# Thread: Unrelative Relativity Part 2

1. ## Unrelative Relativity Part 2

Crisp, James R, Thed, Q, Xev, and anyone else that cares,

I see my thread "Unrelative Relativity" was to complicated for some of you to understand. Here is a simpler problem:

Two metal balls are in outerspace. One weighs one kilogram, while the other one weighs two kilograms. You don't know which ball is moving and which ball is stationairy, all you know is that they are moving towards each other at .90c. Finally, the two balls impact each other, and ALL of the mass of both balls are converted to energy in the impact. Give the total energy output of the collision in both examples listed below.

1) Since you don't know which ball is moving and which ball is stationairy, assume that the 1 kilogram ball is moving at .90c and the 2 kilogram ball is stationairy. If the total mass(relative and rest) of both balls is converted to energy in the collision, how much energy is produced in the collision??

2) Since you don't know which ball is moving and which ball is stationairy, assume that the 2 kilogram ball is moving at .90c and the 1 kilogram ball is stationairy. If the total mass(relative and rest) of both balls is converted to energy in the collision, how much energy is produced by the collision???

Finally, which answer is correct: The answer from example 1 or from example 2, or is the answer somewhere in between the two???

If you do not know the answer, what additional information would you need to obtain the answer??

Hint: Formulas to use:

E=mc^2

m= m0/sqrt(1-(v^2/c^2))

Xev,

You probably already know the answer. I'm interested in the answer that the die-hard relativists come up with.

Tom

2. Tom (May I call you Tom?):

1) Since you don't know which ball is moving and which ball is stationairy, assume that the 1 kilogram ball is moving at .90c and the 2 kilogram ball is stationairy. If the total mass(relative and rest) of both balls is converted to energy in the collision, how much energy is produced in the collision??
E=mc^2
E= (1+3kg)(9*10^10m2/s2)
E= 3.6*10^11 joules

I do hope I did this one correctly, I get a bad feeling that I did not.

2) Since you don't know which ball is moving and which ball is stationairy, assume that the 2 kilogram ball is moving at .90c and the 1 kilogram ball is stationairy. If the total mass(relative and rest) of both balls is converted to energy in the collision, how much energy is produced by the collision???
Same as above....I think.....

3. Xev,

Sure you can call me Tom.

You made a slight error in your calculations. You should have used a calculator. I did.

Example 1:

In the first case the 1 kilogram mass is travelling .90 c. This would mean that the relative mass of the 1 kilogram mass is:

m1=m0/sqrt(1-(v^2/c^2)
m1=1 kg/sqrt(1-(v^2/c^2)
m1=1kg/0.436
m1=2.29 kg

Since the 2kg weight is at rest

m2=2 kg

The sum of both masses is:

m=m1+m2
m=2.29 kg+2 kg
m=4.29 kg

Since all the mass gets converted to energy:

E=mc^2
E=4.29 * c^2
E=386100000000000000 Joules
E= 3.86*10^17 Joules

In the second example:

The 1 kg mass is at rest therefore:

m1=1

The 2 kg mass is traveling at .90c therefore:

m2=m0/sqrt(1-(v^2/c^2)
m2=2 kg/sqrt(1-(v^2/c^2)
m2=2kg/0.436
m2=4.58 kg

The sum of both masses is:

m=m1+m2
m=1 kg+4.58 kg
m=5.58 kg

Since all the mass gets converted to energy:

E=mc^2
E=5.58 * c^2
E=502200000000000000 Joules
E= 5.02*10^17 Joules

As you can see, in the same frame of reference, example 1 gives 3.86*10^17 joules while example 2 gives 5.02*10^17 Joules.

Can you guess which one is the correct??

Tom

4. I see my thread "Unrelative Relativity" was to complicated for some of you to understand.

There was nothing complicated at all with your other thread. You are simply unable to understand the explanations.

As you can see, in the same frame of reference, example 1 gives 3.86*10^17 joules while example 2 gives 5.02*10^17 Joules.

Wrong. You are using two frames of reference:

Here:

Since the 2kg weight is at rest

and here:

The 1 kg mass is at rest therefore:

5. Q,

It was two frames of reference that merged into one.

In the end, when the two metal balls no longer exist, how much energy do you believe the collision produced???

Tom

6. Hi Tom,

This example will not help you disprove special relativity. You will probably claim that the total amount of energy must be the same for both observers, while special relativity (and I can claim this without any calculations) will predict different energies for both observers.

Well, guess what: energy is not an invariant quantity between observers, only the size of the energy-momentum fourvector = (E/c)^2 - p^2 is.

Bye!

Crisp

7. Crisp,

This example will not help you disprove special relativity. You will probably claim that the total amount of energy must be the same for both observers, while special relativity (and I can claim this without any calculations) will predict different energies for both observers.
After the collision, there are no more observers. The observers have been destroyed.

So I ask you just as I asked Q, how much energy is left after the metal balls are gone???

Tom

8. It was two frames of reference that merged into one.

Sorry, you can't do that. Not only does that imply an absolute frame of reference, it also implies there is no difference between an inertial and non-inertial frame.

9. Q,

"It was two frames of reference that merged into one."

Sorry, you can't do that. Not only does that imply an absolute frame of reference, it also implies there is no difference between an inertial and non-inertial frame.
The only frame of reference left after the collision is the frame of reference of the energy. The metal ball's frames of reference have been destroyed. They no longer exist because the metal balls no longer exist.

Tom

10. The only frame of reference left after the collision is the frame of reference of the energy. The metal ball's frames of reference have been destroyed. They no longer exist because the metal balls no longer exist.

Then your question is no worded correctly. You explicitly stated two examples; one example in which the (1) kg. ball was at rest and another example in which the (2) kg. ball was a rest. That is (2) separate reference frames.

Obviously the aftermath of the explosion must be viewed by a third party observer. That again, is another reference frame altogether.

11. Q,

Obviously the aftermath of the explosion must be viewed by a third party observer. That again, is another reference frame altogether.
How much energy does this third party observer detect???

Tom

12. How much energy does this third party observer detect???

That is not the issue. Asking how much energy a third party observer will detect is meaningless. The four vector formula allows you to convert a set of components from one frame to a set of components of another frame. In this case, one is evaluating the components from the FOR of the (1) kg. ball to the FOR of the (2) kg. ball.

13. Q,

"How much energy does this third party observer detect???"

That is not the issue. Asking how much energy a third party observer will detect is meaningless.
It's not meaningless. We know that the two balls collided and annihilated each other. The only question that now remains is how much energy was created from the collision???

Tom

14. Tom:
You made a slight error in your calculations. You should have used a calculator. I did.
More a problem in that I used my T-83 when I am more used to the 86. Thanks for the clarification on the end results.

As you can see, in the same frame of reference, example 1 gives 3.86*10^17 joules while example 2 gives 5.02*10^17 Joules.

Can you guess which one is the correct??
They both are.

The only question that now remains is how much energy was created from the collision???
Okay, for the first collision:

3.86*10^17 J

And for the second:

5.02*10^17 J

So, umm, this may be a stupid question, but, where exactly is the contradiction?

Edit to note:
I do not consider myself a partisan of either side. Relativity seems well verified by experiment, from what I read, and I've seen no reason to consider it flawed.

Incomplete, definitly, but we have yet to find a complete physical theory.

15. Xev,

Okay, for the first collision:

3.86*10^17 J

And for the second:

5.02*10^17 J

So, umm, this may be a stupid question, but, where exactly is the contradiction?
Relativity claims that both collisions are the same. In other words, relativity claims that you do not need to know the absolute motions of two objects in a frame of reference, you only need to know the motion of one object relative to the other in order to calculate the result of any interaction.

However, in the examples I provided, the relative motion of one ball to the other is the same in both cases, but the results are different. This goes against the core of relativity.

As I indicated in the beginning of this thread, you don't know which metal ball is moving and which one isn't, you only know that they are moving towards each other at .90 c. The examples I have provided are only two of an infinite amount of possibilities(for example both balls can be moving towards each other).

The problem is that to get the true answer to this problem, you would have to compare the motion of the balls to the absolute frame of reference in order to obtain the absolute motions of the balls. Only when you know what the absolute motions of the balls are, can you calculate the exact energy resulting from the collision.

Although the relativists on this forum will argue that there is no absolute frame of reference, I'm still waiting for them to answer my question:

How much energy is present after the collision:

a) 3.86*10^17 J
b) 5.02*10^17 J
c) none of the above

Tom

16. Crisp and Q,

Now that I think about it, there is only one frame of reference in the examples I have provided above.

In example 1 the one kilogram ball assumes that it is travelling towards the two kilogram ball.

In example 2 the one kilogram ball assumes that the two kilogram ball is moving towards it.

Notice that in both cases, we are in the one kilogram balls' frame of reference.

After all, there is no rule that say's that the one kilogram ball has to consider itself stationairy in it's frame of reference.

Tom

17. Hi Tom,

There is always an extra frame of reference involved. When you said

"In example 1 the one kilogram ball assumes that it is travelling towards the two kilogram ball."

then you imply that there is a third observer who can confirm that the one kilogram ball is moving towards the 2kg ball. From the point of view of the 1kg ball, it cannot tell (because it will only see the 2kg ball closing in).

Bye!

Crisp

18. Hi Tom,

"The problem is that to get the true answer to this problem, you would have to compare the motion of the balls to the absolute frame of reference in order to obtain the absolute motions of the balls. Only when you know what the absolute motions of the balls are, can you calculate the exact energy resulting from the collision."

First of all, energy is relative to the observer - I'll illustrate this with an example in just a second. But you don't need an absolute frame of reference at all if you think the disappearence of the two frames of reference of the two balls are a problem. You just need a third observer who is watching the entire proces. This observer can take the relative motions of the two balls into account and calculate the energy of the resulting collision from his point of view.

Now, energy is relative to the observer even in classical mechanics. A small example would be a moving object. For an observer witnessing the object fly by, it will have a kinetic energy of mv<sup>2</sup>/2, while for an observer moving along with the object it will have zero kinetic energy (for that observer the object does not seem to move).

So to answer your question: the answer is (c), none of the above, since you did not specify the frame of reference where the energy is to be calculated.

Bye!

Crisp

19. Q:

"That is not the issue. Asking how much energy a third party observer will detect is meaningless. "

CRISP:

"But you don't need an absolute frame of reference at all if you think the disappearence of the two frames of reference of the two balls are a problem. You just need a third observer who is watching the entire proces. This observer can take the relative motions of the two balls into account and calculate the energy of the resulting collision from his point of view. "

It seems that the Einstein fanclub is in contradiction with each other. I've noticed this posts with contradictions more than once. It reminds me of neo-darwinian discussions.

20. Crisp,

If the energy is released as photons(like a matter-antimatter reaction) I'm pretty sure that all observers would detect the same amount of photons. To assume that one observer would only see five photons while the other observer would see one million photons, sounds very illogical.

As I have indicated in a another thread, I don't consider kinetic energy, energy at all. Kinetic energy is something that physicists use to balance there formulas. Therefore, I wouldn't be surprised if it was relative(considering that it doesn't exist in my frame of reference )

Your problem is that you assume everthing is relative. There are some things that have to be absolute. Science is based on the assumption that something is real if everyone can percieve it(either directly or indirectly). If Einstein is correct, and something that exists in one frame of reference doesn't exist in another, then how can science ever prove that anything exists.

Ive noticed that the more I hear about relative frames of reference, the more I see flaws in the theory. I've found that it is impossible to prove that relativity is wrong when dealing with frames of reference where there are two observers. However, I found that when there are three observers, one observer in both of the other two observers frames of reference, the concept of relativity becomes illogical.

Let me give you an example:

There are three different clocks in one location. Two of the clocks head in opposite directions, both at .90c, from each other while the third clock remains stationairy. According to relativity for every one second of the stationairy clock .44 seconds pass for each of the moving clocks. Logic would dictate:

1 second(stationairy clock)=0.44 seconds(clock a)
and
1 second(stationairy clock)=0.44 seconds(clock b)

Therefore:

0.44 seconds(clock a)=1 second(stationairy clock)=0.44 seconds(clock b)

Therefore:

0.44 second(clock a)=0.44 seconds(clock b)

Therefore:

1 second(clock a)= 1 second(clock b)

Mathematically it would be:

If a=c and b=c then a=b.

In other words logic dictates that time slowed down equally for both clocks, and therefore they are synchronized.

However, Einstein would argue that they can't be synchronized because, in each of their frames of reference, the other clock is moving at a high speed.

This is the kind of problem where I have to choose between logic and Einstein. And for me, the choice will always be logic.

Tom

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