Thread: relativity busted from the get-go?

1. relativity busted from the get-go?

Okay. This will probably sound like everything everyone has heard in ATM before, but I'll post it anyway.

In the MM experiment, two beams of light were separated perpendicular to each and brought back together so their difference in times could be measured between that in the line of motion through the ether and that cross-wise to it. That in the line of motion is found to be t=d/(c-v)+d/(c+v)=2dc/(c^2-v^2) for the sum of the back and forth times and that cross-wise is found to be t=2dc/sqrt[c^2-v^2]. Lorentz noticed that if the distance were to shrink in the line of motion by the ratio of these two times, then that would produce a null result, as was found to be the case. Einstein also used a thought experiment with the same setup to come to the same conclusion, whereby gamma=sqrt[1-(v/c)^2] in relativity was born.

Now, there are two things about this. First, finding the speed of light in each direction should not have been c-v and c+v as were used for this, but c*(c+vo)/(c-vs) and c*(c-vo)/(c+vs) for Doppler, where vo is the velocity of the observer and vs is the velocity of the source. For some reason, the velocity of the source was not considered for this. This only applies for after the point of separation and before they reunite, however, but as long as all of the mirrors used are travelling at the same speed in the same direction in respect to one another, one will recieve a null result every time. So that causes relativity to become unsubstantiated right off the bat.

But I want to go a little further with this. Both the actual experiment and thought experiment used beams that travelled perpendicular to each other. So what I want to know is if we use a different angle between the beams, will we still find that the component of the distance in the line of motion must still be contracted by a factor of gamma in order for the times to match, or would it be something else. The experiment would still work fine and should still produce a null result, otherwise we could gain positive results this way. If it is gamma every time, then that would strengthen the case for relativity. But if not, then it is probably blown out of the water, and back to Doppler we go.

Of course, I had to do it in exactly the same way they did, using c+v and c-v for the velocities, to see if I come to the same results, and then compare for different angles. So we designate some angle we are finding for, figure the light is travelling over the resultant angle, so the distance there is ct, (sin L)d+vt along the line of motion, x along the original angle, and therefore (cos L)x for the base. From this we get

C=cos L, S=sin L
d=sqrt[(ct)^2-(Sx+vt)^2]/C

Separating the component for that which contracts along the line of motion with G (some value for gamma), then

x=sqrt[(GSd)^2+(Cd)^2]=d*sqrt[(GS)^2+C^2]

Factoring in for d, we now have

x^2=(GSct/C)^2-(GS*(Sx+vt))^2+(ct)^2-(Sx+vt)^2

Expanding this and solving for the exponents of x using the quadratic formula, this becomes

x=t*[-Sv+sqrt((Sc)^2+c^2/z-v^2/z)]/(1/z+S^2), where z=1+(GS/C)^2

Finally, solving for t, we get

t=(x/c)(1/z+S^2)/[-S(v/c)+sqrt(S^2+(1-(v/c)^2)/z)]

Alright. But that is only the time travelling to the reflecting mirror. We still need the time for coming back. By going through the same method as before, one finds this to be

t(back)=(x/c)(1/z+S^2)/[S(v/c)+sqrt(S^2+(1-(v/c)^2)/z)]

which turns out to be almost the same formula, since only the value for S changes sign. Adding these two times together and reducing a little gives us the total time for a beam travelling along some angle. So this is

t=t1+t2=2(x/c)*sqrt[S^2+(1-(v/c)^2)/z]/(1-(v/c)^2)

Finally, since we are actually looking for the formula in terms of d instead of x here, we can substitute back in for x and get

t=2(d/c)*sqrt[(GS)^2+C^2]*sqrt[S^2+(1-(v/c)^2)/z]/(1-(v/c)^2)

Okay. It's looking good. There's a couple of gammas already in there if they will only hold up. So let's find out. For the cross-wise beam, the angle is zero degree, so S=0 and C=1, so z=1. That gives us

t=2(d/c)*sqrt[1-(v/c)^2]/(1-(v/c)^2)=2(d/c)/sqrt[1-(v/c)^2], so that works out.

For along the line of motion, the angle is 90 degrees, so S=1, C=0, so z=infinity, and we have

t=2(d/c)*sqrt[1-(v/c)^2]/(1-(v/c)^2)=2(d/c)/sqrt[1-(v/c)^2], when G=gamma=sqrt[1-(v/c)^2], so that works out also.

Now let's try it for some angle in between, say 45 degrees, using the same G=gamma=sqrt[1-(v/c)^2] in the line of motion for the Lorentz contraction and v=.6c. So

C=S=sqrt(.5)
G=sqrt[1-(v/c)^2]=.8
z=1+(GS/C)^2=1+G^2=1+(1-(v/c)^2)=2-(v/c)^2=1.64
t=2(d/c)*sqrt[.5*.64+.5]*sqrt[.5+.64/1.64]/(.64)=2.67(d/c)

Finding for the same values at zero or 90 degrees, which come out to the same thing either way, we find

t=2(d/c)/sqrt[1-(v/c)^2]=2(d/c)/(.64)=3.125(d/c)

It is not the same time for other angles! That means the value of gamma must be changed depending on the angle between the beams. Therefore, the value of gamma is not always G=sqrt[1-(v/c)^2] for the contraction in the line of motion, and relativity does not hold for this. Since this is what relativity is based on from the get-go, it looks like we might be looking back at Doppler to explain the effects we observe.

Of course, before I make such a conviction, and pass the point of no turning back for relativity, starting to treat it as Doppler instead, I will need every person that can to run through these equations and make sure I haven't made any errors. After all, it was starting to look good there for a minute, with gamma actually carrying through in the calculations themselves, but it is the end result that matters. If they are difficult to follow, maybe somebody can re-perform the same calculations their own way to see if they come to the same conclusions and post it here.
Last edited by grav; 2007-Jan-31 at 03:31 PM. Reason: fixed typos

2. I didn't understand it, but I assure you, it doesn't sound like everything in ATM. For one, that really looked like math! (Or even if it isn't, it's closer than most others get.)

3. Established Member
Join Date
Jan 2007
Posts
927
Originally Posted by grav
Now, there are two things about this. First, finding the speed of light in each direction should not have been c-v and c+v as were used for this, but c*(c+vo)/(c-vs) and c*(c-vo)/(c+vs) for Doppler, where vo is the velocity of the observer and vs is the velocity of the source. For some reason, the velocity of the source was not considered for this. This only aplies for after the point of separation and before they reunite, however, but as long as all of the mirrors used are travelling at the same speed in the same direction in respect to one another, one will recieve a null result every time. So that causes relativity to become unsubstantiated right off the bat.
Are you saying that (x+y)/(x-y) = (x-y)/(x+y)?

As to the rest, you will actually have to go out and do the experiment at a different angle in order to show that there is a problem with special relativity.

4. Originally Posted by Kwalish Kid
Are you saying that (x+y)/(x-y) = (x-y)/(x+y)?
Sorry. I meant to include that. If you are referring to Doppler, the velocity is positive for when one approaches the other and negative when moving away. In the case where they are stationary to each other while moving along together, vo=-vs, so (c+vo)/(c-vs)=(c-vo)/(c+vs)=1. I wrote it that way because that is the way it is usually written, but I would personally prefer it if they were written as just (c+vo)/(c+vs) and (c-vo)/(c-vs), where vo=vs along the same line of motion. One can visualize it much easier that way.

Originally Posted by Kwalish Kid
As to the rest, you will actually have to go out and do the experiment at a different angle in order to show that there is a problem with special relativity
Ah, but that's the beauty of mathematics. The Lorentz contraction and relativity were based upon these very same mathematics. Mathematics does not lie, although it's possible I made an error somewhere, as the procedure is very lengthy and complicated, so I'll definitely need it checked. If there are no errors, then the value of gamma is not always sqrt[1-(v/c)^2], but only for an angle of 90 degrees between the beams, so matter does not contract in the line of motion, and we are back to using the Doppler effect.
Last edited by grav; 2007-Jan-31 at 03:48 PM.

5. Member
Join Date
Jan 2007
Posts
41
Following the nonstandard math format is somewhat slow but it appears that what you are proposing is the LET interpretation.

Originally Posted by grav
Finding for the same values at zero or 90 degrees, which come out to the same thing either way, we find

t=2(d/c)/(1-v2/c2)=2(d/c)/(.64)=3.125(d/c)
As you are assuming that gamma g = (1-v2/c2) is valid here. This does in fact lead the the same result for the MM experiment. This was actually how it was interpreted by Lorentz and the rest of the community prior to Einstein. However this interpretation gives preference to a particular frame of reference even though the same rules are equally valid for all frames of reference. You can read what Einstein said about it here. Einstein was a positivist and based his entire framework only on what was observable and not on any interpretation of what was observed. If you can define a situation where the two interpretations lead to different observable results then you can claim an empirical difference.

Originally Posted by grav
It is not the same time for other angles! That means the value of gamma must be changed depending on the angle between the beams. Therefore, the value of gamma is not always G=(1-v2/c2) for the contraction in the line of motion, and relativity does not hold for this.
You are assuming that the whole experiment exist on a single line of motion such that v = v' in both mirror frames relative to the beam splitter. You used the same v in g in both cases but what v represents is the rate of change of distance between the mirrors relative to the splitter. To see this imagine a rock 50 km from you traveling to a point 1 m to your left at 1 m/s. Initially the rocks velocity toward you is very nearly 1 m/s. When the rock reaches a point where your position is at a right angle to its direction of motion its velocity relative to you becomes zero for that instant in time. This is why Minkowski's spacetime is considered superior to Einstein's. They are mathematically equivalent.

It might be said that the distance between the mirrors and the splitter do not change under either situation as they are mounted on the same base. However by assuming the speed of light is relative to a medium which transforms according to g then by changing orientation in this medium it is physically equivalent because g as defined by the medium changes the relative distance between the points on the apparatus.

When english speakers are told wednesdays meeting is moved forward two days they are split 50/50 on wether that means monday or friday. Why is that?

6. Established Member
Join Date
Jan 2007
Posts
927
(c+vo)/(c-vs)=(c-vo)/(c+vs)=1 can only be the case if vo=-vs.

7. Originally Posted by my wan
You are assuming that the whole experiment exist on a single line of motion such that v = v' in both mirror frames relative to the beam splitter. You used the same v in g in both cases but what v represents is the rate of change of distance between the mirrors relative to the splitter. To see this imagine a rock 50 km from you traveling to a point 1 m to your left at 1 m/s. Initially the rocks velocity toward you is very nearly 1 m/s. When the rock reaches a point where your position is at a right angle to its direction of motion its velocity relative to you becomes zero for that instant in time. This is why Minkowski's spacetime is considered superior to Einstein's. They are mathematically equivalent.
Right. That is what I'm working on now. I'm trying to find what the effect would be using the actual formula for Doppler shift. That just depends on how each reflecting mirror and the half mirror are moving in respect to each other while the beams are split. If there is no relative motion between the mirrors, then one produces a null result. The velocity of Earth must also be figured in for along the line of motion, so we could potentially find our speed through an ether this way, but again, if there is no relative velocity between the mirrors, the overall effect cancels out. The angle at which two bodies are moving in respect to each other is important also, as you said. There would also be some effects produced by the time of propagation that would produce about the same results as that for relativity, at least in the line of motion, with just a different way of thinking about it, as mere optics.

Originally Posted by my wan
It might be said that the distance between the mirrors and the splitter do not change under either situation as they are mounted on the same base. However by assuming the speed of light is relative to a medium which transforms according to g then by changing orientation in this medium it is physically equivalent because g as defined by the medium changes the relative distance between the points on the apparatus.
Yes. I figured Lorentz wouldn't have made such a claim if the value of gamma changed depending on the orientation of the apparatus, so I didn't bother with that. So just to be clear on this, it is not the orientation of the apparatus I am changing, but the angle between the beams themselves.

8. Originally Posted by Kwalish Kid
(c+vo)/(c-vs)=(c-vo)/(c+vs)=1 can only be the case if vo=-vs.
That's right. It is rather confusing, but think of two bodies moving together along the same line of motion at the same speed. The one in front is basically attempting to move away from the one behind it, so it's velocity is negative. At the same time and at the same rate, the one in back is attempting to catch up to the one in front, so is positive. That is the way the Doppler effect is usually stated, but I personally think it only makes it more complicated than necessary to put it that way.

9. Established Member
Join Date
Jun 2006
Posts
2,445

Wrong

Originally Posted by grav
In the MM experiment, two beams of light were separated perpendicular to each and brought back together so their difference in times could be measured between that in the line of motion through the ether and that cross-wise to it. That in the line of motion is found to be t=d/(c-v)+d/(c+v)=2dc/(c^2-v^2) for the sum of the back and forth times and that cross-wise is found to be t=2dc/sqrt[c^2-v^2]. Lorentz noticed that if the distance were to shrink in the line of motion by the ratio of these two times, then that would produce a null result, as was found to be the case. Einstein also used a thought experiment with the same setup to come to the same conclusion, whereby gamma=sqrt[1-(v/c)^2] in relativity was born.

Now, there are two things about this. First, finding the speed of light in each direction should not have been c-v and c+v as were used for this, but c*(c+vo)/(c-vs) and c*(c-vo)/(c+vs) for Doppler, where vo is the velocity of the observer and vs is the velocity of the source. For some reason, the velocity of the source was not considered for this. This only applies for after the point of separation and before they reunite, however, but as long as all of the mirrors used are travelling at the same speed in the same direction in respect to one another, one will recieve a null result every time. So that causes relativity to become unsubstantiated right off the bat.

You completely miss the point. The velocity of light, as measured by any observer, regardless of the velocity of the observer relative to the velocity of the source of the light, is always c. And this has been confirmed again and again. There is no (c+v) or (c-v) for source or observer for c. Yes, you will get Doppler shift, but c is not affected. Doppler shift is a change in the wavelength, it is not a change in the velocity. All observers everywhere – and probably in all times since the big bang – measure the velocity of light in a vacuum in their inertial frame as c. Even if they are in an accelerating reference frame and measuring the speed of light from a source in a differently accelerating reference frame, they will still get c as the result!

Read the reference below. Remember, your GPS system includes relativistic corrections for satellite motion. We use Einstein’s postulates because they demonstrably work.

http://en.wikipedia.org/wiki/Special...ity#Postulates

10. Originally Posted by John Mendenhall
You completely miss the point. The velocity of light, as measured by any observer, regardless of the velocity of the observer relative to the velocity of the source of the light, is always c. And this has been confirmed again and again. There is no (c+v) or (c-v) for source or observer for c. Yes, you will get Doppler shift, but c is not affected. Doppler shift is a change in the wavelength, it is not a change in the velocity. All observers everywhere – and probably in all times since the big bang – measure the velocity of light in a vacuum in their inertial frame as c. Even if they are in an accelerating reference frame and measuring the speed of light from a source in a differently accelerating reference frame, they will still get c as the result!

Read the reference below. Remember, your GPS system includes relativistic corrections for satellite motion. We use Einstein’s postulates because they demonstrably work.

http://en.wikipedia.org/wiki/Special...ity#Postulates
I agree. That's a point I made in another thread, that if it is always c, then how could Einstein have used c+v and c-v for the relative velocities in his original paper in order to find gamma? It is really only an exact replicate of the original MM experiment and the Lorentz contraction hypothesized to explain the null results. So if you are arguing that c+v and c-v shouldn't be used, then you are not arguing against me, but Einstein himself. The thing is, though, that it is incorrect for both relativity and the MM experiment as far as Doppler is concerned. But Einstein just uses exactly the same equations as used in the experiment to find it to begin with. Have a look at this. Then compare it to this (sections 1-3). Look familiar? He does precisely the same thing, by considering the back and forth motion for light in the direction of propagation, then (about the middle of section 3), considers it for the cross-wise direction as well and compares them, the same as with the Michelson-Morley experiment, using the same equations, and finding the same results.

11. Established Member
Join Date
Apr 2006
Posts
1,031

Nereid Can Do It

Originally Posted by grav
...

, I will need every person that can to run through these equations and make sure I haven't made any errors. ...
I imagine: "This looks like a job for Nereid."!!

Read that with the familiar "voice" from the quote: "This looks like a job for Superman."

12. Order of Kilopi
Join Date
Mar 2004
Posts
13,441
Originally Posted by Squashed
I imagine: "This looks like a job for Nereid."!!

Read that with the familiar "voice" from the quote: "This looks like a job for Superman."
What can I say?

Reports of my superpowers are greatly exaggerated ...

13. Established Member
Join Date
Apr 2006
Posts
1,031

Excellent Come Back

Thanks for the laugh.

14. I've been making some diagrams using Doppler for the MM experiment and I've noticed three very important things about it.

One, if a person positions themself stationary to the medium and runs the same experiment as M&M did for the paths of the light as they reflect off of the mirrors, one gains the same results they did for the times. That is to say, if one takes the same paths as they occur for when the observer is stationary to the apparatus, and apply that to a velocity of v in the line of motion, while the observer remains stationary to the medium, then all results come out exactly as they say for the paths and times for the light. Apparently, though, and unfortunately, this is as far they go in their considerations, which make all of the difference in the world, as we shall see.

Okay. So the paths and times are exactly as predicted according to Doppler. But what else might be going on with this? Well, take a look at these diagrams. The first is for an observer travelling with the apparatus. The second is for an observer that remains stationary with the medium. Everything appears to be in order at first glance. But look a little closer. Notice anything odd? That's right, when considering the second diagram, the angles of incidence are not equal to the angles of reflection. In fact, if they were, the two beams could not realign at all. Believe me, I have tried. So this is the way the light really reflects off of the mirrors. The angle of reflection is only equal to the angle of incidence to an observer that is stationary with the mirror, but changes with if the observer and mirror have a relative velocity. For instance, if the apparatus is travelling through the medium at .5c, then the path relative to the medium itself is 30 degrees. That means that the light bounces off the cross-wise mirror, again at 30 degrees, but in the opposite direction, and reaches the half silvered mirror at 30 degrees of the regular 45, or only 15 degrees off the normal. But it is still reflected back in the line of motion at 45 degrees, tripling its ratio for the angles. I think that might be very important, not just in regard to relativity, but for optics as well.

Alright, so the angle of incidence and reflection change with relative velocities between mirror and observer. But that would not change the times, right? True, but the frequency would have already had to have changed upon reflection. To see this, imagine the light reflecting off of just one of the mirrors using Doppler shift. Let's say the cross-wise mirror is moving away from the apparatus at the same time. As we travel with the apparatus, we notice that as the light catches up to the mirror along the normal, since the mirror is moving away, let's say at .5c, then the pulses take longer to reach the mirror before they are reflected, twice as long for .5c . Because of this, by the time they are reflected back again, the wavelength has increased two-fold. So the frequency is cut in half. Now let's look at that the way the MM experiment does. The light travels to the same mirror at an angle, but only the point of reflection is considered, and that path back. It is not considered for where the frequency might have changed at the point of reflection for the stationary observer, only the paths and angles.

So what happens when we apply this? Well, the two reflecting mirrors don't actually have a relative velocity to the apparatus and so the angles of reflection do equal the angles of incidence for them , and the frequency does not change. But let's look at the the one where the angles do change, the half silvered mirror. If the angles become different for different relative velocities of the observer, then the observed frequency must change as well. This, then, would be where the real Lorentz transformation takes place, but nothing like an actual contraction or anything. I haven't worked out the details for this, yet, but I will. That would also be important in optics, I'm sure. But what this means is that if the frequency changes for the stationary observer upon reflection on the half silvered mirror, then the points of reflection and times of travel are not the only considerations to be made here.

I have also figured out why gamma=sqrt[1-(v/c)^2] followed us all the way through the equations, but not at the end, depending on the angle between the beams. It is related to the time of travel in the line of motion, that's true, but is not a contraction or anything. For true Doppler, it would simply mean that if the apparatus is moving at greater than c through the medium, then the pulses, which travel at constant c through the medium themselves, can never catch up in order to be reflected. So the apparatus can never travel faster than c if the pulses are to be reflected in the first place. So [1-(v/c)^2] just represents an physical impossibilty, not that the apparatus can never travel faster than c through the medium, but that the pulses can never be reflected if it does.

15. Established Member
Join Date
Jun 2006
Posts
2,445

Originally Posted by grav
Have a look at this. Then compare it to this
Folks, these are really good links that Grav provided.

In the famous words of the Fonz, I could be "Wr . . . wro . . . You said it!"

16. Originally Posted by Nereid
What can I say?

Reports of my superpowers are greatly exaggerated ...
Not to me...

17. Aha! I've got it. As usual, it's always so simple in the end. In reference to my last post, I've been trying to figure out how the light could reflect off of the mirrors at a different incident angle and still come by the same angle of reflection as for the same mirror when stationary. It turns out that it really is the times we consider here, but not for light at an angle, but for the light when an observer is stationary to the apparatus, because of the motion of the source after all. Consider a source that moves with the apparatus. The apparatus and source both move at .5c through the medium. That means the light must travel at a 30 degree angle to get to the first mirror, right? That's where the difference in times in the MM experiment comes from, since after splitting, one beam moves in a straight line back and forth in the line of motion and the other continues to move at an angle.

But look at what is really happening here. Yes, the pulses move at an angle. But it is not a straight-line angle for the light. That is to say, the pulses of light are not spaced along a line, one behind the other, on that angle. So how are they spaced? Let's find out. Let's say a pulse is emitted by the source at t=0 while the apparatus moves through the medium at .5c . At t=1, that pulse travels .02 meters, say. Since the light is travelling at c and the apparatus at .5c, then if the pulse is to meet the half mirror, then it is travelling at an angle of 30 degrees. So if it travels .02 meters, it has gone .01 meters in the forward direction and .01732 horizontally. At this point, the source emits another pulse, but it has moved ahead .01 meters in the same time. So the first and the second pulses are at the same position in the forward direction. After two time intervals, the first pulse is now at coordinates (.034641, 0+2*.01=.02), the second at (.01732, .01+1*.01=.02) at the third at (0, .02+0*.01=.02). See a pattern here? Even though the overall path of each individual pulse is at a 30 degree angle to the apparatus through the medium, the sum of the times of emission and the times of travel of each individual pulse puts them on the same horizontal line, side by side. If an observer that is stationary to the medium were watching this, they wouldn't say the path is at an angle at all, any more than the observer moving with the apparatus would! Instead, they would say that the line of the pulses, one directly behind the other, lie upon a horizontal line at all times, just as the moving observer would, and they would see that entire horizontal line moving with the apparatus as it travels forward at .5c, and with the same difference between the frequencies and times! There is absolutely no difference between what a stationary and moving observer will see in respect to the motion of the apparatus itself when the apparatus and source are moving together, and so there is no difference in the measurements as compared to when the apparatus is stationary to the medium or moving through it when the source moves with it as well!
Last edited by grav; 2007-Feb-02 at 05:27 PM.

18. Originally Posted by grav
Aha! I've got it.
Are you still thinking that relativity is busted from the get-go, or no?

19. Originally Posted by hhEb09'1
Are you still thinking that relativity is busted from the get-go, or no?
Yes. At least as far as a real spatial contraction in the line of motion in terms of gamma and so forth goes, but not as far as redshift and time dilation and other effects are concerned which are caused solely by the propagation time of light through purely geometrical means in an otherwise unchanging space-time.

The MM experiment never should have figured for light travelling at an angle. The individual pulses do, but if the source travels with the apparatus, the overall path of the light is straight across, same as for when the entire device is stationary. The measured frequency as emitted along this line is different than it would be when stationary, by f/sqrt[1-(v/c)^2], that's true, but that is just the frequency of the pulses along the straight line because of the path the pulses take, which becomes f/[cos(sin-1(v/c))] in both directions being considered, so the measured difference in the two beams remains the same, zero. In other words, it is the wavelengths (distance between the pulses) as measured along this line that appear shorter than they should otherwise be if travelling one behind the other along the angle, not an actual contraction of the device itself in the line of motion or anything, because of the difference between the pulses travelling at an angle, but the line through the pulses, one behind the other, still being that of a horizontal line.
Last edited by grav; 2007-Feb-02 at 06:40 PM.

20. Established Member
Join Date
Apr 2006
Posts
1,031

Using Light to Measure Things

Originally Posted by grav
Yes. At least as far as a real spatial contraction in the line of motion in terms of gamma and so forth goes, but not as far as redshift and time dilation and other effects are concerned which are caused solely by the propagation time of light through purely geometrical means in an otherwise unchanging space-time.
grav, I speculate that once you have completed your investigation you'll find that the length contraction is an artifact of using light to measure things - the length contraction is not real - but it all works out in the end because the relevant calculations null out the artifact.

21. Originally Posted by grav
The individual pulses do, but if the source travels with the apparatus,
Quick question, which I was not able to answer just by going back through this thread. Why "if the source travels with the apparatus"? Isn't the source part of the apparatus, or am I misunderstanding something?

22. Originally Posted by hhEb09'1
Quick question, which I was not able to answer just by going back through this thread. Why "if the source travels with the apparatus"? Isn't the source part of the apparatus, or am I misunderstanding something?
Yes. If the source is travelling with the apparatus, then it can be considered part of it, or attached. The only way we would actually see the light travelling at an angle, then, is if the source remained stationary while the apparatus travelled away from it, but that angle will then constantly change with time as the device travels further and further away from the source, in order to strike the mirror. If it is travelling with it, the laser would have to be aimed forward, "ahead of the target", in order to strike it in the first place. But since the source and device are moving together, the overall path of the light will just be that of a straight line from source to mirror.
Last edited by grav; 2007-Feb-02 at 07:19 PM.

23. Established Member
Join Date
Apr 2006
Posts
1,031

Relativistic Beaming

Originally Posted by grav
Yes. If the source is travelling with the apparatus, then it can be considered part of it, or attached. The only way we would actually see the light travelling at an angle, then, is if the source remained stationary while the apparatus travelled away from it, but that angle will then constantly change with time as the device travels further and further away from the source, in order to strike the mirror. If it is travelling with it, the laser would have to be aimed forward, "ahead of the mirror", in order to strike it in the first place. But since the source and device are moving together, the overall path of the light will just be that of a straight line from source to mirror.
The "aiming" that you refer to is automatically done by relativistic beaming. This post investigates the phenomenon.

Maybe a better name is Relativistic Aberration.

24. Order of Kilopi
Join Date
Nov 2002
Posts
6,235
Originally Posted by grav
Yes. At least as far as a real spatial contraction in the line of motion in terms of gamma and so forth goes,
Grav,
You are aware that the spatial contraction is only for the one dimensional object used to illustrate the concept, right? An actual three dimensional object would appear to be rotated, not contracted, the result of light propogation time.

25. Originally Posted by Squashed
The "aiming" that you refer to is automatically done by relativistic beaming. This post investigates the phenomenon.

Maybe a better name is Relativistic Aberration.
Yes. That all sounds similar, and it would work the same with a light reflecting clock as well, as you wrote. We would always see the beams travelling straight up and down along the same straight-line path in relation to the mirrors, no matter how the clock was moving relative to us.

26. Established Member
Join Date
Apr 2006
Posts
1,031

Angles

Originally Posted by Tensor
Grav,
You are aware that the spatial contraction is only for the one dimensional object used to illustrate the concept, right? An actual three dimensional object would appear to be rotated, not contracted, the result of light propogation time.
Tensor, are you talking about what they present in Section 4. An Application: The Rotation of Bodies in Transverse Motion?

27. Originally Posted by Tensor
Grav,
You are aware that the spatial contraction is only for the one dimensional object used to illustrate the concept, right? An actual three dimensional object would appear to be rotated, not contracted, the result of light propogation time.
Are you referring to the work of Minkowski? I haven't had a chance to look into that completely yet, but if it is based solely on the time of light propagation, then there should be no problem with it, as it is then purely optics.

28. Order of Kilopi
Join Date
Mar 2004
Posts
13,441
Originally Posted by grav
Are you referring to the work of Minkowski? I haven't had a chance to look into that completely yet, but if it is based on the time of light propagation, then there should be no problem with it, as it is then purely optics.
(my bold)

Given your history of working through relativity problems grav, I'd counsel a little caution when it comes to qualifiers like "purely" ...

29. Originally Posted by Nereid
(my bold)

Given your history of working through relativity problems grav, I'd counsel a little caution when it comes to qualifiers like "purely" ...
Advice taken. I edited my post to say that if it is based solely on the time of light propagation, then it is purely optics. (Of course, that refers only to a propagation through flat space-time as well, so I see your point)

30. Order of Kilopi
Join Date
Nov 2002
Posts
6,235
Originally Posted by Squashed
Tensor, are you talking about what they present in Section 4. An Application: The Rotation of Bodies in Transverse Motion?