Why nothing can go faster than speed of light.

[Robittybob1]

 

Time dilation and length contraction are a consequence of the speed of light being invariant, so your question doesn't make much sense.

 

 

Again, I don't really understand what you are asking. "distance per sec" is known as speed and so all observers will see light moving at the same "distance per sec".

Remember we agreed that in the forward direction the craft moved at half the speed of light, but light shown to the rear would separate from the craft at 0.5c plus c.

To the front they say it is because of time dilation and length contraction they can account for this apparent slower moving wave front (0.5c) but how do they account for the same at the rear?

Can you see where the video is in error?

Edited March 8, 2016 by Robittybob1

[Mordred]

its not 0.5 plus c the craft is 0.5c to outside observer at rest the speed of light is 1 c to outside observer at rest. same with observer on the ship. the length and time dilation cdifferences between observers allows c to remain invarient to all observers. The video isnt wrong you had to listen to it correctly

[StringJunky]

One method has a spinning mirror, which rotates a certain amount between the first bounce and the return. The deflection and rotation speed tell you the elapsed time.

Do you have any links that might elaborate on the apparatus setup and how it works? Or even the name of that technique would help me to search myself.

[Robittybob1]

its not 0.5 plus c the craft is 0.5c to outside observer at rest the speed of light is 1 c to outside observer at rest. same with observer on the ship. the length and time dilation cdifferences between observers allows c to remain invarient to all observers. The video isnt wrong you had to listen to it correctly

In #43 Strange says "But, yes the maximum separation speed is 2c - but it is important to note that no one, in any frame of reference, sees anything moving at more than c." I accept this but if we look at the original example the craft is going at 0.5c not nearly 1c, so the separation will be 0.5c + 1c (for the light and the craft are going in opposite directions).

Just as in the video when the light is going in the same direction as the ship the light advances twice as fast as the ship so it appears to have only gone half as far in front of the ship compared to the stationary ship. They then say time dilation accounts for some of the change, so there is more time to go the same distance but it didn't account for all the difference, and the implication is that length contraction accounts for the rest.

This next bit is not in the video:

If they fired their lasers to the rear can you use length contraction and time dilation to account for the even greater apparent distance traveled by the light?

The arguments used in one direction seem inadequate in the opposite direction.

 

There isn't "time contraction behind you" or "length dilation behind you" is there? When the direction of travel is opposite to the light travel it seems like you need the opposite of time dilation and length contraction to account for the rate that the light wavefront separates from the source.

The EO has no problem accounting for the change, it is only those onboard the moving ship that need time dilation and length contraction explanations hence allowing c to be invariant. yet the ones onboard aren't aware of either of these two effects so they can't use them in their arguments.

 

I tend to think the arguments in the video are wrong, for speed of light measurements are always based on the two-way speed of light. You can only sense the light when it comes back to you. Like it has to be reflected back and you measure the time interval for light to travel the distance and back.

Do you have any links that might elaborate on the apparatus setup and how it works? Or even the name of that technique would help me to search myself.

https://en.wikipedia.org/wiki/Fizeau%E2%80%93Foucault_apparatus is this a start?

Mordred - I think you might have pasted your reply into the wrong thread

http://www.scienceforums.net/topic/93472-gravitational-lens-and-gravitational-waves-question/page-3#entry909902

 

 

hooh boy there is a big difference between seperation speed and actual speed of travel.

 

lets run a basic analogy.

 

you have a ball, a wall and a thrower. the thrower throws the ball at the wall and runs in the opposite direction. ball is moving 20 m/s in one direction, boy is moving 20 m/s in the opposite direction. The vector sum of the two ACTUAL speeds is the seperation speed. 40 /m/s in this case. The ball still moves at the same speed 20 m/s regardless if the thrower moving at 20 m/s is running away or towards the ball and same is true for the ball to the thrower.

 

now apply that to your spaceship analogy. where the actual speed of light is always c ( PERIOD.)

 

I can't see a connection to the topic sorry.

If one was use the physics of the situation of the ball thrower throwing the ball and running toward the wall, then there might be a connection. But I don't see that. So throwing the ball is a separate event to the running toward the wall or away from the wall.

Edited March 8, 2016 by Robittybob1

[Mordred]

spacecraft a is moving at 75c . (right) spacecraft B is moving 75c to the left. velocity of seperation from an external observer is 1.8 c

 

however observer on craft a looking at spacecraft b will get a value less than c from observer a on the craft looking at craft b he will say craft b is moving at 96c.

Edited March 8, 2016 by Mordred

[Robittybob1]

oh man come on think. OK lets simplify it further.

 

spacecraft a is moving at 90c . (right) spacecraft B is moving 90c to the left. Neither craft is moving faster than 9c but if Observer in craft A measures the speed of craft B. (via doppler shift) he will get a Doppler shift of 1.8c. If craft a is going the same direction as craft b and the same speed he will measure a speed of 0 m/s via doppler shift.

 

The light does not advance twice as fast. LIGHT ALWAYS MOVES AT C>>>>> You can however get different doppler red/blue shifts.

I'll correct the errors first:

"spacecraft A is moving at 0.9c to the right spacecraft B is moving 0.9c to the left. Neither craft is moving faster than 0.9c but if Observer in craft A measures the speed of craft B. (via doppler shift) he will get a Doppler shift of 1.8c. If craft A is going the same direction as craft B and the same speed he will measure a speed of 0 m/s via doppler shift.

The light does not advance twice as fast. LIGHT ALWAYS MOVES AT C>>>>> You can however get different doppler red/blue shifts."

 

That seems nearly right but in these relative motion situations one observer is at rest and the other will be in motion. I'm not sure how Doppler and relativity relate to each other. Do we add the two motions before you do the doppler calculations IDK.

[Mordred]

I fixed that post I had the wrong observer lol.

 

speration is literally how fast two objects are moving away or toward each other. It is not a measure of the individual speeds

Edited March 8, 2016 by Mordred

[Janus]

In #43 Strange says "But, yes the maximum separation speed is 2c - but it is important to note that no one, in any frame of reference, sees anything moving at more than c." I accept this but if we look at the original example the craft is going at 0.5c not nearly 1c, so the separation will be 0.5c + 1c (for the light and the craft are going in opposite directions).

Just as in the video when the light is going in the same direction as the ship the light advances twice as fast as the ship so it appears to have only gone half as far in front of the ship compared to the stationary ship. They then say time dilation accounts for some of the change, so there is more time to go the same distance but it didn't account for all the difference, and the implication is that length contraction accounts for the rest.

This next bit is not in the video:

If they fired their lasers to the rear can you use length contraction and time dilation to account for the even greater apparent distance traveled by the light?

The arguments used in one direction seem inadequate in the opposite direction.

There is one more important aspect of SR that you are missing: The Relativity of simultaneity.

 

Let's assume that our spaceship has three clocks, one in the front, one in the back and one in the middle, and according to the ship, they are all in sync with each other. The ship is long (one light sec long as measured by the ship). A laser in fired from the middle clock towards both the front and back and reflected back to the middle. The time on the front and rear clocks upon arrival is noted, along with the time on the middle clock upon the light's return.

The laser's leave the middle clock when it reads 0, strikes both of the other clocks when they read 0.5 and returns to the middle clock when it reads 1 sec. In the ship this is explained by the fact that light travels at c and the round trip distance is 1 light sec.

 

Now consider what happens to each laser according to an observer watching the ship traveling at 0.5 c. One laser leaves the middle clock when it reads 0, and heads toward the front of the ship. The ship is length contracted, so the front for the ship is 0.433 light sec away, and due to the relative difference between the ship's velocity and the speed of light, arrives at the front of the ship after .866 sec. it reflects back towards the middle. Because the light is moving in the opposite direction from the ship, it will take 0.2887 sec to meet up with the middle clock for a total round trip time of 1.1547 sec. Due to time dilation,the middle clock ticks off 1 sec in this time.

The laser going in the opposite direction takes 0.2887 sec to reach the rear clock and 0.866 sec to return to the middle clock. This is the same 1.1547 sec round trip time.

Now we just need to consider the front and back clocks. In the time that it takes the forward facing laser takes to reach the front clock, 0.75 sec passes on the ship's clocks. The front clock still reads 0.5 sec when the light strikes it. This means that when the light left the middle clock it read -0.25 sec.

In the time it takes the rearward facing laser to reach the the clock tick off 0.25 sec. this means that when the light left the middle clock it already read 0.25 sec.

Thus while an observer in the ship will say that all three clocks read the same at all times, according to the observer for which the ship is moving at 0.5c, the front clock always reads 0.25 sec behind the middle clock and the rear clock always reads 0.25 sec ahead of the middle clock. This is the Relativity of simultaneity.

So even though the ship measures the light as traveling at c relative to itself in both directions, and the other observer measures the speed difference between ship and light to be different in the two directions, they both still agree as to the readings on all the clocks when the light strikes them due to the effects of time dilation, length contraction and Relativity of Simultaneity..

[Mordred]

Thanks Janus, I'm a bit too busy to concentrate atm. Making too many stupid mistakes lol

[Robittybob1]

@Janus - That is rather difficult to understand but I'll have a go at it.

 

The bit that really confused me was this:

 

 

....means that when the light left the middle clock it read -0.25 sec.
In the time it takes the rearward facing laser to reach the the clock tick off 0.25 sec. this means that when the light left the middle clock it already read 0.25 sec.

.Is that physically possible when you have set the center clock to fire at 0.00 sec? That sounds like a malfunction if it fires on two occasions.

Edited March 8, 2016 by Robittybob1

[Mordred]

One handy idea would be to read and study this low math textbook on SR.

 

 

http://www.lightandmatter.com/sr/

It's a full open source textbook.

[Strange]

In #43 Strange says "But, yes the maximum separation speed is 2c - but it is important to note that no one, in any frame of reference, sees anything moving at more than c." I accept this but if we look at the original example the craft is going at 0.5c not nearly 1c, so the separation will be 0.5c + 1c (for the light and the craft are going in opposite directions).

 

But that separation speed is from the point of view of someone "stationary". From the point of view of people of the craft, there is no difference between light going in any direction.

"spacecraft A is moving at 0.9c to the right spacecraft B is moving 0.9c to the left. Neither craft is moving faster than 0.9c but if Observer in craft A measures the speed of craft B. (via doppler shift) he will get a Doppler shift of 1.8c. If craft A is going the same direction as craft B and the same speed he will measure a speed of 0 m/s via doppler shift.

 

This is wrong. The observers will measure a speed of 0.995c for the other craft receding from them.

 

No one can ever measure anything moving at more than c.

Edited March 8, 2016 by Strange

[Robittybob1]

 

But that separation speed is from the point of view of someone "stationary". From the point of view of people of the craft, there is no difference between light going in any direction.

 

This is wrong. The observers will measure a speed of 0.995c for the other craft receding from them.

 

No one can ever measure anything moving at more than c.

You need to relate it to the situation in the video.

It might be wrong but it was what Mordred who was saying this and I rewrote his post since he had made so many mistakes. I was looking into how the relativistic Doppler should be handed.

And I think you have handled it correctly. Cheers.

Edited March 8, 2016 by Robittybob1

[swansont]

Do you have any links that might elaborate on the apparatus setup and how it works? Or even the name of that technique would help me to search myself.

 

 

"rotating mirror speed of light" gives several pdfs and other links.

Such as https://www.phys.ksu.edu/personal/rprice/SpeedofLight.pdf

 

This one has animations

http://www.pas.rochester.edu/~pavone/particle-www/teachers/demonstrations/FoucaultDemonstration.htm

[StringJunky]

 

 

"rotating mirror speed of light" gives several pdfs and other links.

Such as https://www.phys.ksu.edu/personal/rprice/SpeedofLight.pdf

 

This one has animations

http://www.pas.rochester.edu/~pavone/particle-www/teachers/demonstrations/FoucaultDemonstration.htm

Thanks.

[Mordred]

 

It might be wrong but it was what Mordred who was saying this and I rewrote his post since he had made so many mistakes. I was looking into how the relativistic Doppler should be handed.

And I think you have handled it correctly. Cheers.

 

 

yeah yeah rub it in, remind me sometime I hate graveyard shift...

 

particularly since I'm doing lab tests every half hour grrr

[Janus]

@Janus - That is rather difficult to understand but I'll have a go at it.

 

The bit that really confused me was this:

.Is that physically possible when you have set the center clock to fire at 0.00 sec? That sounds like a malfunction if it fires on two occasions.

The lasers only fire once, when the middle clock reads 0. The point is that while according to the ship observers, the front and rear clocks also read 0 at this moment, according to the "external" observer, they do not; the front clock reads 0.25 sec shy of 0 and the rear clock is already reading 0.25 at the moment the the middle clock reads 0 and the lasers are fired.. The two observers do not agree that the clocks are synchronized.

This is a very important idea in Relativity, just as much so as length contraction or time dilation. Observers in motion with each other do not only disagree on the rate of time flow, or their respective length along the line of motion, but also to what events are simultaneous. The ship says the events of the rear clock, middle clock and front clock reading 0 are simultaneous, while the external observer says that it is the events of the rear clock reading 0.25, the middle clock reading 0 and the front clock reading -0.25 that are simultaneous..

Edited March 8, 2016 by Janus

[Robittybob1]

The lasers only fire once, when the middle clock reads 0. The point is that while according to the ship observers, the front and rear clocks also read 0 at this moment, according to the "external" observer, they do not; the front clock already reads 0.25 sec and the rear clock is reading 0.25 sec shy of 0 at the moment the the middle clock reads 0 and the lasers are fired.. The two observers do not agree that the clocks are synchronized.

This is a very important idea in Relativity, just as much so as length contraction or time dilation. Observers in motion with each other do not only disagree on the rate of time flow, or their respective length along the line of motion, but also to what events are simultaneous. The ship says the events of the rear clock, middle clock and front clock reading 0 are simultaneous, while the external observer says that it is the events of the rear clock reading -0.25, the middle clock reading 0 and the front clock reading 0.25 that are simultaneous..

I thought I was replying to Mordred sorry.

Janus I'll read it again in light of what you've just told me.

You confirm the laser fires when the center clock reads 0:00 and at no other time.

You also confirm it only fires once.

 

On what basis does this happen "the external observer says that it is the events of the rear clock reading -0.25, the middle clock reading 0 and the front clock reading 0.25 that are simultaneous"? Is that from evidence or just from his calculations? Or was it from his own synchronised clocks at those positions?

 

 

 

yeah yeah rub it in, remind me sometime I hate graveyard shift...

 

particularly since I'm doing lab tests every half hour grrr

I feel for you. Don't over do it.

Edited March 8, 2016 by Robittybob1

[Janus]

I thought I was replying to Mordred sorry.

Janus I'll read it again in light of what you've just told me.

You confirm the laser fires when the center clock reads 0:00 and at no other time.

You also confirm it only fires once.

 

On what basis does this happen "the external observer says that it is the events of the rear clock reading -0.25, the middle clock reading 0 and the front clock reading 0.25 that are simultaneous"? Is that from evidence or just from his calculations? Or was it from his own synchronised clocks at those positions?

First off, I must apologize, along the line between my last posts I inadvertently switched the order of the clock readings. I've gone back and edited the posts to correct this. But I will note it it here also. It is the front clock that reads -0.25 sec and the rear that reads 0.25 sec when the middle clock reads 0.

 

A to the question. Assume that the external observer has clocks strung along the path of the ship that according to him, are all in sync. The clocks are spaced so that at a given moment, three of them line up with the three clocks in the ship.

Further assume that when the middle clock of these three external clock lines up with the middle clock of the ship, they both read 0. Then, according to the external frame, all three external clocks read zero at that moment, and the forward external clock will read zero while it is next to the front ship clock when it reads -.25 sec, and the rearward external clock will read 0 when it is next to the rear ship clock when it reads 0.25 sec. So yes by comparing the readings of its own clocks to the readings of the ship's clocks, the external frame can measure that when the middle ship's clock reads 0, the rear clock reads 0.25 and the front clock reads -.25

Again,this is just as much a consequence of Relativity as time dilation and length contraction. In fact, it is only by taking all three into account that you can make sense of what is going on.

 

For example. We will start at the point where the front ship clock reads -.25 sec while it is next to the external forward clock when it reads 0, and work out the respective readings on the middle ship clock and forward external clocks when they meet.

 

First from the external frame. The three ship and three external clocks start lined up. In order for this to be true the external clocks must be 0.433 light sec apart (the three clocks in the ship are 0.5 light sec apart in the ship frame, but as measured from the external frame this distance is length contracted to 0.433 light sec.)

It takes .866 sec for the middle clock to travel 0.433 light sec at 0.5c so the external clocks will tick off .866 sec. Since the forward external clock started at 0, it will read 0.866 sec when the middle ship clock reaches it. The middle ship clock, being time dilated will tick off 0.75 sec and since it also started at zero, this is what it will read when it meets up wit the forward external clock.

Thus when the two clocks meet, the middle ship clock reads 0.75 sec and the forward external clock reads 0.866 sec.

 

Now the ship frame. Front ship clock reads -.25 sec and forward external clock reads zero when they are nest to each other. The distance between front and middle ship clocks is 0.5 light sec. At .5 c this means its takes 1 sec for the middle ship clock and forward external clock to meet in the ship frame. Since the front ship clock starts at -.25 sec and all three ship clocks are in sync in this frame, the middle ship clock also reads -.25 sec when we start so reads -0.25+1 = .75 sec when it meets up with the forward external clock. The forward external clock starts at zero and runs at a time dilated rate of 0.866, so it reads 0.866 when it meets up with the middle ship clock.

Thus when the two clocks meet, the middle ship clock reads 0.75 sec and the forward external clock reads 0.866 sec; the same as we got for the external frame.

 

Now consider one more thing, above we said that from the external frame, the three external and three ship clocks lined up at the same moment. This does not happen according to the ship frame. In the ship frame, the ship clock separation is 0.5 light sec and the it is the external clock separation that is length contracted. Sine this proper distance is .433 light sec in the external frame, it will be 0.375 light sec in the ship frame.

 

Thus this means that when the front ship clock is aligned with the forward external clock, then the middle ship clock and middle external clock are not aligned. The middle ship clock will be 0.125 light sec short of the middle external clock. It will not be until .125/.5 = .25 sec later that the two will meet. Adding 0.25 sec to the present reading of -.25 sec of the middle ship clock means its will read 0 when it lines up with the middle external clock. (again agreeing with the external frame.)

The external middle clock will read -0.10825 sec when the two forward clocks align and will advance 0.125*.866= 0.10825 and also read 0 when it aligns with the middle ship clock.

 

The point is that if you take any pair of clocks, one from each frame, everyone agrees as to what they read as they pass each other, However you compare those same two clock's reading when they are separated from each other, the two frames will disagree as to their respective readings.

[Robittybob1]

It is confusing to say the least how all three clocks on the craft can be synchronized and all three clocks on the exterior can be synchronized yet when you compare them 1:1 they are reading different times.

 

You have covered this before but I may not have got my head around it yet. I'll work on it.

Method to overcome Relativity of simultaneity.

 

Each frame of reference synchronize their clocks using the (2 way time) of travel method.

as used in this video:

using light beams and spacetime diagrams.

 

When the frames of reference pass each other a laser fires when the center clocks align.

One half of the beam is split again and half goes along the craft in both directions.

The return time is compared and the average is used to synchronize the distant clocks.

The other half of the original beam is sent to the platform and there split again and half goes along the platform in both directions.
The return time is compared and the average is used to synchronize all the many distant clocks along the platform.

 

Will the amount of time that the various clocks are out on the second attempt to synchronize them inform us which way (direction) and how much the relative motion has changed?

[Robittybob1]

IFoR = inertial frames of reference.

Situation 1. If two IFoR are moving at the same rate light reflected off one will return in the same time period to the sender.

Situation 2 If one IFoR is moving away slower the light reflected off one will return in shorter time period to the sender.

 

Situation 3 If one IFoR is moving away faster the light reflected off one will return in a longer time period to the sender.

 

How to synchronize their clocks.

Return time = time period between start of outgoing signal from A and the reflected return signal.

Situation 1. If two IFoR are moving at the same rate light reflected off one will return in the same time period to the sender.

step 1. send a signal with the time plus half the return time, and get second person (B) to set clock to this time.

Edited March 10, 2016 by Robittybob1

[Klaynos]

IFoR = inertial frames of reference.

If two IFoR are moving at the same rate

 

Relative to what?

light reflected off one will return in the same time

In what frame?

period to the sender.

If one IFoR is moving away slower

Relative to what?

the light reflected off one will return in shorter time

In what frame?

period to the sender.

 

If one IFoR is moving away faster

Relative to what?

the light reflected off one will return in a longer time

In what frame?

period to the sender.

Etc...

[swansont]

IFoR = inertial frames of reference.

If two IFoR are moving at the same rate light reflected off one will return in the same time period to the sender.

If one IFoR is moving away slower the light reflected off one will return in shorter time period to the sender.

 

If one IFoR is moving away faster the light reflected off one will return in a longer time period to the sender.

 

How to synchronize their clocks.

Return time = time period between start of outgoing signal from A and the reflected return signal.

1. If two IFoR are moving at the same rate light reflected off one will return in the same time period to the sender.

step 1. send a signal with the time plus half the return time, and get second person (B) to set clock to this time.

 

 

 

No, that won't work.

 

Your time will depend on both your speed and your location with respect to a point in another frame. Many examples rely on the clocks becoming co-located for comparison to avoid this issue.

 

https://en.wikipedia.org/wiki/Lorentz_transformation#Coordinate_transformation

 

Note the equation for transforming time from t to t' (from frame F to F')

It depends on the location of the event in F. Events happening at the same time in F but at different location will be seen at different times in F'. It looks like Janus has been giving you an example of this.

 

Relativity is well-tested. You can't synchronize clocks in different frames. Any new idea you come up with which disagrees with this is going to be wrong.

[Robittybob1]

 

 

 

No, that won't work.

 

Your time will depend on both your speed and your location with respect to a point in another frame. Many examples rely on the clocks becoming co-located for comparison to avoid this issue.

 

https://en.wikipedia.org/wiki/Lorentz_transformation#Coordinate_transformation

 

Note the equation for transforming time from t to t' (from frame F to F')

It depends on the location of the event in F. Events happening at the same time in F but at different location will be seen at different times in F'. It looks like Janus has been giving you an example of this.

 

Relativity is well-tested. You can't synchronize clocks in different frames. Any new idea you come up with which disagrees with this is going to be wrong.

Did you look at the video? So was that teacher actually wrong too?

The above method was just for frames going the same velocity. But you were definitely right that the place where the light reflected from and the clock receiving the signal needed to be in the same location.

 

I was trying to comprehend Janus's example, and I am looking at how each frame would synchronize their clocks. Can you synchronize clocks in the same frame without bringing them together?

Relative to what?

In what frame?

Relative to what?

In what frame?

Relative to what?

In what frame?

 

Etc...

Relative to each other. All motion is relative to the sender frame, so if a signal returns at same interval the two IFoR are going at the same relative velocity. If the time period is shorter the distance between them is getting shorter or longer the frames are not going the same velocity.

I'm surprised that you question this?

Edited March 9, 2016 by Robittybob1

[swansont]

Did you look at the video? So was that teacher actually wrong too?

The above method was just for frames going the same velocity. But you were definitely right that the place where the light reflected from and the clock receiving the signal needed to be in the same location.

 

I was trying to comprehend Janus's example, and I am looking at how each frame would synchronize their clocks. Can you synchronize clocks in the same frame without bringing them together?

 

No, I did not watch the video. They aren't a substitute for discussion (and arguably your use of them violates the rules).

 

You described more than one frame. You talk of them moving slower or faster, which only makes sense if it's relative to another frame. Frames going the same velocity are the same frame.