Space-time curvature is incorrect

zankat,

The 10% value of the distance is obtained by subtracting the distance the light traveled by the distance the ship traveled in the stationairy frame of reference.

You can't apply the 10% value to the time because the time that the ship travelled and the time that the light travelled are identical. They are both one second. Are you claiming that for every one second of the stationairy clock, 0.0436 seconds pass for the clock in the spaceship???

Tom

 

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Prosoothus:

Whatever percentage you apply must be applied to both the time and distance in a given frame. In the ship’s frame, you applied 10% to the distance but not the time. Here is your story mostly corrected:

There are two clocks, one that is stationary on Earth, and one in my spaceship. I fly towards the stationary clock at a speed of .90c in my spaceship. When the clock in my spaceship is right above the stationary clock they become synchronized, at that exact moment, I turn on my flashlight and point it forward (in the direction of my motion). I continue to fly away at .90c for 1 second of the stationary clock. How fast is the beam of light moving in my frame of reference?

In the stationary clock's frame, I am 270,000 km away from the clock after one stationary second, and the beam of light is 300,000 km away from the stationary clock after one stationary second. That means that after 1 stationary second, the light is 30,000 km away from me in the stationary clock's frame. To summarize:

t=1 second
L=30,000 km

Now let's find out the time and distance in my frame:

Because I measure the stationary clock as running slow by 43.6% and vice versa, every second in the stationary clock’s frame is converted to:

t(me)=0.436 seconds

Because length along my axis of motion is contracted by 43.6%, 30,000 km in the stationary clock's frame is converted to:

L(me)=13,076.7 km

But t(me) and L(me) are apples-to-oranges, because in my frame:

t=1 second
L=300,000 km

That is, as measured by me in my frame, the light is 300,000 km away from me after 1 second, which is 10 times further than the distance measured by the stationary clock after 1 second by its clock. But we said above “Now let's find out the time and distance in my frame.” In 1 second in either frame, the L=30,000 km that is measured by the stationary clock is only 10% of the distance that I measure. In my frame, L=300,000 km. So to make the time and distance apples-to-apples, I must multiply L(me) by a factor of 10:

L(me)=130,767

Therefore, the light is 130,767 km away from me in .436 seconds in my frame. To calculate the speed that the light is traveling relative to me I divide the length by the time:

v=L(me)/t(me)
v=130,767/0.436
v=300,000 km/s

 

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zankat,

This thought experiment doesn't focus on how far the light is from the stationairy clock, it focuses on how far the light is from the spaceship. And the fact is that the light is 30,000 km away from the spaceship after 1 second in the stationairy clock's frame of reference.

I agree with you that to the stationairy clock the speed of the light is equal to c. However, relativity asserts that not only is the speed of light equal to c in a "relatively" stationairy frame of reference, but that it is also c in any inertial frame of reference, such as in the spaceship in my experiment.

To repeat myself, the light is 30,000 km away from the spaceship after 1 second in the stationairy clock's frame of reference. There is no need to apply 10% to the distance or the time, all that is necessary is to apply the length contraction and time dilation ratios to those values.

Of course there is another way to calculate the speed, but it leads to the same result:

t=1 second
L1=300,000 km (the distance traveled by the light)
L2=270,000 km (the distance traveled by the spaceship)

After time dilation and length contraction:

t(me)=.436 seconds (the dilated time)
L1(me)=130,800 km (the contracted distance traveled by the light)
L2(me)=117,720 km ( the contracted distance traveled by the spaceship)

But you still have to find out how far the light is from the spaceship so that we can find the speed of the light RELATIVE to the spaceship:

L=L1-L2
L=130,800 km-117,720 km
L=13,080 km

When you divide this number by the contracted time, you get:

v=13,080 km/.436 seconds
v=30,000 km/s

Tom

 

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The relativistic explanation

Hi Tom,

Here's the explanation using Special Relativity. First it is a good idea to clear out some misunderstandings:

• The formula for time dilatation that you use, t' = <font face="symbol">g</font>t is not valid for this scenario. It is only valid in what you call a "roundtrip scenario": the two events, i.e. shining the lightbeam and letting it return, have to happen at x = 0 for the observer before this formula can be used (and that would hence require a roundtrip).
• The length contraction formula is also dubious, I'd have to look up the exact conditions where you can use the formula... I think the only restruction is that the measurements of the start and the end of the length to measure have to happen simultaneously (obviously).

To relate physical quantities between observers, you need special transformations. In this case we will use the so called "Boost Transformation", which is a special case of the Lorentz-transformation. A Boost transformation is a Lorentz Transformation where one observer moves along the x-axis of the other observer. Denoting O the "stationary" observer, and O' the moving observer, then we have for the general form of the Boost transformation:

t' = <font face="symbol">g</font>(t - vx/c²)
x' = <font face="symbol">g</font>(x - vt)
y' = y
z' = z

where v is the relative speed between the observers.

Note that the formula for the time transformation reduces to the formula you use in the case that x = 0 (i.e. the two events happen at the same location for the observer). In our scenario, the start event is the shining of the lightbeam, which occurs for the moving observer at x' = 0, but after one second, the light beam will NO longer be at x' = 0, and hence your formula is not valid!

The (hopefully

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) correct calculation would be as follows, assuming that the speed of light is c = 300.000 km / s in the stationary frame of reference:


STATIONARY FRAME OF REFERENCE - O:

Event 1: Turning on the light
t₁ = 0 s
x₁ = 0 km

Event 2: After one second in the stationary FOR
t₂ = 1 s
x₂ = 300.000 km


MOVING FRAME OF REFERENCE - O':

Event 1: Turning on the light
t'₁ = <font face="symbol">g</font>( t₁ - vx₁/c²) = 0 s
x'₁ = <font face="symbol">g</font>(x₁ - vt₁) = 0 km

Event 2: After one second in the stationary FOR
t'₂ = <font face="symbol">g</font>(t₂ - vx₂/c²) = <font face="symbol">g</font>(1s - (0.9c)*(c km) / c² ) = 0.1*<font face="symbol">g</font> s

x'₂ = <font face="symbol">g</font>(x₂ - vt₂) = <font face="symbol">g</font>(c km - (0.9c)*(1s) ) = <font face="symbol">g</font>*(0.1c) km

This yields the following speed of light:

v'_light = (x'₂ - x'₁) / (t'₂ - t'₁) = x'₂ / t'₂ = <font face="symbol">g</font>(0.1c km) / <font face="symbol">g</font>(0.1 s) = c km/s.

Note that I have used "c km" to indicate "300.000 km".


HOW TO INTERPRET THE RESULTS

First of all, we calculate what <font face="symbol">g</font> is:
<font face="symbol">g</font> = 1 / ( 1 - (v/c)² )^1/2 = 1 / (0.19)^1/2 = 2,294

During the 1s in the stationary frame of reference, (0.1*<font face="symbol">g</font>) s = 0.229 s pass in the moving frame of reference. This is in accordance with the idea that "time passes more slowly for a moving observer".

For the moving observer, after 1s in the stationary frame (or 0.229 s in his frame of reference), the beam of light is at a distance x₂ = <font face="symbol">g</font>*(0.1c) km = 68820 km.


CONCLUSION

Using the special theory of relativity, we have calculated that the speed of light is c for every observer. Not really suprising since this is one of the postulates of the theory and that until today, it has been proven to be mathematically consistent

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.

Note that I have nowhere used that there has to be a "roundtrip" of light of some sort. The Boost transformation does not require that. Also note that the derivation of the boost transformation is solely done on the assumption that the speed of light is the same in all frames of reference.


For more information on the Boost transformation:

<A HREF="http://www.physics.nyu.edu/hogg/sr/sr.pdf">Derivation of the Boost transformation</A>, p20 (Princeton course document, quite technical)
<A HREF="http://www.beyondweird.com/einstein/contents/ap01.htm">Simple Derivation of the Lorentz transformation</A> (actually it's a boost transformation

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).



Hope this helps,

Bye!

Crisp

[edited to fix some sub/sup formatting errors]

 

Crisp,

Thanks for your explanation. These are the same formulae that James demonstrated previously.

Although the formulae indicate that the one way trip of light is always equal to c, in all inertial frames of reference, have you considered what you are suggesting by implying that they are correct??

You are claiming that the spaceship's clock is influenced by the speed of other objects in the spaceship's frame of reference. In other words, it the spaceship was travelling at .90c in empty space, it's clock will read .436 seconds, but if it is travelling in the same direction as a beam of light, it's clock will indicate 0.229 seconds.

If that fact isn't illogical enough, what happens if you have multiple beams of light travelling in multiple directions? Will the clock blow up because it can't show all the different times at the same time?

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If I were truly travelling at .90c, what time would my clock indicate considering all the objects in the universe that are moving at different speeds and in different directions?

Because of these serious questions, I have to assume that the formulae you provided are wrong simply because they're illogical. If you can answer my questions so that there is some logic to these formulae, I may reconsider my beliefs.

Tom

 

Prosoothus:

Agreed. But your story says “Now let's find out the time and distance in my frame of reference.” In your frame, the ship’s frame, the fact is that the light is 300,000 km away from the ship after 1 second. In the ship’s frame, L=300,000 km.

In this alternate story, L=L1-L2 is true only in the stationary clock’s frame. In your original story, the question is “How fast is the beam of light moving in my frame of reference?” And you say “Now let’s find out the time and distance in my frame of reference.” In your frame, the ship’s frame, L=L1=300,000 km, so v=300,000 km/s.

 

Hi Tom,

" You are claiming that the spaceship's clock is influenced by the speed of other objects in the spaceship's frame of reference. In other words, it the spaceship was travelling at .90c in empty space, it's clock will read .436 seconds, but if it is travelling in the same direction as a beam of light, it's clock will indicate 0.229 seconds."

This is not correct: the clock onboard the spaceship will always tick at the same rate, i.e. 1s = 1s. What we are considering here are events that are independent of the observer, and we are looking at how the different observers perceive them.

Switching on the light, and then off, takes 1s for the stationary observer. The moving observer however, will say it took 0.229s. We are talking about those two events now, other events might have different effects. It could very well be that the stationary observer measures 1s between switching on the light and turning it off, and 1s between dropping a penny and it falling on the floor, while the moving observer perceives 0.229s for the lightflash and 0.834s for the penny. (the last number was typed randomly). It is NOT the observer we are talking about and how he perceives things, but we are talking about EVENTS and the time seperation inbetween events. Think about it for a while: one stationary meter will always be 1m for an observer, and one second will always be 1s if measured on a stationary clock. From the moment we throw a stick of 1 m away at 0.9c from the observer, he will measure it and say that it is no longer 1m, but a contracted length. The contraction is not something that happens at the observer's side, but at the object's/event's side!

(Small thought on that last sentence: ok, it is the observer that sees the contraction and measures the time dilatation - note that I wrote MEASURES and not "experiences" or something - but it is independent of the observer's motion relative to another observer, it is only dependent on the object's motion relative to the observer).

"I have to assume that the formulae you provided are wrong simply because they're illogical. If you can answer my questions so that there is some logic to these formulae, I may reconsider my beliefs."

Once you bend your mind into seeing that we are talking about events, and it are the distances and times between events that observers disagree on, then I think we are a whole lot closer to resolving your relativity disbelief

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.

Bye!

Crisp

 

Crisp,

If the clock on the spaceship always ticks at the same rate (1 stationairy second = 0.436 spaceship seconds) how can you say that a 1 second stationairy event can occure at more or less time than the clock on the ship indicates??

Where is the proof??? As far as I know, the only way of measuring time is with clocks, but you are claiming that the clock in the spaceship is wrong. If you have another way of measuring "real" time, please share.

Is there anywhere that I can put my clock in my spaceship where it will indicate that 0.229 seconds instead of 0.436? And if there isn't, how can I even assume that 0.229 seconds passed instead of the 0.436 seconds that my clock is indicating??

Tom

 

Hi Tom,

I have a feeling we are getting somewhere

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.

"If the clock on the spaceship always ticks at the same rate (1 stationairy second = 0.436 spaceship seconds) how can you say that a 1 second stationairy event can occure at more or less time than the clock on the ship indicates??"

First of all, forget about the 0.436s, it does not occur anywhere in this problem.

A clock always ticks at the same rate. If someone on the spaceship looks at the clock, he will see one second passing as one second, there is nothing wrong with his clock. When comparing to how fast time goes for other people/observers from his point of view, things start to go funny with Special Relativity.

To explain how you can compare times, let's for a minute forget about the spaceship/light example, and take the example of a penny dropping on the floor. Imagine you stand next to the penny and you measure it to drop on the floor 1s after letting it go - let's denote this time t = 1s.

Now let's drag in another observer: if he lets a penny drop on the floor, then for him, it will also take t' = 1s, regardless of what speed he is moving at with respect to the other observer. The same kind of events take the same time in each frame of reference (sidenote: we are assuming that gravitational attraction of the penny is the same for everybody

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).

The transformation between the observers tells you something on how one observer sees what happens at the OTHER observer's location. If observer O, that measured t = 1s for his penny dropping on the floor, sees observer O' rushing by at 0.9c, dropping HIS penny on the floor, then observer O will conclude: "hey, my penny takes 1s to drop on the floor, but his penny takes 2.294 seconds to drop on the floor!".

Observer O' on the other hand, would have noticed nothing special about his penny, and would have measured it to took 1s for the penny to drop.

That's the entire point of the discussion: an observer would see nothing go wrong or funny, it are only the people that look at him see things going wrong.

Sorry for the ridiculous example, but I hope it illustrated what is meant by times going slower or faster. Back to our spaceship/light example now:

The beam of light travels 1s for the stationary observer. After that time, the beam of light stops there (ok, let's assume we freeze the situation at that moment). The stationary observer would have measured 1s on his clock. The moving observer, at that time, would have only measured 0.229s for the lightbeam to move after it departed from his point of view. For him, the lightbeam would still have travelled at lightspeed c, but it would only hasn't travelled that long to reach the point were we frooze everything. (Note that for him the light also hasn't travelled that far because of length contraction!).

Hope this clears something up

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...

Bye!

Crisp

 

Crisp,

Unfortunately, it doesn't.

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Let's simplify the situation. First let's get rid of all the observers. We only have 2 clocks: one that is relatively stationairy and one on the spaceship that is travelling at .90c.

1) When the "stationairy" clock says 1 second what time does the clock in the spaceship indicate?

2) Is the clock in the spaceship influenced by anything else besides the speed of the spaceship??

3) If the answer to 2 is no, how can you assume that the time is anything else than what the clock in the spaceship indicates?

You keep saying that two seperate events that both take 1 second to occur in the stationairy clock's frame of reference, can take two different times to occur in the spaceship's frame of reference. However, you seem to be forgetting that the clock in the spaceship can only indicate ONE time. And since the clock in the spaceship is the only reliable way an observer in the spaceship can measure time, it's the ONLY device that offers any proof of the actual time. All other values, regardless of the formulae used to obtain them, are fictious until there's a clock that confirms them.

Tom

 

Hi Tom,

"Let's simplify the situation. First let's get rid of all the observers. We only have 2 clocks: one that is relatively stationairy and one on the spaceship that is travelling at .90c. "

You can't get rid of the observer, the clock is the observer in this case (whether somebody reads it or not is irrelevant for the physical reality ofcourse).

"1) When the "stationairy" clock says 1 second what time does the clock in the spaceship indicate?"

In relativity (or any other physical theory) you have to speak in reference to what 1 second of time is. We'll assume that it is relative to t₁ = 0 where the light is turned on. Then the time-interval t = t₂ - t₁ = 1 s corresponds to t' = t'₂ - t'₁ = 0.229 s. Here t₂ and t'₂ refer to the same physical event (event number 2: the light beaming reaching the position such that it is at 300.000 km away from the starting point for the stationary observer/clock). The TIME DIFFERENCE between the two events is different for both observers.

"2) Is the clock in the spaceship influenced by anything else besides the speed of the spaceship??"

No, the clock itself is not influenced, the time between two events is dilatated, and the "space between events" (which is a distance or length) is contracted.

"3) If the answer to 2 is no, how can you assume that the time is anything else than what the clock in the spaceship indicates?"

That's exactly the core of relativity: there is no way to say at what time something happened, it is relative to the observer. But what your question probably refers to is how you can measure time if one second remains one second... [see below]

"You keep saying that two seperate events that both take 1 second to occur in the stationairy clock's frame of reference, can take two different times to occur in the spaceship's frame of reference."

Yes, entirely correct.

"However, you seem to be forgetting that the clock in the spaceship can only indicate ONE time."

Ofcourse, but you are forgetting that something that happens simultaneously for one observer (eg. the light beam arrives and the penny drops on the floor, both took one second for the stationary observer) is not necessarily simultaneous for the other observer. Simultanity is an absolute concept, which does not fit into the theory of relativity.

You have to make a clear distinction between events on the one side - those are considered to be absolute: if one observer sees a penny hitting the floor, then sooner or later, another observer will also see this happen. On the other side you have the observers that witness the events. Both are considered to be completely independent: the observer cannot react with the events, and is assumed to be passive.

Each observer has to give a time label and space coordinate to each event, and it happens to be that different observers give different labels. For example, in our spaceship-light problem, both observers label the start of the lightbeam with 0 (t₁ = 0s and t'₁ = 0s), but the arrival of the lightbeam has different labels: t₂ = 1s and t'₂ = 0.229s.

By seperately labeling each event, you avoid confusion about simultanity: an event in this concept would be simultaneous if t₁ = t'₁ and t₂ = t'₂, which is not the case here. This means that for the moving observer the light beam will arrive before it does for the stationary observer!

(Note that this has nothing to do with roundtrips of light: the event really happens sooner for the moving observer, nowhere we took into account that for the moving observer to know that the lightbeam did in fact arrive, there needed to be some sort of information transfer, i.e. time dilatation is more than just the finite transmission of signals).

Bye!

Crisp

 

Crisp,

Your confusing me. First I asked:

You answered:

You are indicating that the atomic clock in the spaceship says that .229 seconds passed. What bothers me is that you calculated this figure using the speed of the light from my flashlight. What if I didn't turn on the flashlight? Would my clock say something different? What if I pointed my flashlight in the opposite direction? Would this influence the clock as well? But then you say:

Which is contrary to what you said above. The figure .229 seconds you obtained was influenced by my flashlight, but then you turned around and said that the flashlight shouldn't influence the clock.

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I will give you two hypothetical situations to try to understand what you are saying. Please indicate which one is correct:

1) Two events in the stationairy frame of reference take 1 second to occur. These two events take two different times to occur in the spaceship's frame of reference. The atomic clock in the spaceship is influenced by these two events.

2) Two events in the stationairy frame of reference take 1 second to occur. These two events take two different times to occur in the spaceship's frame of reference. The atomic clock in the spaceship IS NOT influenced by these events. The atomic clock is only influenced by the speed of the spaceship.

Telling me which one is correct, 1 or 2, will really help me understand your point of view.

One more question I would like you to answer regardless of whether you chose answer 1 or 2:

Are you sure that you aren't confusing the time that two events occur WITH when those same events are "observed"? After all, you don't have to travel at relativistic speeds to observe two events at the same time that actually occurred at different times. This is simply the result of the finite speed of light.

Tom

 

zankat,

If I wanted to find out how far the light was in my frame of reference, why wouldn't I just take the 30,000 km (the distance between me and the light in the stationairy frame of reference), and apply the length contraction factor to it?

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Tom

 

Prosoothus:

Because that method ignores a key piece of information, that the distance the light pulse is ahead of you, as a percentage of the light's elapsed distance, is 10% in the stationary clock's frame, but is 100% in your frame. You must take into account this percentage difference in addition to the distortion factor if you will use the 30,000 km distance in your frame.

The reason another percentage enters the fray, in addition to the distortion factor, is because you are measuring the distance that light travels, as opposed to a material object. Loosely speaking, when you attempt to apply a distortion factor to light, it "breaks free" to ensure that c remains constant. Another percentage will always crop up to make that happen. (Of course this really happens because the distortion formula itself is designed to ensure that c remains constant.)

A roundabout way to see that your method excludes key information is to imagine that the stationary clock has the flashlight instead of you. What the stationary clock observed about you--namely you being 10% behind the light's elapsed distance--now you would so observe about the stationary clock. If you were to use this information to determine the speed of light in your frame, how would you go about it? The easiest way would be to ignore the 10% distance value altogether and just measure the speed of the light; after all, you care only about the speed of the light, not the speed of the stationary clock or what distance it elapsed. Measuring the speed of the light entails dividing 100% of the distance the light traveled by 100% of the time it took. If you insisted on using the 10% distance value as you did in your story problem, you'd have to take 10% of the time as well, as I did in my first solution above in this thread.

 

Hi Tom,

"You are indicating that the atomic clock in the spaceship says that .229 seconds passed. What bothers me is that you calculated this figure using the speed of the light from my flashlight. What if I didn't turn on the flashlight? Would my clock say something different? What if I pointed my flashlight in the opposite direction? Would this influence the clock as well?"

You are too fixated on the idea that the clock itself ticks more slowly when going at a large speed. It doesn't, from the point of view of the clock, nothing, and I mean nothing, happens: whether it is moving or standing still, it always ticks at the same rate.

To measure a time interval, you need two events: one event that you label the "start" of the time interval, and one that you label "stop". One observer will measure a time (stop - start = 1s) while another will measure (stop - start = 0.229s). In order for this to happen, nothing strange has to happen with the clock, it's reality that changes for the time dilatated observer.

So to answer your questions:

1. If you didn't turn on the flashlight, then there would be nothing to measure. All you could do is look at your clock and say "hey, it just keeps on ticking". Note once again that for someone moving along with the clock, nothing strange would happen, or he wouldn't notice anything strange.

2. The opposite direction also doesn't matter for the clock's ticking rate. If you'd shine the light in the opposite direction, then the start event would be the same, but the "stop" event would be perceived differently by the moving observer.

"Please indicate which one is correct:"

Situation 2 is correct: the moving clock is not influenced for the observer moving along with the clock. I can't stress enough that an observer moving along with the clock will experience nothing strange with times or lengths moving along with him. It are only other people that look at him that see his clock go funny (or events have different time intervals). Different observers will measure different time INTERVALS (= stop - start) for different events.

"Are you sure that you aren't confusing the time that two events occur WITH when those same events are "observed"? After all, you don't have to travel at relativistic speeds to observe two events at the same time that actually occurred at different times. This is simply the result of the finite speed of light."

I'd have to think on this one to be sure actually, a few days ago I thought I had figured it out but now you're making me doubt again

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. I don't think the finite transmission speed of light is taken into account, that would have to be added additionally... Perhaps anybody can comment on this ?

Bye!

Crisp

 

If I understand the question right, you're talking about the difference between what is measured and what is seen. Prosoothus's story problem deals with measurements and ignores the Doppler effect (the signal transmission time delay due to the finite speed of light) that affects what is seen, as do most relativity story problems. Crisp's replies also ignore the Doppler effect.

A "measurement" or "observation" typically implies that the Doppler effect is ignored. In practice or imagination, you can derive a measurement from what is seen by compensating for the Doppler effect. Alternatively, you can employ a host of assistant observers "flying" in formation with you (at rest with respect to you). The assistant observer who is local to what is being measured makes the measurement (sans the Doppler effect by virtue of being local), and sends the information to you.

 

Hi zanket,

"If I understand the question right, you're talking about the difference between what is measured and what is seen. Prosoothus's story problem deals with measurements and ignores the Doppler effect (the signal transmission time delay due to the finite speed of light) that affects what is seen, as do most relativity story problems. Crisp's replies also ignore the Doppler effect."

Yes, this is also what I thought: nowhere in the derivation of the Boost transformation one uses the fact that light (or any other signal) has to be transmitted in order to actually witness the event. Hence the Boost transformation (and Lorentz transformations in general) describe how the observer experiences reality, ignoring the finite transmission speed of signals. I suddenly recall all those nice lightcone figures that had to be taken into account to calculate when the event happens and when the observer witnesses the event

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Bye!

Crisp

 

zanket,

The "delay" you are referring to is not the Doppler effect. The Doppler effect is the change in frequency of electromagnetic radiation (or sound) as a result of the motion of the source of that radiation.

Tom

 

Hi Tom,

Nomenclature aside, I think the point was clear

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Bye!

Crisp