# Speed of light becomes Infinity and Zero

by RiddlerA
Tags: speed of light
P: 86
 Quote by ghwellsjr But this is not right.
why not?what i meant is this (an example) consider A moving at 0.86c and B is stationary. Now B has to wait for 2 years in order to have A 1 year old but he himself ages 1 year in 1 year (of course B is FoR)
 Quote by ghwellsjr Also, in Special Relativity, the term "event" has a specific meaning. It refers to an instant in time at a specific location. It shouldn't be used to refer to a duration or an interval of time. An event is the four coordinates making up a "point" in spacetime.
P: 260
 Quote by Snip3r why not?what i meant is this (an example) consider A moving at 0.86c and B is stationary. Now B has to wait for 2 years in order to have A 1 year old but he himself ages 1 year in 1 year (of course B is FoR)
Because from A's rest frame, B is the one who ages slower.
P: 86
 Quote by elfmotat Because from A's rest frame, B is the one who ages slower.
yes... but i mentioned B is my FoR
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....
P: 260
 Quote by Snip3r yes... but i mentioned B is my FoR

 Quote by Snip3r If an event takes Δt in S it would appear to take longer when happens in S'(seen from S) hence if Δt' is the time taken in S' Δt'=$\frac{Δt}{\sqrt{1-V^{2}/C^{2}}}$
Now tell me whether or not it makes sense given the fact that a person at rest in S' ages slower when viewed from S.
P: 58
 Quote by elfmotat Because from A's rest frame, B is the one who ages slower.
Is this the twin paradox is all about? It is a hole in the special theory of relativity right?
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....
P: 86
 Quote by elfmotat Read what you wrote: Now tell me whether or not it makes sense given the fact that a person at rest in S' ages slower when viewed from S.
i never denied a person at rest ages slowly when seen by a person in motion. i just took one of the 2 cases. may be i should have mentioned Δt' is the time taken for the event in S' according to the S clock.
P: 260
 Quote by Snip3r i never denied a person at rest ages slowly when seen by a person in motion. i just took one of the 2 cases. may be i should have mentioned Δt' is the time taken for the event in S' according to the S clock.
Your equation should have looked like this: $\Delta t' = \Delta t \sqrt{1-v^2/c^2}$.
P: 15,319
 Quote by RiddlerA I mean in reality Only one of them is gonna age slower than the other,.
There is no "in reality"; that would imply a privileged point from which it can be viewed. There are only frames of reference.

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.
P: 11
RiddlerA, the way you defined the times...
 Quote by RiddlerA t = time duration for the desired frame of reference to = time duration for the relative frame of reference for example: if a rocket is moving with some speed, then 't' is the time duration for the rocket which we want to calculate and 'to' is the time duration for the stationary frame of reference from which we are observing the rocket....
... it means your formula is incorrect, like the other guys said. Your t is the proper time (which someone in the rocket measures), while your to is the coordinate time (which someone on Earth measures, the observer, "us").

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
$\Delta t_{o} = \Delta t \sqrt{1-v^2/c^2}$, it's the same as $\Delta t = \frac{\Delta t_{o}}{\sqrt{1-v^2/c^2}}$, which you take from Wikipedia.

$\Delta t = \Delta t_{o} \sqrt{1-v^2/c^2}$, it's the same as $\Delta t_{o} = \frac{\Delta t}{\sqrt{1-v^2/c^2}}$, 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, to 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 $v^2/c^2$ increases as well, which means $1-v^2/c^2$ decreases towards zero (also $\sqrt{1-v^2/c^2}$ 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 $\sqrt{1-v^2/c^2} = 0.8$: when the clock on the observer from Earth (to) 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 (to = 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 = to / 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.
P: 86
 Quote by elfmotat Your equation should have looked like this: $\Delta t' = \Delta t \sqrt{1-v^2/c^2}$.
no you dont get me. Consider this a stationary observer S and a moving observer S' having identical light clocks. Now let S measure the time Δt between 2 consecutive ticks of the clock in S. Now how does he see the same clock to tick in S' (time difference between 2 ticks in S' as seen from S)it is Δt'=$\frac{Δt}{\sqrt{1-V^{2}/C^{2}}}$its also true for S'(as FoR and seeing S) i m just taking one of the case
 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.
P: 260
 Quote by MihaiM Your t is the proper time (which someone in the rocket measures), while your to is the coordinate time (which someone on Earth measures, the observer, "us").
You have them reversed. t0 is proper time and t is coordinate time.

 Quote by MihaiM The correct formula is this, as elfmotat wrote: $\Delta t_{o} = \Delta t \sqrt{1-v^2/c^2}$, it's the same as $\Delta t = \frac{\Delta t_{o}}{\sqrt{1-v^2/c^2}}$, which you take from Wikipedia.
If you use my above definitions, then this is correct.

 Quote by MihaiM So t is the time of the moving clock, to is the time of the static clock (ours).
to should be the moving clock's time.

 Quote by MihaiM 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 $v^2/c^2$ increases as well, which means $1-v^2/c^2$ decreases towards zero (also $\sqrt{1-v^2/c^2}$ decreases).
Correct so far.

 Quote by MihaiM 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").
This is incorrect. The way you had your equations the stationary clock would experience less time. Also, the two observers DO NOT synchronize their clocks - this is impossible.

 Quote by MihaiM Imagine what happens if v is so great that $\sqrt{1-v^2/c^2} = 0.8$: when the clock on the observer from Earth (to) 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 guy in the rocket would not experience time dilation in his rest frame. From his perspective, the guy on Earth's clock is running slower.

 Quote by MihaiM 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 (to = 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 = to / 0.8 = 75 minutes - dilated time).
Neither of the clocks actually show these calculated times. The person in the rocket thinks the person on Earth's time is being dilated. The person in the rocket thinks the exact opposite.

 Quote by Snip3r no you dont get me. Consider this a stationary observer S and a moving observer S' having identical light clocks. Now let S measure the time Δt between 2 consecutive ticks of the clock in S. Now how does he see the same clock to tick in S' (time difference between 2 ticks in S' as seen from S)it is Δt'=$\frac{Δt}{\sqrt{1-V^{2}/C^{2}}}$its also true for S'(as FoR and seeing S) i m just taking one of the case
You still don't have it right. It should be $\Delta t' = \Delta t \sqrt{1-v^2/c^2}.$
Mentor
P: 17,329
 Quote by RiddlerA Is this the twin paradox is all about? It is a hole in the special theory of relativity right?
not even close! It is essentially a famous homework problem for the one of the first few lectures in an introductory SR class. It is a useful problem because it challenges students to think about a few concepts that are difficult and a little unintuitive. It is not a hole in the theory.

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.
P: 11
 Quote by elfmotat You have them reversed. t0 is proper time and t is coordinate time. If you use my above definitions, then this is correct.
Damn! You are right that my equations were wrong based on my notations, however I prefer to stick to mine (even if they don't respect established conventions or yours) in order to accomodate RiddlerA's definition:
 Quote by RiddlerA for example: if a rocket is moving with some speed, then 't' is the time duration for the rocket which we want to calculate and 'to' is the time duration for the stationary frame of reference from which we are observing the rocket....
I corrected the equations in my post.
 Quote by RiddlerA;3698911t[SUB o[/SUB] should be the moving clock's time. ... This is incorrect. The way you had your equations the stationary clock would experience less time.
Now it is ok, based on the notations I established. I insist on keeping his definition in order for RiddlerA to easier understand what's going on.
 Quote by RiddlerA Also, the two observers DO NOT synchronize their clocks - this is impossible. The guy in the rocket would not experience time dilation in his rest frame. From his perspective, the guy on Earth's clock is running slower. Neither of the clocks actually show these calculated times. The person in the rocket thinks the person on Earth's time is being dilated. The person in the rocket thinks the exact opposite.
Let's say that is what the guy in the static reference frame sees it happening and imagines how the guy in the rocket would react, not necessarily that it actually happens, ignoring "contracted time" as well. Is that fine for you?
P: 86
Quote by elfmotat
 Quote by Snip3r no you dont get me. Consider this a stationary observer S and a moving observer S' having identical light clocks. Now let S measure the time Δt between 2 consecutive ticks of the clock in S. Now how does he see the same clock to tick in S' (time difference between 2 ticks in S' as seen from S)it is Δt'=$\frac{Δt}{\sqrt{1-V^{2}/C^{2}}}$its also true for S'(as FoR and seeing S) i m just taking one of the case
You still don't have it right. It should be $\Delta t' = \Delta t \sqrt{1-v^2/c^2}.$
an apparent misunderstanding between us. can you explain your stand?
P: 260
 Quote by Snip3r an apparent misunderstanding between us. can you explain your stand?
Alright:

 Quote by Snip3r Consider this a stationary observer S and a moving observer S' having identical light clocks. Now let S measure the time Δt between 2 consecutive ticks of the clock in S.
No problems so far.

 Quote by Snip3r Now how does he see the same clock to tick in S' (time difference between 2 ticks in S' as seen from S)
When viewed from S the clock at rest in S' should undergo time dilation, correct? So the clock in S' should age less than the one in S. Let's see whether or not this fact is reflected in your equation:

 Quote by Snip3r it is Δt'=$\frac{Δt}{\sqrt{1-V^{2}/C^{2}}}$
This equation says that the clock in S' should age more than the clock in S. Based on what we just established this is clearly wrong. The correct equation would be $\Delta t' = \Delta t \sqrt{1-v^2/c^2}$.
P: 86
 Quote by elfmotat This equation says that the clock in S' should age more than the clock in S. Based on what we just established this is clearly wrong. The correct equation would be $\Delta t' = \Delta t \sqrt{1-v^2/c^2}$.
ok lets restrict ourselves to frame S and dont jump to S' when we discuss.Let S(stationary) and S' (say velocity=0.8c) have identical light clocks and time between consecutive ticks is 1 second(meaning in any inertial frame, the frame ages by 1 second for each tick in a clock placed in that frame).Going by your equation Δt'=0.6 s meaning S' registers 1 second for every 0.6 s in S. Do you think thats correct?

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