the mass. At that point you can choose if you want to calculate the rest mass or the relativistic mass.
The only ways I can think of to measure mass require an apparatus at rest relative to the mass and therefore measure rest mass.
lightarrow said:For which reason should it suggest ftl travels? It doesn't suggest it at all. They've given you the formula:
V = (v1 + v2)/(1 + v1*v2/c^2)
which always gives a value < c, even if you re-iterate it as many times as you like.
Don't confuse what can be called the "3rd party separation rate" with relative speed.
Imagine this situation: Spaceship A is 1 light year to the west of me and travels at a speed of 0.9999c towards me; Spaceship B is 1 light year to the east of me and travels at a speed of 0.9999c towards me. What's their distance apart according to me? 2 light-years. When do they reach me? In about 1 year. So the separation distance decreases at a rate of about 2c according to me.
Of course the relative speed of spaceship B as measured by spaceship A is still less than c.
What we actually see and measure rather than calculate using reference frames is that the pilot is going faster than us when the spaceship approaches us and going slower than us when the spaceship passes us. One can use the relativistic Doppler formula to calculate the exact ratios.DaveC426913 said:If we were to look in the porthole of the spaceship as it passed Earth, we would see the pilot moving so slowly that he would appear to be virtually frozen in time
matheinste said:The speed of an observer relative to another observer never exceeds c in SR. However their speed of separation, that is the rate of increase of the distance between them, as seen by a third observer, may be up to nearly 2c. But to these two, whose separation distance is increasing at up to 2c, each will measure the other as going at less than c.
Matheinste.
Look, I can make mathematical symbols do what they wants, I can add c as many times as I want, but it doesn't produce a *physical* velocity. Physics IS NOT mathematics. You only have to understand this. If you want to talk about physics, then you have to talk about things which have a physical meaning, not something else, otherwise you can post on the mathematics forum.Nickelodeon said:This formula is fine(ish) for calculating relative observed speeds but not much use for un-observable speeds.
This following text comes from another post on this forum which I thought illustrates the point I am trying to make.Don't confuse what can be called the "3rd party separation rate" with relative speed.
Imagine this situation: Spaceship A is 1 light year to the west of me and travels at a speed of 0.9999c towards me; Spaceship B is 1 light year to the east of me and travels at a speed of 0.9999c towards me. What's their distance apart according to me? 2 light-years. When do they reach me? In about 1 year. So the separation distance decreases at a rate of about 2c according to me.
Of course the relative speed of spaceship B as measured by spaceship A is still less than c.
What part exactly of what he is saying is physically nonsensical?lightarrow said:Look, I can make mathematical symbols do what they wants, I can add c as many times as I want, but it doesn't produce a *physical* velocity. Physics IS NOT mathematics. You only have to understand this. If you want to talk about physics, then you have to talk about things which have a physical meaning, not something else, otherwise you can post on the mathematics forum.
This is, on its face, a true statement. However, IF the universe behaves logically THEN the universe can be described in mathematical terms. Math is the language of logic.lightarrow said:Physics IS NOT mathematics.
That's wrong. Two space ships moving away from each other, each at .999 C, wrt a 3rd observer, will be able to see each other. Since their speed wrt each other is below C, a light beam sent from one to the other will catch the other, bounce off, and go back to the first ship.Nickelodeon said:Now if we go for these two spaceships going passed the Earth and continuing on their way they will not be able to see each other although we still will.
What do you mean by unobservable speed? Give an example.Nickelodeon said:This formula is fine(ish) for calculating relative observed speeds but not much use for un-observable speeds.
That brilliant quote about "3rd party separation rate" is from a post of mine. Reread it carefully, especially the last line.This following text comes from another post on this forum which I thought illustrates the point I am trying to make.
Don't confuse what can be called the "3rd party separation rate" with relative speed.
Imagine this situation: Spaceship A is 1 light year to the west of me and travels at a speed of 0.9999c towards me; Spaceship B is 1 light year to the east of me and travels at a speed of 0.9999c towards me. What's their distance apart according to me? 2 light-years. When do they reach me? In about 1 year. So the separation distance decreases at a rate of about 2c according to me.
Of course the relative speed of spaceship B as measured by spaceship A is still less than c.
Why can't they see each other?Now if we go for these two spaceships going passed the Earth and continuing on their way they will not be able to see each other although we still will.
Ah... I think I see what you're doing. You are misinterpreting that 'rate of separation'. As it would apply to this example, somebody at rest with respect to the ship would observe the Earth moving one way at some speed close to light speed and the pilot moving at some speed in the opposite direction. (For fun, let's have the pilot move through the ship at half the speed of light.) So, according to the stationary observers in the ship (the "third party"), the Earth and the Pilot are separating at a rate of about 1.5 c. So what? With respect to everyone, the pilot still moves at a speed less than the speed of light. In particular, the people on Earth will see the pilot moving at some speed less than c. (Very close to c, since the ship is almost traveling at light speed.)Going back to the original post on this thread. Noone on Earth will see the pilot walking towards the front of the spaceship assuming his 'rate of separation' from us exceeds c. We will see him again when his sits down at the controls.
Doc Al said:What do you mean by unobservable speed? Give an example.
That brilliant quote about "3rd party separation rate" is from a post of mine. Reread it carefully, especially the last line.
Why can't they see each other?
How is that relevant to this thread?Nickelodeon said:Just guessing here but dark matter would probably fall into this category as would whatever is going on within the event horizon of a black hole.
Again, it seems like you are still thinking that somehow the pilot's speed with respect to the Earth is greater than c. Not so.I think it's red-shifting taken to extreme. After c the wavelength goes infinite. Here again the spaceship has to be going away from the observer not across his line of vision.
matheinste said:Quote:-
----I think it's red-shifting taken to extreme. After c the wavelength goes infinite. Here again the spaceship has to be going away from the observer not across his line of vision. -----
There is no after c.
Matheinste.
Define an object's velocity and you'll have the answer.DaveC426913 said:What part exactly of what he is saying is physically nonsensical?
Relativistic mass doesn't exist. Only invariant mass = mass. If then you want to say that accelerating m from 0 to 1000 m/s requires less energy than accelerating it from 1000 m/s to 2000 m/s, that's another story: F = m*a doesn't hold in relativity.DaveC426913 said:No. Unless I'm mistaken though, an external observer looking at the craft as it whizzes by would be measuring its relativistic mass.
lightarrow said:Relativistic mass doesn't exist. Only invariant mass = mass. If then you want to say that accelerating m from 0 to 1000 m/s requires less energy than accelerating it from 1000 m/s to 2000 m/s, that's another story: F = m*a doesn't hold in relativity.
Because it's completely meaningless, unless you call it "Energy". The only meaningful mass is invariant mass, so that's the only mass you should talk about.Nickelodeon said:Surely relativistic mass = invariant mass plus the mass equivalence of its relative kinetic energy so how can you say it doesn't exist?