The Underlying Problem With Some Science Is Interpretation.

[A-wal]

You seem to be catching on that your velocity of 1.5c is meaningless.

There's nothing to catch on to. This is just another pathetic and transparent attempt to cover your mistakes.

 

So, let me ask you this, you are sitting in your lab frame and the two objects (call them protons) are approaching each other at the relativistic velocity of 0.96c. Do you agree that is the velocity they will collide with each other?

IN THEIR FRAME! They collide at 1.5c in the lab's frame. Are you seriously trying to claim that the lab's frame is invalid? You can work it out from any frame and the energy of the collision will be the same in all of them because all frames are equally valid. This is the basic fundamental principle of relativity.

 

Go away and learn it properly if can and then you might be be able to make an actual contribution to the discussion instead of trying to find fault with everyone else because you can't fully grasp the discussion. You even thought that Galilean relativity included a constant speed of light.

 

Why do you think that objects moving relative to each faster than c from the perspective of a third object implies a preferred frame of reference?

 

Now, you in your frame make a doodle that they are really approaching each other at 1.5 c (that is what you are claiming) To test that “theory” you open the inspection hatch on the beam accelerator and stick your head in right at the point of collision. Your head is attached to the rest of your body so it is at rest in the lab frame.

 

Questions:

 

1). Does your head being present in the accelerator change the velocity at which the particles collide?

 

2). Does the presence of your head, at rest in the lab frame, change the relative velocities between the two particles at all?

 

Maybe now you can see that what you doodle in your lab frame has no significance at all to the two particles that are moving at relativistic speeds to one another, it is a meaningless “pseudovelocity” and is of no use to you or to science. It is rubbish!

For the last time, if they're moving away from a central object at 0.75c from the frame of the central object then they're moving away from each other at 1.5c in that frame. You only apply the Lorenz transformations when you change frames. You're claim that they aren't moving away from each other in the frame of the central object (that was clearly what you claiming) shows a total lack of understanding about the most basic aspects of SR.

 

As much you you want to save face, the frame in which they are moving away from each other at 1.5c is every bit as valid as their own frames where they're moving away from each at 0.9c because all frames are equally valid. Deal with it.

 

Now, that is established, I would like to call you out on your other claim that you can derive E=mc^2 with just algebra and common sense.

 

As per the forum rules, which Craig just reminded us about, either put up or shut up, or at least post a link where somebody has done this.

From before:

 

B accelerates to a quarter of the speed of light from A's frame of reference. Work out the time dilation and length contraction from B's frame. Apply that to get the difference.

 

B then accelerates to half the speed of light from A's frame of reference. From A's perspective B accelerated by the same amount the second time but B's own perspective they accelerated by a greater amount the second time from and therefore more energy was used the second time.

 

From A's frame a greater amount of energy was used the second time to produce the same amount of acceleration therefore the mass of B was greater the second time B accelerated than it was the first time.

 

The greater the relative velocity of an object, the greater the object's mass because of time dilation and length contraction. Work out the difference in the amount of energy required to accelerate an object by the same amount at different relative velocities and you should be able to get E=mc^2.

 

What you are referring to is the gap velocity of two very distant objects where there is no restriction on the rate of expansion of space, because the local velocity in the vicinity of the objects still obeys the speed limit of c. There is no restriction at all on how fast space can expand, but that has nothing to do with this discussion. No mention has been made about how far apart the objects are. As long as they are in the same locality, they cannot move apart faster than c.

What gives you that impression. I think what he's referring to is two objects that are moving away from each other from the perspective of a third observer.

 

Now, you are free to add the velocities any way you like and get whatever number you like, such as 1.5c, but is has no meaning at all in science. The velocities that have meaning are 0.75c between you and each of the objects, and 0.96c between the two objects. That is all.

Utter BS! The frame in which they're moving at 1.5c is every bit as valid as their own frames.

 

Fair enough, but you did several things that fall into the category of "special pleading": making up your own word, your own special speedometer to justify a velocity of 260,000 km/s in violation of SR, for example. So, I feel justified in pointing that out.

 

As for the velocity addition formula, yes, that is what I posted, but not what others here have been doing.

 

And A-Wal says he CAN do the math! He claims he can derive E=mc^2 with just algebra and common sense, and says that any higher math is "nonsense".

 

OK, as per site rules, let him do this! If he does, I will post my derivation of E=mc^2 from first principles using differential calculus.

 

More BS. Using proper time to get a velocity of 260,000 km/s is in no way in violation of SR.

 

I never said I could do the math. I said it could be done with just algebra and common sense.

 

I think it would be very interesting and valuable for us non-specialist spectators to see you and A-Wal provide your rival derivations of this. 

 

I am intrigued by A-Wal's claim that is easy, as I had thought one needed GR (tensors and that) to derive it. But my training is in chemistry, so I easily get out of my depth with relativity. (I am a bit sounder on QM). 

It would be a mission. You definitely wouldn't need GR, that came later. You could do it by working out the difference in acceleration using the same amount of energy at different relative velocities to work out the the difference in mass.

 

The particle accelerator example would work as well because the lab frame in which they're colliding at 1.5c is equally as valid as they're own frame where they're colliding at 0.9c but the amount of energy released in the collision doesn't change when you change reference frames so the overall mass involved in the collision must be lower in the lab's frame because the relative velocity of the particles is higher.

 

I'm sure you could just play around with those to get E=mc2, something OceanBreaze thinks would be impossible with using tensors.

[OceanBreeze]

 

I'm sure you could just play around with those to get E=mc2, something OceanBreaze thinks would be impossible with using tensors.

 

Ha! No answer then, just more BS. Just as I thought.

 

A rigorous solution would involve at least vector calculus, but I did work it out using only differential calculus and a Lorentz transform.

Edited December 9, 2016 by OceanBreeze

[A-wal]

I made my derivation as simple as I could, no tensor or vector calculus, just using the Lorentz transform and basic differential calculus. A purist would not approve, but it is mathematically correct, but perhaps not rigorous. I am a marine engineer, not a physicist or a mathematician, but SR and GR interest me.

But actually learning how it works doesn't interest you apparently.

 

Why do you think that objects moving relative to each faster than c from the perspective of a third object implies a preferred frame of reference?

 

You completely skipped over the refutations of all your misunderstandings and mistakes and focused on the one part that doesn't really matter. All I said was this:

Then you just have to get the square root of what the TD or LC would need to be on their own because velocity is distance over time so the effect of TD and LC are multiplied together. Because they're equal it should just be a case of what the square root of one of them on its own would need to be. I don't see why the SR equations are so complex. Surely that's a really cack handed way of doing it?

It even has a question mark at the end, I wasn't even making a statement.

 

You can get time dilation and length contraction with only the knowledge that the speed of light is the same in all inertial frames so you can get the velocity addition formula and from that you could work out the difference in mass at different relative velocities by the difference in the amount of acceleration that occurs at those relative velocities using a consistent burst of energy each time.

Edited December 9, 2016 by A-wal

[OceanBreeze]

 

 

 

For the last time, if they're moving away from a central object at 0.75c from the frame of the central object then they're moving away from each other at 1.5c in that frame. You only apply the Lorenz transformations when you change frames. You're claim that they aren't moving away from each other in the frame of the central object (that was clearly what you claiming) shows a total lack of understanding about the most basic aspects of SR.

 

 

 

In the interest of maintaining peace, I will grant that you can add the relative velocities the way you are doing, but it is a meaningless number.

 

You can see that in the particle collider example I presented. No matter what you do, or what frame you choose to view the experiment from, the two particles will collide at 0.96c

 

In the example where the two objects are moving away from you at 0.75c in different directions, you and each of the observers will agree on these things:

 

Object 1 is moving away from you at 0.75 c and both you and the observer on object 1 agree on this relative velocity.

 

Object 2 is moving away from you at 0.75 c and both you and the observer on object 2 agree on this relative velocity.

 

Object 1 and Object 2 are moving away from each other at 0.96 c and both observers on these objects and you will agree on this relative velocity between the two objects.

 

You may add the velocities of Object 1 and 2 in a non-relativistic way, if you wish, and arrive at 1.5 c. Neither of the other observers will agree with you on this relative velocity, making it a frame-specific velocity, something that should be avoided.

 

That is as far as I can go towards agreeing with you.

[OceanBreeze]

 

 

You can get time dilation and length contraction with only the knowledge that the speed of light is the same in all inertial frames so you can get the velocity addition formula and from that you could work out the difference in mass at different relative velocities by the difference in the amount of acceleration that occurs at those relative velocities using a consistent burst of energy each time.

 

If you think so, why don't you do it? Anyway, what you are saying is that you can use the principles of special relativity to derive special relativity. That is a circular argument.

[A-wal]

In the interest of maintaining peace, I will grant that you can add the relative velocities the way you are doing, but it is a meaningless number.

 

You can see that in the particle collider example I presented. No matter what you do, or what frame you choose to view the experiment from, the two particles will collide at 0.96c

I disagree. All frames are equally valid. Your particle collider is a good example. Do you agree with this:

They collide at 1.5c in the lab's frame. Are you seriously trying to claim that the lab's frame is invalid? You can work it out from any frame and the energy of the collision will be the same in all of them because all frames are equally valid. This is the basic fundamental principle of relativity.

and this:

You can get time dilation and length contraction with only the knowledge that the speed of light is the same in all inertial frames so you can get the velocity addition formula and from that you could work out the difference in mass at different relative velocities by the difference in the amount of acceleration that occurs at those relative velocities using a consistent burst of energy each time.

?

Edited December 9, 2016 by A-wal

[A-wal]

If you think so, why don't you do it? Anyway, what you are saying is that you can use the principles of special relativity to derive special relativity. That is a circular argument.

No I'm saying you can derive the exact model that special relativity describes without using any advanced mathematics with only the knowledge that the speed of light is the same in all inertial frames of reference.

[OceanBreeze]

No I'm saying you can derive the exact model that special relativity describes without using any advanced mathematics with only the knowledge that the speed of light is the same in all inertial frames of reference.

 

You keep saying that, but you have not even attempted to actually do it.

 

Anyway, I am tired of arguing with you. Peace!

 

I will post my derivation as soon as I have it in a presentable format and I figure out how to use LaTex on this site.

 

After that, we can discuss it if you want.

[A-wal]

I fully accept your peace treaty and look forward to the derivation to see if I can follow it.

 

Do you now agree that the lab's frame with the particles colliding at 1.5c with a lower overall mass is just as valid as the frame's of either particle with a 0.9c collision and higher overall mass?

[A-wal]

Step 1

LV = Lab frame collision velocity (1.5c).
PV = Particle frame collision velocity (0.9).

3 x 1.5 = 4.5
5 x 0.9 = 4.5
LV = PV x 0.6

The mass in the lab frame is three fifths of what it is in the particles' frames.


LV = Lab frame collision velocity (1.2c).
PV = Particle frame collision velocity (0.8 ).

2 x 1.2 = 2.4
3 x 0.8 = 2.4
LV = PV x 0.6r

The mass in the lab frame is two thirds of what it is in the particles' frames.

 

 

Step 2

?

 

 

Step 3

E=Mc^2 :)

Edited December 9, 2016 by A-wal

[OceanBreeze]

I fully accept your peace treaty and look forward to the derivation to see if I can follow it.

 

 

 

Great. Maybe if you and I can start to agree, there might be a chance of getting through to xyz.

 

Do you now agree that the lab's frame with the particles colliding at 1.5c with a lower overall mass is just as valid as the frame's of either particle with a 0.9c collision and higher overall mass?

 

 

To answer that, I would like to first quote the answer posted in this link: plus, it gives me a chance to test out LaTex

 

Quote:

 

What is the relative speed of two near-light speed particles headed towards each other?

 

Classically we would add the speeds to get 1.8c , which is obviously not allowed. In relativity you simply use the relativistic velocity addition formula:

 

 

 

[math]V = \frac{u + v}{1 + uv/c^2}[/math]

 

(LaTex here is buggy. I had to insert extra spaces that I don't need anywhere else)

 

Anyway,  I prefer not to use any local velocity (and that includes relative velocities) that is greater than c, as that is a violation one of the postulates of relativity.

 

However, keeping in the spirit of peaceful cooperation, I will reluctantly agree you can use the lab frame and 1.5c if you really insist on doing it that way, but I would much prefer that you didn’t.

 

Edited December 9, 2016 by OceanBreeze

[OceanBreeze]

Step 1

LV = Lab frame collision velocity (1.5c).

PV = Particle frame collision velocity (0.9).

 

3 x 1.5 = 4.5

5 x 0.9 = 4.5

LV = PV x 0.6

 

The mass in the lab frame is three fifths of what it is in the particles' frames.

 

 

LV = Lab frame collision velocity (1.2c).

PV = Particle frame collision velocity (0.8 ).

 

2 x 1.2 = 2.4

3 x 0.8 = 2.4

LV = PV x 0.6r

 

The mass in the lab frame is two thirds of what it is in the particles' frames.

 

 

Step 2

?

 

 

Step 3

E=Mc^2 :)

 

Can you expand on step 2 please?

[A-wal]

Anyway,  I prefer not to use any local velocity (and that includes relative velocities) that is greater than c, as that is a violation one of the postulates of relativity.

 

However, keeping in the spirit of peaceful cooperation, I will reluctantly agree you can use the lab frame and 1.5c if you really insist on doing it that way, but I would much prefer that you didn’t.

That makes it very awkward when comparing frames. I always like to stick a central object between any two objects that are in motion relative to each other as a reference frame because it puts the two other objects on an equal footing.

 

It's not a violation of the postulate. The postulate is that no object can move at or over the speed of light relative to the inertial observer, at least that's how it should be worded. I've just looked it up and that's not one of the postulates, it doesn't need to be. The only two are that the laws are the same in all inertial frames and that the speed of light is the same in all of them.

 

Can you expand on step 2 please?

Sure. The formula is whatever it needs to be to keep the collision energy constant despite the varying velocity of the particles relative to each other in all inertial frames of reference. :)

Edited December 9, 2016 by A-wal

[OceanBreeze]

That makes it very awkward when comparing frames. I always like to stick a central object between any two objects that are in motion relative to each other as a reference frame because it puts the two other objects on an equal footing.

 

 

 

It isn't awkward for me, and I have never heard of anyone inserting another object into a two-body problem, but whatever floats your boat.

 

It's not a violation of the postulate. The postulate is that no object can move at or over the speed of light relative to the inertial observer, at least that's how it should be worded. I've just looked it up and that's not one of the postulates, it doesn't need to be. The only two are that the laws are the same in all inertial frames and that the speed of light is the same in all of them.

 

 

It violates the implication of the postulate that is c is the maximum velocity in the universe. As the source I quoted said. But again, if you want to do things differently, who am I to argue?

 

 

Sure. The formula is whatever it needs to be to keep the collision energy constant despite the varying velocity of the particles relative to each other in all inertial frames of reference. :)

 

 

Thanks for clearing that up!

 

Now that I have found the LaTex bug, maybe I can get my derivative posted. It is Effort though!

[A-wal]

It isn't awkward for me, and I have never heard of anyone inserting another object into a two-body problem, but whatever floats your boat.

Try it, it really helps me because it's the perfect starting point to start playing around with other frames because it creates perfect symmetry. I suppose that symmetry is already there but it helps to use it as the central frame. It's more for describing the scenario than thinking about it, it's not really necessary in my head but I used to use it as a go to until I got used to switching frames.

 

It violates the implication of the postulate that is c is the maximum velocity in the universe. As the source I quoted said. But again, if you want to do things differently, who am I to argue?

The postulate is that the speed of light is the same in all inertial frames. The speed limit is derived from that but a better way of looking at it is that no object can move at or over the speed of light relative to any other object in their frame of reference, therefore no object can move at or over twice the speed of light relative to any other object in any reference frame.

[OceanBreeze]

OK, to derive E = mc² I choose as the starting point the Lorentz Factor

Which is: 

[math] \frac { 1 }{ \sqrt { 1\quad -\quad \frac { { v }^{ 2 } }{ { c }^{ 2 } }  }  }[/math]

 

I was going to show how this can be derived using simple algebra and trigonometry, but here is a YouTube video that does it very nicely and save me a lot of time. If you are already familiar with the Lorentz factor just skip the video.

https://www.youtube.com/watch?v=67xr6EZEYV8

 

The next step is to let this Lorentz factor operate on mass, (instead of time as in the video) so that we have:

 

[math]\frac { m }{ { m }_{ o } } =\frac { 1 }{ \sqrt { 1\quad -\quad \frac { { v }^{ 2 } }{ { c }^{ 2 } }  }  }[/math]

 

By simple algebraic manipulation, we solve this expression for v²:

 

[math]\frac { { v }^{ 2 } }{ { c }^{ 2 } } =\quad 1-{ \left( \frac { { m }_{ o } }{ m }  \right)  }^{ 2 }[/math]

 

And:

 

[math]{ v }^{ 2 }=\quad { c }^{ 2 }\left\lceil 1-{ \left( \frac { { m }_{ o } }{ m }  \right)  }^{ 2 } \right\rceil[/math]

 

And that expression is the first important building block in the derivation, and a good place to pause and see if anyone has a question.

[OceanBreeze]

No questions, then? Great! We can proceed :weather_rain:

 

Next, we need to use a bit of calculus to differentiate the expression for v² with respect to m:

 

[math]d{ v }^{ 2 }=\quad 2{ c }^{ 2 }\left( \frac { { m }_{ o }^{ 2 } }{ { m }^{ 3 } }  \right) dm[/math]

 

That is the second important building block. One thing to make note of is the m in the denominator is cubed, not squared.

 

So now we have expressions for v² and dv², what are we going to do with them? One approach is to find an expression for Energy and see if those other expressions can be plugged in.

 

We can make use of the work energy principle in that Work = Force x Distance and the derivative of that is just dW = Fdx, so:

 

[math]dE\quad =\quad \frac { d }{ dt } \left( mv \right) dx[/math]

 

[math]dE\quad =\quad { v }^{ 2 }dm\quad +\quad mv\quad dv[/math]

 

[math]dE\quad =\quad { v }^{ 2 }dm\quad +\quad \frac { 1 }{ 2 } m\quad d{ v }^{ 2 }[/math]

 

In case that last part throws you off, just remember that

[math]\frac { 1 }{ 2 } \quad \frac { d }{ dt } { v }^{ 2 }=\quad v\quad dv[/math]

And everything should be crystal.

 

But look what we now have, an expression for dE that contains both a v² term and a dv² term!

 

All we need to do now is substitute in the two expressions that were derived earlier:

 

[math]dE\quad =\quad { c }^{ 2 }\left\lceil 1-{ \left( \frac { { m }_{ o } }{ m }  \right)  }^{ 2 } \right\rceil dm\quad +\quad \frac { 1 }{ 2 } m\quad 2{ c }^{ 2 }\left( \frac { { m }_{ o }^{ 2 } }{ { m }^{ 3 } }  \right) dm[/math]

 

 

 

 

By very simple algebraic manipulation you can see the two expressions in the parentheses cancel each other out, leaving only :

 

[math]dE\quad =\quad { c }^{ 2 }dm[/math]

 

Integrating both sides:

 

[math]E\quad =\quad m{ c }^{ 2 }[/math]

 

Aaaaannd we are done!

I don’t know if anyone will gain any deep insight into this equation from this simple derivation, but I hope it was at least interesting.