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

[cfrogue]

The general equation is given by the Lorentz transform

$$\Delta t' = \gamma ( \Delta t - \frac {v}{c^2} \Delta x )$$​

The time dilation formula

$$\Delta t' = \gamma \Delta t$$​

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.

$$x = \gamma(x' - vt') = 1.25 (-1 + 0.6) = -0.5$$

$$t = \gamma(t' - vx') = 1.25 (1 - 0.6) = 0.5$$

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.

$$x = \gamma(x' - vt') = 1.25 (-1 + 0.6) = -0.5$$

$$t = \gamma(t' - vx') = 1.25 (1 - 0.6) = 0.5$$

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:
$$\Delta t' = \gamma (\Delta t - v \Delta x /c^2)$$

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

$$\Delta t' = \gamma ( \Delta t - \frac {v}{c^2} \Delta x )$$​

The time dilation formula

$$\Delta t' = \gamma \Delta t$$​

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:
$$\Delta t' = \gamma (\Delta t - v \Delta x /c^2)$$

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

 

[JesseM]

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.

You didn't answer my question of what two events you're thinking of, though.

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?

Yes, this is the physical basis for the time coordinate assigned to any event in an inertial frame--you assume that each frame has a row of clocks attached to different positions on a measuring-rod at rest in that frame, with the clocks synchronized according to the Einstein synchronization convention, and then any event that occurs alongside the rod can be assigned a time-coordinate by looking at the reading on the clock that was right next to the event at the moment it happened.

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

The elapsed time between two events on a worldline of a clock at rest in O?

 

[cfrogue]

You didn't answer my question of what two events you're thinking of, though.

I am thinking about the endpoint of the O' rod.

Please look for this in section 4
Between the quantities x, t, and , which refer to the position of the clock, we have, evidently, x=vt
http://www.fourmilab.ch/etexts/einstein/specrel/www/

It appears that Einstein derives time dilation based on the fact that x=vt or based on the motion of the rod.

Yes, this is the physical basis for the time coordinate assigned to any event in an inertial frame--you assume that each frame has a row of clocks attached to different positions on a measuring-rod at rest in that frame, with the clocks synchronized according to the Einstein synchronization convention, and then any event that occurs alongside the rod can be assigned a time-coordinate by looking at the reading on the clock that was right next to the event at the moment it happened.

OK, that is what I thought.

The elapsed time between two events on a worldline of a clock at rest in O?

Yes, I am only talking about rest clocks

 

[JesseM]

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

No, it's not a single clock C' at rest in the moving frame O' that measures gamma*t for the clock C in O to tick forward by t, it's the coordinate time in O' that's gamma*t, or alternately the difference in readings between two clocks at rest in O', the first measuring the time t1 in O' of the event of clock C showing its first reading, the second measuring the time t2 in O' of the event of clock C showing its second reading. If the separation between the two readings of C was t, then (t2 - t1) will equal gamma*t.

Take a look at my thread an illustration of relativity with rulers and clocks. In this thread I show how each frame assigns coordinates using local readings on a ruler with synchronized clocks attached to different markings, exactly as discussed in the quote from Einstein you provided earlier. Suppose we treat frame O in your example as synonymous with frame A in my diagrams, and frame O' as synonymous with frame B. And suppose the clock C at rest in frame A is the one at position x=692.3 meters on A's ruler, just to the right of the green clock. Finally, suppose the two events on C's worldline are the event of it showing a time of 2 microseconds, and the event of it showing a time of 3 microseconds, so t = 3 - 2 = 1 microsecond.

Now if you look at the top part of the diagram, you can see that the first event of clock C reading 2 microseconds happens right next to the clock at the 346.2 meter mark on ruler B, when that clock reads 0 microseconds. And if you look at the bottom part of the diagram, you can see that the second event of clock C reading 3 microseconds happens right next to a different clock on ruler B, namely the one at the -173.1 meter mark on ruler B, when that clock reads 2 microseconds.

So, the time between the readings of these two different clocks on ruler B is 2 microseconds, and that is t', the time in B's frame between the event of clock C reading 2 microseconds and the event of clock C reading 3 microseconds. Since gamma=2 in my diagram, this does satisfy the time dilation equation t' = gamma*t.

 

[Hans de Vries]

The OP's subject resembles the spherical mirror clock which
is a very nice example to thread all three aspects of SR:
Lorentz contraction, Time dilation and Non-simultaneity.

4.13 Simultaneity from the Spherical Mirror clock
http://physics-quest.org/Book_Chapter_Non_Simultaneity.pdf

Regards, Hans

 

[JesseM]

I am thinking about the endpoint of the O' rod.

That's not an event though. Do you want to pick two events which both happen on the worldline of the clock at the endpoint of the O' rod?

Please look for this in section 4
Between the quantities x, t, and , which refer to the position of the clock, we have, evidently, x=vt
http://www.fourmilab.ch/etexts/einstein/specrel/www/

It appears that Einstein derives time dilation based on the fact that x=vt or based on the motion of the rod.

Yes, but he was looking at the time in the stationary frame between two events on the worldline of the moving clock--the event of it reading 0, and the event of it reading t. The origins coincide so the event of this clock reading 0 happens at a time coordinate of tau=0 in the stationary frame, while the event of it reading t happens at a time coordinate of tau=t*sqrt(1 - v^2/c^2) in the stationary frame. So, the time interval in the stationary frame is just the second time coordinate minus the first, which of course is tau=t*sqrt(1 - v^2/c^2).

The elapsed time between two events on a worldline of a clock at rest in O?

Yes, I am only talking about rest clocks

OK, if we pick two events on the worldline of a clock C at rest in O that have a time separation of t in frame O, then the time separation t' between these events in O is gamma*t.

 

[cfrogue]

No, it's not a single clock C at rest in the moving frame O' that elapses gamma*t, it's the coordinate time in O' that's gamma*t, or alternately the difference in readings between two clocks at rest in O', the first measuring the time t1 in O' of the event of clock C showing its first reading, the second measuring the time t2 in O' of the event of clock C showing its second reading. If the separation between the two readings of C was t, then (t2 - t1) will equal gamma*t.

Take a look at my thread an illustration of relativity with rulers and clocks. In this thread I show how each frame assigns coordinates using local readings on a ruler with synchronized clocks attached to different markings, exactly as discussed in the quote from Einstein you provided earlier. Suppose we treat frame O in your example as synonymous with frame A in my diagrams, and frame O' as synonymous with frame B. And suppose the clock C at rest in frame A is the one at position x=692.3 meters on A's ruler, just to the right of the green diagram. Finally, suppose the two events on C's worldline are the event of it showing a time of 2 microseconds, and the event of it showing a time of 3 microseconds, so t = 3 - 2 = 1 microsecond.

Now if you look at the top part of the diagram, you can see that the first event of clock C reading 2 microseconds happens right next to the clock at the 346.2 meter mark on ruler B, when that clock reads 0 microseconds. And if you look at the bottom part of the diagram, you can see that the second event of clock C reading 3 microseconds happens right next to a different clock on ruler B, namely the one at the -173.1 meter mark on ruler B, when that clock reads 2 microseconds.

So, the time between the readings of these two different clocks on ruler B is 2 microseconds, and that is t', the time in B's frame between the event of clock C reading 2 microseconds and the event of clock C reading 3 microseconds. Since gamma=2 in my diagram, this does satisfy the time dilation equation t' = gamma*t.

This is excellent and I completely agree with all of this.

The coordinate time interval, yes, that is will elapse slower than an interval in the stationary frame from the POV of the stationary frame.

I guess now, we agree I can apply time dilation to describe the motion of the rod to illustrate the elapsed coord time in O'.

Do you agree with the following, when the center of the rod is located at the origin of O, then the left endpoint of the rod is located at xstart = d/(2λ)?

After any time t in O, the endpoint is located at xt = d/(2λ)+vt.

Thus, ∆x = vt.

Also, I asked DR Greg, but, when will O think O' sees the endpoints of the rods struck simultaneously?

I come up with t = d/(2*λ*(c-v)).

 

[JesseM]

This is excellent and I completely agree with all of this.

The coordinate time interval, yes, that is will elapse slower than an interval in the stationary frame from the POV of the stationary frame.

I guess now, we agree I can apply time dilation to describe the motion of the rod to illustrate the elapsed coord time in O'.

Do you agree with the following, when the center of the rod is located at the origin of O, then the left endpoint of the rod is located at xstart = d/(2λ)?

Normally I'd think of the coordinates increasing from left to right so that xstart = -d/(2*gamma), but this is just a matter of convention so sure, I agree.

After any time t in O, the endpoint is located at xt = d/(2λ)+vt.

Assuming the rod is traveling to the left, yes.

Thus, ∆x = vt.

If you pick two events on the worldline of the endpoint which have a time separation of t in frame O, then yes, their spatial separation ∆x in frame O will be ∆x = vt.

Also, I asked DR Greg, but, when will O think O' sees the endpoints of the rods struck simultaneously?

You mean, if the ends of the rod are struck simultaneously in O', what will be the time interval between the strikes in frame O?

I come up with t = d/(2*λ*(c-v)).

Let's say in frame O' the left end is struck at x'=d,t'=0 and the right end is struck at x'=-d,t'=0. Then the time of the left end being struck in frame O is:

gamma*(t' + vx'/c^2) = gamma*v*d/c^2

And the time of the right end being struck is:

gamma*(t' + vx'/c^2) = -gamma*v*d/c^2

So, the time interval between strikes in O would be 2*gamma*v*d/c^2. What was the reasoning behind your answer?

 

[cfrogue]

Normally I'd think of the coordinates increasing from left to right so that xstart = -d/(2*gamma), but this is just a matter of convention so sure, I agree.

Assuming the rod is traveling to the left, yes.

If you pick two events on the worldline of the endpoint which have a time separation of t in frame O, then yes, their spatial separation ∆x in frame O will be ∆x = vt.

You mean, if the ends of the rod are struck simultaneously in O', what will be the time interval between the strikes in frame O?

Let's say in frame O' the left end is struck at x'=d,t'=0 and the right end is struck at x'=-d,t'=0. Then the time of the left end being struck in frame O is:

gamma*(t' + vx'/c^2) = gamma*v*d/c^2

And the time of the right end being struck is:

gamma*(t' + vx'/c^2) = -gamma*v*d/c^2

So, the time interval between strikes in O would be 2*gamma*v*d/c^2. What was the reasoning behind your answer?

I calculated x' = ( x - vt )λ, I then put x = ct, but I could not come up with a good reason so that is why I asked.

I see, you applied the translation of t' back to t. You concluded that x'=d/c, I think it should be d/(2c) but that is no deal.

Why do you have a negative sign?

 

[JesseM]

I calculated x' = ( x - vt )λ, I then put x = ct, but I could not come up with a good reason so that is why I asked.

I see, you applied the translation of t' back to t. You concluded that x'=d/c, I think it should be d/(2c) but that is no deal.

Oh yeah, I guess I accidentally gave the rod a length of 2d rather than d.

Why do you have a negative sign?

Well, if the left end was at x'=d at t'=0 in O' (should have been x'=d/2, but never mind), then the right end would have to be at x'=-d, at t'=0, no? Both ends are equidistant from the origin of the x' axis since that's where the center of the rod is located.

 

[cfrogue]

Oh yeah, I guess I accidentally gave the rod a length of 2d rather than d.

Well, if the left end was at x'=d at t'=0 in O' (should have been x'=d/2, but never mind), then the right end would have to be at x'=-d, at t'=0, no? Both ends are equidistant from the origin of the x' axis since that's where the center of the rod is located.

I think we only need to calculate one anyway since they are both the same time.

It looks good to me.

 

[cfrogue]

Oh yeah, I guess I accidentally gave the rod a length of 2d rather than d.

Well, if the left end was at x'=d at t'=0 in O' (should have been x'=d/2, but never mind), then the right end would have to be at x'=-d, at t'=0, no? Both ends are equidistant from the origin of the x' axis since that's where the center of the rod is located.

OK, I wrote it in this format.

t = d/(2c) ( λ/c )

Is this OK?

 

[Dale]

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?

Sure, as always
$$\Delta t' = \gamma (\Delta t - v \Delta x/c^2)$$.

In this case Δx = ±1, c=1, v=0.6, Δt = 1, so
Δt' = 1.25 (1 - 0.6 (+1)/1²) = 0.5
and
Δt' = 1.25 (1 - 0.6 (-1)/1²) = 2

 

[JesseM]

OK, I wrote it in this format.

t = d/(2c) ( vλ/c )

Is this OK?

If the length is d then it should just be (d/c)*(v*gamma/c). The strike at the left end happens at t' = +gamma*v*d/(2*c^2) and the strike at the right end occurs at t' = -gamma*v*d/(2*c^2), so the time between them is gamma*v*d/c^2.

 

[cfrogue]

Sure, as always
$$\Delta t' = \gamma (\Delta t - v \Delta x/c^2)$$.

In this case Δx = ±1, c=1, v=0.6, Δt = 1, so
Δt' = 1.25 (1 - 0.6 (+1)/1²) = 0.5
and
Δt' = 1.25 (1 - 0.6 (-1)/1²) = 2

What do you get for the time dilation for this?

x = λvt/ (1 + λ)

Note, y > 0 for this case and z = 0.

 

[cfrogue]

If the length is d then it should just be (d/c)*(v*gamma/c). The strike at the left end happens at t' = +gamma*v*d/(2*c^2) and the strike at the right end occurs at t' = -gamma*v*d/(2*c^2), so the time between them is gamma*v*d/c^2.

I cannot see that.

They are struck at the same time in O', no?