Deriving Equations for Light Sphere in Collinear Motion - O and O' Observers

[cfrogue]

But you apparently don't understand the most basic concept of the relativity of simultaneity, namely that if clocks are synched in one frame they are out-of-sync in another frame in motion relative to the first. So if the rod clock is synched so that it reads 0 at the moment the light was emitted according to the definition of simultaneity in the rod's rest frame, that means it will read some time other than 0 at the moment the light was emitted according to the definition of simultaneity in frame O (specifically, it will read rv/c^2 at the moment the light was emitted).

Yea I have this part figured out.

But, time dilation may be a problem for this concept.

Did we figure out the time dilation yet?

 

[cfrogue]

Yes, it is a fact, and we are not ignoring it. The only difference is I'm saying it's not a problem, whereas you are saying it is?

show me the math of one light sphere performing these gymnastics.

 

[Dale]

so is
t' = t*λ

No, you had the correct expression way back in post 5:
t' = ( t - vx/c^2 )γ

Are we moving off of the center topic and onto some different topic now?

 

[cfrogue]

No, you had the correct expression way back in post 5:
t' = ( t - vx/c^2 )γ

Are we moving off of the center topic and onto some different topic now?

Wiki has the following

t' = tλ
http://en.wikipedia.org/wiki/Time_dilation

Is this wrong?

Yes, I am going to bring it all together.

I think some have noticed what I am doing.

 

[atyy]

show me the math of one light sphere performing these gymnastics.

Didn't you say there are multiple centres?

 

[cfrogue]

It seems folks want to abandon time dilation. I refuse to.

I have a light timer on the left end of a rod of length r and a light source in the center of the rod. A light timer records the time when light strikes it. This will be the moving frame at v.

Now, when the center of the rod is at the origin of O, all clocks are synchronized to 0 in each frame.

Now, light will strike the left point of the rod at t = r/(λ(c+v)) in O.

What will the light timer in O' read, t' = t*λ, for time dilation?

Reference

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

 

[cfrogue]

Didn't you say there are multiple centres?

yes...

 

[JesseM]

Yea I have this part figured out.

But, time dilation may be a problem for this concept.

Did we figure out the time dilation yet?

Yes, time dilation says the time interval elapsed on the clock at the end of the rod should be t/gamma in frame O, and if you look at the math I presented, it is. But the time reading when the light hits it is not t/gamma, because in frame O it didn't start out reading zero at the moment the flash was set off. What is your issue here?

 

[cfrogue]

Yes, time dilation says the time interval elapsed on the clock at the end of the rod should be t/gamma in frame O, and if you look at the math I presented, it is. But the time reading when the light hits it is not t/gamma, because in frame O it didn't start out reading zero at the moment the flash was set off. What is your issue here?

Yes, I did the normal sync of SR.

If you refuse my thing then you must refuse Einstein's.

I thought I showed you this.

 

[cfrogue]

Yes, time dilation says the time interval elapsed on the clock at the end of the rod should be t/gamma in frame O, and if you look at the math I presented, it is. But the time reading when the light hits it is not t/gamma, because in frame O it didn't start out reading zero at the moment the flash was set off. What is your issue here?

I did show you this.

We now imagine space to be measured from the stationary system K by means of the stationary measuring-rod, and also from the moving system k by means of the measuring-rod moving with it; and that we thus obtain the co-ordinates x, y, z, and , , respectively. Further, let the time t of the stationary system be determined for all points thereof at which there are clocks by means of light signals in the manner indicated in § 1; similarly let the time of the moving system be determined for all points of the moving system at which there are clocks at rest relatively to that system by applying the method, given in § 1, of light signals between the points at which the latter clocks are located.
http://www.fourmilab.ch/etexts/einstein/specrel/www/

 

[atyy]

yes...

So your problem is you believe there are multiple centres, but one light sphere?

 

[cfrogue]

So your problem is you believe there are multiple centres, but one light sphere?

Yes...

 

[Greg Bernhardt]

Reminder to keep things friendly

 

[JesseM]

Yes, I did the normal sync of SR.

The "normal sync of SR" is exactly why the rod clock doesn't read zero in frame O at the moment the light flash is set off, because it was synched to read zero when the light flash was set off according to the rod frame's definition of simultaneity.

Do you understand that according to the relativity of simultaneity, clocks which are synchronized in one frame are out-of-sync in another? Answer yes/no

If you refuse my thing then you must refuse Einstein's.

Einstein talked a lot about the relativity of simultaneity, see here for example. I can't help it if you lack the intellectual humility to consider that there might be giant gaps in your understanding, and therefore ignorantly tell people that they are "refusing Einstein" or "abandoning time dilation" when they try to explain the basics of how SR works.

 

[Jorrie]

I have a light timer on the left end of a rod of length r and a light source in the center of the rod. A light timer records the time when light strikes it. This will be the moving frame at v.

Now, when the center of the rod is at the origin of O, all clocks are synchronized to 0 in each frame.

There are just two events (which you specified). In both frames the emission event is at coordinates (0,0).

In O', the reception event coordinates are given by you (geometric units, c=1)

x'=-r

t' = r

In O, the reception event coordinates can be found by using the LTs.

<br /> x = \gamma(x&#039;+vt&#039;) = \gamma(-r+vr)<br />

<br /> t = \gamma(t&#039; +vx&#039;) = \gamma(r-vr) <br />

If you use a different method, numerically, your answers must be the same as given by the LTs.

It is much easier using a Minkowski diagram! ;)

 

[Dale]

Wiki has the following

t' = tλ
http://en.wikipedia.org/wiki/Time_dilation

Nobody is "abandoning" time dilation, you are just accidentally misapplying it. That is not exactly what Wikipedia has. Wikipedia has
Δt' = γ Δt
and Wikipedia also points out that the Δt is the time between two co-local events. This is an important distinction. Time dilation is part of the Lorentz transform, but not the whole thing. Let me show you how the time dilation formula is derived from the Lorentz transform:

Given the Lorentz transform of some arbitrary initial and final event
t&#039;_i = \gamma (t_i - v x_i/c^2)
t&#039;_f = \gamma (t_i - v x_f/c^2)
and given
\Delta t&#039; = t&#039;_f - t&#039;_i
\Delta t = t_f - t_i
\Delta x = x_f - x_i

We obtain
\Delta t&#039; = \gamma (\Delta t - v \Delta x /c^2)

This reduces to the time dilation formula only in the special case where \Delta x = 0 (the events are co-local in the unprimed frame), which is not the case here.

By the way, you can clearly see time dilation in the https://www.physicsforums.com/showpost.php?p=2464800&postcount=88". Note, as you follow the line x'=0 away from the origin you cross the line t=1 before you cross the line t'=1, so the primed clock is slow in the unprimed frame. Similarly, as you follow the line x=0 away from the origin you cross the line t'=1 before you cross the line t=1, so the unprimed clock is slow in the primed frame. Not only does the diagram correctly include time dilation, it let's you see how it is reciprocal for each frame and it let's you understand graphically that the time dilation formula only applies as written when \Delta x = 0.

 

[Dale]

Now, when the center of the rod is at the origin of O, all clocks are synchronized to 0 in each frame.

I covered this in post 233 already. The relativity of simultaneity prevents this.

 

[cfrogue]

The "normal sync of SR" is exactly why the rod clock doesn't read zero in frame O at the moment the light flash is set off, because it was synched to read zero when the light flash was set off according to the rod frame's definition of simultaneity.

Do you understand that according to the relativity of simultaneity, clocks which are synchronized in one frame are out-of-sync in another? Answer yes/no

Einstein talked a lot about the relativity of simultaneity, see here for example. I can't help it if you lack the intellectual humility to consider that there might be giant gaps in your understanding, and therefore ignorantly tell people that they are "refusing Einstein" or "abandoning time dilation" when they try to explain the basics of how SR works.

Do you understand that according to the relativity of simultaneity, clocks which are synchronized in one frame are out-of-sync in another? Answer yes/no

Yes.

But, that has nothing to do with what I was talking about. I did not claim to synchronize between frames.

Let's say we have a proper time interval in a stationary frame of t for an object to move from x1 to x2.

What is the elapsed proper time in the time of the moving object?

 

[cfrogue]

Nobody is "abandoning" time dilation, you are just accidentally misapplying it. That is not exactly what Wikipedia has. Wikipedia has
Δt' = γ Δt
and Wikipedia also points out that the Δt is the time between two co-local events. This is an important distinction. Time dilation is part of the Lorentz transform, but not the whole thing. Let me show you how the time dilation formula is derived from the Lorentz transform:

Given the Lorentz transform of some arbitrary initial and final event
t&#039;_i = \gamma (t_i - v x_i/c^2)
t&#039;_f = \gamma (t_i - v x_f/c^2)
and given
\Delta t&#039; = t&#039;_f - t&#039;_i
\Delta t = t_f - t_i
\Delta x = x_f - x_i

We obtain
\Delta t&#039; = \gamma (\Delta t - v \Delta x /c^2)

This reduces to the time dilation formula only in the special case where \Delta x = 0 (the events are co-local in the unprimed frame), which is not the case here.

By the way, you can clearly see time dilation in the https://www.physicsforums.com/showpost.php?p=2464800&postcount=88". Note, as you follow the line x'=0 away from the origin you cross the line t=1 before you cross the line t'=1, so the primed clock is slow in the unprimed frame. Similarly, as you follow the line x=0 away from the origin you cross the line t'=1 before you cross the line t=1, so the unprimed clock is slow in the primed frame. Not only does the diagram correctly include time dilation, it let's you see how it is reciprocal for each frame and it let's you understand graphically that the time dilation formula only applies as written when \Delta x = 0.

Nice lex.

What is the formula for time dilation for a frame moving v.

you had Δt' = λΔt.

Therefore, when O elapses Δt, then O' elapses λΔt.

Is this correct?

 

[JesseM]

Do you understand that according to the relativity of simultaneity, clocks which are synchronized in one frame are out-of-sync in another? Answer yes/no

Yes.

But, that has nothing to do with what I was talking about. I did not claim to synchronize between frames.

Let's say we have a proper time interval in a stationary frame of t for an object to move from x1 to x2.

What is the elapsed proper time in the time of the moving object?

"Proper time interval in a stationary frame" does not make sense as a phrase, "proper time" always refers to the time between events on a specific worldline as measured by a clock moving along that worldline, so it's a frame-invariant quantity. Do you mean there is a coordinate time of t in the stationary frame for an object to move from x1 to x2? In this case, in the object's own rest frame, the coordinate time between the object passing markers at these two points would be t/gamma (since the object is at rest in this frame, this would also be the unique proper time along the object's worldline between these two passing-events). I don't really see what this has to do with the example of the clock at the end of the rod though, since we weren't talking about the time for that clock to move between two positions, we were talking about the time that would be displayed on that clock when the light from the flash reached it. Do you agree that if the clock was synchronized so that it read a time of zero simultaneously with the flash in its own rest frame, then it would read a time of r/c when the light reached it?

 

[cfrogue]

There are just two events (which you specified). In both frames the emission event is at coordinates (0,0).

In O', the reception event coordinates are given by you (geometric units, c=1)

x&#039;=-r

t&#039; = r

In O, the reception event coordinates can be found by using the LTs.

<br /> x = \gamma(x&#039;+vt&#039;) = \gamma(-r+vr)<br />

<br /> t = \gamma(t&#039; +vx&#039;) = \gamma(r-vr) <br />

If you use a different method, numerically, your answers must be the same as given by the LTs.

It is much easier using a Minkowski diagram! ;)

Could you calculate the time dialation?

Is there not time dilation for a rest frame and a moving frame?

So, given an elapsed time in O, what is the elapsed time in O' with relative motion v?

 

[cfrogue]

"Proper time interval in a stationary frame" does not make sense as a phrase, "proper time" always refers to the time between events on a specific worldline as measured by a clock moving along that worldline, so it's a frame-invariant quantity. Do you mean there is a coordinate time of t in the stationary frame for an object to move from x1 to x2? In this case, in the object's own rest frame, the coordinate time between the object passing markers at these two points would be t/gamma (since the object is at rest in this frame, this would also be the unique proper time along the object's worldline between these two passing-events). I don't really see what this has to do with the example of the clock at the end of the rod though, since we weren't talking about the time for that clock to move between two positions, we were talking about the time that would be displayed on that clock when the light from the flash reached it. Do you agree that if the clock was synchronized so that it read a time of zero simultaneously with the flash in its own rest frame, then it would read a time of r/c when the light reached it?

I think I need to take this slowly.

Time dilation if O is synched at 0 is t' = λt for the moving frame O'.

Is this correct?

 

[JesseM]

I think I need to take this slowly.

Time dilation if O is synched at 0 is t' = λt for the moving frame O'.

Is this correct?

The question is too vague. What do t' and t refer to? If they're time-intervals, you need to pick specific events you're measuring the time intervals between in each frame. For example, if we have Event 1A = the light flash, and Event 2 = the light reaching the clock, then no, your equation is not correct, in this case t would be r/(gamma*(c+v)) while t' would be r/c. On the other hand, if we have Event 1B = the event on the clock's worldline that is simultaneous with the light flash in frame O, and the same choice of Event 2, then t' = t/gamma.

 

[cfrogue]

The question is too vague. What do t' and t refer to? If they're time-intervals, you need to pick specific events you're measuring the time intervals between in each frame. For example, if we have Event 1A = the light flash, and Event 2 = the light reaching the clock, then no, your equation is not correct, in this case t would be r/(gamma*(c+v)) while t' would be r/c. On the other hand, if we have Event 1B = the event on the clock's worldline that is simultaneous with the light flash in frame O, and the same choice of Event 2, then t' = t/gamma.

OK, I am simply talking about a general Δt in O.

What is the elapsed time in O'?

In the special theory of relativity, a moving clock is found to be ticking slowly with respect to the observer's clock.
http://en.wikipedia.org/wiki/Time_dilation

Can you confirm or deny this logic?

Further, do you have the equation for this time dilation?

 

[JesseM]

OK, I am simply talking about a general Δt in O.

Sorry but this is meaningless. There is no "general Δt", in SR it only makes time to talk about a time interval between a specific pair of events.

In the special theory of relativity, a moving clock is found to be ticking slowly with respect to the observer's clock.
http://en.wikipedia.org/wiki/Time_dilation

Can you confirm or deny this logic?

Further, do you have the equation for this time dilation?

Yes, of course I confirm it, but you have to understand what types of events it applies to. The time dilation equation says that if you have two events 1 and 2 which happen at the same position in frame A (like two readings on a clock that is at rest in that frame), and the time between these events in frame A is t_A, then if you look at the time between the same pair of events in a different frame B moving at speed v relative to A, and if the time interval between them in B is t_B, the two times will be related by t_B = t_A*gamma.

So if you want to apply the time dilation equation, you have to pick a specific pair of events which are colocated (happen at the same position coordinate) in one of the two frames, otherwise you're misusing the equation. For example, I mentioned the option of picking the following two events:

Event 1B = the event on the clock's worldline that is simultaneous with the light flash in frame O
Event 2 = the light reaching the clock

Since both these events happen along the clock's worldline, obviously they are colocated in frame O' where the clock is at rest. So, plugging in t' for t_A in my above notation, we have t = gamma*t', or equivalently t' = t/gamma, which is exactly what I said would be true for this pair of events at the end of my last post.

 

[cfrogue]

Sorry but this is meaningless. There is no "general Δt", it only makes time to talk about a time interval between a specific pair of events.

Yes, of course I confirm it, but you have to understand what types of events it applies to. The time dilation equation says that if you have two events 1 and 2 which happen at the same position in frame A (like two readings on a clock that is at rest in that frame), and the time between these events in frame A is t_A, then if you look at the time between the same pair of events in a different frame B moving at speed v relative to A, and if the time interval between them in B is t_B, the two times will be related by t_B = t_A*gamma.

So if you want to apply the time dilation equation, you have to pick a specific pair of events which are colocated (happen at the same position coordinate) in one of the two frames, otherwise you're misusing the equation. For example, I mentioned the option of picking the following two events:

Event 1B = the event on the clock's worldline that is simultaneous with the light flash in frame O
Event 2 = the light reaching the clock

Since both these events happen along the clock's worldline, obviously they are colocated in frame O' where the clock is at rest. So, plugging in t' for t_A in my above notation, we have t = gamma*t', or equivalently t' = t/gamma, which is exactly what I said would be true for this pair of events at the end of my last post.

Well, since I said Δt, that means I have selected two timing events in O. That means I have a general start point and a general endpoint. How I pick these may be important later.

But, for the general equation they are not.

So, according to wiki we have,

t' = t*λ

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



You said t = gamma*t', why is that?

 

[Dale]

you had Δt' = γ Δt.

Therefore, when O elapses Δt, then O' elapses γ Δt.

Is this correct?

If Δx=0 then yes.

 

[cfrogue]

If ?x=0 then yes.

what do you mean?

what is the time dilation equation for what I gave?

 

[DrGreg]

what is the time dilation equation for what I gave?

The general equation is given by the Lorentz transform

\Delta t&#039; = \gamma ( \Delta t - \frac {v}{c^2} \Delta x )​

The time dilation formula

\Delta t&#039; = \gamma \Delta t​

applies only in the special case when \Delta x = 0, which is what the Wikipedia article says: "\Delta t is the time interval between two co-local events (i.e. happening at the same place)"

 

[cfrogue]

The general equation is given by the Lorentz transform

\Delta t&#039; = \gamma ( \Delta t - \frac {v}{c^2} \Delta x )​

The time dilation formula

\Delta t&#039; = \gamma \Delta t​

applies only in the special case when \Delta x = 0, which is what the Wikipedia article says: "\Delta t is the time interval between two co-local events (i.e. happening at the same place)"

Do you mean we can no longer apply time dilation between frames?

 

[cfrogue]

The general equation is given by the Lorentz transform

\Delta t&#039; = \gamma ( \Delta t - \frac {v}{c^2} \Delta x )​

The time dilation formula

\Delta t&#039; = \gamma \Delta t​

applies only in the special case when \Delta x = 0, which is what the Wikipedia article says: "\Delta t is the time interval between two co-local events (i.e. happening at the same place)"

what if I said, co-local events means events in the same frame.

 

[Jorrie]

I have a light timer on the left end of a rod of length r and a light source in the center of the rod. A light timer records the time when light strikes it. This will be the moving frame at v.

Could you calculate the time dilation?

The equations for your original scenario have been given a few times already, but if you want numerics, let's make the length of your rod r = 2 light seconds, so that x' = -1 light second and hence the time t' = 1 second. Let the relative speed be v = -0.6c (O relative to O'), giving gamma = 1.25.

<br /> <br /> x = \gamma(x&#039; - vt&#039;) = 1.25 (-1 + 0.6) = -0.5<br /> <br />

<br /> <br /> t = \gamma(t&#039; - vx&#039;) = 1.25 (1 - 0.6) = 0.5<br /> <br />

As JesseM told you before, when there are both time and space increments in one frame, you cannot simply work out 'time dilation' by applying gamma to the time coordinate alone. You have to use the LTs, where the space increment also influences the coordinate time of the other frame.

As a valuable exercise, I urge you to sketch your scenario on a Minkowski spacetime diagram. You will probably immediately understand the above result! ;-)

So, given an elapsed time in O, what is the elapsed time in O' with relative motion v?

This is just the other way around, but the answer is also known from the above calculations.

 

[cfrogue]

The equations for your original scenario has been given a few times already, but if you want numerics, let's make the length of your rod r = 2 light seconds, so that x' = -1 light second and hence the time t' = 1 second. Let the relative speed be v = -0.6c (O relative to O'), giving gamma = 1.25.

<br /> <br /> x = \gamma(x&#039; - vt&#039;) = 1.25 (-1 + 0.6) = -0.5<br /> <br />

<br /> <br /> t = \gamma(t&#039; - vx&#039;) = 1.25 (1 - 0.6) = 0.5<br /> <br />

As JesseM told you before, when there are both time and space increments in one frame, you cannot simply work out 'time dilation' by applying gamma to the time coordinate alone. You have to use the LTs, where the space increment also influences the coordinate time of the other frame.

As a valuable exercise, I urge you to sketch your scenario on a Minkowski spacetime diagram. You will probably immediately understand the above result! ;-)



This is just the other way around, but the answer is also known from the above calculations.

Can you please tell me the restrictions to the time dilation equation?

I did not know I am not allowed to apply it given a frame and given an elapsed time in that frame vs a frame moving with collinear relative motion.

Can you explain this?

 

[JesseM]

Well, since I said Δt, that means I have selected two timing events in O. That means I have a general start point and a general endpoint. How I pick these may be important later.

But you need a specific pair of events in order to decide the time between them in another frame. Suppose we have an event 1 that occurs at t1 in O, and another event 2 that occurs at t2 in O. Now suppose we pick a different pair of events, event 3 that also occurs at t1 in O, and event 4 that also occurs at time t2 in O. Obviously in frame O, the time interval between 1 and 2 is the same as the time interval between 3 and 4, in both cases it would be Δt=(t2 - t1). Now, do you understand that in frame O', the time between 1 and 2 may be different than the time between 3 and 4? Again, please answer yes or no.

But, for the general equation they are not.

But that's where you're wrong, the general equation can only be used when you have picked a pair of events that are colocated in one of the frames.

So, according to wiki we have,

t' = t*λ

http://en.wikipedia.org/wiki/Time_dilation
You said t = gamma*t', why is that?

The equation can be written in different ways depending on which is the frame where the two events are co-located. Note that they define Δt as:

the time interval between two co-local events (i.e. happening at the same place) for an observer in some inertial frame

However, in my notation t' was the time between two events which were co-located in frame O':

Since both these events happen along the clock's worldline, obviously they are colocated in frame O' where the clock is at rest.

So you can see that there is no inconsistency, we both agree that if t_colocated is the time between two events in the frame where they are colocated, and t_noncolocated is the time between the same events in the frame where they are not colocated, then the equation has the form:

t_noncolocated = gamma*t_colocated

 

[Jorrie]

Can you please tell me the restrictions to the time dilation equation?

I can do no better than what JesseM and DrGreg has already done. :)

In the end, my final advice is: learn the basics and learn it well. And, do learn Minkowski Spacetime diagrams...

 

[cfrogue]

But you need a specific pair of events in order to decide the time between them in another frame. Suppose we have an event 1 that occurs at t1 in O, and another event 2 that occurs at t2 in O. Now suppose we pick a different pair of events, event 3 that also occurs at t1 in O, and event 4 that also occurs at time t2 in O. Obviously in frame O, the time interval between 1 and 2 is the same as the time interval between 3 and 4, in both cases it would be Δt=(t2 - t1). Now, do you understand that in frame O', the time between 1 and 2 may be different than the time between 3 and 4? Again, please answer yes or no.

I already said yes.

But, I have two events exclusively in one frame.

So, why can't I apply time dilation?

But that's where you're wrong, the general equation can only be used when you have picked a pair of events that are colocated in one of the frames.

Do you mean the time dilation equations depends on a location?

So, if I have a time interval in O, there must be a location of the same position for two events?

The events, MUST occur at the same location in frame O or time dilation is not applicable.

Is this true?

Therefore, we must have only events at the same location in a frame or there is no time dilation period.

Please confirm or deny this.

Otherwise, you are simply talking about two events in the same frame in which case my example holds.

 

[cfrogue]

I can do no better than what JesseM and DrGreg has already done. :)

In the end, my final advice is: learn the basics and learn it well. And, do learn Minkowski Spacetime diagrams...

I sure do hope this helps me understand when I am allowed to apply time dilation.

 

[cfrogue]

So you can see that there is no inconsistency, we both agree that if t_colocated is the time between two events in the frame where they are colocated, and t_noncolocated is the time between the same events in the frame where they are not colocated, then the equation has the form:

t_noncolocated = gamma*t_colocated

You and I were applying time dilation in a twins thread.

In addition, there is a paper that applies the integral of time dilation.

Can you prove and explain why the paper's logic had colocated events during acceleration?



Otherwise, the logic of the paper is false.

 

[JesseM]

I already said yes.

How could you? I never asked you that question before, I asked a different question about simultaneity.

But, I have two events exclusively in one frame.

Do you? What are they? Anyway, the phrase "two events exclusively in one frame" doesn't seem to make sense, events are physical things that are assigned coordinates by all frames, they don't "belong" to one frame or another.

Do you mean the time dilation equations depends on a location?

So, if I have a time interval in O, there must be a location of the same position for two events?

The events, MUST occur at the same location in frame O or time dilation is not applicable.

It would also be applicable if the events occurred at the same location in frame O' (as in my example with events 1B and 2 from post 275)...the point is just that you need to pick a pair of events that occur at the same location in one of the two frames you're using, it doesn't matter which one. One way or another, the equation will then take the form t_noncolocated = gamma*t_colocated

Therefore, we must have only events at the same location in a frame or there is no time dilation period.

If by "there is no time dilation period" you just mean "you can't plug the time intervals into the time dilation equation and expect it to be valid", then that's correct. You can only plug two time intervals into the time dilation equation and expect it to be valid when the following conditions are met:

1. Both time intervals represent the time between a specific pair of events, as measured in two different inertial frames
2. The two events occurred at the same spatial location in one of those frames

 

[Dale]

what do you mean?

what is the time dilation equation for what I gave?

What I mean is that the time dilation formula, Δt' = γ Δt, applies only if Δx=0 (that is what it means for two events to be co-located). Please see my derivation in https://www.physicsforums.com/showpost.php?p=2472240&postcount=266". Since Δx is not 0 in this case you need to use the full Lorentz transform equation:
\Delta t&#039; = \gamma (\Delta t - v \Delta x /c^2)

IMO, it is a bad idea (especially for beginners) to use the length contraction or time dilation formulas at all, they are too easy to mess up as you have seen. Instead it is best to always use the Lorentz transform, and the time dilation and length contraction formulas will automatically pop out whenever they are appropriate.

 

[atyy]

I can do no better than what JesseM and DrGreg has already done. :)

In the end, my final advice is: learn the basics and learn it well. And, do learn Minkowski Spacetime diagrams...

I sure do hope this helps me understand when I am allowed to apply time dilation.

events are physical things that are assigned coordinates by all frames, they don't "belong" to one frame or another.

IMO, it is a bad idea (especially for beginners) to use the length contraction or time dilation formulas at all, they are too easy to mess up as you have seen. Instead it is best to always use the Lorentz transform, and the time dilation and length contraction formulas will automatically pop out whenever they are appropriate.

All the above is very good advice. Whether or not you use the the full spacetime view, you must use some spacetime view. The basic idea is not relativity of simultaneity, not time dilation, not Lorentz contraction.

The basic ideas begin with a conception of reality: objects move about in spacetime (worldlines of objects), and their spacetime paths intersect each other (events). For example, a photon is an object which travels in spacetime, so there is a photon worldline. The endpoint of a rod is another object which also travels in spacetime, so there is the rod endpoint worldline. The point in spacetime when the photon meets the endpoint of the rod is an event, which is where the photon and rod endpoint worldlines intersect.

Then to describe reality, we use objects such as photons and clocks to assign spatial and temporal coordinates to events. Different observers, who are themselves objects with worldlines, assign different numbers to the same underlying reality, ie. assign different coordinates to the same events.

It turns out to be an experimental fact that there are some observers who assign spacetime coordinates for the intersections of photon worldlines with their rods, such that the speed of light is c in all directions. These special observers are called inertial observers.

It turns out to be yet another experimental fact that any inertial observer A will assign spacetime coordinates for the intersection of the wordlines of any other inertial observer B with A's rods, such that the velocity of any other inertial observer B is constant.

If there is one underlying reality and the different coordinates are merely different descriptions of the same events, then we expect to have some way of relating the the different coordinates assigned by different observers to each event. The Lorentz transformations are the rule for relating the coordinate assignments by different inertial observers to an event. Time dilation is just a special case of the application of the Lorentz transformations to two events in spacetime that one particular inertial observer describes as "colocal".

 

[DrGreg]

The general equation is given by the Lorentz transform

\Delta t&#039; = \gamma ( \Delta t - \frac {v}{c^2} \Delta x )​

The time dilation formula

\Delta t&#039; = \gamma \Delta t​

applies only in the special case when \Delta x = 0, which is what the Wikipedia article says: "\Delta t is the time interval between two co-local events (i.e. happening at the same place)"

what if I said, co-local events means events in the same frame.

Two events are co-local, relative to a frame, if the distance between the events is zero in that frame. "Events in the same frame" is meaningless, as every event occurs "in" every frame. If you meant "events that are both stationary relative to one frame", that doesn't make sense either, because an event exists only at a single instant in time, so it's meaningless to ask whether an event is moving or not.

 

[cfrogue]

Two events are co-local, relative to a frame, if the distance between the events is zero in that frame. "Events in the same frame" is meaningless, as every event occurs "in" every frame. If you meant "events that are both stationary relative to one frame", that doesn't make sense either, because an event exists only at a single instant in time, so it's meaningless to ask whether an event is moving or not.

I will think about this.

I have an additional question. If O' has a rod of rest length d, with a light source at the middle and when the light source and O are coincident the light flashes, when in the time of O will O' see the flashes as simultaneous?

I keep calculating t = d/(2*λ*(c-v)).

 

[cfrogue]

All the above is very good advice. Whether or not you use the the full spacetime view, you must use some spacetime view. The basic idea is not relativity of simultaneity, not time dilation, not Lorentz contraction.

The basic ideas begin with a conception of reality: objects move about in spacetime (worldlines of objects), and their spacetime paths intersect each other (events). For example, a photon is an object which travels in spacetime, so there is a photon worldline. The endpoint of a rod is another object which also travels in spacetime, so there is the rod endpoint worldline. The point in spacetime when the photon meets the endpoint of the rod is an event, which is where the photon and rod endpoint worldlines intersect.

Then to describe reality, we use objects such as photons and clocks to assign spatial and temporal coordinates to events. Different observers, who are themselves objects with worldlines, assign different numbers to the same underlying reality, ie. assign different coordinates to the same events.

It turns out to be an experimental fact that there are some observers who assign spacetime coordinates for the intersections of photon worldlines with their rods, such that the speed of light is c in all directions. These special observers are called inertial observers.

It turns out to be yet another experimental fact that any inertial observer A will assign spacetime coordinates for the intersection of the wordlines of any other inertial observer B with A's rods, such that the velocity of any other inertial observer B is constant.

If there is one underlying reality and the different coordinates are merely different descriptions of the same events, then we expect to have some way of relating the the different coordinates assigned by different observers to each event. The Lorentz transformations are the rule for relating the coordinate assignments by different inertial observers to an event. Time dilation is just a special case of the application of the Lorentz transformations to two events in spacetime that one particular inertial observer describes as "colocal".

This is an excellent description.

 

[cfrogue]

How could you? I never asked you that question before, I asked a different question about simultaneity.

Do you? What are they? Anyway, the phrase "two events exclusively in one frame" doesn't seem to make sense, events are physical things that are assigned coordinates by all frames, they don't "belong" to one frame or another.

It would also be applicable if the events occurred at the same location in frame O' (as in my example with events 1B and 2 from post 275)...the point is just that you need to pick a pair of events that occur at the same location in one of the two frames you're using, it doesn't matter which one. One way or another, the equation will then take the form t_noncolocated = gamma*t_colocated

If by "there is no time dilation period" you just mean "you can't plug the time intervals into the time dilation equation and expect it to be valid", then that's correct. You can only plug two time intervals into the time dilation equation and expect it to be valid when the following conditions are met:

1. Both time intervals represent the time between a specific pair of events, as measured in two different inertial frames
2. The two events occurred at the same spatial location in one of those frames

If a clock in t elapses in O by t, what will elapse on the clock of O'?

 

[cfrogue]

What I mean is that the time dilation formula, Δt' = γ Δt, applies only if Δx=0 (that is what it means for two events to be co-located). Please see my derivation in https://www.physicsforums.com/showpost.php?p=2472240&postcount=266". Since Δx is not 0 in this case you need to use the full Lorentz transform equation:
\Delta t&#039; = \gamma (\Delta t - v \Delta x /c^2)

IMO, it is a bad idea (especially for beginners) to use the length contraction or time dilation formulas at all, they are too easy to mess up as you have seen. Instead it is best to always use the Lorentz transform, and the time dilation and length contraction formulas will automatically pop out whenever they are appropriate.

I know how to derive that equation.

I know it is when Δx = 0.

Can you then use LT and tell me the time dilation for the scenerio we are talking about?

 

[JesseM]

If a clock in t elapses in O by t, what will elapse on the clock of O'?

Having trouble interpreting "a clock in t elapses in O by t". Do you mean we have a clock at rest in O, and pick two events on its worldline separated by a time interval of t in O (and also separated by a proper time of t on the clock since its time matches coordinate time in O)?

 

[cfrogue]

Do you? What are they? Anyway, the phrase "two events exclusively in one frame" doesn't seem to make sense, events are physical things that are assigned coordinates by all frames, they don't "belong" to one frame or another.

Yea, I should have said from the POV of O.

It would also be applicable if the events occurred at the same location in frame O' (as in my example with events 1B and 2 from post 275)...the point is just that you need to pick a pair of events that occur at the same location in one of the two frames you're using, it doesn't matter which one. One way or another, the equation will then take the form t_noncolocated = gamma*t_colocated

We now imagine space to be measured from the stationary system K by means of the stationary measuring-rod, and also from the moving system k by means of the measuring-rod moving with it; and that we thus obtain the co-ordinates x, y, z, and , , respectively. Further, let the time t of the stationary system be determined for all points thereof at which there are clocks by means of light signals in the manner indicated in § 1;
http://www.fourmilab.ch/etexts/einstein/specrel/www/

It appears that Einstein had a "clock" located at each point of the stationary measuring-rod synchronized by SR's simultaneity convention.

Do you agree?

 

[cfrogue]

Having trouble interpreting "a clock in t elapses in O by t". Do you mean we have a clock at rest in O, and pick two events on its worldline separated by a time interval of t in O (and also separated by a proper time of t on the clock since its time matches coordinate time in O)?

Yea, I am simply talking about elaspsed time in O only, no other frame.

 

[cfrogue]

Two events are co-local, relative to a frame, if the distance between the events is zero in that frame. "Events in the same frame" is meaningless, as every event occurs "in" every frame. If you meant "events that are both stationary relative to one frame", that doesn't make sense either, because an event exists only at a single instant in time, so it's meaningless to ask whether an event is moving or not.

OK, with the twins solution on Wiki and many other places, when a clock in O elapses by t, then in a moving frame, its clock in that frame will elapse λt.

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