Mark M said:Hi,
The reason we can not go faster than the speed of light is simple. As you accelerate, your energy gets higher and higher. And from E=mc2, we know that this also increases your mass. The higher your mass, the harder it is to increase your speed(just think about how much easier it is to move a basketball rather than a car.). Therefore, as you go faster, it gets harder and harder to continue to accelerate. To accelerate to the speed of light, you would need to exert an infinite amount of energy. So, even if you had a particle collider with a ridiculous amount of energy you could only make a particle going 99.9% of the SoL continue to approach the SoL but never reach it (i.e. 99.9999999...% of the SoL)
The reason light can reach this speed is that it is massless, unlike matter.
Drakkith said:As I said in my above post, this is not true. The mass of an object does not change. If it did, then at a certain speed anything would be transformed into a black hole, which does not happen.
Homesick345 said:Is this bull?
phinds said:Yes, if that is in fact what he said. Perhaps a translation error? Perhaps you misunderstood him? Perhaps he's an idiot? (one of the first two is more likely).
Homesick345 said:Hi - I'm sure this is what he said. I will maybe remember his name. He's kind of famous (in France at least) - he wrote many books "for the public" - (no, not Hubert Reeves, who is Candian I think). French is almost a mother tongue for me - so there is no mistake - or translation, I'm 100% sure this is exactly what he said. This was around 1980 on a famous french TV "litterary" talk show, hosted by the famous Bernard Pivot. But well, I guess it's not really important
Drakkith said:As I said in my above post, this is not true. The mass of an object does not change. If it did, then at a certain speed anything would be transformed into a black hole, which does not happen.
clamtrox said:It's not the mass that causes gravitation, but all energy. However, kinetic energy enters into the equations of GR differently from other (for example thermal or mass-like) energies, as the Einstein equation relates metric to the 4d energy-momentum tensor. I have not done the explicit calculations but I'm sure you are right in that fast moving things don't collapse into black holes. :-)
salvestrom said:Maybe they would. =D
I'm pretty sure the mass of an object at any particular value of v is determined by the lorentz factor for that v. Anything traveling at 99.9%c would therefore be 22 times more massive than in its rest frame. So maybe our sun traveling at 99.9%c would collapse into a black hole, but a tiny little spaceship wouldn't. But that's just an assumption. I couldn't tell you if some unusual counter-effects would apply, or if it even works that literally.
salvestrom said:Maybe they would. =D
I'm pretty sure the mass of an object at any particular value of v is determined by the lorentz factor for that v. Anything traveling at 99.9%c would therefore be 22 times more massive than in its rest frame. So maybe our sun traveling at 99.9%c would collapse into a black hole, but a tiny little spaceship wouldn't. But that's just an assumption. I couldn't tell you if some unusual counter-effects would apply, or if it even works that literally.
If you look up velocity addition in wikipedia, you will get the answer that 0.5c + 0.5c = 0.8c.jreelawg said:I was about to ask for clarification about this issue.
Say you have three adjacent observers, all at rest relative to each , A,B,C.
A stays at rest with respect to itself, B and C each accelerate to .5 c in opposite directions with respect to A.
From A's perspective, would it not be the case that B, and C, are moving away from each other at the speed of light? What is the trick, that allows A to see B moving away in one direction at .5 c, and C the other at .5 c, without seeing C, and B moving away from each other at C.
How about C, and B what would they observe? What would be the difference between the speed A would observe between C, and B, and the speed B, and C would observe between each other and A?
ghwellsjr said:If you look up velocity addition in wikipedia, you will get the answer that 0.5c + 0.5c = 0.8c.
ghwellsjr said:If you look up velocity addition in wikipedia, you will get the answer that 0.5c + 0.5c = 0.8c.
jreelawg said:Does that follow that the sum of the distances traveled away from the object would also be 80% of what you would expect?
Because the observer can say definitively, that each other observer separately has traveled a certain distance based on their speed .5 C, and you would think that two certain distances should ad up without relativistic effects. I'm confused.
Drakkith said:To observer A, both B and C are traveling at 0.5c away. So the distance between them is increasing at 1.0 c which is perfectly ok, because to observer B, observer c is only moving 0.8 c, and observer C see's B receding at 0.8c also. So between any two observers there is no one traveling faster than c. To the moving observers they would experience time dilation and length contraction, so the distance to an object would appear shorter than to observer A.
jreelawg said:But who is to say which object is moving after B, and C reach .5 c and stop accelerating?
According to B, A could be moving at .5 C, and then A would be experiencing the relativistic effects.
jreelawg said:But who is to say which object is moving after B, and C reach .5 c and stop accelerating?
According to B, A could be moving at .5 C, and then A would be experiencing the relativistic effects.
It seams like in order to work, you would have to consider that for every two objects increasing or decreasing in distance from one another, both are moving relativistically at an equal rate. Otherwise, you could say that one or the other is moving any arbitrary speed so long as the sum of the two is less than C. But which speed you assign to which objects makes a difference in which object is experiencing which effects, and thus if and when they meet again, which one will be older than the other.
No, there is length contraction involved as well as time dilation.jreelawg said:Does that follow that the sum of the distances traveled away from the object would also be 80% of what you would expect?
Like I said, there's length contraction but there's also relativity of simultaneity, so the events of B and C arriving 0.5 light-seconds away from A simultaneously does not correspond with A arriving at 0.433 ls at the same time as C arriving at 0.693 ls.jreelawg said:Because the observer can say definitively, that each other observer separately has traveled a certain distance based on their speed .5 C, and you would think that two certain distances should ad up without relativistic effects. I'm confused.
Drakkith said:For one thing, A never experienced an acceleration. This is noticeable in all 3 frames. A, B, and C would all agree on that. Once B and C stop accelerating then it is equally valid to say that any of them are moving. We could say that all 3 observers accelerated to 0.9 c from starting point D before B and C moved away from A.
You get to say who is moving which you did back in post #16 when you said A remained at rest while B and C moved away. Then you wanted to see what things looked like from B and C's frame of reference which means that they were at rest in a second (and third) frame of reference with A now moving and C or B also moving.jreelawg said:But who is to say which object is moving after B, and C reach .5 c and stop accelerating?
According to B, A could be moving at .5 C, and then A would be experiencing the relativistic effects.
Yes, you are correct, if A is at rest and B is moving at some speed, then when B is at rest, A is moving at the negative of that speed. The same is true between A and C and between B and C. It's just that you can't add the speeds that A sees B and C moving away from him to get the speed that B and C will see each other moving away from each other.jreelawg said:It seams like in order to work, you would have to consider that for every two objects increasing or decreasing in distance from one another, both are moving relativistically at an equal rate. Otherwise, you could say that one or the other is moving any arbitrary speed so long as the sum of the two is less than C. But which speed you assign to which objects makes a difference in which object is experiencing which effects, and thus if and when they meet again, which one will be older than the other.
jreelawg said:So if two observers start in about the same frame at rest with each other, and they move apart at some fraction of C, and eventually meet again;
Will it be the case that the only factor effecting an actual difference what their clocks will read once back in the same frame, will be the acceleration and gravitational effects, and their relativistic speeds will not have had any effect on their clocks, other than in the changes in appearance while they were moving apart, or towards each other?
Drakkith said:If both observers experience the same acceleration for the same amount of time, both away and back, then their clocks will read the same. Note that you are creating a scenario in which two observers leave a location at the same speeds for the same amount of time and then return to that same location. If we say the location contains observer E then both ships show the exact same relative effects to E.
jreelawg said:But what if two objects move apart from the third observer at rest, and each experience the same amount of acceleration, and the same gravitational field strengths,
but one just happens to have gone further before turning around. So even though it experienced the exact same amount of acceleration, it was moving at .5 C relative to the third observer for a longer period of time.
When they all meet eventually, will their clocks all read the same?
Mark M said:You are correct in saying that A would say B and C were undergoing relativistic effects, but B would say A and C were undergoing said effects. I'll illustrate where the error comes in your thinking:
Let's first take the role of observer B. We are moving at a constant velocity, so we claim A and C are moving away from us, and hence we claim that they are undergoing time dilation. Observer A makes a similar claim: he is at rest, but we are in motion, so B is experiencing time dilation. As observer B, we wish to demonstrate to A that his clock is slower than ours by flying over to him. And this is where the problem comes in - by changing our frame of reference we relinquish our ability to claim we are at rest, so once we get back to A, B and A agree that B's clock ran slower. Even if we send a signal, this must also travel at finite speed, so the same thing applies.
jreelawg said:But what if two objects move apart from the third observer at rest, and each experience the same amount of acceleration, and the same gravitational field strengths,
but one just happens to have gone further before turning around. So even though it experienced the exact same amount of acceleration, it was moving at .5 C relative to the third observer for a longer period of time.
When they all meet eventually, will their clocks all read the same?
And both A and B will have less time pass than C.Drakkith said:No, their clocks will not read the same. If A moves away from C at 0.5c for 1 hour before returning, and B moves away from C at 0.5c for 2 hours before returning, and both accelerate the same amount, then B will have had less time pass than A.
Cosmo Novice said:C is a very special number. C is the speed of all massless particles in a vacuum, so anything with 0 mass travels at C.
Anything with mass does not and cannot travel at C. As anything with mass accelerates towards C it takes more and more energy to get closer to C, to actually reach the speed of light would take more energy than is in the entire universe.
General Relativity forbids anything reaching C as it is mathematically impossible for mass to go at that speed.
That is a simplified explanation but holds true.
Why do you keep saying subjective?pawprint said:I came to this thread late in it's evolution. Everything seems well sewn up except the subjective experience of the intrepid pilot.
His on-board clocks continue to show subjective elapsed time. He calculates his speed from his known rate of acceleration. After 3 years at at just under 1 g he calculates a velocity of c. Subjectively he is correct. He is one and a half a light years from his departure point.
Meanwhile observers at the pilot's departure point (they have his flight plan) calculate after (their) three years that his vessel is traveling at about .8 c. (That's from memory, I may be wrong as to the exact figure.) He is only 1.2 light years distant.
It looks like a free ride to the pilot. If he keeps accelerating at that rate he'll have traveled 4.5 light years after 6 subjective years, and after 9 he'll have AVERAGED 1 light year per subjective year. But things are not all beer and skittles for him. Unless he has extraordinary shielding he'll soon die from radiation sickness as photons originating from ahead have (subjectively) blue shifted radically. What he thought would be visible light is now hard xrays. Xrays have become gamma rays etc. etc..
So the universe does not impose speed limits on travellers unless they choose to catch up with friends. Our captain now reverses thrust but he'll have been away for 36 years subjective before he returns home. I leave it to the mathematicians to say how long 'home' thinks he's been away.