PDA

View Full Version : Theory of Relativity Question

thunderchicken
2005-Apr-29, 09:37 PM
I guess this is more of a physics question than an astronomy question, but it does involve a spaceship!

I am trying to understand the principals of the theory of relativity as it relates to time. I have seen various explanations using the light clock. The clock consists of a light source and a mirror placed directly above it at some height. A pulse of light is shone vertically from the source, reflected off the mirror and back down to the source. The time this takes is counted as one unit.

The light clock is placed in a spaceship moving at a very high speed. To an observer in the spaceship the light takes one unit of time to complete its trip. To an observer out side the spaceship, however this is not the case. To him, the light appears to travel in a V. The light is traveling a longer distance, and therefore must take a longer amount of time to complete its trip. Have I understood this correctly??

In the spaceship’s frame of reference the light is traveling directly vertical, therefore the vertical component of the velocity is C. In the astronaut’s frame of reference the vertical component of the velocity does not seem to be taken into consideration. The speed of light is taken as traveling on an angle, and therefore taking more time to complete the trip. Why is this, since the light is not actually traveling on an angle, it only ‘appears’ to be?

Sine the light hitting the mirror is a real, visible event, would the astronaut not see it reflect off the mirror at exactly the same moment in time as someone in the spaceship?

When I fire a bullet horizontally from a gun, it falls to the ground in the same amount of time as a bullet dropped vertically from the same height. The amount of extra horizontal distance covered by the fired bullet has nothing to do with the vertical motion. What is the difference between this and the light clock?

Any help would be great!

Ricimer
2005-Apr-29, 10:18 PM
even if you stop it at the mirror (and remember, you only see something when it gets absorbed!) the fact remains: The shipboard people will see it have traveled a shorter distance than the stationary observer. And they will both measure it at the same speed...

and that's where the time and legth effects come in.

George
2005-Apr-29, 11:28 PM
The light clock is placed in a spaceship moving at a very high speed. To an observer in the spaceship the light takes one unit of time to complete its trip. To an observer out side the spaceship, however this is not the case. To him, the light appears to travel in a V. The light is traveling a longer distance, and therefore must take a longer amount of time to complete its trip. Have I understood this correctly??
No. You are thinking logically, which does not work. The important aspect of special relativitiy is light's speed can only be observed to be one speed regardless of any other motion.

You assumed it would take more time for the light to have traveled to the mirror from the viewpoint of the non-spaceship observer. Let's consider first the distance only involved. The distance vertically (as seen by the spaceman) we'll call... d. The distance the light travelled as seen by the observer....d'. The spaceship traveled a certain distance as seen by the observer.....d'2, but this is simply the velocity of the ship times the time the observer saw it move, therefore, it can be stated as .....vt'.

These three distances form a right triangle. d'^2 = (vt')^2 + d^2

But d' = ct' and d = ct. Substitute and solve for t'.

This site is quite helpful.... here (http://instruct1.cit.cornell.edu/courses/astro101/lec21.htm)

This shows that time is not the same for both observers.

Sylas
2005-Apr-30, 12:45 AM
The light clock is placed in a spaceship moving at a very high speed. To an observer in the spaceship the light takes one unit of time to complete its trip. To an observer out side the spaceship, however this is not the case. To him, the light appears to travel in a V. The light is traveling a longer distance, and therefore must take a longer amount of time to complete its trip. Have I understood this correctly??

Yes, this bit is perfect. The outside observer sees a longer amount of time between the sending and return of the photon. As you have described the matter, the light clock is positioned by the pilot to be sending light perpendicular to the direction of motion.

You can calculate the relativistic time dilation with your method. The pilot sees the photons travel for time t. In that time, they travel a distance ct, where c is the speed of light. You call this direction "vertical", while the direction of the ship is "horizontal".

Suppose the ship is seen from outside to be travelling at speed v, horizontally. The photon is now actually moving at an angle, in the V shape you describe, and the outside observer sees it moving for a different amount of time t'. The vertical component is still the same (ct), and the horizontal component is give by the velocity of the ship and the total time of the journey seen from outside, which is vt’.

Pythagorus tells us the photons travelled a total distance sqrt((ct)^2 + (vt’)^2).

Curiously, however, the outside observer still sees the photons travelling at the speed of light. Hence ct’ = sqrt((ct)^2 + (vt’)^2). Divide both sides by ct, and you get:
t’/t = sqrt(1+(v/c)^2))

This is the time dilation factor, and your example so far is perfect, allowing the time dilation to be calculated correctly. t’ is greater than t, so the observer sees the pilot’s clock running slow.

Why is this, since the light is not actually traveling on an angle, it only ‘appears’ to be?

There is no problem with the light traveling on an angle for one person but not another. Neither case is the “actual” case; each perspective is just as good as the other. You get the same thing in conventional Galilean perspectives.

Sine the light hitting the mirror is a real, visible event, would the astronaut not see it reflect off the mirror at exactly the same moment in time as someone in the spaceship?

No. This is a key point in many of the so-called “paradoxes” of relativity. Not only is time and distance relative; so too is simultaneity. Two events which occur at the same instant for one observer may have a time lag between then for another observer.

For example; from the perspective of the spaceship pilot, the outside observer’s clock is running slow. Each one observers the other clock running slow.

How can this be? The explanation deals with relativity of simultaneity, and is best explained with the spacetime diagrams.

Cheers -- Sylas

worzel
2005-Apr-30, 01:58 AM
No. You are thinking logically, which does not work. The important aspect of special relativitiy is light's speed can only be observed to be one speed regardless of any other motion.
LOL

In the spaceship’s frame of reference the light is traveling directly vertical, therefore the vertical component of the velocity is C. In the astronaut’s frame of reference the vertical component of the velocity does not seem to be taken into consideration. The speed of light is taken as traveling on an angle, and therefore taking more time to complete the trip. Why is this, since the light is not actually traveling on an angle, it only ‘appears’ to be?
It is counter intuitive, light doesn't behave like low speed bullets. A photon always travels at c for every inertial observer (one not accelerating). Imagine you fire a photon at c from you and it passes me while I whiz away from you at 0.9c - I will measure the photon as overtaking me at c and that is really bizarre. This is not a result of SR, but an assumption of SR which is backed experimentally and by Maxwell's equations (which is why the assumption was made). Making that counter intuiitive assumption leads to all sorts of wonderful (and counter intuitive) results like time dialation, lorentz contraction and relativity of simultaneity.

Sam5
2005-Apr-30, 09:59 PM
In the spaceship’s frame of reference the light is traveling directly vertical, therefore the vertical component of the velocity is C. In the astronaut’s frame of reference the vertical component of the velocity does not seem to be taken into consideration. The speed of light is taken as traveling on an angle, and therefore taking more time to complete the trip. Why is this, since the light is not actually traveling on an angle, it only ‘appears’ to be?

The earth is such a “spacecraft” moving through space at very high speeds relative to other bodies. The television industry has a large number of satellites that are located about 23,500 miles above the surface of the earth. Since the earth-orbit time for the satellites at this distance is about 24 hours, they keep pace with the rotation of the earth and they each hover over a different spot on the earth’s surface since they are rotating or “revolving” with the earth. These are called “geostationary” satellites. It’s like having a satellite at the very top of a 23,500 mile tall TV tower. The TV signals sent up and back down basically go in a fairly straight line.

If these signals were seen from the side, from the point of view of a distant star or planet that has a sideways or lateral motion relative to the earth of, say, 1,000 mps, while the earth and satellite remain in the center of view, then the signals would appear to move not as a “V” shaped line but as an “I” shaped line that travels up and down vertically off and back to the same place on the surface of the earth, while at the same time this line is seen traveling sideways through space along with the earth. It doesn’t matter if we consider the earth to be moving at that speed, or the distant planet to be moving at that speed or the speed to just be relative between the two.

peter eldergill
2005-Apr-30, 10:51 PM
Doesn't the graviational field also affect time dilation?

I believe that global positioning satellites must use the relativity equations to compensate for a time difference due to the difference in the field strength at the surface of the earth and in orbit. I've been told that if this was not done, you GPS would be off by several kilometers

Is this correct?

Later

Pete

Kaptain K
2005-Apr-30, 11:26 PM
GPS satellites must compensate for both gravity and velocity!

worzel
2005-Apr-30, 11:34 PM
If these signals were seen from the side, from the point of view of a distant star or planet that has a sideways or lateral motion relative to the earth of, say, 1,000 mps, while the earth and satellite remain in the center of view, then the signals would appear to move not as a “V” shaped line but as an “I” shaped line that travels up and down vertically off and back to the same place on the surface of the earth, while at the same time this line is seen traveling sideways through space along with the earth.
And within that inertial frame travelling laterally to the earth at 1,000 mps the up-down motion that you described combined with the sideways travelling through space that you described combine to have the light moving a longer distance in a V, at c, relative to that frame, according to relativity, and all the experiments that back it.

The OP's question was about the theory of relativity. I think you should point out that your views are against the mainstream when you attempt to answer basic questions about relativity with something that sounds decidedly like you're disagreeing with the theory. You wouldn't want people to mistakedly think that your take on SR is the convention, would you?

Sam5
2005-Apr-30, 11:47 PM
And within that inertial frame travelling laterally to the earth at 1,000 mps the up-down motion that you described combined with the sideways travelling through space that you described combine to have the light moving a longer distance in a V, at c, ....

The light doesn’t travel up and down to and from the satellites in a “V”. It travels up and down in an “I”. It doesn’t matter how fast our earth is moving or our galaxy is moving through space or relative to any other object in space.

worzel
2005-May-01, 12:02 AM
The light doesn’t travel up and down to and from the satellites in a “V”. It travels up and down in an “I”. It doesn’t matter how fast our earth is moving or our galaxy is moving through space or relative to any other object in space.
True, in the almost inertial frame where the satellites and the earth are stationary the light travels up and down in an I. But in an inertial frame which has a significant velocity relative to the earth and its satellites, the light moves in a V. According to relativity light always travels at c in an inertial frame. Therefore, a photon that travels between the earth and a satellite must take longer the faster the intertial frame of the observer is to the earth and its satellites due to the longer path the light must travel in that moving frame. Do you disagree that that is what relativity says?

Sam5
2005-May-01, 01:44 AM
But in an inertial frame which has a significant velocity relative to the earth and its satellites, the light moves in a V.

No, the light doesn’t move in a “V”. The earth stays under the satellite all the time and the signal moves up and down relative to the satellite and the earth. You don’t count any apparent sideways motion of the light beam as part of “the speed of light.”

Normandy6644
2005-May-01, 02:24 AM
If you had a tower on a train and the train was moving to the right with a given velocity, and you shot a light beam from the bottom of the tower to the top while the train was moving, a stationary observer would see the light move in a capital lambda (/\). That's one of the basic ways to calculate time dilation.

Sam5
2005-May-01, 02:33 AM
If you had a tower on a train and the train was moving to the right with a given velocity, and you shot a light beam from the bottom of the tower to the top while the train was moving, a stationary observer would see the light move in a capital lambda (/\). That's one of the basic ways to calculate time dilation.

Oh... are you one who believes the beam will exit the vertical laser diagonally instead of going straight up? Like this:

http://users.powernet.co.uk/bearsoft/LtClk.html

Fortis
2005-May-01, 02:36 AM
But in an inertial frame which has a significant velocity relative to the earth and its satellites, the light moves in a V.

No, the light doesn’t move in a “V”. The earth stays under the satellite all the time and the signal moves up and down relative to the satellite and the earth. You don’t count any apparent sideways motion of the light beam as part of “the speed of light.”
I'm picking this one up late, so I may have misunderstood where this one came from, but if you're at rest in an inertial frame that is moving relative to the earth-satellite system, then you must include the "apparent" sideways motion in the speed of light calculation. After all, you are making your measurements in your frame, not anyone elses.

Sam5
2005-May-01, 02:41 AM
But in an inertial frame which has a significant velocity relative to the earth and its satellites, the light moves in a V.

No, the light doesn’t move in a “V”. The earth stays under the satellite all the time and the signal moves up and down relative to the satellite and the earth. You don’t count any apparent sideways motion of the light beam as part of “the speed of light.”
I'm picking this one up late, so I may have misunderstood where this one came from, but if you're at rest in an inertial frame that is moving relative to the earth-satellite system, then you must include the "apparent" sideways motion in the speed of light calculation. After all, you are making your measurements in your frame, not anyone elses.

That’s like saying you are scanning the sky at night with a searchlight. The distant beam of the light seems to move from star to star, sideways, at millions of times faster than the speed of light because that’s what you see from your frame. But that’s just an illusion.

Fortis
2005-May-01, 02:49 AM
But in an inertial frame which has a significant velocity relative to the earth and its satellites, the light moves in a V.

No, the light doesn’t move in a “V”. The earth stays under the satellite all the time and the signal moves up and down relative to the satellite and the earth. You don’t count any apparent sideways motion of the light beam as part of “the speed of light.”
I'm picking this one up late, so I may have misunderstood where this one came from, but if you're at rest in an inertial frame that is moving relative to the earth-satellite system, then you must include the "apparent" sideways motion in the speed of light calculation. After all, you are making your measurements in your frame, not anyone elses.

That’s like saying you are scanning the sky at night with a searchlight. The distant beam of the light seems to move from star to star, sideways, at millions of times faster than the speed of light because that’s what you see from your frame. But that’s just an illusion.
Not true. How do you measure the speed of a photon in your reference frame? You determine the distance that it travelled in your frame, and divide it by the time it took (in your frame) to get there.

Sam5
2005-May-01, 03:10 AM
How do you measure the speed of a photon in your reference frame?

Well I certainly don’t count its speed going sideways even though I know the earth is going sideways in space. Nor would I count the sideways motion of a photon as part of the “speed” of a photon going up to a mirror and back down to the surface of a distant planet.

Fortis
2005-May-01, 03:23 AM
How do you measure the speed of a photon in your reference frame?

Well I certainly don’t count its speed going sideways even though I know the earth is going sideways in space. Nor would I count the sideways motion of a photon as part of the “speed” of a photon going up to a mirror and back down to the surface of a distant planet.
If you couldn't see the earth or the satellite, but just observe the movement of the photons in your frame (this is, of course, a thought experiment), then how would you measure the speed, given that all you know is where and when the photon started on its journey, and where and when it stopped? (All in your rest frame, of course. :) )

Sam5
2005-May-01, 04:00 AM
How do you measure the speed of a photon in your reference frame?

Well I certainly don’t count its speed going sideways even though I know the earth is going sideways in space. Nor would I count the sideways motion of a photon as part of the “speed” of a photon going up to a mirror and back down to the surface of a distant planet.
If you couldn't see the earth or the satellite, but just observe the movement of the photons in your frame (this is, of course, a thought experiment), then how would you measure the speed, given that all you know is where and when the photon started on its journey, and where and when it stopped? (All in your rest frame, of course. :) )

Doh, umm, hmm, well, :D I suppose I’d say that the photon must be going at “c” up and down in its system, and I wouldn’t count any sideways motion that might result from an optical illusion. It’s like when I see telephone poles that seem to get smaller and smaller in the distance along the side of a long straight highway. That’s sure what it looks like to me, but I’m reasonably certain they don’t actually get smaller the further away they are from me. I certainly wouldn’t imagine a photon bent over sideways and exiting a laser beam at a 45 degree angle like the one in this animation (the animation at the top of the java box):

Or in this drawing:

http://galileoandeinstein.physics.virginia.edu/lectures/srelwhat.html

See the slanted photon in the second drawing down on the right? I don’t know of any laser that emits photons at a 45 degree angle and slanted toward the right, like in that drawing or this one:

http://users.powernet.co.uk/bearsoft/LtClk.html

Such a phenomenon would require an old 19th Century “ether wind” to blow the photon toward the right as it exits the laser.

Grey
2005-May-01, 04:02 AM
Oh... are you one who believes the beam will exit the vertical laser diagonally instead of going straight up?
It would have to, of course. Think about the laser beam as it leaves the device for the final time, from a frame of reference in which the laser is pointed vertically, but moving laterally. During the time the light beam moves from the bottom to the top, the laser will have itself moved to the side, so the exit will be displaced from where it was at the beginning. Any light beam that was not travelling diagonally would hit the side of the tube. If you think about the beam gradually building up to laser strength by bouncing back and forth between the internal mirrors of the laser, you can see that it will in fact select light travelling at a slant as the light to be reinforced.

Of course, we see this effect in reverse directly all the time, as the aberration of starlight.

Sam5
2005-May-01, 04:32 AM
Oh... are you one who believes the beam will exit the vertical laser diagonally instead of going straight up?
It would have to, of course. Think about the laser beam as it leaves the device for the final time, from a frame of reference in which the laser is pointed vertically, but moving laterally. During the time the light beam moves from the bottom to the top, the laser will have itself moved to the side, so the exit will be displaced from where it was at the beginning. Any light beam that was not travelling diagonally would hit the side of the tube. If you think about the beam gradually building up to laser strength by bouncing back and forth between the internal mirrors of the laser, you can see that it will in fact select light travelling at a slant as the light to be reinforced.

Of course, we see this effect in reverse directly all the time, as the aberration of starlight.

I think there is something that keeps the laser beam moving straight up and down if the laser is fixed and aimed vertically upward at the surface of the earth, while the earth is moving through space at hundreds and even thousands of miles per second, relative to other galaxies. But I’m not sure what this “something” is. It might be a local inertial effect or some kind of “medium-like” effect with a “medium”, perhaps the earth’s gravity field, traveling through space with the earth.

It would be very interesting to place a laser on a moving vehicle, moving across the surface of the earth and see if the beam would actually hit a mirror fixed directly above the laser that is moving at high speed across the surface of the earth with the laser. I’m not sure if we have anything that moves fast enough to be able to conduct this test. Even the space shuttle, moving at what, about 17,000 mph, might not be fast enough to see if a laser beam aimed at the wall across from it hits a spot on the wall that is directly and exactly across from the laser. That speed is only about 4.7 miles per second, and the walls of the Space Shuttle aren’t very far apart.

I think some of the very old thought experiments need to be updated and modified. I’m not “anti relativity”, but I do think some the relativity concepts should be brought up to date in light of the past hundred years of new discoveries.

Kaptain K
2005-May-01, 05:08 AM
Another thread successfully hijacked by Sam5! :o :roll: #-o ](*,)

Sam5
2005-May-01, 05:40 AM
Another thread successfully hijacked by Sam5!

Excuse me, :D but this thread was dead, way down on the list, and thunderchicken never came back to comment on the confusing stuff the other guys told him. I can’t help it if I make one post directly to thunderchicken and then 8 guys want to challenge me to a thought experiment duel. What am I supposed to do, not respond to their questions? If I get 8 guys talking to me and asking me questions I’m going to be posting responses to all 8 guys. If you’ve got a complaint, complain to the 8 guys for dragging these discussions out. If the 8 guys hadn’t posted messages directly to me and asked me questions, that one post to thunderchicken would have been the only post I would have made on this thread. If they don’t want me to post any more, then tell them to stop asking me questions. :D

worzel
2005-May-01, 09:21 AM
Another thread successfully hijacked by Sam5!

Excuse me, :D but this thread was dead, way down on the list, and thunderchicken never came back to comment on the confusing stuff the other guys told him.
I resent that.

Firstly, I am in no way as eloquent or knowledgable as many of the other posters on BA, but being one who struggled not so long ago to understand exactly the sort of question the OP asked, I feel that sometimes I may have something useful to offer. I also remember posting these sort of questions and appreciating the wide range of answers attempting to describe the same thing from different points of view and trying to deal with my particular misconceptions.

Secondly, all of the answers provided in this thread other than yours are in agreement.

Thirdly, the OP asked a question about the theory of relativity, not about Sam5's inability to understand the most basic of SR expositions. The fact that you never point out that you are against the mainstream when you reply to these sorts of questions with what appears to me be a deliberate attempt to confuse the issue is intellectually dishonest.

I can’t help it if I make one post directly to thunderchicken and then 8 guys want to challenge me to a thought experiment duel. What am I supposed to do, not respond to their questions?
How about starting your own thread in ATM clearly stating what part of relativity you don't agree with. Or at least being honest enough to state that you are ATM when responding to threads like this.

Sam5
2005-May-01, 11:58 AM
Hi worzel,

I was telling thunderchicken about TV satellite signals that go up and down to the earth, which is similar to the “light clock” thought experiment that has become part of modern science legend, and I explained to him that the signal is actually moving sideways through space in addition to moving up and down. The signal is not moving diagonally. That is true, and as far as a can tell it doesn’t refute any of “relativity”, and this is not an ATM theory. Ask any TV satellite engineer. :D

Grey
2005-May-01, 04:25 PM
...and I explained to him that the signal is actually moving sideways through space in addition to moving up and down. The signal is not moving diagonally.
This doesn't make sense to me. Can you show me a drawing of a path that has both vertical and horizontal components, but isn't diagonal?

Fortis
2005-May-01, 04:34 PM
Hi worzel,

I was telling thunderchicken about TV satellite signals that go up and down to the earth, which is similar to the “light clock” thought experiment that has become part of modern science legend, and I explained to him that the signal is actually moving sideways through space in addition to moving up and down. The signal is not moving diagonally. That is true, and as far as a can tell it doesn’t refute any of “relativity”, and this is not an ATM theory. Ask any TV satellite engineer. :D
This would be your satellite engineer who is at rest in the same rest frame (ignoring for the moment the higher order effects of GR) as the satellite and ground station? ;)

Sam5
2005-May-01, 04:50 PM
...and I explained to him that the signal is actually moving sideways through space in addition to moving up and down. The signal is not moving diagonally.
This doesn't make sense to me. Can you show me a drawing of a path that has both vertical and horizontal components, but isn't diagonal?

Here’s one that is easier to imagine: A highly directional sound emitter is on the floor of a Concorde, and a directional receiving microphone is directly above it on the ceiling. (The sound is picked up and re-emitted by another highly directional speaker back down to the floor.) The sound signal goes up at 1,100 fps and back down in a straight line to the floor, while the signal is traveling sideways relative to the surface of the earth at 2,200 fps, but we don’t add this sideways speed to the speed of the sound inside the Concorde, and the sound signal travels in a straight line up and down, not a diagonal one.

Edit, one sentence added: (The sound is picked up and re-emitted by another highly directional speaker back down to the floor.)

worzel
2005-May-01, 05:02 PM
The signal is not moving diagonally.
Yes it is, in the frame of the observer moving relative to the earth and satellite.

Here’s one that is easier to imagine: A highly directional sound emitter is on the floor of a Concorde, and a directional receiving microphone is directly above it on the ceiling. The sound signal goes up at 1,100 fps and back down in a straight line to the floor, while the signal is traveling sideways relative to the surface of the earth at 2,200 fps, but we don’t add this sideways speed to the speed of the sound inside the Concorde, and the sound signal travels in a straight line up and down, not a diagonal one.
Well that would be travelling diagonally relative to a someone moving relative to it. But unlikle light, so would the medium that it is moving in, according to conventional theory. And unlike light, the speed of sound isn't constant for all (intertial) observers, according to conventional theory. So while you might find this easier to imagine, it is of little help for someone trying to understand special relativity.

Sam5
2005-May-01, 06:07 PM
So while you might find this easier to imagine, it is of little help for someone trying to understand special relativity.

The waves move sideways at the surface of a planet, not diagonally. Light doesn’t come out of a laser at a 45 degree angle. To do so would require a 19th Century “ether wind”. In galaxy after galaxy, no matter how fast they are moving relative to our own, the light comes out of lasers in a straight line. The galaxies move relative to us at high speeds, like the Concorde moving relative to the surface of the earth. The common “light clock” illustration is a modern invention. I can’t find it in any of Einstein’s works. Some modern thought experiments are just not correct.

worzel
2005-May-01, 06:18 PM
So while you might find this easier to imagine, it is of little help for someone trying to understand special relativity.

The waves move sideways at the surface of a planet, not diagonally.
But from another frame that is moving relative to the planet they do move diagonally.

Light doesn’t come out of a laser at a 45 degree angle.
It does if you've moving fast enough relative to the laser. This particular point was already answered comprehensively by Grey in this thread. Rather than refute what he said you instead started woffling on about your aether theory. Whatever you personally believe, you cannot deny that SR says that the light leaves the laser diagonally and therefore covers a greater distance in the moving frame, and therefore takes longer to do so.

To do so would require a 19th Century “ether wind”.
Why?

The common “light clock” illustration is a modern invention. I can’t find it in any of Einstein’s works. Some modern thought experiments are just not correct.
So what. If you understood SR you would understand that the light clock thought experiment is simply a particularly good demonstration of SR's time dialation. If you don't think it is then explain why.

Sam5
2005-May-01, 06:27 PM
But from another frame that is moving relative to the planet they do move diagonally.
:D It doesn’t matter how fast you are moving while looking at a distant laser, the light doesn’t come out of the laser diagonally. You are thinking of an optical illusion, like this one:

You are riding in a car a mile away from flashing lights that you see out your car’s side window, as you pass the lights in the distance. In the distance, a laser sends a brief signal directly upward where it hits a detector and when it does, a bright strobe light flashes on the detector, which you see. A second later the detector sends a brief laser signal downward where it hits a detector at the bottom laser, where a bright strobe light flashes. These lasers are not moving relative to the surface of the earth, but you are, so the bright strobe flashes move across your field of view. You will see a light flash pattern in a zig-zag manner, as a series of “moving” dots. If you connect all the dots you will see this pattern: ΛΛΛΛ

But the light is not going out of the lasers diagonally.

Normandy6644
2005-May-01, 06:47 PM
The light never comes out of the laser diagonally, but its position changes relative to a stationary observer due to the velocity of the train. Therefore the total path of the light is a diagonal, since at each instant it is transmitted upwards as the train moves.

Grey
2005-May-01, 06:51 PM
Here’s one that is easier to imagine: A highly directional sound emitter is on the floor of a Concorde, and a directional receiving microphone is directly above it on the ceiling. (The sound is picked up and re-emitted by another highly directional speaker back down to the floor.) The sound signal goes up at 1,100 fps and back down in a straight line to the floor, while the signal is traveling sideways relative to the surface of the earth at 2,200 fps... and the sound signal travels in a straight line up and down, not a diagonal one.
Um, no. Try drawing this from the perspective of someone on the ground, and I think you'll see that the path is definitely a diagonal one. I maintain that it's nonsensical to describe a path as having both a vertical and a horizontal component, but not being diagonal.

...but we don’t add this sideways speed to the speed of the sound inside the Concorde...
Of course, that's because the speed of sound is always measured relative to the medium in which it's moving, and in this case, the medium is moving horizontally, too.

Sam5
2005-May-01, 06:53 PM
The light never comes out of the laser diagonally, but its position changes relative to a stationary observer due to the velocity of the train. Therefore the total path of the light is a diagonal, since at each instant it is transmitted upwards as the train moves.

That’s ok. But you don’t add the speed of the train to the speed of the light. And time doesn’t slow down on the train as it moves forward.

Sam5
2005-May-01, 07:04 PM
I maintain that it's nonsensical to describe a path as having both a vertical and a horizontal component, but not being diagonal.

I think you could probably say that the “path” of the sound waves traced on a graph by someone on the ground will be diagonal from the point of view of the ground observer, but the sound waves will not come out of the speaker at a sharp angle, they will not be slanted in the direction of travel of the Concorde, and time will not “slow down” on the Concorde just because you see the waves moving both upward and sideways at the same time.

Normandy6644
2005-May-01, 09:36 PM
The light never comes out of the laser diagonally, but its position changes relative to a stationary observer due to the velocity of the train. Therefore the total path of the light is a diagonal, since at each instant it is transmitted upwards as the train moves.

That’s ok. But you don’t add the speed of the train to the speed of the light. And time doesn’t slow down on the train as it moves forward.

That's not what I'm saying at all. Because the train is moving, the light from the laser is emitted at different points along the axis, so it makes a diagonal.

Fortis
2005-May-01, 10:03 PM
How do you measure the speed of a photon in your reference frame?

Well I certainly don’t count its speed going sideways even though I know the earth is going sideways in space. Nor would I count the sideways motion of a photon as part of the “speed” of a photon going up to a mirror and back down to the surface of a distant planet.
If you couldn't see the earth or the satellite, but just observe the movement of the photons in your frame (this is, of course, a thought experiment), then how would you measure the speed, given that all you know is where and when the photon started on its journey, and where and when it stopped? (All in your rest frame, of course. :) )

Doh, umm, hmm, well, :D I suppose I’d say that the photon must be going at “c” up and down in its system, and I wouldn’t count any sideways motion that might result from an optical illusion.
If you can't observe the original source and the intended receiver, how do you know that they are moving with respect to you, or how fast they are moving (which you would need to know in order to apply any kind of correction to your measurement)? :)

Sam5
2005-May-01, 10:12 PM
If you can't observe the original source and the intended receiver, how do you know that they are moving with respect to you, or how fast they are moving (which you would need to know in order to apply any kind of correction to your measurement)? :)

Because that’s part of the thought experiment. If they aren’t moving relatively, we’ve got no thought experiment.

If you get right down to it, we can’t see phonons from the side, and we can’t travel fast enough yet for this thought experiment to be tested. (But it's sure fun to talk about.) :D

worzel
2005-May-01, 11:52 PM
But from another frame that is moving relative to the planet they do move diagonally.
:D It doesn’t matter how fast you are moving while looking at a distant laser, the light doesn’t come out of the laser diagonally.
From the moving frame the light moves in a diagonal path before and after it leaves the laser so that the total distance it travels is greater than in the frame where the laser is stationary. If, as Maxwelll's equations and many experiments say, light always travels at c in any intertial frame, then it must have take longer to make the same journey in the moving frame than it did in the stationary frame. Why can't you point out which bit of this standard SR exposition you disagree with rather than keep arguing over the semantics of the word 'diagonal'. Maybe you're being too literal because you don't understand the maths: prove me wrong Sam5!

You are thinking of an optical illusion, like this one:

[ snip pointless rehashing of the same thought experiment ]

But the light is not going out of the lasers diagonally.
Yes it does, in the frame of the observer moving relatiive to the emitter. How is this example different to the other ones involving light in this thread?

If you get right down to it, we can’t see phonons from the side, and we can’t travel fast enough yet for this thought experiment to be tested. (But it's sure fun to talk about.) :D
But it is what special relativity says actually happens. And general relativity has been tested successfully many times. If something as fundamental as special relativity's time dialation were wrong then it would be quite remarkable that GR worked at all.

If you have something intelligent to say about the thought experiment then why don't you say it instead of repeatedly insisting that sideways motion combined with simultaneous up-down motion doesn't constitute diagonal motion.

Sam5
2005-May-02, 12:05 AM
But from another frame that is moving relative to the planet they do move diagonally.
:D It doesn’t matter how fast you are moving while looking at a distant laser, the light doesn’t come out of the laser diagonally.
From the moving frame the light moves in a diagonal path before and after it leaves the laser so that the total distance it travels is greater than in the frame where the laser is stationary. If, as Maxwelll's equations and many experiments say, light always travels at c in any intertial frame, then it must have take longer to make the same journey in the moving frame than it did in the stationary frame. Why can't you point out which bit of this standard SR exposition you disagree with rather than keep arguing over the semantics of the word 'diagonal'.

If we have a laser stationary on earth, aimed up to the mirror, the light is moving in the earth frame, no matter who is looking at it. It’s going to do what its supposed to do on earth, and it will not go out from the laser at a diagonal angle, and it will not take any kind of “longer path” just because someone is watching it from afar.

If you want to talk about “optical illusions” then make that clear. But the light doesn’t leave the laser diagonally and the photon is not sloped at an angle like in this U. of Virginia drawing:

http://galileoandeinstein.physics.virginia.edu/lectures/photclk0.gif

I think some guys get relativity illusions mixed up with reality. It’s ok to see an illusion, but it is not ok to claim that reality, on earth or any other specific place, is different for different observers.

The part I disagree with are the modern drawings that show the light beam going up and out of the laser at an angle and the photon tilted at an angle. This is not an Einstein thought experiment. I can’t find it in his papers or in any old relativity books. This is a modern thought experiment that is misleading and not accurate.

Oh, and PS: The whole idea that students should give up their “common sense” and “logic” is an absurd thing to tell students. That’s what cultists tell their followers, but that’s not what teachers should tell their students. You might tell them that “things work differently at high speeds” or “on the very small scale”, but the students shouldn’t be told to “give up thinking” and just “trust and believe” in every relativity thought experiment.

:D

Sam5
2005-May-02, 12:19 AM
If you get right down to it, we can’t see phonons from the side, and we can’t travel fast enough yet for this thought experiment to be tested. (But it's sure fun to talk about.) :D
But it is what special relativity says actually happens.

Where does it say that? That’s what you say, but I don’t see the SR paper saying that. Where does it say that light is going to come out of a laser (or flashlight or spotlight) diagonally?
:D

worzel
2005-May-02, 12:36 AM
If we have a laser stationary on earth, aimed up to the mirror, the light is moving in the earth frame, no matter who is looking at it. It’s going to do what its supposed to do on earth, and it will not go out from the laser at a diagonal angle, and it will not take any kind of “longer path” just because someone is watching it from afar.
Your mistake is in saying that "the light is moving in the earth frame, no matter who is looking at it". While it is true that any observer will reckon that the light is moving vertically in the earth's frame, it is not moving only vertically in the observer's frame.

If you want to talk about “optical illusions” then make that clear. But the light doesn’t leave the laser diagonally and the photon is not sloped at an angle like in this U. of Virginia drawing:
If you want to pretend that fundamental SR results are just optical illusions then be honest and stop pretending that you're talking about SR.

I think some guys get relativity illusions mixed up with reality. It’s ok to see an illusion, but it is not ok to claim that reality, on earth or any other specific place, is different for different observers.
But it's ok to completely not understand SR but nevertheless pretend to be representing it with your misunderstandings because you've failed to understand more papers and books on the subject than most of us have read, right?

The part I disagree with are the modern drawings that show the light beam going up and out of the laser at an angle and the photon tilted at an angle. This is not an Einstein thought experiment. I can’t find it in his papers or in any old relativity books. This is a modern thought experiment that is misleading and not accurate.
It is a direct consequence of Einstein's theory of special relativity. From "Relativity, Albert Einstein, Routeledge" page 39:

As a consequence of its motion the clock goes more slowly than when at rest.

Notice that he says "goes", not "seems to go". Do you disagree with him?

Oh, and PS: The whole idea that students should give up their “common sense” and “logic” is an absurd thing to tell students. That’s what cultists tell their followers, but that’s not what teachers should tell their students. You might tell them that “things work differently at high speeds” or “on the very small scale”, but the students shouldn’t be told to “give up thinking” and just “trust and believe” in every relativity thought experiment.
Now you're just being ridiculous. When one says to forget about common sense and logic to a newbie to relativity it is meant light heartedly because much of relativity is counter intuitive. One must let go of the natural intuitions (which one may initially think are perfectly logical) we have about space and time if one is to understand it. But it is perfectly logical and consistent even though it may not seem so at first.

PS I'm boerd, I'm going tot bed.

thunderchicken
2005-May-02, 01:22 AM
Hi guys, thanks for all the replies and other stuff......lol.

So, let me see if I understand this correctly:

To the stationary observer outside the spaceship, the light beam IS traveling on an angle (it does not have a vertical velocity of C). Because the speed of light is always the same (regardless of the frame of reference and direction) it takes longer to complete this diagonal trip.

The light pulse is not observed to hit the mirror simultaneously in both frames of reference.

The stationary person observes the light clock in the spaceship to be ticking slower than his identical, stationary clock.

So, if the above is true I've heard it said that the person in the spaceship will not age as fast as the stationary person. BUT, if you look at it from the other way around, wouldn't the person in the spaceship see the stationary person's light clock as ticking just as slowly as he zips past???

Sam5
2005-May-02, 01:42 AM
Hi guys, thanks for all the replies and other stuff......lol.

-snip-

So, if the above is true I've heard it said that the person in the spaceship will not age as fast as the stationary person. BUT, if you look at it from the other way around, wouldn't the person in the spaceship see the stationary person's light clock as ticking just as slowly as he zips past???
Yes, and that is called the “clock paradox,” which is unresolvable in SR theory. You can read about it here in Professor Herbert Dingle’s 1972 book. He was a well-known British professor in the 1930s-60s.

http://russamos.narod.ru/dingle/9.htm

And see this, written in 1996, by Louis Essen, the inventor of the first Cesium atomic clock:

http://www.btinternet.com/~time.lord/Relativity.html

:D

Sylas
2005-May-02, 02:22 AM
To the stationary observer outside the spaceship, the light beam IS traveling on an angle (it does not have a vertical velocity of C). Because the speed of light is always the same (regardless of the frame of reference and direction) it takes longer to complete this diagonal trip.

Right.

The light pulse is not observed to hit the mirror simultaneously in both frames of reference.

"Simultaneous" usually refers to equality of time co-ordinate for two different events, measured in the one frame. I think you mean to say that the time from leaving the laser to hitting the mirror is different in the two frames of reference. This is true.

The stationary person observes the light clock in the spaceship to be ticking slower than his identical, stationary clock.

Yes... although there is no absolute notion of stationary. Everyone is stationary in their own frame, by definition of what we mean by "their own frame".

So, if the above is true I've heard it said that the person in the spaceship will not age as fast as the stationary person. BUT, if you look at it from the other way around, wouldn't the person in the spaceship see the stationary person's light clock as ticking just as slowly as he zips past???

Yes, that is true. In the frame of each observer, the clocks for the other observer are running more slowly.

The solution to this apparent dilemma is relativity of simultaneity. The observer from outside the ship marks an interval of time of length t, being the time from their perspective of how long it took the ship's clock to cycle. Suppose the outside observer raises a little red flag at this instant t. From the perspective of the outside observer, they raise the red flag simultaneously with the end of the clock cycle.

The outside observer also confirms simultaneity by noting that they SEE the event at a time (v/c)t seconds later, since it took place at a distance of vt.

But from inside the ship, the inside observer sees the red flag being raised at an instant t', significantly AFTER the clock completed its cycle! They confirm this, because they see the flag being raised after a further (v/c)t' seconds.

Let d be the duration of the clock cycle in its own frame. Let g = 1/sqrt(1-(v/c)^2)) be the Lorentz contraction factor.

The outside observer raises a flag at time t = g*d in their own frame. In their frame, this is simultaneous with the end of the clock cycle.

In the ship frame, the clock completed its cycle at time d, and the red flag was raised at time t' = g^2*d.

Cheers -- Sylas

Fortis
2005-May-02, 02:32 AM
If you can't observe the original source and the intended receiver, how do you know that they are moving with respect to you, or how fast they are moving (which you would need to know in order to apply any kind of correction to your measurement)? :)

Because that’s part of the thought experiment. If they aren’t moving relatively, we’ve got no thought experiment.

If you get right down to it, we can’t see phonons from the side, and we can’t travel fast enough yet for this thought experiment to be tested. (But it's sure fun to talk about.) :D
I think that you mean photons. ;)

As it's a thought experiment, let's have a pulse of photons, so that we can position detectors at two locations that the pulse will pass through. (Let's say that the detectors are only 50% efficient, so that they let ~50% of the pulse through un-disturbed.)

Now these detectors are stationary in the rest frame of my observer. As an added bonus, my observer is sitting inside of a sealed room, and only knows the distance between the detectors and the only additional input that he receives will be the times at which the two detectors detect the pulse.

So, what will he measure for c? :)

thunderchicken
2005-May-02, 02:44 AM
The solution to this apparent dilemma is relativity of simultaneity. The observer from outside the ship marks an interval of time of length t, being the time from their perspective of how long it took the ship's clock to cycle. Suppose the outside observer raises a little red flag at this instant t. From the perspective of the outside observer, they raise the red flag simultaneously with the end of the clock cycle.

The outside observer also confirms simultaneity by noting that they SEE the event at a time (v/c)t seconds later, since it took place at a distance of vt.

But from inside the ship, the inside observer sees the red flag being raised at an instant t', significantly AFTER the clock completed its cycle! They confirm this, because they see the flag being raised after a further (v/c)t' seconds.

Let d be the duration of the clock cycle in its own frame. Let g = 1/sqrt(1-(v/c)^2)) be the Lorentz contraction factor.

The outside observer raises a flag at time t = g*d in their own frame. In their frame, this is simultaneous with the end of the clock cycle.

In the ship frame, the clock completed its cycle at time d, and the red flag was raised at time t' = g^2*d.

It seems to me that the clock on the spaceship is ticking at exactly the same rate as the stationary clock. Problems only occur when you observe either clock from another reference frame. Does it matter what each of them perceive the others clock to be doing? What they are seeing is not the reality of the situation, is it?

When a car passes me on the road, it appears to become smaller the farther away it gets. I also appear to get smaller in the rearview mirror of that car. The reality of the situation, however, is that neither of us actually got smaller. OK.....I know that that is probably a very bad example, but couldn’t this clock paradox be a similar sort of thing?

Sylas
2005-May-02, 03:15 AM
It seems to me that the clock on the spaceship is ticking at exactly the same rate as the stationary clock. Problems only occur when you observe either clock from another reference frame. Does it matter what each of them perceive the others clock to be doing? What they are seeing is not the reality of the situation, is it?

The whole point of relativity is that there is no one "reality" of the situation. That's what "relativity" means. All views are equally valid, and the mathematics lets you translate between them.

Each clock does tick at a fixed rate when measured by someone who is moving with the clock. This works for accelerated motions as well, and you can use special relativity to solve problems involving accelerated motions just fine. (It is a common mistake to think that general relativity is required for accelerated motions.)

Differences arise when you consider what the clocks are doing relative to a different frame. These are "problems" in the sense that "what is 4 cubed?" is a problem. You can calculate the answers.

The answers are consistent; the claims for "paradoxes" in special relativity always arise from forgetting some part of the mathematical formalism, or insisting on conclusions that actually don't fit reality. Some of the conclusions are unintuitive, but the results are unambiguous, and conform to observations.

For example, what happens when spaceship turns around and comes back? In this case, once the ship clock is back with another clock at the same location, you can certainly tell unambiguously which one has the greater reading. It is possible to solve this in special relativity, but you can't use an inertial frame for an accelerated observer. And only one of the clocks is accelerated.

There are many pages where this is worked out correctly. Here is one of them. (http://www.phys.unsw.edu.au/~jw/twin.html)

When a car passes me on the road, it appears to become smaller the farther away it gets. I also appear to get smaller in the rearview mirror of that car. The reality of the situation, however, is that neither of us actually got smaller. OK.....I know that that is probably a very bad example, but couldn’t this clock paradox be a similar sort of thing?

Similar in some respects and not in others. Analogies are never the whole story. Relativity has some definite consequences, that are quite different from the consequences of conventional Newtonian assumptions. And reality happens to fit the relativistic consequences more accurately.

General relativity is only required if we also have to take gravity into account.

Cheers -- Sylas

A Thousand Pardons
2005-May-02, 03:18 AM
General relativity is only required if we also have to take gravity into account.
What about if we want to use non-inertial reference frames? :)

Sylas
2005-May-02, 03:32 AM
General relativity is only required if we also have to take gravity into account.
What about if we want to use non-inertial reference frames? :)

No problem. You can apply the theory of special relativity to non-inertial frames. You just need a bit of calculus, rather than using a straight Lorentz transform and co-ordinate mapping.

Cheers -- Sylas

Tensor
2005-May-02, 03:58 AM
General relativity is only required if we also have to take gravity into account.
What about if we want to use non-inertial reference frames? :)

No problem. You can apply the theory of special relativity to non-inertial frames. You just need a bit of calculus, rather than using a straight Lorentz transform and co-ordinate mapping.

Cheers -- Sylas

SR can also handle gravity. Remember the equivalence between gravity and accelerations? I think that was ATP's point.

Sylas
2005-May-02, 04:10 AM
SR can also handle gravity. Remember the equivalence between gravity and accelerations? I think that was ATP's point.

SR does not handle gravity, which is why there was a need to develop GR. The equivalence principle is that the effects of gravitation are indistinguishable from an acceleration, but only over a sufficiently small region.

Based on this, general relativity was derived as a consequence; and general relativity goes beyond small regions where you can use special relativity.

Hence, for example, you NEED general relativity to calculate time dilation for a clock moved into a gravity well and then returned to the starting point. But you don't need general relativity to calculate a time dilation for a clock accelerated away, and then returned to the starting point.

General relativity really is different to special relativity, and you do need it for gravity. You don't need all the GR machinery for accelerated motions outside of gravitational fields.

Cheers -- Sylas

Tensor
2005-May-02, 04:23 AM
SR can also handle gravity. Remember the equivalence between gravity and accelerations? I think that was ATP's point.

SR does not handle gravity, which is why there was a need to develop

snip.....

You don't need all the GR machinery for accelerated motions outside of gravitational fields.

Cheers -- Sylas

Check out chapter 6 of MTW. The first section, of that chapter, decribes how SR can handle gravity. It wouldn't be pretty, but it can be done.

Sylas
2005-May-02, 04:58 AM
Check out chapter 6 of MTW. The first section, of that chapter, decribes how SR can handle gravity. It wouldn't be pretty, but it can be done.

Alas, I tried to look it up, but it is borrowed from the library and does not return for three weeks. I'm obviously missing something, so give me a clue. Why was GR needed at all?

I know how to calculate the time dilation in a space ship that accelerates away and later returns to a reference clock. I don't know how to do it if the space ship moves away into a deep gravitational well and returns.

Cheers -- Sylas

worzel
2005-May-02, 10:12 AM
It seems to me that the clock on the spaceship is ticking at exactly the same rate as the stationary clock. Problems only occur when you observe either clock from another reference frame. Does it matter what each of them perceive the others clock to be doing? What they are seeing is not the reality of the situation, is it?
We have to be careful about what we mean when we say "see". In most of these descriptions it is assumed that one could mark off distances in ones own frame, place clocks and cameras at regular intervales, and examine, after the event, what actually happened at each location at each instance of time.

If the observers are moving apart they will see each other's clocks run slow due the doppler effect (like a siren dropping in pitch as it goes away from you), and if they are moving towards each they will see each others clocks running fast for the same reason. If they take this into account, that is, calculate the rate of each other's clock after taking into account how long each pulse of light would take to reach them, then they would both conclude that each other's clock is running slow, whether they are moving away from, or towards each other.

So in the twin paradox you can work out what actually happens in the frame of each observer, and on top of that you can work out what they would actually see.

SeanF
2005-May-02, 02:49 PM
Check out chapter 6 of MTW. The first section, of that chapter, decribes how SR can handle gravity. It wouldn't be pretty, but it can be done.

Alas, I tried to look it up, but it is borrowed from the library and does not return for three weeks. I'm obviously missing something, so give me a clue. Why was GR needed at all?

I know how to calculate the time dilation in a space ship that accelerates away and later returns to a reference clock. I don't know how to do it if the space ship moves away into a deep gravitational well and returns.

Cheers -- Sylas
Sylas,

If acceleration and gravity are indistinguishable, then doesn't it stand to reason that any theory which can handle acceleration can also handle gravity?

Conversely, if you're right, then any situation which can be dealt with in SR could be said to be acceleration and not be gravity, which would thus disprove the "indistinguishable" principle.

:)

Sylas
2005-May-02, 07:09 PM
If acceleration and gravity are indistinguishable, then doesn't it stand to reason that any theory which can handle acceleration can also handle gravity?

Conversely, if you're right, then any situation which can be dealt with in SR could be said to be acceleration and not be gravity, which would thus disprove the "indistinguishable" principle.

My understanding was given earlier in the thread: gravitation and acceleration are locally indistinguishable, and that GR gives SR as a special case over a sufficiently localized region.

If you are in a small box (elevator), you cannot distinguish acceleration and gravity. But if the box is sufficienctly large, and gravity is from a point source, you can distinguish, or so I had been lead to believe. But I'm not a GR expert.

Cheers -- Sylas

SeanF
2005-May-02, 07:38 PM
If acceleration and gravity are indistinguishable, then doesn't it stand to reason that any theory which can handle acceleration can also handle gravity?

Conversely, if you're right, then any situation which can be dealt with in SR could be said to be acceleration and not be gravity, which would thus disprove the "indistinguishable" principle.

My understanding was given earlier in the thread: gravitation and acceleration are locally indistinguishable, and that GR gives SR as a special case over a sufficiently localized region.

If you are in a small box (elevator), you cannot distinguish acceleration and gravity. But if the box is sufficienctly large, and gravity is from a point source, you can distinguish, or so I had been lead to believe. But I'm not a GR expert.

Cheers -- Sylas
Yes, that's true. The relativistic effect on two clocks or observers at different heights in a uniformly accelerating box would be different than on two clocks at different heights in a mass-generated (non-uniform) gravitational field.

But I'm not sure why that leads you to the conclusion that SR can't handle gravity.

Sylas
2005-May-02, 08:38 PM
Yes, that's true. The relativistic effect on two clocks or observers at different heights in a uniformly accelerating box would be different than on two clocks at different heights in a mass-generated (non-uniform) gravitational field.

But I'm not sure why that leads you to the conclusion that SR can't handle gravity.

I'm missing something here. GR goes to SR in the limit; or SR applied independently over lots of sufficiently small regions and combined gives GR.

By GR, I mean the general case, in which you effectively integrate over those many smaller regions in which SR applies. By SR, I mean the special case, where you can set up as a reference an inertial frame that covers all space, and use that as a base for calculating effects of accelerated motions.

Is my usage of these terms "special" and "general" improper or non-standard? How do you guys distinguish "using special relativity" and "using general relativity"?

Cheers -- Sylas

SeanF
2005-May-02, 09:03 PM
I think we're using the terminology the same way, Sylas. My point is that when you say:

SR applied independently over lots of sufficiently small regions and combined gives GR.
That holds true regardless of whether you're dealing with acceleration or gravity in GR.

In fact, because of equivalence, it must be true in both cases if it's true in either.

Tensor
2005-May-02, 09:13 PM
Yes, that's true. The relativistic effect on two clocks or observers at different heights in a uniformly accelerating box would be different than on two clocks at different heights in a mass-generated (non-uniform) gravitational field.

But I'm not sure why that leads you to the conclusion that SR can't handle gravity.

I'm missing something here. GR goes to SR in the limit; or SR applied independently over lots of sufficiently small regions and combined gives GR.

By GR, I mean the general case, in which you effectively integrate over those many smaller regions in which SR applies. By SR, I mean the special case, where you can set up as a reference an inertial frame that covers all space, and use that as a base for calculating effects of accelerated motions.

Ahhhhhh, Ok, I think (in my case) it was just a misunderstanding as to what you meant by "SR does not handle gravity". A lot of people out there don't understand that you can use just SR by integrating over all those small areas without resorting to tensors or forms. It now appears you do know. Sorry.

Is my usage of these terms "special" and "general" improper or non-standard? How do you guys distinguish "using special relativity" and "using general relativity"? Cheers -- Sylas

I don't think your use is really improper or non-standard. Over all, I would say we use GR for accelrating frames and gravity, and SR for inertial frames. Just wanted to point out that SR can deal with gravity and accelerations just by breaking the problem into a lot of little frames.

Sylas
2005-May-02, 09:29 PM
OK. I made a statement about gravity and relativity in this post (http://www.badastronomy.com/phpBB/viewtopic.php?p=464012&amp;highlight=#464012), and it was queried by A Thousand Pardons in this post (http://www.badastronomy.com/phpBB/viewtopic.php?p=464014&amp;highlight=#464014). The statement was: "General relativity is only required if we also have to take gravity into account."

I think I understand the objection now, but I don't think it really damages my intended point. How about if I phrase as follows:

Special relativity works fine on accelerated motions, but motions in a gravitational field are more difficult, since we can't identify an inertial frame in which the motions take place. Such cases are handled by general relativity, which is effectively a way of applying special relativity in many localised regions and integrating over them all.

I claim that there is a difference between special and general relativity, that general relativity is more difficult, and that the motivation for developing the general theory was to provide a way of handling gravity. The principle of equivalence is sufficient to allow derivation of the general theory from the special theory.

Cheers -- Sylas

PM. Thanks to ATP, Sean and Tensor, and to Tensor for a PM. I'm still learning this stuff and dealing with this objection has, I think, helped clarify a few matters for me, assuming I have this right now.

Sylas
2005-May-02, 10:00 PM
I don't think your use is really improper or non-standard. Over all, I would say we use GR for accelrating frames and gravity, and SR for inertial frames. Just wanted to point out that SR can deal with gravity and accelerations just by breaking the problem into a lot of little frames.

Addendum. Actually, my point is that SR can deal with accelerations without breaking up the problem into many little frames.

I solve an accelerated motion problem by first fixing some inertial frame in which the motion takes place. I can then calculate the velocity at any point in the locus, apply a time dilation, and integrate over the path for proper time of the accelerated observer. I'm not sure if that really coresponds to breaking up the problem into frames; I guess it does in some sense. But I have no idea how to do something similar with gravitation.

Example in one spatial dimension.

Suppose that the position of an accelerated particle in some inertial frame is given by x = sqrt(1 + t^2), in units where the speed of light is 1.

Then taking a time derivative, x' = t/sqrt(1 + t^2)

The gamma factor, g = 1/sqrt(1-x'^2) = sqrt(1 + t^2)

(Can you tell I picked an easy example?)

Let proper time for the particle be s. By SR, ds/dt = 1/g

Integrate to give s = ASinh(t)

This is the reading of a clock on the particle at time t in the frame where we observe its accelerated motion.

Put another way, when the particle's clock reads s, then the outside time is sinh(s), and the location is cosh(s).

Perhaps that integration step is effectively taking many little frames. I had not thought of it in such terms before. I think of it as working in one frame to define locations and times, and treating the reading on the particle's clock as a number that depends on time. The time derivative of this reading is given by SR, and I can integrate to get the reading as a function of time in the inertial frame. I'm not sure how to do anything similar with gravity.

Cheers -- Sylas

A Thousand Pardons
2005-May-03, 09:33 AM
The principle of equivalence is sufficient to allow derivation of the general theory from the special theory.
By that, do you mean, special relativity plus the principle of equivalence directly implies Einstein's general relativity? I don't think that's true--just look at all the alternative theories.

Sylas
2005-May-03, 01:09 PM
By that, do you mean, special relativity plus the principle of equivalence directly implies Einstein's general relativity? I don't think that's true--just look at all the alternative theories.

Yes; it is my understanding that SR + equivalence => GR.

For an alternative theory to refute this claim, we would need an alternative to GR that is equivalent to SR in the absence of gravitation; and that retains the principle of equivalence for acceleration and gravitation.

I am not aware of any such alternatives, always excepting those that are internally inconsistent. (Just look around this forum for examples!) As far as I am aware, alternative notions of gravitation generally violate equivalence in one way or another. But I'm open to counter examples.

Cheers -- Sylas