Distance & Time Light Travels in Approaching Mirrors Problem

[stevmg]

Say that two mirrors are 3 ltyr apart and perfectly facing each other. Say they approach at 0.6c (relative to a "stationary" observer)
M1---------------->0.6 0.6<----------------M2

If a lightbeam is flashed at M1 towards M2 and bounces back and forth until the mirrors cross each other, how far will the light beams travel and how long will it take for the light beams to travel until the mirrors do cross.

This is a variant of the hummingbird-approaching trains problem but now we have relativity mixed in.

Maybe this belongs in the homework section, so, if so, please move it there.

My answer is 2.72 years... is that correct? If we were Newtonian or Galilean it would be 2.5 years.

stevmg

 

[Doc Al]

My answer is 2.72 years... is that correct?

How did you arrive at this answer? (I presume you are trying to find the time according the 'stationary' observer.)

 

[stevmg]

1) Closing speed is not 1.2c because that's impossible but is 0.8824c (Velocity Addition formula)
2) By length contraction distance = 3 (1 - .8824^2)^(1/2) = 1.411 ltyr
3) time = 1.411/0.88241.60 = 1.60 year
4) If we were Galilean, it would be 3.0/(0.6 + 0.6) = 2.5 years
I think I screwed up before. Hope I am right now.

stevmg

 

[yossell]

stevmg - it's late so I'm confused - but why is relativity even relevant to this question? It looks as though all the information and the calculation to be done wholly from the stationary observer's point of view and I can't see any funny stuff going on that calls for anything particularly relativistic.

 

[Doc Al]

1) Closing speed is not 1.2c because that's impossible but is 0.8824c (Velocity Addition formula)

Here's your mistake. As seen by the stationary observer, the closing speed is 1.2c. Why do you think that's impossible?

The velocity addition formula is only relevant if you wanted to know the speed of one mirror in the frame of the other.

 

[starthaus]

1) Closing speed is not 1.2c because that's impossible but is 0.8824c (Velocity Addition formula)

Arrrgh, you are loosing all your previous gains. The distance between the two mirrors is covered at the closing speed 1.c in 2.5 years. End.

2) By length contraction distance = 3 (1 - .8824^2)^(1/2) = 1.411 ltyr

There is no length contraction from the perspective of an observer that notices the mirrors "closing" the 3ly distance at a closing speed of 1.2c

3) time = 1.411/0.88241.60 = 1.60 year

No.

4) If we were Galilean, it would be 3.0/(0.6 + 0.6) = 2.5 years
I think I screwed up before. Hope I am right now.

stevmg

Closing speed applies exactly the same way in SR as in galilean kinematics. I am quite sure I told you this early on when we got started on the subject of closing speeds.

 

[stevmg]

S-o-o-o-r-r-y

OK, looking from mirror 1 would I be right?

Steve

 

[Doc Al]

OK, looking from mirror 1 would I be right?

No. For one thing, the length contraction would be based on 0.6c, not the relative speed of the mirrors.

 

[stevmg]

Now, here's we I need a class on this subject as I am getting lost.

Doc Al. You would agree that the relative speed from M2 onto a "stationary" M1 would be 0.8824. Is that correct?

starthaus - I am really confused about closing speed.

In the MMX .pdf file you referred me to, you explained that the lack of difference in roundtrip times for the reflecting beams of light (head on versus perpendicular) was due to a contraction factor in the direction of motion of the Earth. But that contraction factor was explained by Lorentz and Fitzpatrick by a later 1898 experiment. At the time of the MMX-3 experiment, I thought that the apparent lack of difference in roundtrip time for these light beams was unexplained using just closing velocities. By closing velocity assumption, there should have been a difference in roundtrip times. At least, that's what I got out of your paper. the time dilation part at the end was more icing on the cake or another way of proving it.

You would have to theorize a length contraction in the direction of the Earth orbital velocity to account for the similarity of the roundtriptimes.

 

[starthaus]

Now, here's we I need a class on this subject as I am getting lost.

Doc Al. You would agree that the relative speed from M2 onto a "stationary" M1 would be 0.8824. Is that correct?

starthaus - I am really confused about closing speed.

In the MMX .pdf file you referred me to, you explained that the lack of difference in roundtrip times for the reflecting beams of light (head on versus perpendicular) was due to a contraction factor in the direction of motion of the Earth. But that contraction factor was explained by Lorentz and Fitzpatrick

You mean FitzGerald.

At the time of the MMX-3 experiment, I thought that the apparent lack of difference in roundtrip time for these light beams was unexplained using just closing velocities.

No, the FitzGerald contraction has been simply replaced by the "length contraction" of SR.

 

[Doc Al]

Doc Al. You would agree that the relative speed from M2 onto a "stationary" M1 would be 0.8824. Is that correct?

Yes. The relative speed of the mirrors is 0.8824c.

 

[stevmg]

You mean FitzGerald.

Never knew the man.

No, the FitzGerald contraction has been simply replaced by the "length contraction" of SR.

So what? What I was saying that it took the Fitzgerald contraction in the direction of the Earth's motion to explain the LACK of difference in roundtrip time for the light bouncing between the mirrors. Is that right?

 

[stevmg]

Doc Al -

The relative speed of the mirrors is 0.8824c. Why would we use 0.6c as you stated?

If we assume that the two mirrors are 3 ltyr apart in the M1 FR (that wasn't the original problem, but let's use that), would it take 3/0.8824 = 3.40 year in the M1 FR or do we length contract the 3 ltyr. I know this is getting basic but I have to get out of my "black hole" of misunderstanding somehow.

If the answer is yes, then I have a little more.

 

[Doc Al]

Doc Al -

The relative speed of the mirrors is 0.8824c. Why would we use 0.6c as you stated?

Because the 3 LY distance was given in the stationary frame. (Until you changed the problem!)

If we assume that the two mirrors are 3 ltyr apart in the M1 FR (that wasn't the original problem, but let's use that), would it take 3/0.8824 = 3.40 year in the M1 FR or do we length contract the 3 ltyr. I know this is getting basic but I have to get out of my "black hole" of misunderstanding somehow.

This is a different problem. Since the distance is now 3 LY in the M1 frame, and we are interested in the time according to the M1 frame, no length contraction is required. 3.4 years would be correct.

 

[stevmg]

Because the 3 LY distance was given in the stationary frame. (Until you changed the problem!)

This is a different problem. Since the distance is now 3 LY in the M1 frame, and we are interested in the time according to the M1 frame, no length contraction is required. 3.4 years would be correct.

Doc Al, in the stationary frame, wouldn't one use the 0.6 + 0.6 = 1.2 for the time then (in other words, one would add up the times that the two mirrors would need to travel 1.5 ltyr each to = 3 ltyr.) In other words, it would take 1.5/0.6 = 2.5 yr for each mirror to move the necessary 1.5 ltyr to "meet." Thus, the 2.5 years would be the answer for that scenario. True? Kind of a round about way of looking at it but conceptually is clearer to me.

Now, with regards to the Michelson-Morley experiment, if we did not use length contraction which is not part of Galilean mechanics, there WOULD be a difference in roundtrip time for the perpendicular and parallel light beams, right? See the figure below:

Because of the mathematics of inverses, the total travel time back and forth parallel to the relative motion of an "ether" would be greater than the back and forth travel time for light perpendicular to the "ether." That would be true in a Galilean or Newtonian Universe, right?

 

[Doc Al]

Doc Al, in the stationary frame, wouldn't one use the 0.6 + 0.6 = 1.2 for the time then (in other words, one would add up the times that the two mirrors would need to travel 1.5 ltyr each to = 3 ltyr.) In other words, it would take 1.5/0.6 = 2.5 yr for each mirror to move the necessary 1.5 ltyr to "meet." Thus, the 2.5 years would be the answer for that scenario. True? Kind of a round about way of looking at it but conceptually is clearer to me.

Nothing wrong with that.

Now, with regards to the Michelson-Morley experiment, if we did not use length contraction which is not part of Galilean mechanics, there WOULD be a difference in roundtrip time for the perpendicular and parallel light beams, right?

If there was an ether in which light traveled, then there would be an expected time difference. One way to explain the null result is with Fitzgerald's length contraction proposal.

Because of the mathematics of inverses, the total travel time back and forth parallel to the relative motion of an "ether" would be greater than the back and forth travel time for light perpendicular to the "ether." That would be true in a Galilean or Newtonian Universe, right?

Yes.

 

[stevmg]

If there was an ether in which light traveled, then there would be an expected time difference. One way to explain the null result is with Fitzgerald's length contraction proposal.

Fitzgerald's (as I told starthaus after calling him Fitzpatrick - "I never knew the man") length contraction is NOT part of Galilean or Newtonian mechanics or physics, thus, a new concept must be introduced to explain the null difference in order to justify the use of closing velocity (c + u) or (c - u). You have to shorten the distance to make the elapsed roundtrip times come out the same by, coincidently, the same gamma factor of later relativity fame.

Now, where does "closing velocity" (c + u) or (c-u) fit in? In other words, either the "u" doesn't exist (i.e., no ether) and closing velocity just remains a mathematic theoretical entity which has no place here. Light always approaches at c and there is no closing velocity - at least not with light unless one calls c the universal closing velocity for light, no pluses or minuses.

By "your" reasoning, one can look at two objects approaching each other, say two trains steaming at each other one from the west and one from the east. The total closing velocity would be <2c but each individual's speed would be <c (because of so-called "relativistic mass increase" - an outmoded term these days.)

One farmer once saw such an event in his field - two trains on the same track blasting towards each other at way subrelativistic speeds, but fast, never-the-less.

A reporter asked, "What did you do?"

The old codger replied, "W-e-l-l - I thought 'What a hell of a way to run a railroad!'"

 

[yuiop]

Say that two mirrors are 3 ltyr apart and perfectly facing each other. Say they approach at 0.6c (relative to a "stationary" observer)
M1---------------->0.6 0.6<----------------M2

If a lightbeam is flashed at M1 towards M2 and bounces back and forth until the mirrors cross each other, how far will the light beams travel and how long will it take for the light beams to travel until the mirrors do cross.

This is a variant of the hummingbird-approaching trains problem but now we have relativity mixed in.

Maybe this belongs in the homework section, so, if so, please move it there.

My answer is 2.72 years... is that correct? If we were Newtonian or Galilean it would be 2.5 years.

stevmg

This is my initial stab at the problem, but it seems to easy and maybe I am missing something.

In the frame of the stationary observer one mirror travels at 0.6c and travels 1.5 light years to arrive at the centre. The time is then 1.5/0.6= 2.5 years and of course the other mirror arrives at the centre at the same time. In 2.5 years the light signal travels 2.5 lightyears. That answers both your questions.

 

[stevmg]

This is my initial stab at the problem, but it seems to easy and maybe I am missing something.

In the frame of the stationary observer one mirror travels at 0.6c and travels 1.5 light years to arrive at the centre. The time is then 1.5/0.6= 2.5 years and of course the other mirror arrives at the centre at the same time. In 2.5 years the light signal travels 2.5 lightyears. That answers both your questions.

Thanks, kev. Doc Al and starthaus have hammered my poor brain with, essentially, what you have said, but please look at what I wrote about closing velocities in my post above
https://www.physicsforums.com/showpost.php?p=2824889&postcount=17
to Doc Al. Any comments would be appreciated!

BTW - what DO they call "relativistic mass increase" these days?

Thanks,


stevmg

 

[yuiop]

Fitzgerald's (as I told starthaus after calling him Fitzpatrick - "I never knew the man") length contraction is NOT part of Galilean or Newtonian mechanics or physics, thus, a new concept must be introduced to explain the null difference in order to justify the use of closing velocity (c + u) or (c - u). You have to shorten the distance to make the elapsed roundtrip times come out the same by, coincidently, the same gamma factor of later relativity fame.

Fitzgerald contraction and the length contraction of SR are mathmatically the same thing. They only differ philosophically in how "real" the length contraction is.

..and closing velocity just remains a mathematic theoretical entity which has no place here. Light always approaches at c and there is no closing velocity - at least not with light unless one calls c the universal closing velocity for light, no pluses or minuses.

By "your" reasoning, one can look at two objects approaching each other, say two trains steaming at each other one from the west and one from the east. The total closing velocity would be <2c but each individual's speed would be <c

We are not talking about the closing velocity of light. Closing velocity applies to two objects seen by one observer. In your OP example, the closing speed of the two mirrors is simply 0.6+0.6 = 1.2c (yes, even in relativity). At a closing velocity of 1.2c the mirrors collide in time 3/1.2 = 2.5 years. Relativistic velocity addition is not required for closing velocities. The velocity addition formula is used when we wish to find the velocity of a single object according to one observer, when we know the velocity of the object according to a different observer (and we know the velocity of the second observer relative to the first). The important thing is that no single object exceeds the speed of light relative to any given inertial observer. If we want to know the speed of m2 from the point of view of m1 then that is a different thing to closing velocity and that is when we would use the relativistic velocity addition formula. This of course means that the total travel time and distance of the light signal is different according to an observer comoving with one of the mirrors, but of course we expect that in relativity.

 

[yuiop]

BTW - what DO they call "relativistic mass increase" these days?

They don't. Its use has been banned! Sometimes it is called the quantity that is obtained when you divide the the momentum of an object by its velocity (p/v) or the enrgy increase divided by c^2, when its use is unavoidable in polite circles.

 

[Doc Al]

Fitzgerald's (as I told starthaus after calling him Fitzpatrick - "I never knew the man") length contraction is NOT part of Galilean or Newtonian mechanics or physics, thus, a new concept must be introduced to explain the null difference in order to justify the use of closing velocity (c + u) or (c - u). You have to shorten the distance to make the elapsed roundtrip times come out the same by, coincidently, the same gamma factor of later relativity fame.

Now, where does "closing velocity" (c + u) or (c-u) fit in? In other words, either the "u" doesn't exist (i.e., no ether) and closing velocity just remains a mathematic theoretical entity which has no place here. Light always approaches at c and there is no closing velocity - at least not with light unless one calls c the universal closing velocity for light, no pluses or minuses.

I think kev already made the point, but I don't see what this has to do with closing speed. Please read his comments.

By "your" reasoning, one can look at two objects approaching each other, say two trains steaming at each other one from the west and one from the east. The total closing velocity would be <2c but each individual's speed would be <c (because of so-called "relativistic mass increase" - an outmoded term these days.)

Why do you mean by 'your' reasoning? The closing speed is a simple matter of kinematics--nothing to do with "relativistic mass increase" or even relativity at all.

 

[stevmg]

I think kev already made the point, but I don't see what this has to do with closing speed. Please read his comments.

Why do you mean by 'your' reasoning? The closing speed is a simple matter of kinematics--nothing to do with "relativistic mass increase" or even relativity at all.

"Your reasoning" doesn't mean anything pejorative or "out of thin air." "Your reasoning" refers to accepted state-of-art current modern physics concepts or kinematics, as is stated.

Both your comments and kev's comments have cleared up my confusion immensely.

I APOLOGIZE for the use of "relativistic mass increase." I hope this doesn't mean that I get banished to some dark corner of the universe. It is mentioned by kev that sometimes it is used in "polite circles." I also run around in circles, but unfortunately, my "circle" doesn't have any people in it.

So, again, as far as Michelson-Morley III is concerned there is no closing velocity and c is the only velocity (in a vacuum) that light either approaches or eminates. There is no need to adjust with the 1/\gamma correction factor to make-up for lost time by shortening the distance because light going parallel to the Earth's tangental orbit and light going perpendicular to it are going at the same speed relative to the MM apparatus.

That apparatus bangs the light back and forth for 22 miles (35 km), doesn't it?

 

[starthaus]

So, again, as far as Michelson-Morley III is concerned there is no closing velocity and c is the only velocity (in a vacuum) that light either approaches or eminates. There is no need to adjust with the 1/\gamma correction factor to make-up for lost time by shortening the distance because light going parallel to the Earth's tangental orbit and light going perpendicular to it are going at the same speed relative to the MM apparatus.

In the frame of the lab, there is no "closing speed" but in any frame at rest wrt the lab, you need both closing speed and length contraction in order to explain the null result.

That apparatus bangs the light back and forth for 22 miles (35 km), doesn't it?

No, it doesn't. This would be a huge distance. The longest light trip in MMX is about 32m, three orders of magnitude smaller than what you are thinking. See here

 

[yuiop]

Fitzgerald's (as I told starthaus after calling him Fitzpatrick - "I never knew the man") length contraction is NOT part of Galilean or Newtonian mechanics or physics, thus, a new concept must be introduced to explain the null difference in order to justify the use of closing velocity (c + u) or (c - u). You have to shorten the distance to make the elapsed roundtrip times come out the same by, coincidently, the same gamma factor of later relativity fame.

Now, where does "closing velocity" (c + u) or (c-u) fit in?

By now you have probably figured this out, but I will show you another way to do it without explicitly using closing velocity that might help.

Imagine we have a rod AB of length L going to the right with velocity u.


|--ut--><-------ct------|
A----------L------------B

The mirror at A travels to the right with velocity u and the light signal from B travels to the left at velocity c. We can obtain the time t for the "collision" of the light particle with the mirror by dividing the distance L by the "closing velocity" (c+u) so that t = L/(c+u). Note that the closing velocity is not the speed of the light, but the speed with which the light and the mirror are approaching each other. However if you are dubious about the notion of closing speeds, then you can reason like this. The distance ut+ct must equal L as can be seen in the diagram, so we can say ut+ct=L => t(u+c)=L => t = L/(u+c).

After the light reflects off the mirror

|-------------------ct-------------------->|
A----------L------------B--------ut------->|

the light chases after the B mirror which has a head start of L and we can use similar reasoning to conclude that ct=L+ut => t(c-u)=L => t=L/(c-u). (or we could just say the distance is L and the closing velocity is (c-u) and t=L/(c-u).)

It turns out that when you work out the diagonal path of the signal traveling along the vertical arm, using good old Pythagorous, that the signal traveling the vertical path returns before the signal traveling the horizontal path. This contradicts what was actually measured in the MM experiment. Now if the speed of light is constant and independent of the velocity of the source and since the time is measured by a common clock at the fulcrum, the only variable left that can explain the null result is if L is shorter by a factor of gamma when it moving. The only other explanation is a ballistic theory of light in which the velocity of light depends on the velocity of the source (and this case the velocity of light really would be c+/-u) and then length contraction would not be required. Since length contraction was a pretty radical concept at the time it shows the high degree of confidence that the likes of Lorentz and Fitzgerald had in Maxwell's equations (and the constant speed of light) and ruling out the ballistic theory.

Hope that helps some.

Note that I have been sloppy with velocity signs and just use magnitudes, but it works out the same if you do it properly.

 

[starthaus]

if you are dubious about the notion of closing speeds, then you can reason like this. The distance ut+ct must equal L as can be seen in the diagram, so we can say ut+ct=L

Actually this the way closing speed is derived, so you are using closing speed.

 

[stevmg]

Actually this the way closing speed is derived, so you are using closing speed.

Thus, "relative speed" which is c and closing speed, which is c + u are two different entities. One is "real world" (relative speed) while "closing speed" is a mathematical concept.

I see how Einstein did use the closing speed to get the gamma factor when he related it to the "real world" in which c is invariant.

I suppose an example using your MMX.pdf as a source could be this;

Say I was sitting on the Sun head pointed North and looking at the Earth moving to the West at 30 km/sec. A light source to my left ("west" of the Earth) would approach me at c. The closing speed is c with respect to me and the relative speed with respect to me is likewise c.. The same light source, from my time frame, shining on the westward moving Earth has a closing speed of c + 30 km/sec. The relative speed to the Earth is still c.

The max closing speed of any two bodies is just shy of 2c with each body having an actual speed of slightly less than c. The relative velocity of one body to the other (either way) would be slightly less than c.

Is that the right idea?

 

[matheinste]

Thus, "relative speed" which is c and closing speed, which is c + u are two different entities. One is "real world" (relative speed) while "closing speed" is a mathematical concept.

Closing speed is a real rate of decrease/increase of the spatial distance between two inertially "moving" objects when measured by an inertial observer at rest in a frame relative to which the two velocities are given. It is a perfectly well defined concept which does not depend on any relativistic concepts.

Matheinste

 

[stevmg]

Closing speed is a real rate of decrease/increase of the spatial distance between two inertially "moving" objects when measured by an inertial observer at rest in a frame relative to which the two velocities are given. It is a perfectly well defined concept which does not depend on any relativistic concepts.

Matheinste

That is what I did say. Look above. Now, with regards to light, look at the example of the man standing at the Sun and seeing the Earth move to the left at 30 km/sec. Say a beam of light from the man's left is coming to the man. It will be coming at c. That would be the closing speed and the relative speed with respect to the man at the Sun.

Now, as he looks out to see the Earth moving across him at 30 km/sec to the left, does the beam of light from the Earth's left (the same light source that hits the man standing at the Sun described above) have a closing speed of c + 30 with the Earth?. The relative speed of the light beam to the Earth would be still be c.

Is this right?

 

[stevmg]

kev did answer my question above (post 29) here:
https://www.physicsforums.com/showpost.php?p=2825141&postcount=25
light speed can be part of a closing speed (c + u) or (c - u)

I just didn't see it before...

It seems that this explains the length contraction part of relativity. You need
\gamma = SQRT[1 - u²/c²] to make the elapsed times of the roundtrips parallel and perpendicular to the motion of the Earth to come out the same otherwise the roundtrip times over the 32 meters wouldn't come out the same.

Actually, if I remember the derivation of this correctly, the equations never establish any relationship between the relative elapsed times of the round trips other than that they are equal. By equation (1.2) in starthaus's blog on the MMX,
t = \gammal₁/c which means that when v = 0, \gamma = 1, so t₀ likewise "expands" to t₁ = t₀\gamma as v > 0. Maybe I have the t's backwards but you get the idea.

starthaus - the 22 mile distance I was referring to was the 22 miles between Mt Wilson and Lookout Mountain used in the 1920s by Michelson to determine the speed of light (actually through air.) He did it with a rotating octagon mirror sending reflected light to a mirror which reflected it back. I don't know which mountain had which mirror but that's the gist of it. I am sure you already knew of that experiment. I didn't know Michelson either.

 

[starthaus]

Now, for the time dilation portion...

starthaus - the 22 mile distance I was referring to was the 22 miles between Mt Wilson and Lookout Mountain used in the 1920s by Michelson to determine the speed of light (actually through air.) He did it with a rotating octagon mirror sending reflected light to a mirror which reflected it back.

I am pretty familiar with the experiment. It has nothing to do with time dilation and/or with MMX.

 

[stevmg]

I am pretty familiar with the experiment. It has nothing to do with time dilation and/or with MMX.

Absolutely true... I was just correcting myself about the incorrect distances I stated earlier for the MMX apparatus mirrors, that's all.

When I first heard of that experiment in high school physics I thought that was really neat and what an engineering feet - to get those mirrors within 1/4" of true considering they did it with triangulation and the long arm of the triangle was 22 miles. I can't even get a 6-inch ruler to come out that well when I'm measuring it directly.

 

[yuiop]

That is what I did say. Look above. Now, with regards to light, look at the example of the man standing at the Sun and seeing the Earth move to the left at 30 km/sec. Say a beam of light from the man's left is coming to the man. It will be coming at c. That would be the closing speed and the relative speed with respect to the man at the Sun.

Now, as he looks out to see the Earth moving across him at 30 km/sec to the left, does the beam of light from the Earth's left (the same light source that hits the man standing at the Sun described above) have a closing speed of c + 30 with the Earth?. The relative speed of the light beam to the Earth would be still be c.

Is this right?

That seems about right.

Lets take a slightly more complicated situation to make the differences between closing velocities and relative velocities a bit clearer. Let's call the velocity of the Earth in the mans's frame (-0.5c) and just for today the Earth is going to move in a straight line I have given the velocity a negative sign to indicate it is going to the left in the man's frame. To the left of the Earth is an attack spaceship that is a distance of one light year away from the Earth in the man's frame. The spaceship lauches a laser pulse towards the Earth and the pulse is moving at c of course. The spaceship also launches a missile moving at 0.8c in the man's frame.

The equation for closing speed is s = | v1-v2 | where | | means we are only concerned with the magnitude of the velocity. In the man's frame the closing speed of the light with the Earth is | c-(-0.5c) | = 1.5c in the man's frame. The closing speed of the missile with the Earth is | 0.8c-(-0.5c) | =1.3c in the man's frame. The closing speed of the man with the Earth is | 0c-(-0.5c) | = 0.5c in the man's frame.

The equation for relative velocity is v = (w+u)/(1+w*u/c^2) where u is the velocity of an object as measured in a frame which has velocity w relative to the frame that measures the velocity to be v. Now we can look at everything from the Earth's point of view. The velocity of the man in the Earth frame is (-0.5c+0)/(1+0c*-0.5c/c^2) = -0.5c. The velocity of the missile in the Earth frame is (-0.5c+(-0.8c))/(1+(-0.5c*-0.8c)/c^2) = -0.9286c. The velocity of laser pulse in the Earth frame is (-0.5c+(-1c))/(1+(-0.5c*-1c)/c^2) = -c.

The closing speeds of the three "objects" with the Earth, in the Earth frame are | 0c-(-0.5c) | = 0.5c for the man, | 0c-(-0.9286c) | = 0.9286c for the missile and | 0c-(-1c) | = c for the laser pulse respectively.

 

[stevmg]

That seems about right.

Lets take a slightly more complicated situation to make the differences between closing velocities and relative velocities a bit clearer. Let's call the velocity of the Earth in the mans's frame (-0.5c) and just for today the Earth is going to move in a straight line I have given the velocity a negative sign to indicate it is going to the left in the man's frame. To the left of the Earth is an attack spaceship that is a distance of one light year away from the Earth in the man's frame. The spaceship lauches a laser pulse towards the Earth and the pulse is moving at c of course. The spaceship also launches a missile moving at 0.8c in the man's frame.

The equation for closing speed is s = | v1-v2 | where | | means we are only concerned with the magnitude of the velocity. In the man's frame the closing speed of the light with the Earth is | c-(-0.5c) | = 1.5c in the man's frame. The closing speed of the missile with the Earth is | 0.8c-(-0.5c) | =1.3c in the man's frame. The closing speed of the man with the Earth is | 0c-(-0.5c) | = 0.5c in the man's frame.

The equation for relative velocity is v = (w+u)/(1+w*u/c^2) where u is the velocity of an object as measured in a frame which has velocity w relative to the frame that measures the velocity to be v. Now we can look at everything from the Earth's point of view. The velocity of the man in the Earth frame is (-0.5c+0)/(1+0c*-0.5c/c^2) = -0.5c. The velocity of the missile in the Earth frame is (-0.5c+(-0.8c))/(1+(-0.5c*-0.8c)/c^2) = -0.9286c. The velocity of laser pulse in the Earth frame is (-0.5c+(-1c))/(1+(-0.5c*-1c)/c^2) = -c.

The closing speeds of the three "objects" with the Earth, in the Earth frame are | 0c-(-0.5c) | = 0.5c for the man, | 0c-(-0.9286c) | = 0.9286c for the missile and | 0c-(-1c) | = c for the laser pulse respectively.

kev -

You are a gentleman and a scholar. When I ask a question, as naive as it may be, you give me an answer at my naive level.

I am not being facetious here. I really appreciate your effort.

Is my later comment post 30 about the necessity of length contraction and concomitant time dilation also true (see above?)

stevmg

the .pdf file attached is from starthaus

 

[yuiop]

Actually, if I remember the derivation of this correctly, the equations never establish any relationship between the relative elapsed times of the round trips other than that they are equal.

The MM experiment can be completely explained by length contraction alone. Once you know the length contraction factor you soon work out the time dilation factor by calculating light travel times parallel to rod with relative motion. If you workout the light travel time for the zig zag taken by the light moving along the transverse arm you can obtain the time dilation gamma factor without involving length contraction. This is essentially Einstein's light clock thought experiment. In fact Einstein never mentions the MM experiment in his 1905 paper.

I am not being facetious here. I really appreciate your effort.

Is my later comment post 30 about the necessity of length contraction and concomitant time dilation also true (see above?)

Your welcome. My posts are as much for my benefit as they are for you or anyone else. I had to think real hard about the signs in the velocity addition equations (and I am still not 100% sure I have it right ) Why is that getting positive and negative signs right seems to be one the hardest things in relativity??

 

[Austin0]

The MM experiment can be completely explained by length contraction alone. Once you know the length contraction factor you soon work out the time dilation factor by calculating light travel times parallel to rod with relative motion. If you workout the light travel time for the zig zag taken by the light moving along the transverse arm you can obtain the time dilation gamma factor without involving length contraction. This is essentially Einstein's light clock thought experiment. In fact Einstein never mentions the MM experiment in his 1905 paper.

Your welcome. My posts are as much for my benefit as they are for you or anyone else. I had to think real hard about the signs in the velocity addition equations (and I am still not 100% sure I have it right ) Why is that getting positive and negative signs right seems to be one the hardest things in relativity??

I am not sure if you are kidding or not here but...Shouldn't it more properly be called the Subtractions of Velocities formula ? The formula itself is subtractive and it is only when the motions are in opposition +x <--->-x that it becomes additive as the signs change.? v-(-v) and v*(-v)

 

[yuiop]

I am not sure if you are kidding or not here but...Shouldn't it more properly be called the Subtractions of Velocities formula ? The formula itself is subtractive and it is only when the motions are in opposition +x <--->-x that it becomes additive as the signs change.?

Seems you are equally as confused as me so pull up a chair and have a beer with me while we work this out.

The relativistic velocity addition formula is normally given as something like:

v = \frac{v&#039;+u}{1+v&#039;u}

(in units of c=1)

Let us say we have a mothership going to the right with velocity u=0.7 and it fires a missile with velocity v' = 0.9 also to the right (by their measurements) then the total velocity in our reference frame is (0.7+0.9)/(1-0.7*0.9) = 0.9816c which is conveniently less than c in our frame. Put like that, it seems to be reasonable to call it an addition formula althouth it is a weird mix of adding the velocity of the mothership (as measured in our frame) to the velocity of the missile (as measured in their frame).

Now let's say we stop being being lazy and actually measure the velocity (v) of the missile in our frame (finding it is 0.9186c) and ask what the velocity (v') of the missile is in their frame, (which we can not directly measure) then we use the inverse formula:

v&#039; = \frac{v-u}{1-vu}

and obtain (0.7-0.9186)/(1-0.7*0.9186) = 0.9c

which is the analogue of (v-u) in Newtonian/Galilean physics and I supose you could call it a subtraction formula when used like this.

Now the bit that was giving me giving me a headache was if the velocity of the spaceship is +0.7c according to me, what is my velocity according to the spaceship (plus or minus 0.7c?). Turns out it is -0.7c as can be seen by the following example. Let us say I fire a bullet to the left at -0.7c the velocity of the bullet according to the spaceship using the addition formula is (-0.7+(-0.7))/(1+(-0.7)*(-0.7)) = -0.9396c which seems reasonable, while if I had assumed my velocity was +0.7c then I would of got the velocity of the bullet in the spaceship frame to be (-0.7+0.7)/(1+(-0.7)*(0.7)) = 0 which does not make sense. With hindsight it seems obvious now... maybe too much beer...

 

[stevmg]

Just use common sense and keep the cart before the horse... It'll come out. If you get a weird answer, try it the other way and it may make sense. Sort of like being married - same logic, trust me.

kev -

You are quite correct in that the MMX-3 can be explained on length contraction OR time dilation. Here's where I get messed up...isn't one of them logical correlative of the other?

But in the MMX-3 experiment if there were a true "closing speed" (c + v) or (c - v) and had no relativity other than length contraction (1898 Lorentz-Fitzgerald experiment), if you had length contraction this would "cover it" so that trivial (zero) difference in parallel and perpendicular light roundtrip times would be explained.

If you used relative speed, there is no addition to c or subtraction from c and naturally the parallel and perpendicular roundtrip times are the same. Also, by using relativity one gets a length contraction AND a time dilation. They go together, right?

Am I adding more to everyone's confusion or just my own. But that's OK, I'm used to it because of marriage and knowng the rules of "logic."

 

[Austin0]

Seems you are equally as confused as me so pull up a chair and have a beer with me while we work this out...

Hi kev ...well if I wasn't confused before, I certainly am now after listening to you , so I will definitely go for that beer

The relativistic velocity addition formula is normally given as something like:

v = \frac{v&#039;+u}{1+v&#039;u}

(in units of c=1) ...

This a specific version for projectiles.
I think you will find that this is not the essential formula which is as below. I offered the Subtraction idea as a simple mneumonic device which I found useful to keep it straight.

Let us say we have a mothership going to the right with velocity u=0.7 and it fires a missile with velocity v' = 0.9 also to the right (by their measurements) then the total velocity in our reference frame is (0.7+0.9)/(1-0.7*0.9) = 0.9816c which is conveniently less than c in our frame. Put like that, it seems to be reasonable to call it an addition formula althouth it is a weird mix of adding the velocity of the mothership (as measured in our frame) to the velocity of the missile (as measured in their frame). ...

In this instance it is actually ...0.9 -(-0.7)/ 1-(0.9*(-0.7)) It is not a mix of adding the mothership velocity as measured in our frame. WHich is +0.7 It is subtracting the velocity of our frame as measured in the mothership.WHich is -0.7
The weird mix would not just be weird but incorrect in principle , as the formula relates to velocities measured in a single frame. It just seems like that is what's going on. because the projectile equation already reversed the signs for conveinience.

Now let's say we stop being being lazy and actually measure the velocity (v) of the missile in our frame (finding it is 0.9186c) and ask what the velocity (v') of the missile is in their frame, (which we can not directly measure) then we use the inverse formula:

v&#039; = \frac{v-u}{1-vu}

and obtain (0.7-0.9186)/(1-0.7*0.9186) = 0.9c

which is the analogue of (v-u) in Newtonian/Galilean physics and I supose you could call it a subtraction formula when used like this...

Yep

Now the bit that was giving me giving me a headache was if the velocity of the spaceship is +0.7c according to me, what is my velocity according to the spaceship (plus or minus 0.7c?). Turns out it is -0.7c as can be seen by the following example. Let us say I fire a bullet to the left at -0.7c the velocity of the bullet according to the spaceship using the addition formula is (-0.7+(-0.7))/(1+(-0.7)*(-0.7)) = -0.9396c which seems reasonable, while if I had assumed my velocity was +0.7c then I would of got the velocity of the bullet in the spaceship frame to be (-0.7+0.7)/(1+(-0.7)*(0.7)) = 0 which does not make sense. With hindsight it seems obvious now... maybe too much beer...

You don't need to give any thought to this question or formulas. It is just a convention for convenience. Velocity of the primed frame, in the unprimed frame, is handled as positive or negative and for ease it is considered that the primed frame has the x-axis in the same orientation, so the relative velocity must be the opposite sign.
Another brew?

 

[Doc Al]

Why is that getting positive and negative signs right seems to be one the hardest things in relativity??

Keeping signs straight is much easier--at least for me--if I write the velocity addition formula like this:

V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}

Where V_{a/c} is the velocity of a relative to c, and so on. V_{a/c} = - V_{c/a}, of course. Here b is the 'common' reference frame.

 

[stevmg]

Keeping signs straight is much easier--at least for me--if I write the velocity addition formula like this:

V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}

Where V_{a/c} is the velocity of a relative to c, and so on. V_{a/c} = - V_{c/a}, of course. Here b is the 'common' reference frame.

I like that one as it makes intuitive sense (the a's -> b's -> c's)

Also, a prior attachment MMX and length contraction.pdf by starthaus literallly explains the Einsteinian Relativity Equations in reverse - i.e., if you assume the constancy of c in relativistic terms, the length contraction and correlative time dilation are a result of the null difference in elapsed time because it forces the parallel distance to be the perpendicular distance divided by gamma. In relativity, you cannot add to c or subtract from it so length contraction is the consequence. If you use closure speeds, you get the null difference ONLY if you apply length contraction to it which is a hybrid universe. But that's how Einstein derived the Lorentz equations - by using closure speeds (c + v and c - v) and then assuming the difference in relativistic speeds of light was zero, a "leap of faith" (non-religious meaning here) which turned out to be true.

stevmg

 

[Austin0]

Keeping signs straight is much easier--at least for me--if I write the velocity addition formula like this:

V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}

Where V_{a/c} is the velocity of a relative to c, and so on. V_{a/c} = - V_{c/a}, of course. Here b is the 'common' reference frame.

Hi Doc AL how do you use this additive formulation with

{V_{a/b}= 0.4 and V_{b/c} = 0.5} both positive??

 

[Doc Al]

Hi Doc AL how do you use this additive formulation with

{V_{a/b}= 0.4 and V_{b/c} = 0.5} both positive??

Just plug in the values and turn the crank!

<br /> V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2} = \frac{0.4c + 0.5c}{1 + (0.4c\times 0.5c)/c^2} = (.9/1.2) c = .75c<br />

This might represent, for example, the relative velocity of two rockets where rocket A is traveling east at .4c with respect to the Earth (frame B) and rocket C is traveling west at .5c with respect to the earth. Note: If I let east be positive and west negative, then V_c/b = -.5c and V_b/c = +.5c.

 

[Austin0]

Hi Doc AL how do you use this additive formulation with

{V_{a/b}= 0.4 and V_{b/c} = 0.5} both positive??

Just plug in the values and turn the crank!

<br /> V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2} = \frac{0.4c + 0.5c}{1 + (0.4c\times 0.5c)/c^2} = (.9/1.2) c = .75c<br />

This might represent, for example, the relative velocity of two rockets where rocket A is traveling east at .4c with respect to the Earth (frame B) and rocket C is traveling west at .5c with respect to the earth. Note: If I let east be positive and west negative, then V_c/b = -.5c and V_b/c = +.5c.

Well... Uh, I would say that .75c is not actually correct for both heading in the same direction east or west .
It would be correct for your example of east and west but then they would not both be positive as I specified , yes??
What happens if you plug in equal values like +0.5 and +0.5 into the additive form you are using?? It looks to me like you get 0.8 when clearly the answer is 0c
That is why I suggested it might be better to consider the basic formula as subtractive.
Then you just plug in the actual values +,+... -,- or -,+ and it always comes out.
Maybe?...:-)

 

[Doc Al]

Well... Uh, I would say that .75c is not actually correct for both heading in the same direction east or west .
It would be correct for your example of east and west but then they would not both be positive as I specified , yes??

You mixed up your signs. Pay attention to the meaning of V_a/b and V_b/c, and the distinction between V_b/c and V_c/b.

If both rocket A and rocket C are moving east, then V_a/b = +0.4c and V_c/b = +0.5c. That makes V_b/c = -0.5c. Which gives you V_a/c = -0.125c.

What happens if you plug in equal values like +0.5 and +0.5 into the additive form you are using?? It looks to me like you get 0.8 when clearly the answer is 0c

My formula gives 0c, of course, when both rockets are traveling at the same speed in the same direction with respect to the third frame (frame B).

(I answered the question you actually asked, not the one you meant to ask. )

That is why I suggested it might be better to consider the basic formula as subtractive.
Then you just plug in the actual values +,+... -,- or -,+ and it always comes out.
Maybe?...:-)

My formula gives the correct answer all the time with no guesswork.

 

[Austin0]

You mixed up your signs. Pay attention to the meaning of V_a/b and V_b/c, and the distinction between V_b/c and V_c/b.

If both rocket A and rocket C are moving east, then V_a/b = +0.4c and V_c/b = +0.5c. That makes V_b/c = -0.5c. Which gives you V_a/c = -0.125c.

My formula gives 0c, of course, when both rockets are traveling at the same speed in the same direction with respect to the third frame (frame B).

(I answered the question you actually asked, not the one you meant to ask. )


My formula gives the correct answer all the time with no guesswork.

Yeah well My formula gives the correct answer all the time with no guesswork too.
But you are right I didn't get your notation and am now curious as to why you would have the velocity of the rest frame as observed in one of the relative frames in question?
Isn't the formula based on velocities relative to a single frame??

Thanks

 

[Doc Al]

But you are right I didn't get your notation and am now curious as to why you would have the velocity of the rest frame as observed in one of the relative frames in question?

No frame is really 'at rest'. There are three frames involved in that formula; you can choose anyone to be the 'rest' frame.

Isn't the formula based on velocities relative to a single frame??

Sure. It's useful when you have the velocities of A and C relative to a common frame B.

 

[Austin0]

No frame is really 'at rest'. There are three frames involved in that formula; you can choose anyone to be the 'rest' frame.

Sure. It's useful when you have the velocities of A and C relative to a common frame B.

It is understood that no frame is really at rest. But in this case I question whether you can choose anyone of the three to analyze from . In A you have only a velocity for B so you can't use the formula from there can you?
Likewise from C??

But hey anything that works , right? :-)

 

[Doc Al]

It is understood that no frame is really at rest. But in this case I question whether you can choose anyone of the three to analyze from . In A you have only a velocity for B so you can't use the formula from there can you?

Sorry, but I really don't know what you're talking about. The formula given is just the relativistic extension of the Galilean velocity addition formula:

<br /> V_{a/c} = V_{a/b} + V_{b/c}<br />

Note that 'b' is the common frame linking a and c.

 

[yuiop]

Well... Uh, I would say that .75c is not actually correct for both heading in the same direction east or west .
It would be correct for your example of east and west but then they would not both be positive as I specified , yes??

They both would be positive in this case.
Rocket A has a velocity of +0.4c (going East) relative to the Earth (frame b) so Va/b= +0.4c.
Rocket C has a velocity of -0.5c (going West) relative to the Earth (frame b) so Vc/b= -0.5c.

The velocity of Earth from the point of view of Rocket C is therefore Vb/c = +0.5c.
The velocity of Rocket A from the point of view of Rocket C is then given by Va/c = (Va/b+Vb/c)/(1+Va/b*Vb/c) = (0.4+0.5)/(1+0.4*0.5) = 0.75c as claimed by DrGreg.

Now if we take the next example you want analysed, when both Rocket A and C are going East relative to the Earth with a velocity of 0.5c, then Va/b = +0.5c and Vc/b = 0.5c (so Vb/c = -0.5) and the velocity of A from the point of view of C is then Va/c = (Va/b+Vb/c)/(1+Va/b*Vb/c) = (0.5+(-0.5))/(1+0.4*(-0.5)) = 0.0c as you would expect.