# Thread: MacM's Claims

1. Originally Posted by cato
you are an idiot. here is your problem:
You really should slow down and think before you start spouting off calling somebody an idiot.

you are launching D with a velocity of .366 WRT C, not A. that is the only time you would need to use the VAF.
Now I can return the favor. YOU IDIOT. "B" and "D" accelerate side by side consuming the same fuel, creating the same thrust and yielding the same acceleration. "B" also accelerated 0.366c from the inertial period and you have no problem with claiming 0.866c wrt "A" in B's case.

Now if you are not an idiot you will be able to give some justification to claim that since "B" and "D" underwent the same acceleration cycles for the same periods and run side by side the entire trip just how you justify them achieving differnt velocities.

Give a rational explanation why "B" re-accelerating is physically any different what-so-ever than "D" accelerating. Remember they flew the entire trip side by side.

The only differance is in the case of "D" you have two objects which can have velocity "C" and "D" and Special Relativity cannot allow 0.9c + 0.5C to be > c, so the velocity addition is applied to compound velocities.

In the scenario I kept the compound velocity below v = c so that the problem would be more obvious.

Come on, give a valid explanation. This is a challenge you idiot. (just kidding you are just confused).

your objects cannot fly side by side because you have chosen them to have the same speed but WRT 2 different, moving, observers. if you pick any one observer and look at the speed of both objects, they will not be the same.
Not true at all. One can write "B" re-accelerates wrt "C" until it has 0.866c wrt "A". Where is the differance now for the flight of "D". Come on idiot. Answer the question.

here is your argument in a nutshell "two spacecraft, traveling at different speeds must stay side by side, and if they don't then SR is wrong"
Where in the hell have you come to the conslusion they have different speeds. That is nuts. They use the same fuel, create the same thrust and achieve the same velocity via acceleration because they are identical and they flew identical acceleration cycles the entire trip.

Come on idiot. We are waiting.

2. Mac, you really don't get it do you? C does not agree with A that the velocity differential between C and B (and between C and D) is .366c. .366c is A's measurement. But u in the formula is C's measurement. Get it? As you originally state the problem w and v are the known's and u is the _unknown_.

3. Originally Posted by kevinalm
Mac, you really don't get it do you?
I am afraid it is not me that has his ears full of wax.

C does not agree with A that the velocity differential between C and B (and between C and D) is .366c. .366c is A's measurement.
I know damn well what the claim is. You are not listening since you have not given any justification for that claim.

But u in the formula is C's measurement. Get it? As you originally state the problem w and v are the known's and u is the _unknown_.
You are damn right. But you ignore the fact that "C" also sees the SAME accleration and velocity for "B" that it does "D".

If not please explain why you think physics cares if "D" is launched from "C" or was merely another ship like "B" that re-accelerates from the same ordinate point simultaneously and to the same magnitude, such that they remain side by side upto where "B" is now 0.866c wrt "A".

You have not answered the physics issue. You are merely reciting the theory which I know quite well.

4. Mac, SR predicts B and D remain in the same frame. At zero velocity with each other. This is for all observers. A sees them as moving together, C sees them as moving together. The entire alphabet. A and C (et al) will have disagreements as to what those number are, But A's numbers are self consistent, C's are self consistent, etc.

Case 1 (your original problem)
According to A this works out: .866c, .5c, .366c.
According to C this works out: 1.146c, .5c, .646c (apprx.)

Case 2 (What you accidently do by inserting .336c into u.)

Now reword the problem slightly. C accelerates to .5c wrt A, B and D accelerate together to .366c wrt C.

According to A: .732c, .5c, .232c
According to C: .866c, .5c, .366c

Pretty sure I got those numbers right. Haven't done these calculations in a while. Anyway A will have a different set of numbers in I than in II. Likewise for C.

In case 1 B and D both burn greater (and identical) quantities of fuel. A, B and C say so. I intended to make that clear in a previous post. Case 1 and Case 2 are different problems, situations, orders to your pilot's. I don't know how else to put it.

5. You are damn right. But you ignore the fact that "C" also sees the SAME acceleration and velocity for "B" that it does "D".
yes, from C's perspective, they both leave to a speed of 0.732c. if you look from A's perspective they go 0.866. I don't know why you cant get it.

6. perhaps you are thinking that since they accelerate .366 from A, then, since they use the same fuel no matter what frame you are in, that they must reach the same speed no matter what frame you are looking from. that is simply not true.

7. Originally Posted by kevinalm
Mac, SR predicts B and D remain in the same frame. At zero velocity with each other. This is for all observers. A sees them as moving together, C sees them as moving together. The entire alphabet. A and C (et al) will have disagreements as to what those number are, But A's numbers are self consistent, C's are self consistent, etc.

Case 1 (your original problem)
According to A this works out: .866c, .5c, .366c.
According to C this works out: 1.146c, .5c, .646c (apprx.)

Case 2 (What you accidently do by inserting .336c into u.)

Now reword the problem slightly. C accelerates to .5c wrt A, B and D accelerate together to .366c wrt C.

According to A: .732c, .5c, .232c
According to C: .866c, .5c, .366c

Pretty sure I got those numbers right. Haven't done these calculations in a while. Anyway A will have a different set of numbers in I than in II. Likewise for C.

In case 1 B and D both burn greater (and identical) quantities of fuel. A, B and C say so. I intended to make that clear in a previous post. Case 1 and Case 2 are different problems, situations, orders to your pilot's. I don't know how else to put it.
Your answer fails to respond to the physics. We all know SRT claims C and A see different velocities for D. But C sees the same velocities for B and D.

There is simply no physical justification to claim A sees D any differently than it sees B.

There is no physical differance in B and D other than the fact that B is a single object with two cycles of acceleration and D is the result of a the same two acceleration cycles but one is masked by riding inside of C.

I can physically just as well eliminate "C" and accelerate "B" and "D" through "B's" cycle. Can you give any physical reasoning why because "D" piggy backed the first 0.5c acceleration inside of "C" that it makes any differance what-so-ever in the terminal velocity?

I think not. All I have done is mask "D's" acceleration cycle by having it inside of "C" to create the Velocity Addition scenario but clearly B and D undergo the exact same acceleration cycle as though "C" never existed and B and D simply ran co-moving the entire trip.

Now justify the use of Velocity Addition.

8. Mac, please do try to completely read my posts. A sees B and D as having the same velocity. The whole alphabet sees B and D as having the same velocity. What you do by inserting .366c in u is to contramand your order for them to accelerate to .866c wrt A and order them to .732c wrt A.

9. Originally Posted by kevinalm
Mac, please do try to completely read my posts. A sees B and D as having the same velocity. The whole alphabet sees B and D as having the same velocity. What you do by inserting .366c in u is to contramand your order for them to accelerate to .866c wrt A and order them to .732c wrt A.
No you do. I don't. You have NO justification to claim BC is any different than DC. I just explained to you that B and D have undergone the exact same acceleration cycles.

The only differance is that "D" did the first 0.5c segment contained inside of "C". But it still accelerated precisely as did "B" and "B" and "D" accelerted precisely the same in the second segment of the overall acceleration cycle.

At no point is there one iota of differance in the acceleration cycles and velocities of B and D. It is ONLY the AD HOC application of the velocity Addition Formula by SRT because part of D's trip was concealed inside of C that the formula is applied.

That application cannot be justified by any physics since there is in reality NO differance in the flight schedules. It is an absurd and AD HOC band aid to achieve nothing more than keep SR mathematically whole and to satisfy a preconcieved desired end result.

Now if you don't agree please post your JUSTIFICATION for applying the VAF to "D"?

Keep in mind not only what I have shown you here (that B and D have physically identical acceleration and velocity histories) but that to invoke the VAF causes you to advocate that A see two different spatial contractions and time dilations.

Why because "A" always computes "C" without use of the VAF and then computes DC using the VAF. Can you give any rational explanation as to why and how physics can have a kink in its processes.

You are making a desperate effort to claim SRT valid but you are not doing it with physics but merely by reciting the theory. I am asking you to justify the theory and the application of the VAF to D when B and D are defacto identical and have undergone identical acceleration and velocity cycles.

Lets see if I can crack the haze another way.

Forget my first scenario of just A and B. Now A launches "C" and accelerate it until it has reached 0.5c. It goes inertial and then launces "B" from "C" and "B" accelerates until it has a relative velocity wrt "A" of 0.866c.

As I surely hope you can see the acceleration cycle for "B" is precisely the same as it was in the first scenario. And its terminal velocity is 0.866c.

Do you want to claim a different fuel consumption, different time dilation, different spatial contraction in this case vs the first case where the same acceleration cycle was done? I should hope not.

What you are failing to see is that your quest to insure nothing exceed v = c is to cause physics to change with velocity. That is fuel consumption increases with velocity. This has nothing to do with any relavistic mass change.

It means you are claiming my space ship accelerates differently for every observer. I dare say my physics are not subject to change as a consequence of who is observing.

10. Originally Posted by MacM
No you do. I don't. You have NO justification to claim BC is any different than DC. I just explained to you that B and D have undergone the exact same acceleration cycles.

The only differance is that "D" did the first 0.5c segment contained inside of "C". But it still accelerated precisely as did "B" and "B" and "D" accelerted precisely the same in the second segment of the overall acceleration cycle.

At no point is there one iota of differance in the acceleration cycles and velocities of B and D. It is ONLY the AD HOC application of the velocity Addition Formula by SRT because part of D's trip was concealed inside of C that the formula is applied.

That application cannot be justified by any physics since there is in reality NO differance in the flight schedules. It is an absurd and AD HOC band aid to achieve nothing more than keep SR mathematically whole and to satisfy a preconcieved desired end result.

Now if you don't agree please post your JUSTIFICATION for applying the VAF to "D"?

Keep in mind not only what I have shown you here (that B and D have physically identical acceleration and velocity histories) but that to invoke the VAF causes you to advocate that A see two different spatial contractions and time dilations.

Why because "A" always computes "C" without use of the VAF and then computes DC using the VAF. Can you give any rational explanation as to why and how physics can have a kink in its processes.

You are making a desperate effort to claim SRT valid but you are not doing it with physics but merely by reciting the theory. I am asking you to justify the theory and the application of the VAF to D when B and D are defacto identical and have undergone identical acceleration and velocity cycles.
Who's treating b and d differently. I'm treating them as a single entity from the intermission on.

11. Originally Posted by kevinalm
Case 1 According to C this works out: 1.146c, .5c, .646c (apprx.)

Case 2 (What you accidently do by inserting .336c into u.)
w = 0.866c, u = .336c

w = v + u / (1 + vu / c^2) = v+u / (1 +vu)

(1 + vu) = (v + u) /w

1 + vu = (v + u) /w = (v/w) + (u/w)

vu - v/w = u/w - 1

v(u-1/w) = (u/w) - 1

v = [(u/w) - 1]/(u - 1/w)

v = 0.7759???

12. Originally Posted by cato
yes, from C's perspective, they both leave to a speed of 0.732c.
What the hell?

if you look from A's perspective they go 0.866. I don't know why you cant get it.
I know what the theory claims. I get that. What you don't get and have not done is justify that conclusion physically. Considering the described acceleration cycles.

13. Sorry, that was a typo. .366c. Damn new bifocals. Arms weren't long enough anymore.

14. Originally Posted by kevinalm
Who's treating b and d differently. I'm treating them as a single entity from the intermission on.
No you don't you claim either A sees B as 0.866c and D as 0.732c OR you claim A sees both as 0.866c but that C is 0.5c and DC is 0.232c when B and D accelerated side by side and B increase 0.366c from his inertial break point which can be viewed as C'.

If you are wanting to claim that BC' is also 0.232c and it is only 0.366c to A then I must ask why things changed when I could have (omitting the spatial ordinate point shift due to the inertial period) I would have been at the same location and velocity had I merely accelerated upto 0.866c with no inertial break.

Give it up it doesn't work guys. Mathematically perhaps it makes you and SRT happy but it defies any physical justification.

15. kevinalm & cato,

You both seem to want to argue theory and not physics. Perhaps I can show you what I mean when I say I expect a physical justification.

I ran my scenario past a NASA, Phd Physicist, Dr E. Dowdye. Here is his response and my e-mail to him:

************************************************** **
Dear Dan:

Guess what Professor Dan, we are in total agreement here on scenarios you presented. You see, the catch here is what the observers work out if relativity is accepted.

The velocity addition scheme used by Relativity and everybody that wants to exact the so-called velocity dependent mass m(v) = m_0 / root(1-v2/c2), the solutions for the resultant velocities must be such that the kinetic energies must work out agreeing with the laws of conservation of energies.

Recall that there are many particle physicists who don't accept the velocity dependent mass m = m(v), where v approaches c, and the mass goes to infinity.

Some believe that the actual mass is unchanging. According to Relativity, the Energy is m_0 c2/root(1-v2/c2) approx m_0 c2 + 0.5 m_0 v2.

The energy gain is still 0.5 m_0 v2. See, this is how these Relativists get away with this. The finagle the thing out so that the math comes out. If you linearly add the velocities using Galilean Transformations, the m(v) mass fails. However, if the mass is unchanging, its OK to add velocities.

The effective mass I calculation M_eff has the same equation as relativity, but for a different reason. Only the effective mass is a function of velocity, the actual mass is unchanging.

The relative velocity of D is only 0.732c because Relativity snow balls the hell out of the mass. (smaller velocity, bigger mass), to save the energy conservation.

That's my spill on that.

EHDowdye

At 09:22 PM 8/26/2005, you wrote:

Dr Dowdye,

Once again I think we agree. I have found your ExtinctionShift Theory to be a most helpful concept.

I believe the following scenario that I use when debating Velocity Addition supports the ES view.

************** Gendankin ******************

Given a mothership "A" and a shuttlecraft "B" which are initally at inertial rest.

"B" accelerates away until it reaches 0.5c and then goes inertial for 5 minutes (B's clock) and then reaccelerates another 0.366c to reach 0.866c wrt "A".

Nobody seems to object to this statement.

Now add to this scenario that another larger shuttlecraft "C" launches from "A" simultaneously with "B" and accelerates equally, and for the same duration, so as to be co-moving with no relative velocity wrt "B", upto the 0.5c and inertial period. When "B" reaccelerates "C" simultaneously launches another shuttlecraft "D" that is identical to "B" and as such consumes the same fuel, generates the same thrust and achieves the same acceleration as "B" so that "B" and "D" accelerate side by side until "B" and "D" once again go inertial when "B" reaches the 0.866c velocity.

Again the physics here seem rather straight forward and nobody has a problem with "B" having a 0.866c relative velocity to "A".

BUT Special Relativity claims that since "D" was launched from "C" that one must use the Velocity Addition Formula to compute "D's" relative velocity to "A" (w) where: v is the velocity of "C" wrt "A" and "u" is the relative velocity of "D" wrt "C".

w = v + u / (1 + vu / c^2); which claims that "D" has a relative velocity of only 0.732c whereas "B" has a relative velocity of 0.866c while both "B" and "D" have expereinced the same exact series of accelerations and consumed the same energy and have flown side by side over the entire trip .

There is no physical justification to claim Velocity Addition must be applied except to keep Special Relativity consistant. It violates basic physics by making thrust (acceleration) vs energy input disproportionate between the two cases which are physically identical.
************************************************** ******

I would like your comments on this proof, if you don't mind..

Thanks.

Dan
************************************************** **

Now this is interesting since I already have stated many times that the correct physical view is that there is no change in mass and that data is being miss interpreted that mass increase is an illusion caused by a decrease in energy transfer efficiency. So Dr Dowdye and I see things the same because we see physics the same way.

So do you have ANY physics basis to reject the concepts I put forth (other than they are different than relativity). I think not.

16. Let see if I can explain this, it's late and I need to call it a night. Think of the solution set for the "3 frame problem" as a 3x3 matrix of measurements. The rows can hold the numbers for one observer(A,B or C) and the columns for a particular frame pair, Vab, Vbc, or Vac. Any physically real situation(problem,whatever you want to call it) has a unique set of entries. If you specify just two entries that are not in the same column you have "locked in" the entire matrix. I think can figure out why yourself. Suffice it to say that there are a number of constraints. Now initially you satisfy the locked in condition with Vab wrt a and Vac wrt a, all the numbers are fixed, we only need to fill in the blancs. The way you are using the vaf is changing the matrix to an unreal state, as it is putting Vbc wrt a into Vbc wrt c's spot and then recalculating Vab wrt A, restoring a real state. (I think. It's late. There may be some other recalculation necessary but I don't think so). But importantly it's a completely different matrix. Basically I'm saying that you can tell B and D to dock together and accel to Vab wrt .866c or tell them to dock and accel to Vac wrt c .366c, but if you try to tell them to do both they throw you in the brig and convene a court martial to relieve you of command.

It's late and I'm getting punchy. Time for bed. We can continue this sometime tomorrow if you like.

17. Originally Posted by kevinalm
Let see if I can explain this, it's late and I need to call it a night. Think of the solution set for the "3 frame problem" as a 3x3 matrix of measurements. The rows can hold the numbers for one observer(A,B or C) and the columns for a particular frame pair, Vab, Vbc, or Vac. Any physically real situation(problem,whatever you want to call it) has a unique set of entries. If you specify just two entries that are not in the same column you have "locked in" the entire matrix. I think can figure out why yourself. Suffice it to say that there are a number of constraints. Now initially you satisfy the locked in condition with Vab wrt a and Vac wrt a, all the numbers are fixed, we only need to fill in the blancs. The way you are using the vaf is changing the matrix to an unreal state, as it is putting Vbc wrt a into Vbc wrt c's spot and then recalculating Vab wrt A, restoring a real state. (I think. It's late. There may be some other recalculation necessary but I don't think so). But importantly it's a completely different matrix. Basically I'm saying that you can tell B and D to dock together and accel to Vab wrt .866c or tell them to dock and accel to Vac wrt c .366c, but if you try to tell them to do both they throw you in the brig and convene a court martial to relieve you of command.

It's late and I'm getting punchy. Time for bed. We can continue this sometime tomorrow if you like.
I am amazed how quickly you move. I have spent considerable time trying to get you to justify the physics but you jump to launching some other form of arguement.

Are you going to ignore the physics? Does that mean you don't have a valid reply?

18. your problem comes from the VAF. could you tell me again why you must use it? I don't really see any justification for using it. I think it is either unnecessary, or being used wrong.

19. Originally Posted by cato
your problem comes from the VAF. could you tell me again why you must use it? I don't really see any justification for using it. I think it is either unnecessary, or being used wrong.
I believe your comments show you are not that familiar with the VAF but it also shows you seem to see it my way. that is there is no justifiction for its application.

However, the facts are it is mandated by Special Relativity. There is the problem. It isn't my problem.

20. I think you misinterpret SR, I don't think you really need the VAF in your situation. links/explanations/rules would be nice to explain why we need it in this situation. I have a gut feeling that it is being applied wrong here.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•