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

[cfrogue]

LOL, I wonder who is doing the "not understanding" (or perhaps the "not writing clearly") here!

I wrote to you:

The common light cone has a static spacetime origin at the vertex in all frames. As viewed from reference frame O, the light spheres have apparent spatial centers (or as atyy has written: "assigned" centers), one for each v AND one for each t. They do not have spacetime origins.

Look at https://www.physicsforums.com/showpost.php?p=2467732&postcount=182" again... ;)

The common light cone has a static spacetime origin at the vertex in all frames

This is artificial, the center moves vt in the moving frame.

OK, I have another way to look at it, tell me what you think.

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?

 

[cfrogue]

The centre of the light sphere in each frame is just an "assigned" centre - no event actually happens there, except when the light is emitted, and the sphere has zero radius - at this point the origins coincide and both frames agree on the centre of the light sphere. So it doesn't matter that the frames disagree about it, just as they don't agree about what is simultaneous.

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.

 

[cfrogue]

No arguments what exactly is wrong with my statements, just an ASCII art that has nothing to do with SR? Well here is my diagram:

+----------+
|  PLEASE  |
|  DO NOT  |
| FEED THE |
|  TROLLS  |
+----------+
    |  |    
    |  |    
  .\|.||/..

Cute art.

Sorry, I misread your statement to mean O and O' disagree with the center of O'.

After rereading it, you are saying the same as me.

I agree with you if you are saying the center of the light sphere is moving with the moving frame.

Thus, O will see it at 0 and O' will see it at vt and they disagree.

 

[JesseM]

cfrogue, can you answer my questions from post 193?

At any given time coordinate in frame O, there is only one light sphere (assuming we are talking about light emitted in all directions from a single event in the past), and it is always centered at the origin of O and spherical in shape. The different light spheres seen by O and O' are just different ways of slicing up a single light cone, based on their different definitions of simultaneity. Do you disagree?

Also, did you look at my diagram in post 182? If so did you understand it?

 

[A.T.]

I agree with you if you are saying the center of the light sphere is moving with the moving frame.

No, there is no frame in which the center of the light sphere is moving.

Thus, O will see it at 0 and O' will see it at vt and they disagree.

This sounds right, assuming "0" and "vt" refer to space points given in O-coorindantes: Each frame observes a different physical center of the light sphere. But nobody observes a moving center, two centers, or two spheres.

 

[cfrogue]

cfrogue, can you answer my questions from post 193?

What I am finding is the following.

One light sphere has two different origins.

I am not ready to say what it really is yet though.

So, no it is not the same light sphere.

 

[cfrogue]

No, there is no frame in which the center of the light sphere is moving.

This sounds right, assuming "0" and "vt" refer to space points given in O-coorindantes: Each frame observes a different physical center of the light sphere. But nobody observes a moving center, two centers, or two spheres.

We are in complete agreement

up to this point

"But nobody observes a moving center, two centers, or two spheres"

O knows the center of the light sphere for O' is located at vt. That therefore, is a moving light sphere.

 

[A.T.]

O knows the center of the light sphere for O' is located at vt. That thereforen is a moving light sphere.

Yes O knows / can calculate that the center of the light sphere observed by O' is located at vt, but O doesn't observe the center of the sphere there.

O uses his simultaneity to determine the center, and doesn't have to care about the simultaneity in O' or any other of the infinite number of frames.

 

[Dale]

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.

No, it does not. The light flash moves spherically from the origin of the flash regardless of the subsequent motion of the source. That is the http://en.wikipedia.org/wiki/Postulates_of_special_relativity" : "As measured in an inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body."

 

[cfrogue]

No, it does not. The light flash moves spherically from the origin of the flash regardless of the subsequent motion of the source. That is the http://en.wikipedia.org/wiki/Postulates_of_special_relativity" : "As measured in an inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body."

You have failed to realize in the moving frame, it thinks it is at rest.

Light proceeds spherically in that frame from the light emission point in the frame.

Now, from the rest frame, that center point is at vt at any time t.

We must be scientific.

 

[cfrogue]

Is anyone going to tackle this?



OK, I have another way to look at it, tell me what you think.

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?

 

[JesseM]

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.

No, it does not. The light flash moves spherically from the origin of the flash regardless of the subsequent motion of the source. That is the http://en.wikipedia.org/wiki/Postulates_of_special_relativity" : "As measured in an inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body."

You have failed to realize in the moving frame, it thinks it is at rest.

Light proceeds spherically in that frame from the light emission point in the frame.

Now, from the rest frame, that center point is at vt at any time t.

We must be scientific.

No, your reasoning is incorrect. Suppose that the source is at rest in the moving frame. Then it will be true that in the moving frame, the light proceeds spherically from the position of the source, and since the source is at rest in the moving frame the light sphere always remains centered on the source in this frame. It is also true that in the stationary frame, the source is moving with speed vt. However, this does not mean that in the stationary frame, the light is centered on the source--it's not, in the stationary frame it's centered on the position the source was when it emitted the light. If you would actually do the math instead of relying on vague verbal arguments you'd see this (I showed you the math in post 189, where I started by assuming the light had the equation of an expanding sphere in the coordinates of the moving frame, and showed that if you then apply the Lorentz transformation to the coordinates the light passes through in the moving frame to find the coordinates it passes through in the stationary frame, you see that it still has the equation of an expanding sphere in the coordinates of the stationary frame). You'd also be able to see this conceptually if you bothered to learn how Minkowski diagrams work. As long as you continue to rely only on words you're going to be confused, though.

 

[cfrogue]

No, your reasoning is incorrect. Suppose that the source is at rest in the moving frame. Then it will be true that in the moving frame, the light proceeds spherically from the position of the source, and since the source is at rest in the moving frame the light sphere always remains centered on the source in this frame. It is also true that in the stationary frame, the source is moving with speed vt. However, this does not mean that in the stationary frame, the light is centered on the source--it's not, in the stationary frame it's centered on the position the source was when it emitted the light. If you would actually do the math instead of relying on vague verbal arguments you'd see this (I showed you the math in post 189, where I started by assuming the light had the equation of an expanding sphere in the coordinates of the moving frame, and showed that if you then apply the Lorentz transformation to the coordinates the light passes through in the moving frame to find the coordinates it passes through in the stationary frame, you see that it still has the equation of an expanding sphere in the coordinates of the stationary frame). You'd also be able to see this conceptually if you bothered to learn how Minkowski diagrams work. As long as you continue to rely only on words you're going to be confused, though.

Suppose I told you I am relying on the math.

Now, the sphere expands in the rest frame.

But, it is expanding in the moving frame as well.

Confirm or deny this.

 

[A.T.]

Is anyone going to tackle this?

It is trivial:

You confuse the center that is observed by an observer, with the some center that he can calculate for some moving observer. You can calculate a lot of moving coordinates in your frame, that doesn't move the light sphere, or any other physical object.

Similar to the Length contraction thread you entire argument is based on confusing two different things, by using inaccurate language, and ignoring all attempts to clarify what is meant.

 

[cfrogue]

It is trivial:

You confuse the center that is observed by an observer, with the some center that he can calculate for some moving observer. You can calculate a lot of moving coordinates in your frame, that doesn't move the light sphere, or any other physical object.

Similar to the Length contraction thread you entire argument is based on confusing two different things, by using inaccurate language, and ignoring all attempts to clarify what is meant.

I cannot find an answer.

Is time dilation applicable yes or no.

 

[JesseM]

Suppose I told you I am relying on the math.

Now, the sphere expands in the rest frame.

But, it is expanding in the moving frame as well.

Confirm or deny this.

Yes, that is true, but it's perfectly spherical and centered on the origin in both frames. The correct way to do the math is to figure out which x',y',z',t' coordinates the light passes through in one frame, then apply the Lorentz transformation to find the x,y,z,t coordinates the light must therefore pass through in the other frame (this is exactly what I did in post 189). Does your version of "the math" involve figuring out the coordinates the light passes through in one frame and applying the Lorentz transformation? If not it is probably based on mistaken classical assumptions (but it would help if you would actually write out whatever math you are basing your statements on, so people could identify the flaw).

 

[cfrogue]

Yes, that is true, but it's perfectly spherical and centered on the origin in both frames. The correct way to do the math is to figure out which x',y',z',t' coordinates the light passes through in one frame, then apply the Lorentz transformation to find the x,y,z,t coordinates the light must therefore pass through in the other frame (this is exactly what I did in post 189). Does your version of "the math" involve figuring out the coordinates the light passes through in one frame and applying the Lorentz transformation? If not it is probably based on mistaken classical assumptions (but it would help if you would actually write out whatever math you are basing your statements on, so people could identify the flaw).

Yea, I am glad you finally confessed one light sphere has two origins.

(but it would help if you would actually write out whatever math you are basing your statements on, so people could identify the flaw)

You wrote the above to me.

Now, can you provide the math in the context of a light sphere without frames on how this is realizable in the real external world?

 

[JesseM]

Yea, I am glad you finally confessed one light sphere has two origins.

No I did not, because the two frames are not measuring "one light sphere", they are measuring different light spheres which represent different cross-sections of the same light cone (the light cone being the set of all points in spacetime that the light passes through). Each frame defines a "light sphere" as a set of points on the light cone that all occur at the same time-coordinate in that frame, so due to the relativity of simultaneity, the set of events that one frame defines to be contained on a single light sphere would actually occur at many different times and thus be part of many different light spheres in the other frame.

 

[Dale]

You have failed to realize

Why do you put in pointless little snide side comments like this? You should realize by now that I do not have any knowledge deficit or comprehension problem on this topic. I understand it well and fully, and I have been nothing but courteous and patient with you. Frankly, it is tiring to try to help you out when you are so repetitively insulting to myself and others. I certainly have done nothing to provoke this nor deserve it. Such behaviour is very immature.

I don't even think that you misunderstand the material, I think that you just deliberately write sloppily and avoid standard terminology in order to purposefully provoke argument and you add such extraneous insults in order to deliberately inflame the argument.

 

[cfrogue]

Why do you put in pointless little snide side comments like this? You should realize by now that I do not have any knowledge deficit or comprehension problem on this topic. I understand it well and fully, and I have been nothing but courteous and patient with you. Frankly, it is tiring to try to help you out when you are so repetitively insulting to myself and others. I certainly have done nothing to provoke this nor deserve it. Such behaviour is very immature.

I don't even think that you misunderstand the material, I think that you just deliberately write sloppily and avoid standard terminology in order to purposefully provoke argument and you add such extraneous insults in order to deliberately inflame the argument.

Sorry to offend you.

I will not do it again.

But, I have some material here.

No, I do not misunderstand any more than you do.

I am glad you figured that out.

 

[cfrogue]

No I did not, because the two frames are not measuring "one light sphere", they are measuring different light spheres which represent different cross-sections of the same light cone (the light cone being the set of all points in spacetime that the light passes through). Each frame defines a "light sphere" as a set of points on the light cone that all occur at the same time-coordinate in that frame, so due to the relativity of simultaneity, the set of events that one frame defines to be contained on a single light sphere would actually occur at many different times and thus be part of many different light spheres in the other frame.

Your light cone is based on forcing the origins to be the same.

The M diagrams place the origin of the light emission at a common point.

Are you claiming this is true in reality?

 

[cfrogue]

No I did not, because the two frames are not measuring "one light sphere", they are measuring different light spheres which represent different cross-sections of the same light cone (the light cone being the set of all points in spacetime that the light passes through). Each frame defines a "light sphere" as a set of points on the light cone that all occur at the same time-coordinate in that frame, so due to the relativity of simultaneity, the set of events that one frame defines to be contained on a single light sphere would actually occur at many different times and thus be part of many different light spheres in the other frame.

Say, Post #211, what do you think?

 

[JesseM]

Your light cone is based on forcing the origins to be the same.

The M diagrams place the origin of the light emission at a common point.

Are you claiming this is true in reality?

Of course you can have multiple flashes of light from different points, which will result in multiple light cones. But the whole problem we are considering is what happens when a single source let's out a single flash of light in all directions, and different frames look at how the light from this flash expands over time, from the perspective of their own coordinates. There aren't two separate flashes, one for each frame, it's two frames analyzing the same single flash.

 

[Dale]

I will not do it again.

Thanks, I appreciate that.

But, I have some material here.

Let's summarize the material:

1) you agree with my post 2, that the equation of the light cone in the unprimed is ct = ± x
2) you agree with my post 4, that the equation of the light cone does not violate relativity of simultaneity
3) you agree with my post 191, that the "multiple centers" and elongation are due to the relativity of simultaneity

As far as I can tell the remaining discussion is not substantive and is simply due to lingering communication problems.

 

[cfrogue]

Of course you can have multiple flashes of light from different points, which will result in multiple light cones. But the whole problem we are considering is what happens when a single source let's out a single flash of light in all directions, and different frames look at how the light from this flash expands over time, from the perspective of their own coordinates. There aren't two separate flashes, one for each frame, it's two frames analyzing the same single flash.

You know I am only talking about one light flash.

In the rest frame it proceeds spherically from 0 and in the moving frame it proceeds spherically from vt all from the view of O.

 

[JesseM]

OK, I have another way to look at it, tell me what you think.

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.

You originally said the rod had a rest length of r and the light was emitted from the center, which would mean in the rod's rest frame the distance from the center to the left end was only r/2...but based on this equation, presumably you should have said the distance from the center to the left end was r, and the rod's total rest length was 2r. Assuming that's what you meant, then yes, the time in frame O where the rod is moving to the right at speed v would be t=r/(gamma*(c+v))

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

It depends on how the light timer was synchronized. If the timer was set to read 0 at the moment the flash was emitted according to the definition of simultaneity in frame O, then yes, when the light reached it the timer would read t' = t*gamma = r/(c+v). But if the timer was set to read 0 at the moment the flash was emitted according to the definition of simultaneity in the rod's own rest frame, then no, it wouldn't, it would just read t' = r/c.

 

[cfrogue]

Thanks, I appreciate that.
Let's summarize the material:

1) you agree with my post 2, that the equation of the light cone in the unprimed is ct = ± x
2) you agree with my post 4, that the equation of the light cone does not violate relativity of simultaneity
3) you agree with my post 191, that the "multiple centers" and elongation are due to the relativity of simultaneity

As far as I can tell the remaining discussion is not substantive and is simply due to lingering communication problems.

You and I agree. I see all this and agree.

There is more to explore.

Please consider post #211.

Also, consider the multiple light source origins.

 

[JesseM]

In the rest frame it proceeds spherically from 0 and in the moving frame it proceeds spherically from vt all from the view of O.

Wrong. Again, I showed the math to prove that if the flash proceeds spherically from the origin in one frame, then applying the Lorentz transformation shows that the same flash also proceeds spherically from the origin in the other frame. You have no math to support your own silly claims.

 

[cfrogue]

You originally said the rod had a rest length of r and the light was emitted from the center, which would mean in the rod's rest frame the distance from the center to the left end was only r/2...but based on this equation, presumably you should have said the distance from the center to the left end was r, and the rod's total rest length was 2r. Assuming that's what you meant, then yes, the time in frame O where the rod is moving to the right at speed v would be t=r/(gamma*(c+v))

It depends on how the light timer was synchronized. If the timer was set to read 0 at the moment the flash was emitted according to the definition of simultaneity in frame O, then yes, when the light reached it the timer would read t' = t*gamma. But if the timer was set to read 0 at the moment the flash was emitted according to the definition of simultaneity in the rod's own rest frame, then no, it wouldn't, it would just read t' = r/c.

Geez, another error.

I meant the radius.

Each clock sync occurs only in the frame and not the other.

Are you OK?

 

[JesseM]

Geez, another error.

I meant the radius.

Each clock sync occurs only in the frame and not the other.

If the light timer was synchronized so that it read 0 at the moment of the flash in the rod's rest frame, then it would show a time of r/c when the light reached it. If this wasn't true the second postulate of SR would be violated.

 

[cfrogue]

If the light timer was synchronized so that it read 0 at the moment of the flash in the rod's rest frame, then it would show a time of r/c when the light reached it. If this wasn't true the second postulate of SR would be violated.

You did not understand what I said.

t = r/(λ*(c+v))

What is t'?

t' = t*λ for time dilation?

 

[cfrogue]

Wrong. Again, I showed the math to prove that if the flash proceeds spherically from the origin in one frame, then applying the Lorentz transformation shows that the same flash also proceeds spherically from the origin in the other frame. You have no math to support your own silly claims.

I already know the know the equations show two different origins for one light sphere.

Now that you have admitted this, operating from the light sphere, how do you make this happen?

 

[Dale]

Please consider post #211.

Oops, I did indeed miss this one.

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

Look at the https://www.physicsforums.com/showpost.php?p=2464800&postcount=88" again, note that the lines t=0 and t'=0 intersect only at the origin. You cannot synchronize all of the clocks in each frame, you can only synchronize one clock in each frame. The relativity of simultaneity requires that all of the remaining clocks be out of sync between the frames.

I suspect this is just a communication problem and that you understand that already.

 

[JesseM]

I already know the know the equations show two different origins for one light sphere.

Now that you have admitted this, operating from the light sphere, how do you make this happen?

I already told you in post 218 that the light spheres in different frames are not "one light sphere" but rather different cross-sections of the same light cone, so I guess the taunting "now that you have admitted this" is just one of those "pointless little snide side comments" that DaleSpam referred to in post 219. Not going to keep discussing this stuff with you if you act in this sort of trollish way.

 

[cfrogue]

Oops, I did indeed miss this one.Look at the https://www.physicsforums.com/showpost.php?p=2464800&postcount=88" again, note that the lines t=0 and t'=0 intersect only at the origin. You cannot synchronize all of the clocks in each frame, you can only synchronize one clock in each frame. The relativity of simultaneity requires that all of the remaining clocks be out of sync between the frames.

I suspect this is just a communication problem and that you understand that already.

This diagram does not address the many different origins.

Why??

If it does not, then it is incomplete.

Do you know what that means in mathematical logic?

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

 

[cfrogue]

I already told you in post 218 that the light spheres in different frames are not "one light sphere" but rather different cross-sections of the same light cone, so I guess the taunting "now that you have admitted this" is just one of those "pointless little snide side comments" that DaleSpam referred to in post 219. Not going to keep discussing this stuff with you if you act in this sort of trollish way.

Why do you need to call me names given that I corrected you into accepting the fact that there exists two different light spheres?

How do you make this happen in the context of one light sphere?

 

[JesseM]

Why do you need to call me names given that I corrected you into accepting the fact that there exists two different light spheres?

There are an infinite number of light spheres, one for every single instant (single value of the time coordinate) in any given frame. A light sphere at a single instant in one frame consists of events on the light cone that are spread out over many different times in the other frame, and thus are part of many different light spheres in that frame. If you have trouble understanding what I'm talking about that's fine, but don't accuse me of "admitting" something I didn't admit or I'll assume you're trolling.

You did not understand what I said.

t = r/(λ*(c+v))

What is t'?

t' = t*λ for time dilation?

No, it's you who doesn't understand. t' = r/c, just like I said. If we assume that in frame O the light was emitted at x=0 and t=0, and that the light hit the left end of the rod at x=-rc/(gamma*(c+v)) and t=r/(gamma*(c+v)), then just apply the Lorentz transformation and you'll see that t'=r/c. This is the time that would show on a clock at the left end when the light hit it, assuming the clock was set to show a time of 0 at t'=0 in this frame (the time when the light was emitted in this frame).

 

[cfrogue]

There are an infinite number of light spheres, one for every single instant (single value of the time coordinate) in any given frame. A light sphere at a single instant in one frame consists of events on the light cone that are spread out over many different times in the other frame, and thus are part of many different light spheres in that frame. If you have trouble understanding what I'm talking about that's fine, but don't accuse me of "admitting" something I didn't admit or I'll assume you're trolling.

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

No, it's you who doesn't understand. t' = r/c, just like I said. If we assume that in frame O the light was emitted at x=0 and t=0, and that the light hit the left end of the rod at x=-rc/(gamma*(c+v)) and t=r/(gamma*(c+v)), then just apply the Lorentz transformation and you'll see that t'=r/c. This is the time that would show on a clock at the left end when the light hit it, assuming the clock was set to show a time of 0 at t'=0 in this frame (the time when the light was emitted in this frame).

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.

 

[cfrogue]

There are an infinite number of light spheres, one for every single instant (single value of the time coordinate) in any given frame. A light sphere at a single instant in one frame consists of events on the light cone that are spread out over many different times in the other frame, and thus are part of many different light spheres in that frame. If you have trouble understanding what I'm talking about that's fine, but don't accuse me of "admitting" something I didn't admit or I'll assume you're trolling.

No, it's you who doesn't understand. t' = r/c, just like I said. If we assume that in frame O the light was emitted at x=0 and t=0, and that the light hit the left end of the rod at x=-rc/(gamma*(c+v)) and t=r/(gamma*(c+v)), then just apply the Lorentz transformation and you'll see that t'=r/c. This is the time that would show on a clock at the left end when the light hit it, assuming the clock was set to show a time of 0 at t'=0 in this frame (the time when the light was emitted in this frame).

This is time dilation.

Why does it not apply?

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

 

[Dale]

This diagram does not address the many different origins.

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.

 

[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?