Velocity Addition Theorem

[Vilas Tamhane]

I am in the spaceship in frame A. I burn fuel quantity q and with the help of accelerometer I notice that I gain speed of 0.01c. Now I am in the frame B. I again burn quantity of fuel q and again gain velocity of 0.01c and enter the frame C.

 

Since my instrument cannot measure uniform velocity, I keep a record. Question is, should I consider my velocity as 0.02c in frame C or use relativistic velocity addition theorem (VAT) and consider my velocity lesser, say 0.019c?

 

Note that the frame B is no different from any other inertial frame and for all mechanical laws; velocity of the frame can be considered as 0. This is a fundamental law which no theory has so far refuted. Therefore there is no reason why I should add velocity 0.009c and not 0.01c.

 

So there cannot be any dispute that I have gained velocity of 0.01c while going from frame B to frame C and instead of adding this velocity if I add 0.009c then it is a mathematical process, just to satisfy VAT.

 

But note that if I add velocity 0.009c then I am treating frame B different from frame A. In fact all the succeeding frames will be different (not equivalent) from all the preceding frames.

 

 

Argument that I do gain velocity of 0.01c while accelerating from frame B to C but the velocity in the frame C with respect to that in frame A is 0.019c would be circular. It is wrong because it presupposes correctness of VAT. If this figure is true then I am simply not adding 0.01c but I am adding 0.009c to the velocity in frame B. I cannot do this unless I actually gain velocity of 0.009c while going from frame B to frame C. After all, second burst of fuel cannot be different than effect of first burst of fuel. Velocity in the frame B remains intact. IOW, I must add 0.009c instead of 0.01c just to satisfy VAT.

 

But the law which states that all inertial frames are equivalent is so fundamental that it overrides VAT (either of the two can be correct) and so during 2nd burst, I did acquire velocity of 0.01c. Addition is now just a mathematical process which cannot affect physical processes. IOW, if I must add the figure 0.009c then I must acquire this velocity in reality, while accelerating from frame B to frame C.

 

 

[Janus]

I am in the spaceship in frame A. I burn fuel quantity q and with the help of accelerometer I notice that I gain speed of 0.01c. Now I am in the frame B. I again burn quantity of fuel q and again gain velocity of 0.01c and enter the frame C.

 

Since my instrument cannot measure uniform velocity, I keep a record. Question is, should I consider my velocity as 0.02c in frame C or use relativistic velocity addition theorem (VAT) and consider my velocity lesser, say 0.019c?

 

Note that the frame B is no different from any other inertial frame and for all mechanical laws; velocity of the frame can be considered as 0. This is a fundamental law which no theory has so far refuted. Therefore there is no reason why I should add velocity 0.009c and not 0.01c.

 

So there cannot be any dispute that I have gained velocity of 0.01c while going from frame B to frame C and instead of adding this velocity if I add 0.009c then it is a mathematical process, just to satisfy VAT.

 

But note that if I add velocity 0.009c then I am treating frame B different from frame A. In fact all the succeeding frames will be different (not equivalent) from all the preceding frames.

 

 

Argument that I do gain velocity of 0.01c while accelerating from frame B to C but the velocity in the frame C with respect to that in frame A is 0.019c would be circular. It is wrong because it presupposes correctness of VAT. If this figure is true then I am simply not adding 0.01c but I am adding 0.009c to the velocity in frame B. I cannot do this unless I actually gain velocity of 0.009c while going from frame B to frame C. After all, second burst of fuel cannot be different than effect of first burst of fuel. Velocity in the frame B remains intact. IOW, I must add 0.009c instead of 0.01c just to satisfy VAT.

 

But the law which states that all inertial frames are equivalent is so fundamental that it overrides VAT (either of the two can be correct) and so during 2nd burst, I did acquire velocity of 0.01c. Addition is now just a mathematical process which cannot affect physical processes. IOW, if I must add the figure 0.009c then I must acquire this velocity in reality, while accelerating from frame B to frame C.

 

If you are at rest in frame B and moving at 0.01c with respect to frame A and then accelerate to 0.01c with respect to frame B you will now be moving at 0.019992c with respect to frame A.

 

But this does not mean that you didn't change your velocity by 0.01c. It means that you are now moving at 0.01c with respect to B and 0.019992c with respect to frame A. The difference in velocity between Frame A and B is now 0.009992 according to you in frame C. What this means is that the difference in speed between A and B is different for you in C than it was for you when you were in B. But that makes sense, since B and C have a relative velocity difference and will measure time and distance differently and thus would measure their relative velocity with respect to a third frame differently.

 

.

[swansont]

Note that the frame B is no different from any other inertial frame and for all mechanical laws; velocity of the frame can be considered as 0. This is a fundamental law which no theory has so far refuted. Therefore there is no reason why I should add velocity 0.009c and not 0.01c.

The reason is that all velocities are measured with respect to something. Note that in Janus's explanation, the velocities are stated as being with respect to a particular frame. So the velocity you end up with depends on which frame it is being measured against.

[Vilas Tamhane]

If you are at rest in frame B and moving at 0.01c with respect to frame A and then accelerate to 0.01c with respect to frame B you will now be moving at 0.019992c with respect to frame A.

 

But this does not mean that you didn't change your velocity by 0.01c. It means that you are now moving at 0.01c with respect to B and 0.019992c with respect to frame A. The difference in velocity between Frame A and B is now 0.009992 according to you in frame C. What this means is that the difference in speed between A and B is different for you in C than it was for you when you were in B. But that makes sense, since B and C have a relative velocity difference and will measure time and distance differently and thus would measure their relative velocity with respect to a third frame differently.

 

.

 

 

I very much appreciate your answer. However this answer pre-supposes relativistic effects. You agree that when I move from frame B to frame C, I do acquire velocity 0.01c. I can’t measure uniform velocity and so I must record velocities and the only fundamental law which I am certain about is the equivalence of inertial frames. So if I deviate from the process of addition as 0.01c+0.01c+0.01c…..then I am invalidating the most fundamental law of physics.

 

 

[J.C.MacSwell]

I very much appreciate your answer. However this answer pre-supposes relativistic effects. You agree that when I move from frame B to frame C, I do acquire velocity 0.01c. I can’t measure uniform velocity and so I must record velocities and the only fundamental law which I am certain about is the equivalence of inertial frames. So if I deviate from the process of addition as 0.01c+0.01c+0.01c…..then I am invalidating the most fundamental law of physics.

 

Lets suppose you are on a circle of 5 mile radius at point A. You walk along the circumference until you are at point B, exactly 1 mile from A. You then walk further on the circle until you are at point C, exactly 1 mile from B.

 

However, you are not 2 miles from point A. Does this prove that not all points on a circle are equivalent? That there are preferred positions?

 

Or does it tell you something about your path?

[Vilas Tamhane]

Lets suppose you are on a circle of 5 mile radius at point A. You walk along the circumference until you are at point B, exactly 1 mile from A. You then walk further on the circle until you are at point C, exactly 1 mile from B.

 

However, you are not 2 miles from point A. Does this prove that not all points on a circle are equivalent? That there are preferred positions?

 

Or does it tell you something about your path?

It tells me that the example is unrelated to my post.

 

 

[J.C.MacSwell]

It tells me that the example is unrelated to my post.

 

It's an analogy. I think it is an appropriate one. If you are confident in your logic and arrive at an incorrect answer then you should check your assumptions.

 

Your post only makes sense if you assume or believe velocity is something that it is not. Velocity simply does not add the way you would like it to though the difference is not obvious until relativistic speeds become significant.

Edited June 30, 2011 by J.C.MacSwell

[Vilas Tamhane]

It's an analogy. I think it is an appropriate one. If you are confident in your logic and arrive at an incorrect answer then you should check your assumptions.

 

Your post only makes sense if you assume or believe velocity is something that it is not. Velocity simply does not add the way you would like it to though the difference is not obvious until relativistic speeds become significant.

 

 

I don’t think there is any point in stretching the discussions. I am right considering only classical physics and you are right taking into account special relativity. Some agree with theory of relativity and some don’t. If we discuss validity of the theory itself then we cannot take any prior position about its correctness. I think reply by Janus clarifies SR point of view eminently and is sufficient.

 

 

[between3and26characterslon]

I very much appreciate your answer. However this answer pre-supposes relativistic effects. You agree that when I move from frame B to frame C, I do acquire velocity 0.01c. I can't measure uniform velocity and so I must record velocities and the only fundamental law which I am certain about is the equivalence of inertial frames. So if I deviate from the process of addition as 0.01c+0.01c+0.01c…..then I am invalidating the most fundamental law of physics.

 

 

 

 

Maybe if you had some very accurate clocks you could measure relatavistic effects rather than pre-supposing them.

 

Are you invalidating the most fundamental law of physics?

 

No.

 

You will measure an acceleration to a speed of 0.01c from the frame you are in to the next frame. When you move from that frame to the next one you will again measure an acceleration of the next frame to 0.01c. This conserves the principle of relativity.

 

If you were to stay in your initial frame and measure each sucessive frame from there you will notice they increase in velocity in increments progressively smaller than 0.01c, this is confirmed by comparing their clocks, and this is fundamentaly the special theory of relativity.

 

edited for spelling

Edited June 30, 2011 by between3and26characterslon

[swansont]

I very much appreciate your answer. However this answer pre-supposes relativistic effects. You agree that when I move from frame B to frame C, I do acquire velocity 0.01c. I can’t measure uniform velocity and so I must record velocities and the only fundamental law which I am certain about is the equivalence of inertial frames. So if I deviate from the process of addition as 0.01c+0.01c+0.01c…..then I am invalidating the most fundamental law of physics.

Addition is not a fundamental law of physics. Velocity addition using a Euclidean geometry/Galilean transform is not a fundamental law of physics. It was overturned when relativity was discovered and verified.

 

I don’t think there is any point in stretching the discussions. I am right considering only classical physics and you are right taking into account special relativity. Some agree with theory of relativity and some don’t. If we discuss validity of the theory itself then we cannot take any prior position about its correctness. I think reply by Janus clarifies SR point of view eminently and is sufficient.

No, you are not right. If you cannot cannot presuppose relativity, then you cannot presuppose classical physics, either. You go with what has been experimentally verified, which is relativity.

[baric]

I don't think there is any point in stretching the discussions. I am right considering only classical physics and you are right taking into account special relativity. Some agree with theory of relativity and some don't. If we discuss validity of the theory itself then we cannot take any prior position about its correctness.

 

Are you challenging the validity of GR? If so, then instead of asking the question you should be giving your expected answer and then explaining why it deviates from the answer provided by GR.

 

It tells me that the example is unrelated to my post.

 

 

 

 

This was a good analogy because the space that your ship is traveling through is curved just as the circle in the analogy.

 

Therefore the velocity differential from previous frames is going to change differently. And just as your distance from any other point on the circle cannot exceed 2pi no matter how far you travel along its circumference, your velocity differential with any other frame in space cannot exceed c.

[Vilas Tamhane]

Addition is not a fundamental law of physics. Velocity addition using a Euclidean geometry/Galilean transform is not a fundamental law of physics. It was overturned when relativity was discovered and verified.

 

 

No, you are not right. If you cannot cannot presuppose relativity, then you cannot presuppose classical physics, either. You go with what has been experimentally verified, which is relativity.

 

 

I am curious to know about experimental verification of VAT. Is it possible for you to provide a link to such an experiment? Of course I know about energy-velocity relationship of charged particles. However charged particles radiate energy during acceleration and so they are improper candidates for the experiment. Are there any experiments conducted with uncharged particles?

 

 

[ajb]

I am curious to know about experimental verification of VAT. Is it possible for you to provide a link to such an experiment?

 

The Fizeau experiment is considered to be the classical test of the velocity addiction formula. I do not know what the modern direct tests of the velocity formula are. However, the experimental evidence supporting special relativity is overwhelming (within its domain of applicability). A good review can be found here.

[Vilas Tamhane]

The Fizeau experiment is considered to be the classical test of the velocity addiction formula. I do not know what the modern direct tests of the velocity formula are. However, the experimental evidence supporting special relativity is overwhelming (within its domain of applicability). A good review can be found here.

 

 

Thanks for giving me the link. I will go through it.

[swansont]

I am curious to know about experimental verification of VAT. Is it possible for you to provide a link to such an experiment? Of course I know about energy-velocity relationship of charged particles. However charged particles radiate energy during acceleration and so they are improper candidates for the experiment. Are there any experiments conducted with uncharged particles?

There's the Fizeau experiment

http://en.wikipedia.org/wiki/Fizeau_experiment

 

I suspect there are a bunch in particle physics, from the decay cascades of neutral particles; these will produce charged particles, but you can measure their speeds before they are slowed much by radiating.

 

The velocity addition formula is a straightforward consequence of the Lorentz transforms. If they hold, the velocity addition formula must hold.

[ajb]

The velocity addition formula is a straightforward consequence of the Lorentz transforms. If they hold, the velocity addition formula must hold.

 

Yes and this is how modern papers will phrase the issue. Violation of Lorentz invariance is a well studied topic; modifications of space-time geometry, quantum gravity and so on could lead to violation of Lorentz symmetry and this could be observable.

[Vilas Tamhane]

There's the Fizeau experiment

http://en.wikipedia....zeau_experiment

 

I suspect there are a bunch in particle physics, from the decay cascades of neutral particles; these will produce charged particles, but you can measure their speeds before they are slowed much by radiating.

 

The velocity addition formula is a straightforward consequence of the Lorentz transforms. If they hold, the velocity addition formula must hold.

 

 

Application of VAT to Fizeau’s experiment is no doubt classic and for that matter velocity addition equation itself is sweet. However I have a small problem that must be fixed.

 

 

In the moving frame O’ velocity of the object u’ is the ratio of distance by time duration and so we have,

 

 

u' = L’/dt’ = L*Gamma^2/dt = u*Gamma^2

 

or

 

u=u’(1-v^2/c^2)

 

 

I don’t get velocity addition equation though I don’t see anything wrong in the steps. Please explain.

 

 

 

[csmyth3025]

Application of VAT to Fizeau's experiment is no doubt classic and for that matter velocity addition equation itself is sweet. However I have a small problem that must be fixed.

 

In the moving frame O' velocity of the object u' is the ratio of distance by time duration and so we have,

 

u' = L'/dt' = L*Gamma^2/dt = u*Gamma^2

 

or

 

u=u'(1-v^2/c^2)

 

I don't get velocity addition equation though I don't see anything wrong in the steps. Please explain.

 

Is this the part you're having trouble with?

 

 

Galilean addition of velocities

 

As Galileo observed, if a ship is moving relative to the shore at velocity v, and a fly is moving with velocity u as measured on the ship, calculating the velocity of the fly as measured on the shore is what is meant by the addition of the velocities v and u. When both the fly and the ship are moving slowly compared to light, it is accurate enough to use the vector sum

 

where is the velocity of the fly relative to the shore.

 

 

Special theory of relativity

According to the theory of special relativity, the frame of the ship has a different clock rate and distance measure, and the notion of simultaneity in the direction of motion is altered, so the addition law for velocities is changed. This change isn't noticeable at low velocities but as the velocity increases towards the speed of light it becomes important. The addition law is also called a composition law for velocities. For collinear motions, the velocity of the fly relative to the shore is given by

 

 

(ref. http://en.wikipedia....n_of_velocities )

 

Chris

[Vilas Tamhane]

Is this the part you're having trouble with?

 

 

 

(ref. http://en.wikipedia....n_of_velocities )

 

Chris

 

 

What is your problem? You didn't read the message or you couldn't follow it? It is very simple text.

[swansont]

http://en.wikipedia.org/wiki/Velocity-addition_formula#Derivation

[csmyth3025]

What is your problem? You didn't read the message or you couldn't follow it? It is very simple text.

Well, since I read the message I have to conclude that I couldn't follow it. Is there a simpler explanation suitable for the mathematically challenged?

 

Chris

[Vilas Tamhane]

Well, since I read the message I have to conclude that I couldn't follow it. Is there a simpler explanation suitable for the mathematically challenged?

 

Chris

 

 

In your post you just mentioned Galilean velocity addition and relativistic equation for addition of velocities. Therefore I did not open the link. Please accept my apology.

 

In the derivation of the equation, Einstein used elementary algebra. Even the space-time diagram was introduced later. It appears to me that in the link which you have provided there is complexity which need not be attended.

 

Velocity addition equation can be easily derived by using Lorentz equations. So I think I am justified in using simple length and time duration in the moving frame and then transform the values in the stationary frame O. Therefore it should be easy for you to point out the mistake in my derivation rather than myself visiting the site.

 

 

[Janus]

Velocity addition equation can be easily derived by using Lorentz equations. So I think I am justified in using simple length and time duration in the moving frame and then transform the values in the stationary frame O. Therefore it should be easy for you to point out the mistake in my derivation rather than myself visiting the site.

 

You have to include the relativity of simultaneity also.

[Vilas Tamhane]

You have to include the relativity of simultaneity also.

 

 

Yes and this is automatically included when we use Lorentz equations and not length and time segments in which simultaneity is eliminated. Thanks.

 

 

[Janus]

Yes and this is automatically included when we use Lorentz equations and not length and time segments in which simultaneity is eliminated. Thanks.

 

No,you cannot eliminate simultaneity in this problem.

 

Here's what you did:

 

u' = L’/dt’ = L*Gamma^2/dt = u*Gamma^2

 

or

 

u=u’(1-v^2/c^2)

 

You took the time, t' measured by a clock in the primed frame and divided it into the length L' in the primed frame to get u' the speed relative to the primed frame. So far, so good.

You then applied time dilation and length contraction to t' and L' to arrive at the formula to give u, the speed relative to he primed frame as measured from the unprimed frame. Thus is wrong and I'll explain why.

 

Assume that in the primed frame that there are two synchronized clocks at points A' and B' a distance L' apart. we have an object traveling from A' to B' at speed u' If we take the reading on clock A'(t'0) as the object passes it and the reading on clock B'(t'1) as the clock passes it,and take the difference between t'1 and t'0 we get t', which is the time it takes for the object to pass from A' to B' in the primed frame. If we assume that t'0=0, then t'=t'1.

 

if we divide L' by t' we get u' the relative speed of the object with respect to the primed frame as measured in the primed frame.

 

Now we switch to the non-primed frame, to which the primed frame is traveling at v with respect to. From this frame's perspective, the object will also pass clock A' when it reads t'0(0) and clock B' when it reads t'1 (these are events that both frames must agree upon). However, unlike the primed frame, clock A' and B' are not synchronized; Clock B' will be behind clock A' (by how much is determined by L' and v). This means that we cannot just equate t'1 to the time that passes in the primed frame according to the non-primed frame while the object travels from A' to B'.

 

For example, if t'1= 30 sec and t'0 = 0 and the difference between the readings on Clock A' and Clock B' is 5 sec, then clock A' goes from 0 to 35 sec as the object passes between the clocks and clock B' goes from -5 sec to 30 sec in the same period. IOW, from the non-primed frame, 35 sec pass in the primed frame and not 30 sec (the time measured in the primed frame). It is this 35 sec that you would apply time dilation to in order to get t in the non-primed frame.