Speed of light becomes Infinity and Zero

[RiddlerA]

Speed of light becomes Infinity and Zero!

Hi Guys, I am just new to SR & GR. I learned about Lorentz contraction and time dilation, but they don't make sense when i consider the following situation...

Your(Moving non inertial frame) Time runs slow relative to a stationary observer when your speed increases and it will become infinity(as t = t₀/√(1-v²/c²)) if speed v equals 'c' right?
Since photons travel at speed c, their time relative to our world or the whole universe must be infinity.. Since it takes infinite time to move relative to us, we can safely assume that the photon doesn't move at all(relative to our frame).. i.e. speed of light = 0

Now consider Lorentz contraction... L = L₀√(1-v²/c²))
This means that as the object's speed increases, its length decreases(in the direction of motion)... So if the object's speed equals 'c' ,then its lenth would be 0 (in the direction of motion)... But we know there is no absolute frame(since einstein forbade the ether )
So we can say that photon travels at c relative to universe, but if we consider universe moves at c relative to photon, then the length of universe would be 0 and hence photon can traverse the whole universe in no time.. which means Speed of light = Infinity


I know I am wrong but i don't know where i went wrong.. The above anomalies are the conclusions I am arriving at whenever i think of relativity... Please someone clear that for me...

 

[Snip3r]

Your(Moving non inertial frame) Time runs slow relative to a stationary observer when your speed increases and it will become infinity(as t = t₀/√(1-v²/c²)) if speed v equals 'c' right?
Since photons travel at speed c, their time relative to our world or the whole universe must be infinity.. Since it takes infinite time to move relative to us, we can safely assume that the photon doesn't move at all(relative to our frame).. i.e. speed of light = 0

how can time run slow and become infinity?time ceases at light speed. btw your equation is wrong

if we consider universe moves at c relative to photon, then the length of universe would be 0 and hence photon can traverse the whole universe in no time.. which means Speed of light = Infinity

https://www.physicsforums.com/showthread.php?t=511170

 

[DaveC426913]

Since photons travel at speed c, their time relative to our world or the whole universe must be infinity..

Yup.

This is why photons (and anything else moving at c) do not have a valid reference frame.

One definition of a valid reference frame is that it is a frame in which you are at rest. Since, by definition, photons always move at c in any reference frame, then you have a reference frame in which photons are both stationary and moving at c.

So no, photons do not experience time. Photons do not experience anything at all (since experience requires the passage of time). It is no valid to try to think of a photon's frame of reference.

 

[RiddlerA]

how can time run slow and become infinity?time ceases at light speed. btw your equation is wrong

https://www.physicsforums.com/showthread.php?t=511170

Sorry small correction, i was going to say Time runs slow for moving frame and becomes 0 when speed equals 'c'. Now the time becomes infinity for the stationary observer(relative to moving frame).. which means the observer cannot perceive the event at all... And again I am getting that SPeed of light = Infinity

And please tell me the correct equation for time dilation...
thanks in advance...

 

[ghwellsjr]

t = t₀/√(1-v²/c²)

And please tell me the correct equation for time dilation...

Where did you get that equation from? Did it explain what t₀ is or what t is?

 

[harrylin]

[..] Your(Moving non inertial frame) Time runs slow relative to a stationary observer when your speed increases and it will become infinity(as t = t₀/√(1-v²/c²)) if speed v equals 'c' right?
Since photons travel at speed c, their time relative to our world or the whole universe must be infinity..
[...correction added:] Time runs slow for moving frame and becomes 0 when speed equals 'c'. Now the time becomes infinity for the stationary observer(relative to moving frame).. which means the observer cannot perceive the event at all... And again I am getting that Speed of light = Infinity

Now consider Lorentz contraction... L = L₀√(1-v²/c²))
This means that as the object's speed increases, its length decreases(in the direction of motion)... So if the object's speed equals 'c' ,then its lenth would be 0 (in the direction of motion)... But we know there is no absolute frame(since einstein forbade the ether )

He did no such thing.

So we can say that photon travels at c relative to universe, but if we consider universe moves at c relative to photon, then the length of universe would be 0 and hence photon can traverse the whole universe in no time.. which means Speed of light = Infinity

Well that is quite correct, in a certain way: if an astronaut could travel near the speed of light, he/she would be able to the cross the universe in almost no time according to himself/herself.

Einstein formulated it as follows in 1905:

the velocity of light in our theory plays the part, physically, of an infinitely great velocity.

- see section 4 ("Physical Meaning of the Equations ..") of:
http://www.fourmilab.ch/etexts/einstein/specrel/www/

 

[RiddlerA]

Where did you get that equation from? Did it explain what t₀ is or what t is?

t = time duration for the desired frame of reference
t_o = 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 't_o' is the time duration for the stationary frame of reference from which we are observing the rocket...

 

[RiddlerA]

He did no such thing.

What i meant was that einstein eliminated the thought of ether by introducing relativity...

Well that is quite correct, in a certain way: if an astronaut could travel near the speed of light, he/she would be able to the cross the universe in almost no time according to himself/herself.

Wow that gives me an idea.. photon can be converted to particle/anti-particle pair and vice versa right?.. So if we find a way to photonize(matter annihilation) ourselves then we can travel at light speed.. In the destination we again do the reverse process(pair production) to bring our former self
And hence the mystery of Nightcrawler's teleportation is solved. Case Closed...



And btw thanks for the link.. that cleared lots of doubt... thank you very much..

 

[ghwellsjr]

Where did you get that equation from? Did it explain what t₀ is or what t is?

t = time duration for the desired frame of reference
t_o = time duration for the relative frame of reference

I asked you where you got the equation from and you didn't say.
Are those explanations of t and t_o from the same place where you got the equation from or are you adding your interpretation?

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 't_o' is the time duration for the stationary frame of reference from which we are observing the rocket...

Please note that your example does not follow your previous explanations of the two times and in fact is wrong because it will give you more time for the rocket instead of less time. That's why I'm trying to find out if it is your interpretation of the equation that is wrong or if was stated incorrectly in wherever it was you got the equation.

 

[RiddlerA]

I asked you where you got the equation from and you didn't say.
Are those explanations of t and t_o from the same place where you got the equation from or are you adding your interpretation?

I got that equation from wikipedia.. See the Time dilation article in wikipedia...

Please note that your example does not follow your previous explanations of the two times and in fact is wrong because it will give you more time for the rocket instead of less time. That's why I'm trying to find out if it is your interpretation of the equation that is wrong or if was stated incorrectly in wherever it was you got the equation.

Im totally confused... i guess i will have to learn it from the beginning once again..
And thank you very much for your time...

 

[ghwellsjr]

I got that equation from wikipedia.. See the Time dilation article in wikipedia...

Here's the link to the article I found:

http://en.wikipedia.org/wiki/Time_dilation

I don't see your equation in there. If that is the same article you referred to, could you point out the section, paragraph, etc where it is? If not, could you provide the link to the article you found.

 

[alexg]

Here's the link to the article I found:

http://en.wikipedia.org/wiki/Time_dilation

I don't see your equation in there. If that is the same article you referred to, could you point out the section, paragraph, etc where it is? If not, could you provide the link to the article you found.

The equation is in section 2, Simple inference of time dilation due to relative velocity

http://en.wikipedia.org/wiki/Time_d...nce_of_time_dilation_due_to_relative_velocity

 

[ghwellsjr]

The equation is in section 2, Simple inference of time dilation due to relative velocity

http://en.wikipedia.org/wiki/Time_d...nce_of_time_dilation_due_to_relative_velocity

Are you talking about this equation:

It's not the same as t = t₀/√(1-v²/c²) nor do I find these explanations in the text:

t = time duration for the desired frame of reference
t_o = 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 't_o' is the time duration for the stationary frame of reference from which we are observing the rocket...

RiddlerA, please provide the link to where you got the equation and its explanation.

 

[elfmotat]

Are you talking about this equation:

It's not the same as t = t₀/√(1-v²/c²) nor do I find these explanations in the text:

RiddlerA, please provide the link to where you got the equation and its explanation.

My first physics class in high school used Wilson/Buffa, and i remember them using those symbols in their time dilation equation as well. The way they had it, t₀ represented a change in proper time for a "moving" observer (most of the problems involved a scenario where one person left Earth on a spaceship and returned after some time) and t represented a change in time for the stationary observer.

I also recall hating that textbook .

 

[RiddlerA]

Are you talking about this equation:

It's not the same as t = t₀/√(1-v²/c²) nor do I find these explanations in the text:

RiddlerA, please provide the link to where you got the equation and its explanation.

[/PLAIN]
http://www.thebigview.com/spacetime/timedilation.html

How do you say that the equation from wikipedia and what i mentioned earlier are not the same? it looks the same to me...

 

[Snip3r]

[/PLAIN]
http://www.thebigview.com/spacetime/timedilation.html

How do you say that the equation from wikipedia and what i mentioned earlier are not the same? it looks the same to me...

by convention Δt is the measure of change in time and t is of time at an instant. when a moving clock S' reads t' and stationary S reads t then
t'=t\sqrt{1-V^{2}/C^{2}}(of course FoR is S)
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}}}

 

[ghwellsjr]

Are you talking about this equation:

It's not the same as t = t₀/√(1-v²/c²) nor do I find these explanations in the text:

RiddlerA, please provide the link to where you got the equation and its explanation.

[/PLAIN]
http://www.thebigview.com/spacetime/timedilation.html

How do you say that the equation from wikipedia and what i mentioned earlier are not the same? it looks the same to me...

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:

This is due to the fact that time is stretched by factor 7 at approx. 99% of the speed of light, which means that in the space traveller’s reference frame, one year is equivalent to seven years on earth.

The truth is that in the space traveler's reference frame, one year is equivalent to 1/7^th year on Earth and in the Earth reference frame, one year is 1/7^th 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:

Time in the moving system will be observed by a stationary observer to be running slower by the factor t'

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:

which expresses the fact that for the moving observer the period of the clock is longer than in the frame of the clock itself.

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.

 

[ghwellsjr]

by convention Δt is the measure of change in time and t is of time at an instant. when a moving clock S' reads t' and stationary S reads t then
t'=t\sqrt{1-V^{2}/C^{2}}(of course FoR is S)

That's right.

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}}}

But this is not right. You should have left it the same way with no division. So if something takes Δt in S, it will take less time in S' (moving with respect to S).

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.

 

[Snip3r]

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)

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]

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]

Because from A's rest frame, B is the one who ages slower.

yes... but i mentioned B is my FoR

 

[RiddlerA]

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 1/7^th year on Earth and in the Earth reference frame, one year is 1/7^th 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]

yes... but i mentioned B is my FoR

Read what you wrote:

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.

 

[RiddlerA]

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 going to 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 doesn't make any sense...

 

[Snip3r]

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]

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}.

 

[DaveC426913]

I mean in reality Only one of them is going to 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]

RiddlerA, the way you defined the times...

t = time duration for the desired frame of reference
t_o = 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 't_o' 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 t_o 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, 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 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 (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.

 

[Snip3r]

Your equation should have looked like this: \Delta t' = \Delta t \sqrt{1-v^2/c^2}.

no you don't 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

 

[darkhorror]

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]

Your t is the proper time (which someone in the rocket measures), while your t_o 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.

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.

So t is the time of the moving clock, t_o is the time of the static clock (ours).

t_o should be the moving clock's time.

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.

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.

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 (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 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.

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).

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.

no you don't 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}.

 

[Dale]

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]

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 accommodate RiddlerA's definition:

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 't_o' is the time duration for the stationary frame of reference from which we are observing the rocket...

I corrected the equations in my post.

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.

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]

no you don't 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?

 

[elfmotat]

an apparent misunderstanding between us. can you explain your stand?

Alright:

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.

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:

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}.

 

[Snip3r]

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 let's restrict ourselves to frame S and don't 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 that's correct?

 

[elfmotat]

ok let's restrict ourselves to frame S and don't jump to S' when we discuss.

That's what we've been doing the whole time.

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 that's correct?

If Δt'=0.6 s then the clock in S' registers 0.6 seconds for every one second in S. That's what Δt'=0.6 s means; 0.6 seconds has passed in S' as viewed from S!

 

[Snip3r]

If Δt'=0.6 s then the clock in S' registers 0.6 seconds for every one second in S. That's what Δt'=0.6 s means; 0.6 seconds has passed in S' as viewed from S!

Δt' is NOT the time measured in S'. It is measured in S as the time taken between consecutive ticks for a clock placed in S'.

 

[elfmotat]

Δt' is NOT the time measured in S'. It is measured in S as the time taken between consecutive ticks for a clock placed in S'.

Exactly.

 

[harrylin]

Combining and rearranging several posts:

Wow that gives me an idea.. photon can be converted to particle/anti-particle pair and vice versa right?.. So if we find a way to photonize(matter annihilation) ourselves then we can travel at light speed.. In the destination we again do the reverse process(pair production) to bring our former self
And hence the mystery of Nightcrawler's teleportation is solved. Case Closed...

I don't know the details of particle/anti-particle creation and certainly not about teleportation; but if a Start Trek kind of teleportation would be technologically feasible then according to current theory, then yes it should be possible to travel like that to another part of the universe.

What i meant was that einstein eliminated the thought of ether by introducing relativity...

He thought so, but later he thought differently. The essential point which he showed is that if we accept certain premises based on observation (the postulates), then the Lorentz transformations follow of necessity, without explicitly introducing a physical model of what happens "in reality".

And btw thanks for the link.. that cleared lots of doubt... thank you very much..

You're welcome .

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 going to 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 doesn't make any sense...

Mutual time dilation is not really what the twin paradox is about as it goes one step further (and it is also not a hole in special relativity). However, it does rise a philosophical discussion point which also often comes up in discussions about the twin paradox - see below.

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.

That remark about asymmetry is perfectly correct; however the opinion that "reality" should correspond to "a privileged point" of observation is (positivistic?) philosophy; philosophy about reality is definitely not part of relativity. Relativity theory describes observations of physical phenomena.

 

[DaveC426913]

...the opinion that "reality" should correspond to "a privileged point" of observation is (positivistic?) philosophy;

No, what's philosophy is RiddlerA's assertion that there is any kind of "in reality" in the first place - there is no way to experience it except through a frame of reference chosen arbitrarily.

How does he propose to go about demonstrating or measuring a "reality" that is not subject to a frame of reference?

 

[harrylin]

No, what's philosophy is RiddlerA's assertion that there is any kind of "in reality" in the first place - there is no way to experience it except through a frame of reference chosen arbitrarily.

How does he propose to go about demonstrating or measuring a "reality" that is not subject to a frame of reference?

That is exactly what I meant with "positivistic" philosophy, as opposed to Riddler's "realistic" philosophy. Indeed, special relativity does not relate to "in reality", it only relates to measurable phenomena.

Coincidentally (and as discussed in other threads), Langevin* provided the space traveler example (later "twin paradox") as a demonstration of the existence of such a reality; but it depends on the way one's brain is wired if one agrees or not.

But let's not deviate too much from the topic; the issue is merely if mutual time dilation (etc.) is clear for the OP.

*Langevin was one of the early teachers of special relativity and was also involved in its early conception

 

[RiddlerA]

Sorry I mentioned the equations by accident so don't keep bothering about the equations.. My question arised from the concept, not the equation...

DaveC gave a lot of points that i didnt know before such as that photon don't experience time and they see all the events of universe simultaneously etc etc...
Actually that makes sense as well as makes me think that relativity fails to explain the proper working of the universe..
Please DOnt ask why I am saying like this, its just a feeling that i have...

 

[ghwellsjr]

Sorry I mentioned the equations by accident so don't keep bothering about the equations.. My question arised from the concept, not the equation...

None of the equations you quoted or supplied the links for nor any of the others that have been discussed in this thread were like Einstein's and his simple explanation:

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.

Do you understand this? Doesn't it seem simple to you?

DaveC gave a lot of points that i didnt know before such as that photon don't experience time and they see all the events of universe simultaneously etc etc...

Dave did say that photons don't experience time but he didn't say they see all events simultaneously:

...photons do not experience time. Photons do not experience anything at all (since experience requires the passage of time)...

Actually that makes sense as well as makes me think that relativity fails to explain the proper working of the universe..
Please DOnt ask why I am saying like this, its just a feeling that i have...

I sure hope you aren't giving up on understanding Special Relativity. (And it doesn't explain everything in the universe because it ignores the effects of gravity which is not simple.) Can you please just respond to my question about Einstein's equation and his explanation? Do you understand it? Don't you think it is simple? If you don't think so, please tell me why because I like to explain the simple concepts of Special Relativity to people and if what I think is simple does not seem simple to others, then I need to know why. Thanks for responding.

 

[RiddlerA]

None of the equations you quoted or supplied the links for nor any of the others that have been discussed in this thread were like Einstein's and his simple explanation:

Do you understand this? Doesn't it seem simple to you?

τ = t√(1-v2/c2)

Let me tell u what i understand from this equation..
τ = denotes the time as read in a clock which is moving in a frame(say A)..
t = denotes the time as read in a clock which is at REST in the same frame A..
v = velocity of the clock which is in motion relative to the other which is at rest
c = speed of light in vacuum

Time Dilation:

Now let's consider you and me, where I am moving at speed v and you are at rest.. we both are in the same frame(say on a staircase to heaven ) I am rushing to enter the heaven at speed v while you wait patiently on the staircase... Let's take our heartbeats to be the clocks from which we measure our time... so my heartbeat is read as 'τ' and yours as 't'...
From the equation above, we can come to the conclusion that my heart beats slower than yours(Biologically: Assume everyone's heart beats at the same rate for all physical conditions (meaning fear don't increase heartbeat rate and so on))
For convenience, I am assuming my speed v = √(3/4)th of c
So the eqn becomes τ = t(1/2)

SO the above eqn implies that For every one of my heartbeats, your heart beats twice..

Mass Variation:

And also My mass would increase by the factor 1/√(1-v2/c2)

Length Contraction:

Now consider the same story as mentioned above..

A small change in the story for this concept
τ = time INTERVAL of my clock, t = time INTERVAL of your clock
and ofc τ= half of t

For you, my speed is √(3/4)th of c ...
so from your point of view, i would seem like traveling (√(3/4)th of c)*t metres in 1 second
If space remains absolute regardless of speed
Then from my own point of view, i would travel (√(3/4)th of c)*t metres in half a second(remember 1 second in my frame equals 2 seconds in your frame)
But the above condition cannot occur, it doesn't make sense because my speed is not the same in both our frames...
So we are introducing Length Conraction..
Now since I am moving, the staircase will be contracted by a factor of √(1-v2/c2) in my own point of view...
So as a result i would travel a contracted distance in 1 second in my own frame which would appear as me moving (√(3/4)th of c)*t metres in 1 second in your frame...
And so my speed remains constant in both the frames...
Problem solved...




Phew... Are the above stories correct?

I sure hope you aren't giving up on understanding Special Relativity. (And it doesn't explain everything in the universe because it ignores the effects of gravity which is not simple.) Can you please just respond to my question about Einstein's equation and his explanation? Do you understand it? Don't you think it is simple? If you don't think so, please tell me why because I like to explain the simple concepts of Special Relativity to people and if what I think is simple does not seem simple to others, then I need to know why. Thanks for responding.

Curiosity is one of the most dangerous incurable disease that happens to humans...
So my friend, i will never stop wondering why the universe works the way it works...
-RiddlerA

 

[ghwellsjr]

τ = t√(1-v2/c2)

Let me tell u what i understand from this equation..
τ = denotes the time as read in a clock which is moving in a frame(say A)..
t = denotes the time as read in a clock which is at REST in the same frame A..
v = velocity of the clock which is in motion relative to the other which is at rest
c = speed of light in vacuum

Time Dilation:

Now let's consider you and me, where I am moving at speed v and you are at rest.. we both are in the same frame(say on a staircase to heaven ) I am rushing to enter the heaven at speed v while you wait patiently on the staircase... Let's take our heartbeats to be the clocks from which we measure our time... so my heartbeat is read as 'τ' and yours as 't'...
From the equation above, we can come to the conclusion that my heart beats slower than yours(Biologically: Assume everyone's heart beats at the same rate for all physical conditions (meaning fear don't increase heartbeat rate and so on))
For convenience, I am assuming my speed v = √(3/4)th of c
So the eqn becomes τ = t(1/2)

SO the above eqn implies that For every one of my heartbeats, your heart beats twice..

Excellent. In fact, perfect. I especially like your comment that both of us are in the same frame and you consistently point out that only you are moving and only I am at rest. A lot of people never seem to learn this concept that everyone and everything is in any frame you choose.

Mass Variation:

And also My mass would increase by the factor 1/√(1-v2/c2)

This is what I was taught in school many decades ago but I'm told that this idea is no longer in vogue. If you want to know more, do a search in this forum for "mass" and specify "Search Titles Only".

Length Contraction:

Now consider the same story as mentioned above..

A small change in the story for this concept
τ = time INTERVAL of my clock, t = time INTERVAL of your clock
and ofc τ= half of t

For you, my speed is √(3/4)th of c ...
so from your point of view, i would seem like traveling (√(3/4)th of c)*t metres in 1 second
If space remains absolute regardless of speed
Then from my own point of view, i would travel (√(3/4)th of c)*t metres in half a second(remember 1 second in my frame equals 2 seconds in your frame)
But the above condition cannot occur, it doesn't make sense because my speed is not the same in both our frames...
So we are introducing Length Conraction..
Now since I am moving, the staircase will be contracted by a factor of √(1-v2/c2) in my own point of view...
So as a result i would travel a contracted distance in 1 second in my own frame which would appear as me moving (√(3/4)th of c)*t metres in 1 second in your frame...
And so my speed remains constant in both the frames...
Problem solved...

Phew... Are the above stories correct?

This last part is correct as far as your calculations go but your explanations are not quite right because you are mixing two frames together. That will always introduce confusion. Always make it clear which frame you are discussing.

And you started using terms like "point of view" and "seem like" and "appear". When we are using Einstein's method, we need to emphasize that the times on clocks are what the equations say they are, it's not a mere optical effect or opinion or illusion. In fact, you and I may not even be able to observe the other ones clock as being time dilated like the equation says or we may not be able to detect length contraction but that is what is really happening according to the definition of the chosen frame of reference.

Another point: when you talk about your point of view, and by that, I understand you to mean you are now switching to your frame of reference in which you are at rest, you can't then say that you are moving. Rather you should say that the staircase is moving a contracted distance and you would say that you are moving a certain speed in my rest frame and I am moving the same speed in your rest frame. I think you really mean this and understand this, but I just wanted to clarify the terminology.

Anyway, you've done a really great job and I'm proud of you for seeing this through.