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Speed of light becomes Infinity and Zeroby RiddlerA
Tags: speed of light 
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#19
Jan812, 03:18 AM

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#21
Jan812, 03:24 AM

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#22
Jan812, 03:30 AM

P: 58

Thank you so much for your support.. So long story short, Time runs slower for a moving clock relative to a stationary clock? Mass increases with velocity? And length contracts as the speed increases relative to a stationary FoR? Are the above statements correct? Once again thank you very much for spending time to clear my doubt.... 


#23
Jan812, 03:32 AM

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#24
Jan812, 03:35 AM

P: 58

I mean in reality Only one of them is gonna age slower than the other, but if we interchange FoR then it shows that either of them have to age slower relative to each other, which doesnt make any sense.... 


#25
Jan812, 03:50 AM

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#26
Jan812, 04:03 AM

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#27
Jan812, 09:08 AM

P: 15,319

The key is that, to decide which one is younger, one of them must turn around and go back to meet the other. And in doing so, this causes an asymmetry in the passage of their timelines. By the time they meet up again, they will both agree on who is younger. 


#28
Jan812, 09:34 AM

P: 11

RiddlerA, the way you defined the times...
I think it is important to get it right, because this way you can figure out what happens from the equation. The correct formula is this, as elfmotat wrote:
Spoiler
[itex]\Delta t_{o} = \Delta t \sqrt{1v^2/c^2}[/itex], it's the same as [itex]\Delta t = \frac{\Delta t_{o}}{\sqrt{1v^2/c^2}}[/itex], which you take from Wikipedia.
[itex]\Delta t = \Delta t_{o} \sqrt{1v^2/c^2}[/itex], it's the same as [itex]\Delta t_{o} = \frac{\Delta t}{\sqrt{1v^2/c^2}}[/itex], which you take from Wikipedia. You may take a look at either and figure out what's going on, let's take a look at the former (just notice that the time notations are swapped compared to your equation, based on how you defined them): So t is the time of the moving clock, t_{o} is the time of the static clock (ours). Now v is the speed of the moving clock, and we know it is lesser than c, both in respect to the observer. When v increases, then [itex]v^2/c^2[/itex] increases as well, which means [itex]1v^2/c^2[/itex] decreases towards zero (also [itex]\sqrt{1v^2/c^2}[/itex] decreases). You can see now how, as the speed of the t clock (in the rocket) is increased, its period has to be lesser, in its frame of reference, to synchronize its ticks to the observer's clock (on Earth, "us"). Imagine what happens if v is so great that [itex]\sqrt{1v^2/c^2} = 0.8[/itex]: when the clock on the observer from Earth (t_{o}) rings "one hour" [1], the guy in the rocket says "hey, wtf, based on my clock (t), only 48 minutes has passed!"  which is 60*0.8. The two equations are equivalent, the former is for use by the observer in the rocket to figure out what time the clock on Earth shows when an hour has passed (t_{o} = 0.8 * t = 48 minutes  contracted time), the latter is for use by the static observer to figure out what time the clock in the rocket shows (t = t_{o} / 0.8 = 75 minutes  dilated time). Does this make intuitive sense? (I hope I haven't messed something up, that would confuse things even more :P)  [1]  this is the second tick, the first was when they started the stopwatches, synchronously. 


#29
Jan812, 09:56 AM

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#30
Jan812, 11:39 AM

P: 140

A lot of you are just arguing semantics with which equation is right or wrong. They are right or wrong depending on how you define the variables. This is why you should specifically specify which variable is which so there is no confusion over how you use the formula.



#31
Jan812, 02:09 PM

P: 260




#32
Jan812, 03:29 PM

Mentor
P: 17,293

However, it really isn't the topic of this thread, and there are numerous other threads on the twin paradox. So I wouldn't try to derail the current topic. 


#33
Jan812, 04:13 PM

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#34
Jan812, 11:26 PM

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#35
Jan912, 01:18 AM

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#36
Jan912, 02:33 AM

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