Graduate Why is the speed of light invariant despite relative motion?

[stevmg]

i understand the michelson–morley experiment and its result; but what i don't know yet is the reason.
Example:
A torch in free space is moving at a velocity [v] w.r.t me. Considering the material nature of light, shouldn't the speed of photons emitted from the torch be [v+c] w.r.t me?
According to the experiment, it's not so.
I know how relativistic velocities are formulated (lorentz transformation, einstein's addition, etc.). But all these calculations are based on the accepted norm that light speed in invariant. Why? What's the scientific explanation of this (if any)? Is it still a mystery?

I've seen other posts regarding the speed of light, but couldn't go through them all. So, apologies if this topic already exists.

To elucidate on this point in particular, the invariance of the speed of light in his situation is taken up by time dilation and length contraction. They do work opposite to each other but one offsets the other. What should be an increase in closure speed as you posit is enhanced by the length contraction yet offset by the time dilation so that, relative to you, things "slow down" back to light speed.

stevmg

Using drudkh's example cited above and using Einstein's velocity addition formula, closure speed between a particle moving to the right at v and light coming from the right and moving left at c would still be c.

Why? Let's use v as the velocity of the particle expressed as a fraction of c and therefore c = 1

closure velocity = (c + v)/(1 + vc/c²) = (1 + v)/1 + v) = 1

If, instead of a particle moving to right at v, we substitute a photon or a light wave moving at c or 1 and still light comes to the left at c or 1, the formula becomes:

closure velocity = (1 + 1)/(1 + 1) = 1

Ain't life grand? The max speed attainable is 1 or c.

Can we ever have a closure velocity > c? Not according to the above.

Someone show me and drudkh wrong!

 

[Fredrik]

The velocity addition formula tells you the velocity of that light in the particle's rest frame, but you asked about the closing speed. It's defined as the rate of decrease of the coordinate distance between the two objects, i.e. as minus the time derivative of the coordinate distance. Your two objects are located at a+vt and b-ct for some a and b with b>a, so the coordinate distance between them at time t is (b-ct)-(a+vt)=b-a-ct-vt, and the closing speed is therefore

-d/dt(b-a-ct-vt)=c+v.

 

[matheinste]

Using drudkh's example cited above and using Einstein's velocity addition formula, closure speed between a particle moving to the right at v and light coming from the right and moving left at c would still be c.

Why? Let's use v as the velocity of the particle expressed as a fraction of c and therefore c = 1

closure velocity = (c + v)/(1 + vc/c²) = (1 + v)/1 + v) = 1

If, instead of a particle moving to right at v, we substitute a photon or a light wave moving at c or 1 and still light comes to the left at c or 1, the formula becomes:

closure velocity = (1 + 1)/(1 + 1) = 1

Ain't life grand? The max speed attainable is 1 or c.

Can we ever have a closure velocity > c? Not according to the above.

Someone show me and drudkh wrong!

Closing velocity or mutual velocity describes the rate of change in a single frame of a vector connecting two objects. It is just the normal Newtonian addition of velocities and can be as high as 2c. Of course it is only the distance between the objects that is changing at the closing velocity, nothing material is moving at this velocity.

The relative velocity is the velocity which one of the objects ascribes to the other and is calculated using the relativitydition formula for velocities and cannot exceed c.

Matheinste.

 

[stevmg]

The closing speed is defined as minus the rate of change of the coordinate distance. Your two objects are located at a+vt and b-ct for some a and b>a, so the coordinate distance between them at time t is (b-ct)-(a+vt)=b-a-ct-vt, and the closing speed is therefore

-d/dt(b-a-ct-vt)=c+v.

...0........(a + vt)...>......<...(b - c^-t)

Help me out here... If the origin of one FOR was at A (a + vt and moving right at a steady speed) and the origin for a second frame of reference was at B (b - c^-t) where "c^- means "just below the speed of light but greater than v," then doesn't the closure speed mean that closing speed is c^- + v which is > c.

How is that possible? How do you "leap frog" > c? Relative to the FOR at A any particle moving left at c^- would be booking > c and I thought that was impossible in any FOR. Even looking at it from B being steady and A coming from the left, a particle would be moving to the right at c^- + v which would be > c (both cases if v were large enough but still , c.)

This is a major new concept for me to understand. I understand it from Galilean or Newtonian point of view - that's easy.

 

[stevmg]

Closing velocity or mutual velocity describes the rate of change in a single frame of a vector connecting two objects. It is just the normal Newtonian addition of velocities and can be as high as 2c. Of course it is only the distance between the objects that is changing at the closing velocity, nothing material is moving at this velocity.

The relative velocity is the velocity which one of the objects ascribes to the other and is calculated using the relativity addition formula for velocities and cannot exceed c.

Matheinste.

Matheinste or Fredrik - if you just dropped the velocities of both objects to c^- or speeds just below light speed, say 29,900,000 m/sec, which allows for the movement of "real" particles what is the difference here? How, in the real world which SR does apply, would one ever achieve a closure velocity or relative velocity of > c. I am conflating closure velocity with relative velocity and therein may lie my misconception, but then when do we use closure velocity versus relative velocity?

stevmg

 

[Fredrik]

If the origin of one FOR was at A (a + vt and moving right at a steady speed) and the origin for a second frame of reference was at B (b - c^-t) where "c^- means "just below the speed of light but greater than v," then doesn't the closure speed mean that closing speed is c^- + v which is > c.

Yes. (But you don't need to mention several frames of reference. You can just talk about the positions of two "objects", and you don't have to let that speed be slightly less than c. If one of the "objects" is light, then the speed is exactly c).

How is that possible?

Why wouldn't it be?

Relative to the FOR at A any particle moving left at c^- would be booking > c

No, because if you want to calculate the speed in frame A of that particle, you have to use the velocity addition formula to get (v+c)/(1+vc/c²). If you want to calculate the closing speed, you just add their speeds (assuming that they're moving towards each other) to get v+c.

This is a major new concept for me to understand. I understand it from Galilean or Newtonian point of view - that's easy.

Then you understand it in Minkowski spacetime too. There is no difference.

 

[Fredrik]

How, in the real world which SR does apply, would one ever achieve a closure velocity or relative velocity of > c.

Just have Alice shoot a beam of particles with speed 0.6c towards Bob, and at the same time have Bob shoot a beam of partices with speed 0.6c towards Alice. The closing speed is 1.2c. The velocity of one of the beams in the rest frame of the other is (0.6c+0.6c)/(1+0.6*0.6)=1.2c/1.36<c.

I am conflating closure velocity with relative velocity and therein may lie my misconception,

I agree.

but then when do we use closure velocity versus relative velocity?

I don't know any situation where closing speed is an interesting concept.

 

[stevmg]

Yes. (But you don't need to mention several frames of reference. You can just talk about the positions of two "objects", and you don't have to let that speed be slightly less than c. If one of the "objects" is light, then the speed is exactly c).


Why wouldn't it be?


No, because if you want to calculate the speed in frame A of that particle, you have to use the velocity addition formula to get (v+c)/(1+vc/c²). If you want to calculate the closing speed, you just add their speeds (assuming that they're moving towards each other) to get v+c.

Then you understand it in Minkowski spacetime too. There is no difference.

No, because if you want to calculate the speed in frame A of that particle, you have to use the velocity addition formula to get (v+c)/(1+vc/c²). If you want to calculate the closing speed, you just add their speeds (assuming that they're moving towards each other) to get v+c.

That is where you lose me. I understand your words but not the concept.

 

[starthaus]

I don't know any situation where closing speed is an interesting concept.

It is used routinely in explaining MMX from a frame of reference outside the lab.

 

[Fredrik]

That is where you lose me. I understand your words but not the concept.

I think you're just assuming that there's something tricky here that needs to be understood. There isn't. If you're driving 50 mph relative to the road and the cars going in the opposite direction are also going at 50 mph, the closing speed is 100 mph because it's defined to be the number of miles that the distance to the other car is decreasing each hour. It doesn't involve a second coordinate system in any way. The only coordinate system we're talking about is the one in which the road is stationary. So you have no use for the Lorentz transformation or the velocity addition formula, which both involve two inertial frames.

 

[stevmg]

I don't know any situation where closing speed is an interesting concept.

Fredrik, Here is where I got "messed up." This was a beautiful explanation of the twins paradox solution (just using SR) by JesseM which he wrote to my questions months ago. I have highlighted in magenta where he used the term "closing speed" but on review now, I see he put it in quotes and he did initially use the term "relatavistic velocity addition formula."

No, you can analyze the problem from any inertial frame and all will have the same answer about the age of the inertial twin and the age of the non-inertial twin when they reunite. Let's call the inertial (Earth-bound) twin "Terence" and the traveling twin "Stella", following the Twin Paradox FAQ. First let's look at the numbers in Terence's rest frame. Suppose that in this frame, Stella travels away from Terence inertially at 0.6c for 10 years, at which point she is at a distance of 0.6*10 = 6 light-years from Earth in this frame, then she turns around (i.e. she accelerates, a non-inertial motion which will cause her to experience G-forces that show objectively that she wasn't moving inertially) and heads back towards Terence at 0.6c, finally reuniting with Terence after 20 years have passed since her departure in this frame. Since Terence is at rest in this frame, he has aged 20 years. But since Stella was moving at 0.6c in this frame, the time dilation formula tells us her aging was slowed down by a factor of \sqrt{1 - 0.6^2} = 0.8, so she only aged 0.8*10 = 8 years during the outbound leg of her trip, and another 0.8*10 = during the inbound leg, so she has only aged 16 years between leaving Earth and returning.

Now let's analyze the same situation in a different inertial frame--namely, the frame where Stella was at rest during the outbound leg of her trip (she can't also be at rest during the inbound leg in this frame, since this is an inertial frame while Stella accelerated between the two legs of the trip). In this frame, Terence on Earth is initially moving away from Stella at 0.6c while she remains at rest. In Terence's frame, remember that Stella accelerated when she was 6 light-years away from Earth, so we can imagine she turns around when she reaches the far end of a measuring-rod at rest in Terence's frame and 6 light-years long in that frame, with Terence sitting on the near end; in the frame we're dealing with now, the measuring-rod will therefore be moving along with Terence at 0.6c, so it'll be shrunk via length contraction to a length of only 0.8*6 = 4.8 light-years. So, Stella accelerates when the distance between her and Terence is 4.8 light-years in this frame, and since Terence as moving away from her at 0.6c in this frame, they will be 4.8 light-years apart after 4.8/0.6 = 8 years have passed. During these 8 years, it is Terence's aging that is slowed down by a factor of 0.8, so while Stella ages 8 years during this leg, Terence only ages 0.8*8 = 6.4 years. Then Stella accelerates to catch up with Terence, while Terence continues to move inertially at 0.6c. Using the relativistic velocity addition formula, if Stella was moving at 0.6c in Terence's frame and Terence is moving at 0.6c in the same direction in this frame, then in this frame Stella must be moving at (0.6c + 0.6c)/(1 + 0.6*0.6) = 0.88235c during the inbound leg. And since Terence is still moving at 0.6c in the same direction, the distance between Stella and Terence will be closing at a "closing speed" of 0.88235c - 0.6c = 0.28235c. Since the distance was initially 4.8 light years at the moment Stella accelerated, in this frame it will take 4.8/0.28235 = 17 years for Stella to catch up with Terence on Earth. During this time Terence has aged another 0.8*17 = 13.6 years, so if you add that to the 6.4 years he had aged during the outbound leg, this frame predicts he has aged 20 years between Stella leaving and Stella returning, same as in Terence's frame. And since Stella is traveling at 0.88235c her aging is slowed by a factor of \sqrt{1 - 0.88235^2} = 0.4706, so during those 17 years in this frame she only ages 0.4706*17 = 8 years during the inbound leg. If you add that to the 8 years she aged during the outbound leg, you find that this frame predicts she has aged 16 years between departing and returning, which again is the same as what was predicted in Terence's frame.

As illustrated in his solution, JesseM really used the "relatavistic velocity addition calculation" and NOT closing speed. Now that you have stated there is sort of no use for closure speed, things now fit in place. With reference to starthaus's "MMX"

It is used routinely in explaining MMX from a frame of reference outside the lab.

I have no understanding. As an old Air Force officer he may be referring to a missile closing in on a target which is clearly "outside the lab." I also see MMX as some technology with new chips.

A little help in that area by starthaus or you would be helpful.

 

[stevmg]

MMX - some new Intel microprocessor technology which, I presume, increases speed of "calculations." I guess there are two stacks reaching out towards each other, or something like it and there is a "closure speed," if you will.

If he meant the missile, then that would be an MX missile, not MMX.

If I know starthaus, it is neither, and it will be months before you guys let me in on your secret.

I am clueless and pulseless.

 

[Fredrik]

MMX - some new Intel microprocessor technology which, I presume, increases speed of "calculations." I guess there are two stacks reaching out towards each other, or something like it and there is a "closure speed," if you will.

If he meant the missile, then that would be an MX missile, not MMX.

If I know starthaus, it is neither, and it will be months before you guys let me in on your secret.

If it makes you feel better, I had to look it up too. Wikipedia has a disambiguation page that says:

• MMX (instruction set), a single-instruction, multiple-data instruction set designed by Intel
• MMX Miner, a Brazilian mining company
• Michelson-Morley Experiment, the 1887 experiment attempting to find evidence of the luminiferous aether
• Mega Man X, a series of video games produced by Capcom and the eponymous main character
• Malmö Airport, an airport in Sweden (IATA airport code)
• MMX, an upcoming album by rapper Xzibit
• 2010, in Roman numerals

I'm assuming it's the third one. I don't think I've heard it called "MMX" before.

 

[jtbell]

MMX as an acronym for "Michelson Morley experiment" is sometimes seen on Internet forums like this one that discuss relativity intensively, and maybe on a few Web sites, but nowhere else as far as I know.

 

[stevmg]

If it makes you feel better, I had to look it up too. Wikipedia has a disambiguation page that says:

• MMX (instruction set), a single-instruction, multiple-data instruction set designed by Intel
• MMX Miner, a Brazilian mining company
• Michelson-Morley Experiment, the 1887 experiment attempting to find evidence of the luminiferous aether
• Mega Man X, a series of video games produced by Capcom and the eponymous main character
• Malmö Airport, an airport in Sweden (IATA airport code)
• MMX, an upcoming album by rapper Xzibit
• 2010, in Roman numerals

I'm assuming it's the third one. I don't think I've heard it called "MMX" before.

MMX as an acronym for "Michelson Morley experiment" is sometimes seen on Internet forums like this one that discuss relativity intensively, and maybe on a few Web sites, but nowhere else as far as I know.

Thanks, gentlemen,

That's all I need now - to become another Sherlock Holmes!

stevmg

 

[starthaus]

[*] Michelson-Morley Experiment, the 1887 experiment attempting to find evidence of the luminiferous aether
I'm assuming it's the third one. I don't think I've heard it called "MMX" before.

yes, Michelson-Morley Experiment

BTW, closing speed figures prominently in the Einstein gedank experiment with the train and the track in explaining relativity of simultaneity.

 

[stevmg]

Thus, "closing speed" is a Galilean concept

"relativistic velocities" is the relativity analogue: a light beam shining directly at another light beam still go at each other at c but the closure speed, which has no bearing on reality, is 2c

Nice shot, starthaus, with "MMX" - you had me. I even saw the list in Wikipedia which included Michelson-Morley and blew it before the other folks did get it.

 

[starthaus]

Thus, "closing speed" is a Galilean concept

Not really, it is a general concept, applies equally in SR.

"relativistic velocities" is the relativity analogue: a light beam shining directly at another light beam still go at each other at c but the closure speed, which has no bearing on reality, is 2c

Closing speed has a lot of bearing on reality. I just showed you that, without it, you can't explain Einstein gedank, nor can you explain MMX as viewed from a frame different from the lab.

 

[stevmg]

Not really, it is a general concept, applies equally in SR.




Closing speed has a lot of bearing on reality. I just showed you that, without it, you can't explain Einstein gedank, nor can you explain MMX as viewed from a frame different from the lab.

Getting late... wife's in bed and so will I be.

Will go over your post in detail later.

stevmg

 

[jfy4]

are closure velocities invariant?

 

[Doc Al]

are closure velocities invariant?

No. They depend on the frame in which they are measured.

 

[stevmg]

Of what value is closing speed in SR or GR? According to Fredrik quoted below it is less than worthless. I understand closing spped to be the speed which two objects approach or "close in" on each other - like two trains. Michelson-Morley did not know relativity so all they could do was to calculate closing speeds - but found no difference no matter how they angled their machine or what time of year. After all, the Earth moves at 30,000 m/sec = 0.001c which is a hell of a lot less than 30,000,000 m/sec (light speed). Also, at relatively low speeds like this, it's good enough to get a difference in a velocity in relation to an "ether" which is what they were looking for. Again, no luck.

Just have Alice shoot a beam of particles with speed 0.6c towards Bob, and at the same time have Bob shoot a beam of partices with speed 0.6c towards Alice. The closing speed is 1.2c. The velocity of one of the beams in the rest frame of the other is (0.6c+0.6c)/(1+0.6*0.6)=1.2c/1.36<c.

I don't know any situation where closing speed is an interesting concept.

 

[starthaus]

Of what value is closing speed in SR or GR? According to Fredrik quoted below it is less than worthless.

You will not be able to explain the Einstein train+platform gedank about relativity of simultaneity without closing speeds.
You will be incable to explain MMX in any frame different from the lab.

So, it isn't "worthless"

Michelson-Morley did not know relativity so all they could do was to calculate closing speeds - but found no difference no matter how they angled their machine or what time of year. After all, the Earth moves at 30,000 m/sec = 0.001c which is a hell of a lot less than 30,000,000 m/sec (light speed). Also, at relatively low speeds like this, it's good enough to get a difference in a velocity in relation to an "ether" which is what they were looking for. Again, no luck.

You are very much mistaken. try writing down the MMX explanation from the POV of an external observer and you'll learn that you can't do it w/o closing speed.

 

[Fredrik]

According to Fredrik quoted below it is less than worthless.

I wouldn't go that far. I couldn't think of any examples where it's useful, but apparently Starthaus could.

 

[stevmg]

You will not be able to explain the Einstein train+platform gedank about relativity of simultaneity without closing speeds.
You will be incable to explain MMX in any frame different from the lab.

So, it isn't "worthless"




You are very much mistaken. try writing down the MMX explanation from the POV of an external observer and you'll learn that you can't do it w/o closing speed.

I wouldn't go that far. I couldn't think of any examples where it's useful, but apparently Starthaus could.

I am missing something here. First, back to basics. Closure velocity is the addition of the two velocities of two objects as they approach each other (even from a vector point of view with the three dimensions.)

Just answer that question - no more no less.

 

[Fredrik]

If they're moving along the same line in opposite directions, then yes. Otherwise it's a bit more complicated than that. But it's always the rate at which the distance in a specific inertial frame is decreasing.

 

[stevmg]

If they're moving along the same line in opposite directions, then yes. Otherwise it's a bit more complicated than that. But it's always the rate at which the distance in a specific inertial frame is decreasing.

I know, and depending on their positions, if they are not going in literally opposite directions and aimed at each each other like two approaching trains, there would be an acceleration involved JUST BY THE GEOMETRY of their relative motions, so let's stick with your definition.

Now, when one is dealing with slow speeds as would been involved with MMX as starthaus refers to it - you know, 30,000 m/sec, then the Einstein velocity addition formulas "reduce" to basic physics closure velocity formulas.

True?

 

[starthaus]

Now, when one is dealing with slow speeds as would been involved with MMX as starthaus refers to it - you know, 30,000 m/sec, then the Einstein velocity addition formulas "reduce" to basic physics closure velocity formulas.

True?

Not. I just wrote https://www.physicsforums.com/blog.php?b=1959 for you. Look under "MMX.pdf".

 

[stevmg]

Not.

A little more than that, starthaus...

Either show me the difference or refer me to a specific source where this would be explained. I am not from Missouri but I still want you to show me. Besides, if I understand this concept correctly, I will learn something I didn't know and that's what this forum is all about.

Steve

 

[starthaus]

A little more than that, starthaus...

Either show me the difference or refer me to a specific source where this would be explained. I am not from Missouri but I still want you to show me. Besides, if I understand this concept correctly, I will learn something I didn't know and that's what this forum is all about.

Steve

Look above your post, I put something together for you personally.