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Speed of light becomes Infinity and Zero

by RiddlerA
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Snip3r
#19
Jan8-12, 03:18 AM
P: 86
Quote Quote by ghwellsjr View Post
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 Quote by ghwellsjr View Post
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.
sorry about that!dint know
elfmotat
#20
Jan8-12, 03:21 AM
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Quote Quote by Snip3r View Post
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.
Snip3r
#21
Jan8-12, 03:24 AM
P: 86
Quote Quote by elfmotat View Post
Because from A's rest frame, B is the one who ages slower.
yes... but i mentioned B is my FoR
RiddlerA
#22
Jan8-12, 03:30 AM
P: 58
Quote Quote by ghwellsjr View Post
Here's the equation from your link:
t' = t/√(1-v2/c2)
Well, all the equations have the form of a time on the left side of the equal sign and another form of a time on the right side of the equal sign divided by the square root of something that is less than one which means the time on the left side of the equal sign will be greater than the time on the right side of the equal sign. So the issue is: what do these two times represent?

In your first post, you had t on the left side of the equal sign but in your link, t is on the right side of the equal sign. You don't see this as a difference? This is why I asked you what the references you were looking at were explaining the different forms of t represent.

So now we have a big mess to clean up.

Let's start with the article in your link. This has to be one of the worst explanations of time dilation that I have ever seen. It is flat out wrong. Note the paragraph heading Time expands, space contracts. In order to show that time gets larger, they use this form of the equation were t' on the left side of the equation is larger than the t on the right side of the equation and they even show a plot with t' getting larger with velocity. And to explain it they reference the twin paradox where one twin travels for one year at 99%c. His age increase is represented by t and his earth brother's age increase is calculated by t' as 7 years. What happened to the reciprocal nature of time dilation? Both brothers see the other one as aging less than them self but they don't mention that. What they do mention is:

The truth is that in the space traveler's reference frame, one year is equivalent to 1/7th year on earth and in the earth reference frame, one year is 1/7th year for the traveler, or 7 years is 1 year for the traveler. Do you see the difference? They got it backwards. And yet they seem to realize that it's not quite right because just above the graph they say:


OK, now let's look at the wikipedia article. Their equation is:
Δt' = Δt/√(1-v2/c2)
This is identical to the one in the link you provided except that there are a couple of deltas in front of the t's and their explanation is:

This again is designed to show the dilation or expansion of time.

But now let's look at how Einstein explains time dilation. We'll look at Einstein's 1905 paper that harrylin linked to in post #6:
http://www.fourmilab.ch/etexts/einstein/specrel/www/
Look down near the end of section 4 where you see this equation:
τ = t√(1-v2/c2)
He explains this as t being the time on a clock at rest in a frame and τ (tau) being the time on a clock moving in that frame. It's so simple, but don't overlook the fact that the square root factor is multiplied, not divided, meaning the time dilation results in the moving clock running slower than the stationary clock. That's all there is to it. What ever speed any clock is moving at in a frame determines how much time dilation there is for that clock. If you switch to a frame where the moving clock is now at rest, the time dilation switches to the other clock.

So why is there so much confusion? It's because some people want to have the equation for time dilation be the reciprocal of the one of length contraction because time dilates and length contracts. But what they fail to realize is that it's the interval of time that gets longer resulting in the clock running slower which means it is displaying less time, not more and so we need to use the same form of the equation for both time dilation and length contraction. Look at your first post to see that you have the two equations different. Look at the Lorentz Transform and you will see that the form of the equation for time is identical to the form of the equation for distance with just the t and x interchanged.

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....
elfmotat
#23
Jan8-12, 03:32 AM
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Quote Quote by Snip3r View Post
yes... but i mentioned B is my FoR
Read what you wrote:

Quote Quote by Snip3r View Post
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'=[itex]\frac{Δt}{\sqrt{1-V^{2}/C^{2}}}[/itex]
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.
RiddlerA
#24
Jan8-12, 03:35 AM
P: 58
Quote Quote by elfmotat View Post
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....
Snip3r
#25
Jan8-12, 03:50 AM
P: 86
Quote Quote by elfmotat View Post
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.
elfmotat
#26
Jan8-12, 04:03 AM
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Quote Quote by Snip3r View Post
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: [itex]\Delta t' = \Delta t \sqrt{1-v^2/c^2}[/itex].
DaveC426913
#27
Jan8-12, 09:08 AM
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Quote Quote by RiddlerA View Post
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.
MihaiM
#28
Jan8-12, 09:34 AM
P: 11
RiddlerA, the way you defined the times...
Quote Quote by RiddlerA View Post
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
[itex]\Delta t_{o} = \Delta t \sqrt{1-v^2/c^2}[/itex], it's the same as [itex]\Delta t = \frac{\Delta t_{o}}{\sqrt{1-v^2/c^2}}[/itex], which you take from Wikipedia.

[itex]\Delta t = \Delta t_{o} \sqrt{1-v^2/c^2}[/itex], it's the same as [itex]\Delta t_{o} = \frac{\Delta t}{\sqrt{1-v^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, 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 [itex]v^2/c^2[/itex] increases as well, which means [itex]1-v^2/c^2[/itex] decreases towards zero (also [itex]\sqrt{1-v^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{1-v^2/c^2} = 0.8[/itex]: 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.
Snip3r
#29
Jan8-12, 09:56 AM
P: 86
Quote Quote by elfmotat View Post
Your equation should have looked like this: [itex]\Delta t' = \Delta t \sqrt{1-v^2/c^2}[/itex].
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'=[itex]\frac{Δt}{\sqrt{1-V^{2}/C^{2}}}[/itex]its also true for S'(as FoR and seeing S) i m just taking one of the case
darkhorror
#30
Jan8-12, 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.
elfmotat
#31
Jan8-12, 02:09 PM
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P: 260
Quote Quote by MihaiM View Post
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 Quote by MihaiM View Post
The correct formula is this, as elfmotat wrote:
[itex]\Delta t_{o} = \Delta t \sqrt{1-v^2/c^2}[/itex], it's the same as [itex]\Delta t = \frac{\Delta t_{o}}{\sqrt{1-v^2/c^2}}[/itex], which you take from Wikipedia.
If you use my above definitions, then this is correct.

Quote Quote by MihaiM View Post
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 Quote by MihaiM View Post
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]1-v^2/c^2[/itex] decreases towards zero (also [itex]\sqrt{1-v^2/c^2}[/itex] decreases).
Correct so far.

Quote Quote by MihaiM View Post
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 Quote by MihaiM View Post
Imagine what happens if v is so great that [itex]\sqrt{1-v^2/c^2} = 0.8[/itex]: 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 Quote by MihaiM View Post
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 Quote by Snip3r View Post
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'=[itex]\frac{Δt}{\sqrt{1-V^{2}/C^{2}}}[/itex]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 [itex]\Delta t' = \Delta t \sqrt{1-v^2/c^2}.[/itex]
DaleSpam
#32
Jan8-12, 03:29 PM
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Quote Quote by RiddlerA View Post
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.
MihaiM
#33
Jan8-12, 04:13 PM
P: 11
Quote Quote by elfmotat View Post
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 Quote by RiddlerA View Post
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 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 Quote by RiddlerA View Post
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?
Snip3r
#34
Jan8-12, 11:26 PM
P: 86
Quote Quote by elfmotat View Post
Quote Quote by Snip3r View Post
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'=[itex]\frac{Δt}{\sqrt{1-V^{2}/C^{2}}}[/itex]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 [itex]\Delta t' = \Delta t \sqrt{1-v^2/c^2}.[/itex]
an apparent misunderstanding between us. can you explain your stand?
elfmotat
#35
Jan9-12, 01:18 AM
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Quote Quote by Snip3r View Post
an apparent misunderstanding between us. can you explain your stand?
Alright:

Quote 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 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 Quote by Snip3r
it is Δt'=[itex] \frac{Δt}{\sqrt{1-V^{2}/C^{2}}}[/itex]
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 [itex]\Delta t' = \Delta t \sqrt{1-v^2/c^2}[/itex].
Snip3r
#36
Jan9-12, 02:33 AM
P: 86
Quote Quote by elfmotat View Post
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 [itex]\Delta t' = \Delta t \sqrt{1-v^2/c^2}[/itex].
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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