Distance & Time Light Travels in Approaching Mirrors Problem

[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 :tongue: ) 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 :tongue: 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:

$$v = \frac{v'+u}{1+v'u}$$

(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:

$$v' = \frac{v-u}{1-vu}$$

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 :tongue: 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 :tongue2:

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

$$v = \frac{v'+u}{1+v'u}$$

(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:

$$v' = \frac{v-u}{1-vu}$$

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!

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

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!

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

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:tongue:.
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:

$$V_{a/c} = V_{a/b} + V_{b/c}$$

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.

 

[stevmg]

Gentlemen:

It seems like we are stating here that if the hound runs to the East at 1/2 the speed of light and the rabbit who is in front runs to the East at 1/2 the speed of light, that hound ain't ever going to catch that rabbit!

One does not have to be a damn rocket scientist to figure that one out!

 

[yossell]

It seems like we are stating here that if the hound runs to the East at 1/2 the speed of light and the rabbit who is in front runs to the East at 1/2 the speed of light, that hound ain't ever going to catch that rabbit!

One does not have to be a damn rocket scientist to figure that one out!

But it's nice to have a precise quantitative answer as to just how badly that hound misses out.

 

[stevmg]

Ha! Ha!

I just my two cents in because I started this thread!

But, yossell, don't forget the old adage, "Missing by an inch is as good as a mile, except in horseshoes." So that hare still gets away!

stevmg

 

[Austin0]

Ha! Ha!

I just my two cents in because I started this thread!

But, yossell, don't forget the old adage, "Missing by an inch is as good as a mile, except in horseshoes." So that hare still gets away!

stevmg

I strenuously take exception to this proposition. Clearly this adage applies equally to handgrenades ,no?

 

[stevmg]

Good on all of you!