Fitzgerald -Lorentz contraction

granpa

the light has to catch up with the front of the object which is moving in the same direction. the simplest way to calculate it as to pretend the distance remains the same and sue c-v for one leg of th etrip and c+v for the other.

mich

ok....but why say "pretend"? ;)

But it works out close to being the same, I would guess.
Still the point being that by letting the distance as remaining the same, and "pretending" the light's velocity as being (c+v) (c-v), we still have a difference in time when the light pulse will return relative to the pulse travelling the vertical path. It's excactly, in my opinion,the same case as in the M&M experiment, except here, we have the observer on a moving frame (relative to the experiment).
Therefore, if that's the case, then, another type of Michelson's experiment can be performed, and here, according to Relativity, if I'm not mistaken, we ought to have a light spectral fringe shift; if the result is still nul, then it seems that the theory of length contraction would have been disproven.

Andre

granpa

I'm not going to do the math for you. if you dont want to do it yourself then look it up on the web. you assume length contraction has takes place but you use c-v and c+v. it works out exactly.

RandallB

No Andre you have miss stated the Michelson-Morley problem.
With two observers you do not have one source you have two, and if they are both together when they send their separate signals if they both have the same coordinate measures of distance for light to travel the same speed the signal for moving observer would have to travel ahead of the light sent from the stationary observer in order to reach the same distance away from its source in the moving frame as the light in the stationary frame.
Unless you point the beams “backwards” where the moving source would need to have its light slow down in that direction.
Combining the two effects in the Michelson-Morley experiments would show this difference as the vertical and horizontal light “could not arrive at the same time”!
That was the point of Michelson-Morley they could not detect the changes that had to be there for unchanging distances.

Lorentz offered the solution that would explain the observed results as objects physically change “shape” by becoming “shorter” in the direction of motion. Lorentz Contraction.

In order to get Lorentz Contraction to work with a the fixed value of “c” Michelson-Morley results were showing also requires Time Dilation be applied.
And that is how you get to Special Relativity.

mich

I don't believe that I was arguing against your point, granpa. Would you agree, however, that the light travelling horizontally will not arrive at the same time as the light travelling vertically, as seen by the observer on the moving frame?

It's just a question of relative simultanety.

Andre

granpa

2 events that occur at the same place at the same time do so for all observers regardless of velocity. all observers see the light pulses arrive at the same time. thats the WHOLE POINT.

its 'pretend' because the light isnt moving at c+v its moving at c.

RandallB

Yes, That that was the point made by Michelson-Morley in the classical view where distances do not change in a reference frame. They first showed mathematically ”that the light travelling horizontally could not arrive at the same time as the light travelling vertically” to establish a prediction of what their Michelson-Morley experiment would show. Again based on a either, with frames using the same gauge for distances, both frames able to define relative simultaneity measures that are the same measured from both frames.

You are incorrectly assigning the MM expectations as a SR expectation – this is wrong.

You need to reread your source information on Michelson-Morley experiment – if you do not have that issue clear, you will not be able to understand Michelson-Morley.
You need to be clear on that BEFORE you can understand Lorentz Contraction.

Then you can add Time Dilation to understand Special Relativity. Where two frames will not agree on simultaneity measures!
And that is why by following SR rules “the light travelling horizontally will arrive at the same time as the light travelling vertically” just as Michelson-Morley unexpectedly found to be true in their experiments.

JesseM

I hope you understand that "relativity of simultaneity" does not allow different frames to disagree in their predictions about localized events happening at a single point in space, only about the simultaneity of events with a spatial separation between them. So if one frame predicts that two beams of light arrive at a single point in space at the same moment, all frames must predict that. This is equally true in a Lorentz aether theory as it is in relativity.

mich

Sorry Randall; I wasn't ignoring you. Since I've gone back to work, my time for computer play is kinda limited. My reply is therefore directed not only to you but to granpa as well as others.

I wasn't speaking of two sources for simplicity sake. Just imagine the M&M experiment performed on one frame, where we all know the result will be null.
But now imagine a second observer on a moving frame who also sees the experiment. According to Relativity, what does he observe?
It seems to me obvious that the two light signals (horizontal path and vertical path) will not arrive at the same time since Relativity claims that one path will be forshortened while the other will not.

Granpa mentioned that the light will have a (c+v) (c-v) velocity relative to the frame where the experiment happens. I did tell him that I don't have any problems with this, since, if I assume the light remains c according to my measurements, then it must be going at (c+v) (c-v) relative to the moving frame. However, I personally disagree with his claim that we ought to pretend the distance remains
the same. This is what the whole thread is about, that is, length contractions of moving frames, so, according to Relativity, we ought to leave the moving frame contracted.

Therefore it seems that the moving observer will measure the path as being 2s*SQRT(1-v2/c2). The time t being
[2s*SQRT(1-v2/c2)] / (c+v) (c-v)

As for the vertical path we would have no contraction, so
t2 = 2s/c, however, because of a dilation of time, we would have
2s/c*SQRT(1-v^2/c^2).

So we would have a ratio of t2 / t1 = (c+v) (c-v), or (c^2 -v^2)

Andre

granpa

the shortening of the paths isnt the problem, its the solution. without contraction the pulses of light wont arrive at the same time. I dont really understand how you can have this so messed up but its clear that you havent done the math. the light appears to both observers to be moving at c. af you want to calculate it that way thats fine. its just easier to do it the way I told you. do the math.

granpa

show us some equations and we'll help you. otherwise, I'm done.

mich

Well, you most probably are correct, for I'm not a scientist of any kind whatsoever....I'm mearly trying to understand something which seems confusing to me.

Now, I did try to explain in mathematical terms the problem I had even using the (c-v)(c+v) terms you asked me to use. So, I'm not quite certain as to what you want me to do.
I'll try something else;

For the horizontal path, the moving oberserver measures the light speed as being c, relative to himself, and therefore will measure the light speed as being (c+v) (c-v) relative to the experimental frame. Would you agree? Now what about the length of the path the light will travel? Will it be : s ( length measured at rest), or will it be s* gamma? I would suspect s* gamma, since Relativity predicts a shortening of measuring rods within moving frames.

I will leave you with this and we could go through it in small steps.

I am going to bed now, so I will write you back tomorrow, if you reply.

Andre

granpa

when I said that the distance remains the same I meant something completely different. if you use c+v and c-v then you use one distance. if you use just c then you have to use a different distance. its much easier to do the first.

granpa

if you are calculating the time as measured by an observer then you would use the distance as measured by that observer.

calculate one leg at a time.

the vertical path is more difficult than you are making. to the moving observer the light moves at an angle. you CAN get into sin and cos but you dont have to. a^2 + b^2 = c^2 is all you really need to solve it. its been a long time since I did it and I dont remember exactly what I did but I know that a lot of things cancel out.

Doc Al

Wrong again. Actually, if you understood the principle of relativity, you could immediately conclude that the two light signals must arrive at the same time.
If you insist on using the speed relative to the lab frame as measured by the moving observer (Rather convoluted, wouldn't you say? Why not just stick to measurements with respect to you?), then the speeds will be (c+v) and (c-v). The length of the path will be s/gamma (not s*gamma). Do the calculation and you'll find, as expected, that the time for the round trip path will be the same as the time measured in the lab frame multiplied by the time dilation factor gamma.

RandallB

NO that is not true.
Your problem is not in understanding SR. It is that you do not understand MM in the first place.
You need to get that straight first, and you won’t do that by applying SR when you clearly do not understand either M&M or SR.

M&M did not need to “a second observer” they used one observer making two observations (horizontal & vertical) and changed the direction of motion for that one observer (horizontal or vertical).
MOST important!! M&M did not agree that “we all know the result will be null” as you claim; you need to show why you think M&M expected a “null result”.

As you look at the details of exactly how M&M expected a NON-Null result you should see why the geometry you just provided for your relativistic length and time calculations is just plain wrong.
Don’t start with the relativistic solution – start by understanding the Math and Geometry used to show the M&M expectation of a Non-Null result.
Once your straight on two things
1) how they made the Non-Null Result predictions and
2) ALL their observations showed Null Results
you will then understand the paradox they and science had to deal with.

THEN apply the relativistic solution to the M&M math and geometry to solve the paradox.
(Note: according to Lorentz himself the SR relativistic solution was a more complete solution to the paradox than his Lorentz solution as the SR version introduced time dilation and the simultaneity issue. I.E. Lorentz does not include “simultaneity” only SR re-interpretations of Lorentz can apply “simultaneity”)

When you get your geometry and math correct to follow M&M you will see that only using relativistic SR can you predict a Null Result as observed; in contrast to the classical Non-Null results predicted by M&M that did not match the observations.

As this is the third time I’ve pointed you to this flaw in your approach – I trust you will take the time to research and ruminate on this one point. And then proceed with further application of what you learn about SR from that.
If you do that I’m confident you will need no additional help understanding the foundation of SR correctly. So I’ll unsubscribe from this thread and see you in another as you move on to more advanced topics.

JesseM

Maybe it would be helpful to do an actual calculation here? Suppose we have a Michelson-Morley type apparatus with two arms at right angles, each 20 light-seconds long in the rest frame of the apparatus. Now consider what will happen from the perspective of an observer who measures the apparatus to be moving at 0.6c to the right along the axis of the horizontal arm. In this frame, the vertical arm will still be 20 light-seconds long, but the horizontal arm is shrunk to 16 light-seconds due to Lorentz contraction, because [tex]20*\sqrt{1 - 0.6c^2/c^2} = 16[/tex]

Now suppose at t=0 some light is released at the point where the left end of the horizontal arm meets the bottom end of the vertical arm. Since the light is moving to the right at 1c while the horizontal arm is moving to the right at 0.6c, in this frame the distance between the light beam and the right end will be decreasing at a rate of 0.4c, and since the horizontal arm is 16 light-seconds long, the time for the light to reach the right end will be 16/0.4 = 40 seconds. Then the light is reflected from the right end back at the left end, and now the left end is approaching the light at 0.6c, so the distance between the light and the left end is decreasing at a rate of 1.6c. So, the time for the light to get back to the left end will be an additional 16/1.6 = 10 seconds. So, in this frame the total time for the light to go from the left end to the right end and back is 40 + 10 = 50 seconds.

Now consider the light going up and down the vertical arm. The top end of the vertical arm starts out 20 light-seconds directly above the point where the light is released, then after 25 seconds it is still 20 light-seconds above the release point on the vertical axis, while it has moved (25 seconds)*(0.6c) = 15 light-seconds away from the release point on the horizontal axis. So using the pythagorean theorem, the top end is now a distance of [tex]\sqrt{(20)^2 + (15)^2} = \sqrt{625} = 25[/tex] light-seconds away from the point where the light was released. Since the light travels 25 light-seconds in 25 seconds, this must be the time when the light catches up to the top end. Similar calculations show that 25 seconds after this, the bottom end of the vertical axis will be 25 light-seconds away from the position where the light hit the top end, so this must be the time the light returns to the bottom end. So, you can see that the total time for the light to go from the bottom to the top and back on the vertical arm is 25 + 25 = 50 seconds, exactly the same as the time for the light to go from the left to the right and back on the horizontal arm, so the two light waves will indeed meet at the same point in space and time. As I said before, it is a basic rule in relativity that if one frame says two events happen at the same position and time, this must be predicted in all frames.