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

[cfrogue]

It does.

Observe the intersection of the light cone with the line t=1. This is the unprimed light sphere at t=1. Observe the intersection of the light cone with the line t'=1. This is the primed light sphere at t'=1.

Note that the midpoint of the unprimed light sphere is on the line x=0. Note that the midpoint of the primed light sphere is on the line x'=0. As you can see in the diagram, the two different light spheres have two different centers.

so is
t' = t*λ

where t = r/(λ*(c+v))

 

[JesseM]

I did not see you put these light spheres at an origin. What is the origin?

Look at the diagram:

This diagram is drawn from the perspective of A's frame, with A at rest at the origin of this frame...the horizontal axis is x and the vertical axis is t, so you can say that A's position on the horizontal axis is x=0, and the light is first emitted from this position at t=0. B is at the origin of his own frame, at x'=0, and the tip of the light cone where the light was first emitted is also t'=0. You can see that for A, the light sphere at the moment of the event E on the left also contains the event E1 on the right, and that A is exactly midway between E and E1 at the moment these events occur. You can also see that for B, the light sphere at the moment of the event E on the left contains the event E2 on the right, and that B is exactly midway between E and E2 at the moment these events occur (according to his own definition of simultaneity). You can look at the graph to find the coordinates of all these events if you want to check the math--for example, in the A frame E occurs at (x=-2, t=2) and E2 occurs at (x=6, t=6), while the event on B's worldline that is simultaneous with these events in his own frame occurs at (x=2, t=4). If you apply the Lorentz transformation to all three of these events (using v=0.5c), you find that in B's frame they all happen at the same t' coordinate, and that the event on B's worldline occurs at x'=0 while E and E2 happen at equal distances from B on either side.

OK, does time dilation not apply?

Please explain why.

I am glad to see you knew what I was doing.

But, you will now need to support time dilation.

It does apply. Because of the relativity of simultaneity, if the clock at the left end of the rod reads 0 at t'=0 in the rod's rest frame, in frame O it does not read 0 at t=0, instead it already reads rv/c^2 at t=0. So, if in frame O the clock ticks forward by t/gamma = r/(gamma^2*(c+v)) in the time it takes the light to reach it, as predicted by the time dilation equation, then the time it will show when the light reaches it will be rv/c^2 + r/(gamma^2*(c+v)).

With a little algebra we can simplify this:

rv/c^2 + r/(gamma^2*(c+v)) = rv/c^2 + r*(1 - v^2/c^2)/(c+v) = [rv*(c+v)]/[c^2*(c+v)] + [rc^2*(1 - v^2/c^2)]/[c^2*(c+v)] = [rvc + rv^2 + rc^2 - rv^2]/[c^2*(c+v)] = [rc*(c+v)]/[c^2*(c+v)] = r/c.

So, taking the initial time on the rod clock at t=0 in frame O (which is not zero due to the relativity of simultaneity) and adding the elapsed time in frame O predicted by the time dilation equation yields the correct prediction that the clock will read r/c when the light reaches it.

 

[cfrogue]

Look at the diagram:

This diagram is drawn from the perspective of A's frame, with A at rest at the origin of this frame...the horizontal axis is x and the vertical axis is t, so you can say that A's position on the horizontal axis is x=0, and the light is first emitted from this position at t=0. B is at the origin of his own frame, at x'=0, and the tip of the light cone where the light was first emitted is also t'=0. You can see that for A, the light sphere at the moment of the event E on the left also contains the event E1 on the right, and that A is exactly midway between E and E1 at the moment these events occur. You can also see that for B, the light sphere at the moment of the event E on the left contains the event E2 on the right, and that B is exactly midway between E and E2 at the moment these events occur (according to his own definition of simultaneity). You can look at the graph to find the coordinates of all these events if you want to check the math--for example, in the A frame E occurs at (x=-2, t=2) and E2 occurs at (x=6, t=6), while the event on B's worldline that is simultaneous with these events in his own frame occurs at (x=2, t=4). If you apply the Lorentz transformation to all three of these events (using v=0.5c), you find that in B's frame they all happen at the same t' coordinate, and that the event on B's worldline occurs at x'=0 while E and E2 happen at equal distances from B on either side.

It does apply. Because of the relativity of simultaneity, if the clock at the left end of the rod reads 0 at t'=0 in the rod's rest frame, in frame O it does not read 0 at t=0, instead it already reads rv/c^2 at t=0. So, if in frame O the clock ticks forward by t/gamma = r/(gamma^2*(c+v)) in the time it takes the light to reach it, as predicted by the time dilation equation, then the time it will show when the light reaches it will be rv/c^2 + r/(gamma^2*(c+v)) = rv/c^2 + r*(1 - v^2/c^2)/(c+v) = [rv*(c+v)]/[c^2*(c+v)] + [rc^2*(1 - v^2/c^2)]/[c^2*(c+v)] = [rvc + rv^2 + rc^2 - rv^2]/[c^2*(c+v)] = [rc*(c+v)]/[c^2*(c+v)] = r/c.

Uh, does that mean that time dilation is false?

Here is the deal under SR.

t' = t*λ

where t = r/(λ*(c+v))

I set up a real problem.

 

[JesseM]

Uh, does that mean that time dilation is false?

Here is the deal under SR.

t' = t*λ

where t = r/(λ*(c+v))

I set up a real problem.

No, it doesn't contradict time dilation. The time interval between the light being emitted at t=0 and the light hitting the left end was t = r/(gamma*(c+v)) in the O frame, and the time elapsed on the clock between its reading at t=0 and the its reading when the light hit it was t/gamma = r/(gamma^2*(c+v)), exactly as predicted by the time dilation equation. But since its reading at t=0 was rv/c^2, its reading when the light hit it was rv/c^2 + t/gamma, which worked out to r/c. Maybe if you'd actually pay attention and think a little before writing a knee-jerk dismissive response you'd learn more, and avoid getting everyone around you frustrated at your attitude.

 

[cfrogue]

No, it doesn't contradict time dilation. The time interval between the light being emitted at t=0 and the light hitting the left end was t = r/(gamma*(c+v)) in the O frame, and the time elapsed on the clock between its reading at t=0 and the its reading when the light hit it was t/gamma = r/(gamma^2*(c+v)), exactly as predicted by the time dilation equation. But since its reading at t=0 was rv/c^2, its reading when the light hit it was rv/c^2 + t/gamma, which worked out to r/c. Maybe if you'd actually pay attention and think a little before writing a knee-jerk dismissive response you'd learn more, and avoid getting everyone around you frustrated at your attitude.

Yea, so let's see


t' = t*λ

where t = r/(λ*(c+v))

Thus,

t' = r/(c+v)

Is this false?

Here is the time dilation logic.

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

 

[cfrogue]

I do not understand, why is this not solved?

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

 

[JesseM]

Yea, so let's see


t' = t*λ

where t = r/(λ*(c+v))

Thus,

t' = r/(c+v)

Is this false?

If t' is supposed to represent the time on the clock when the light reaches it, then yes, your logic is false. You are making the trivial mistake of ignoring the difference between time intervals on a clock (which is what the time dilation equation deals with) and time readings. If a clock moves at 0.6c for 10 minutes between t=0 and t=10 in my frame, then the time interval that elapses on the clock should be 8 minutes according to the time dilation equation. But does that mean I should expect the time reading at the end to be t'=8 minutes? No, because the clock may not have started off showing a time of 0 minutes at t=0. If it started off showing a reading of t'=12 minutes at t=0, for example, then the time dilation equation predicts that at t=10 it'll read t'=12+8=20 minutes.

Similarly, although the rod clock shows a time interval of t/gamma (not t*gamma, you are also making the mistake of running the time dilation equation backwards) in frame O between t=0 when the light was emitted and the time the light reaches the rod clock, that does not mean it shows a time reading of t/gamma when the light reaches it, because at t=0 in frame O it was already showing a time reading of rv/c^2 due to the relativity of simultaneity, so its reading when the light reaches it will be rv/c^2 + t/gamma = r/c.

 

[cfrogue]

If t' is supposed to represent the time on the clock when the light reaches it, then yes, your logic is false. You are making the trivial mistake of ignoring the difference between time intervals on a clock (which is what the time dilation equation deals with) and time readings. If a clock moves at 0.6c for 10 minutes between t=0 and t=10 in my frame, then the time interval that elapses on the clock should be 8 minutes according to the time dilation equation. But does that mean I should expect the time reading at the end to be t'=8 minutes? No, because the clock may not have started off showing a time of 0 minutes at t=0. If it started off showing a reading of t'=12 minutes at t=0, for example, then the time dilation equation predicts that at t=10 it'll read t'=12+8=20 minutes.

Similarly, although the rod clock shows a time interval of t/gamma (not t*gamma, you are also making the mistake of running the time dilation equation backwards) in frame O between t=0 when the light was emitted and the time the light reaches the rod clock, that does not mean it shows a time reading of t/gamma when the light reaches it, because at t=0 in frame O it was already showing a time reading of rv/c^2 due to the relativity of simultaneity, so its reading when the light reaches it will be rv/c^2 + t/gamma = r/c.

Yea, I was very careful to to sync te clocks in their own frames.

Note Einstein did this also.

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/

 

[JesseM]

Yea, I was very careful to to sync te clocks in their own frames.

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).

 

[atyy]

Maybe so, but the fact is if you are in a frame and light emits from a light source, light moves spherically from the light source with the light source as the center.

So, this is what to expect in the moving O' or we are ignoring a fact.

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?

 

[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?