An adjustment to Special Relativity

Let me start saying Albert Einstein was a man truly unique. Who else in the world could ever arrive to the conclusion that time must be relative? That is the idea of a genius. That said, today I have to expose you what, if true, could be a useful adjustment to his Theory of Special Relativity.

We will make use of three well proven tools:
1) The time dilation equation
2) The Hafele-Keating experiment (ok, not really a tool, but a reference to a real world evidence)
3) A thought experiment to put it altogether.

Let’s suppose you are the pilot of a spaceship traveling the deep space, far away from any gravitational source. You just wake up from a long hibernation state, so you don’t know what your speed is. All you know is your spaceship is capable of accelerate and decelerate on a straight line, and that the maximum speed reached is far from the speed of light (these assumptions are appropriate to simplify our calculations).

You are bored, so you prepare an experiment to verify (one more time) the validity of the theory of special relativity. Given the spaceship can only travel longitudinally, you plan to go to the middle of the ship and from there throw two high precision clocks, one to fore-end, and one to aft-end. A pile of pillows will catch the clocks at the end, for they don’t get damaged. Of course, both clocks are already synchronized to the spaceship clock.

spaceship-clocks

Being on a spaceship, you throw 2 clocks on opposite directions to measure time dilation.

While doing the calculations, you find there is missing information. You can’t make any prediction about the output of this experiment, because  the time dilation depends not only on the speed of the clocks, but also on the spaceship speed, which is unknown. Anyway, you throw both clocks with the same speed, and then you compare them both with the spaceship clock.

Let me stop here. Have you noticed something rare on the text above? No? We have the equations but we can’t predict the output (so the readings of both clocks will be undefined for us). What will the clocks show? Will they show an undefined value? I really doubt it. They will delay exactly in the appropriate amount predicted by special relativity.

Now return to the clock readings. Given you don’t know what your speed is, the readings you get could be basically of 2 types:

1) Both clocks show exactly the same time dilation (that is, the spaceship is at rest)
or
2) One clock shows more time dilation than the other, that is, one clock speed up and the other one slow down (what means the spaceship has some speed > 0 in the direction of the slower clock)

Please note you now have one of 2 possible outputs, but most important, you already know what is your state of movement, that is, when you read the clocks you will know for sure if you are at movement or at rest. Nothing new here right? Everybody knows that.

Now, let’s check what we have. By the simple fact of reading the clocks, your unknown speed “collapses” (a Quantum Physics term) to a known state. That is, you certainly know if you are at movement or at rest. But moving relative to what? There is not a planet involved, there is not another spaceship either. It’s just your spaceship, and anyway you know what your speed is. If you know your speed, then you know your speed relative to something, maybe relative to rest. But what is rest? We have no point of reference! It doesn’t matter. We don’t know what is it but it is there anyway, as an absolute reference frame for time dilation. And clocks someway “perceive” that frame of reference. That is what experiments tell us (Hafele-Keating experiment, among others).

So what? If this is true, we may need to do a little “conceptual” adjustment to the theory of special relativity to account for this observation. According to Einstein, no frame of reference is absolute. No experiment can let you know what your own speed is. However, clocks disagree, as we have seen above. Please notice this experiment could be done in a fully closed spaceship, with no windows at all, and it will work anyway.

Now we are talking about an absolute frame of reference I must mention that this absolute frame is the real solution to the Einstein Twins Paradox. Previous solutions talking about the asymmetry of the movements of both twins don’t apply, sorry. We will talk about that in a later article.

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The centripetal force work problem – solved!

A few weeks ago, I wrote the article “Does the centripetal force do some work?“. Since then, I keep myself searching for an answer. Last week, the right answer finally arrived to my inbox. And the answer came from Henry Reich, the creator of MinutePhysics. To be honest, when I wrote him the email my chances of get an answer were really low. However, to my surprise, Henry was patient enough to show me where my error was. Here is the transcription (simplified) of the problem I exposed to Henry, and his very clear response.

First, I explain him why I thought the centripetal force does work:

Henry,
Let me start saying I have some formal physics instruction so I am aware of the energy conservation laws. Actually I don’t think this law can be easily broken. I also know that there is no work done when making a circular path, because of the Work=F*d*Cos(angle) formula.

Now, what takes a man aware of the physics laws to think that it requires work to make a circle? The problem of deflecting an asteroid.

The “deflect an asteroid” problem:
Imagine an asteroid, 1 million tons weight, traveling the flat spacetime with velocity Vx (lets think of it as traveling on a x-y plane). Please note there is no string attached to the asteroid. Just an asteroid happily traveling in a straight line.

You can exert a force Fy over the asteroid, and that force will deflect its trajectory, so Fy will do a work Wy=Fy*dy .
or
You can exert a Fx over the asteroid, opposed to its velocity, so you can slow down the asteroid, and the work done will be Wx=-Fy*dx (negative work, by convention).
To this point every physicist agree there is work done in each of these cases. Perfect!

Now, what if we apply both forces simultaneously? For most physics, work done disappears. Why? Because a circle does not require energy to be created. I know that. The force is always perpendicular to the trajectory, so no work is done.

There is a force Fy applied simultaneously with a Fx force, the asteroid deflects (more than when the Fy force was applied alone), and still there is no work done. Don’t you think this sounds odd, just a little?

If this is true, we can actually deflect such a big asteroid in a circular path, and it will use no energy to do that, because no work has been done!
We can even literally stop the asteroid on a 1 meter circle without any energy expended! I don´t think that could happen in a real world.

Of course, many argue that in fact we will require lots of energy to generate the forces, and that can be true. However, the energy required to generate the force is not considered in my calculation, just the net work done over the asteroid by the force applied. If it takes energy to generate the force, that will add to the net energy required to deflect the asteroid. And that means even more energy would be required.

Does this makes sense?

Please note this goes against what we have learned (It took me months to start to believe this could be true). Also note I am aware that there are lots of things rotating forever with zero energy required (I will explain this later, if we find the asteroid case makes sense).
All we have learned to this day say a circle requires no energy, but please, don’t discard this possibility too fast. Let’s imagine the asteroid and ask your self -no equations involved- Can we really deflect an asteroid without work?

Thanks again for your time reading this.

Henry’s answer to this problem was, of course, brilliant. This is the only response that has taken two apparently different realities and generated a coherent response:

Stalin,

I think you are putting too much faith in your intuition and not enough in your logical reasoning.

Also, I think you are confusing what is meant by “work”: you’re totally correct that in certain cases it can take the expenditure of a LOT of energy to keep an object rotating in a circle – moving an asteroid with rockets is one example – but that is not what work is. “Work” in physics parlance, means specifically energy ADDED to the system in question, and while the rockets may expend a lot of energy changing the asteroid’s direction of motion so that it travels in a circle, if they do so entirely by firing perpendicularly then they do not change the asteroid’s speed or kinetic energy – just the direction of the velocity. The point is, if they stopped firing and let the asteroid continue on its path, it would have the exact same energy it started with (though be traveling in a new direction). And in physics parlance, this means that zero work has been done on the asteroid. This doesn’t mean that nothing happened, or that no one expended any energy! It just means that the energy OF THE ASTEROID has not changed.

I think that you are kind of intuitively using a different definition for “work” than what is meant by physicists. All physicists mean by “work” is the change of total energy of an object over time. If an asteroid in free space starts with velocity V and that velocity changes direction but not magnitude (and nothing else changes), then the total energy of the asteroid is still 1/2*m*V^2 and so the change of energy is zero. If you want “work” to mean something different, that’s fine, but you can’t then go argue with physicists that they’re wrong saying “no work happens”, because by their definition, they are right.

This is why I hate the use of the word “work” by physicists. If I could abolish that word forever and just have everyone say “change of energy” then I think all sorts of confusion would be avoided, because the word “work” has many colloquial meanings, many of which are closer to “power” or “force” or other things than “change of energy”.

Henry

Bell’s spaceship thread does not breaks

Is interesting the kind of ideas that arise from relativity. One of the widely accepted ones are about some kind of “relativistic stresses”. As wikipedia says:

In general, it was concluded by Dewan & Beran and Bell, that relativistic stresses arise when all parts of an object are accelerated the same way with respect to an inertial frame, and that length contraction has real physical consequences.

This assertion is correct, as far as this condition is met: ” when all parts of an object are accelerated the same way with respect to an inertial frame”

However, this is a highly unnatural situation, far from the original “spirit” of the Bell’s spaceship paradox wich initial conditions say:

“Consider two identically constructed rockets at rest in an inertial frame S. Let them face the same direction and be situated one behind the other”.

 

Bell's spaceship paradox

Illustration of w:Bell’s spaceship paradox with vertical motion as suggested by Bell himself, (see also File:Dewan-Beran-Bell-Paradox.svg, rocket taken from http://openclipart.org/detail/ 16043/rocket-by-mystica-16043)

It is obvious that, under this conditions, both spaceships will have exactly the same speed for every instant of time. Moreover, having the same speed they will share the same frame of reference, so B will stay at rest relative to C, and this yields to no broken thread. We can not say the same about A. A will stay at rest, so for A, B and C velocities will be different. Let’s check the facts:

Initial conditions:
V_a=0\\  V_b=0\\  V_c=0\\

Now, let’s analyze the situation by dividing the flight time on very little intervals \Delta t, and we will assume that \Delta t is so small that the initial condition holds for the entire interval.

Step 1 (B and C start motors). The initial conditions of this interval say V_a = V_b = V_c = 0, so A, B and C share a common reference frame, so me assume they share the reference frame for the whole \Delta t. This means \Delta t_a = \Delta t_b = \Delta t_c = \Delta t, because if A, B and C share the reference frame, time flows equally for all of them.
At the end of this step, we will have:
V_a=0\\  V_b=a\Delta t\\  V_c=a\Delta t\\  V_b=V_c=V_1=a\Delta t

Step 2. A is a rest, while B and C has a velocity V_1. This makes B and C to share a common reference frame, while A keeps its original reference frame. This means time flows now equal to B and C, but different for A (by time dilation) \Delta t_b = \Delta t_c = \Delta t. \Delta t_a \neq \Delta t. B and C now share the same reference frame, so for all practical purposes they are at rest relative to each other. This of course means the thread doesn’t break during this interval.

V_a=0\\  V_b=v_1+a\Delta t=2a\Delta t\\  V_c=v_1+a\Delta t=2a\Delta t\\  V_b=V_c=V_2=2a\Delta t

Step 3. A keeps at rest, while B and C has a velocity V_2. B and C share the same reference frame, so they both keep ar rest relative to each other, so no thread breaks.

As you can see we can keep doing this analysis for a long time, and the result will not change. B and C keeps always at rest relative to each other, so the thread never breaks. Now the question is: why this was not seen before? Simple: we are humans, and by starting with A, B and C from the same reference frame and by being acceleration the same from the begining, we easily tend to think B and C will keep synchronized to A. By the way, A keeps synchronized to B and C? No. Why? By the relativity of simultaneity. When B and C get a velocity V the line of simultaneity for them get a little inclination relative to A rest frame, so simultaneous events for B and C are now out of phase for A. The phase difference is given by: \Delta t=Vx/c^2, been x the distance from A to B or C, and V the velocity of B or C. This means (folowing our analysis) that B and C get velocity V_2 at the end of step 2 simultaneously for its own frame of reference, but B get that velocity slightly after C from the A frame of reference. This means the minkowski diagrams for this paradox are wrong too, because they assume B and C get velocity V_2 simultaneously as seen from A frame of reference. The right diagram shows below.

Bell-spaceship-paradox-simultaneity-lines

Bell’s spaceship paradox Minkowski diagram showing the right lines of simultaneity.

Do you see of any error in this article? Let me know! I’ll really appreciate it.

 

The misleading perpendicular force

After seing the article Does the centripetal force do some work? some questions arise: why we could’n see it before? Why are there so many respectable books claiming this? I think I have a culprit: a well known figure that clearly explains why there is no work done when a force is perpendicular to displacement. Yes, it’s all about a simple image. However I must admit it is a very convincing one. Haven’t I found accidentally the asteroid problem we still stay under its hypnotic power.

Let me show, your highness, the innocent gilty of 200 years of conceptual darkness:

Work done by a perpendicular force

The “Classical” graphical demonstration of the work formula.

Is a really convincing one, don’t you think? However I must say, to its favor, that the image is clear and has ever tried to trick us. This figure is only guilty of being misunderstood.

First, let’s check what the image is showing us: it is a simple block, moving over the floor, and an arrow showing us the force applied. Of.course, there is the already classical \Theta angle, pointing out that the force is applied at an angle. The block moves only horizontally (what other direction could it take for christ sake?), so it’s obvious that only the horizontal component of the applied force will do work over the mass. That is what we always have seen. Is there any else?

Normal force

The normal force acting against the vertical component.

Yes! There is a normal force, pointing upwards, equal to the vertical component of the applied force. Don’t you see it? Don’t worry, nobody draws it. Can this missing arrow change things? Of course! When this arrow appears is obvious that the vertical component of the applied force is exactly cancelled by the normal force, so the net force will be zero, and zero force do zero work. There is my friend, the source of this error.

Now think for a minute: will the mass move horizontally if no floor is there to stop it? I rest my case, Your Honor.

Does the centripetal force do some work?

Would you like to show to your physics teacher something he doesn’t know about? Here I leave you the demonstration that the centripetal force, contrary to our beliefs, do work.

Before proceeding, let’s remember that although a circle looks like a “natural” trajectory (after all we see lots of things moving on circles), it happen that bodies don’t move ‘spontaneously’ on circular paths, but on rectilinear paths. Making a circle requires a force that changes the mass trajectory. Even more, it requires a continuous application of this force to keep it circular. Once this force disappears, the object stops making circles and start again a rectilinear motion.

What do we need to create a circle? From the cinematic point of view, all we need is a centripetal acceleration. From the dynamics point of view, all we need is a centripetal force. Now the question is: From the energy point of view, does it requires energy to make a circle? Does this centripetal force do any work over the mass moving on a circle? Or is this just a static force, making no displacement, so the total work will be zero?

Let’s see what formulas say about this.

To analyze the possibility of any work done over the mass we are going to decompose the centripetal acceleration on its x-y components, and calculate a \Delta x and \Delta y displacements over a \Delta \Theta angle.

Work done by a uniform circular motion

Work done by the centripetal force.

The horizontal and vertical little displacements are:
\Delta x = R\Delta\Theta Cos \Theta     (1)
\Delta y = -R\Delta\Theta Sin \Theta     (2)
Please note the minus sign, on the y direction, showing \Delta y is decreasing (y_2<y_1, Thanks exchemist).

The acceleration components over the x and y directions are:
a_{cx} = -a_c Sin \Theta     (3)
a_{cy} = -a_c Cos \Theta     (4)

The work done over a delta x and delta y:

\Delta E_x = \Delta xF_x\\ \Delta E_x =\Delta xa_{cx}m     (5)

\Delta E_y = \Delta yF_y\\ \Delta E_y =\Delta ya_{cy}m     (6)

Replacing (1) and (3) in (5):

\Delta E_x = \Delta xa_{cx}m\\ \Delta E_x =(R\Delta\Theta Cos \Theta)(-a_c Sin \Theta)m\\ \Delta E_x =(R\Delta\Theta Cos \Theta)(- w^2R Sin\Theta)m\\ \Delta E_x =-R^2 Cos\Theta Sin\Theta w^2m\Delta\Theta\\ \Delta E_x =-(R^2w^2) m Cos\Theta Sin\Theta \Delta\Theta\\ \Delta E_x =-V^2 m Cos\Theta Sin\Theta \Delta\Theta     (7)

Replacing (2) and (4) in (6):

\Delta E_y = \Delta ya_{cy}m\\ \Delta E_y =(-R\Delta\Theta Sin \Theta)(-a_c Cos \Theta)m\\ \Delta E_y =(-R\Delta\Theta Sin \Theta)(- w^2R Cos\Theta)m\\ \Delta E_y =R^2 Sin\Theta Cos\Theta w^2m\Delta\Theta\\ \Delta E_y =(R^2w^2) m Sin\Theta Cos\Theta \Delta\Theta\\ \Delta E_y =V^2 m Sin\Theta Cos\Theta \Delta\Theta     (8)

Given the equation:
Sin (x) Cos (x) = \frac {1}{2} Sin(2x)      (9)

Replacing (9) in both (7) and (8):

\Delta E_x =-V^2 m Cos\Theta Sin\Theta \Delta\Theta\\ \Delta E_x =- \frac {1}{2}mV^2 Sin 2\Theta \Delta\Theta     (10)

\Delta E_y =V^2 m Cos\Theta Sin\Theta \Delta\Theta\\ \Delta E_y = \frac {1}{2}mV^2 Sin 2\Theta \Delta\Theta     (11)

So, there we have an expression for the work done by the centripetal force over a small \Delta x and a \Delta y. Note the centripetal force is doing a negative work over the x direction, and a positive work over the y direction. Besides, the magnitude of the delta energy is the same in both x and y directions. That explains why the energy stays the same for every instant of the circular movement. This is the main reason why our physics today believe that no work is done by the centripetal force: because energy does not change on uniform circular motion. However, as I have shown above, a lot of work have been done. If you integrate this formulas over 0<\Theta<\frac{\pi}{2}, then you have the work done over 90° is:

E_x =- \frac {1}{2}mV^2 (12)

E_y = \frac {1}{2}mV^2 (13)

Does those results look familiar to you?

Yes! They are exactly the expressions for the energy of a mass m with velocity V. What do this means?

This results show that in the x direction the centripetal force has done negative work (to stop the mass from moving), and that it has done work equal to the initial energy carried by the mass. In short, that means the centripetal force has stopped the mass in the x direction. At the same time, the centripetal force has done work equal to the initial energy of the mass on the y direction. This in turn means that the centripetal force has accelerated the mass in the y direction from the rest to velocity V, so at every point of the 90° arc the energy of the mass remains constant, even when a work E_x =- \frac {1}{2}mV^2 was done in the x direction and a work of E_y = \frac {1}{2}mV^2 was done on the y direction. Given that energy is required to do work, this also means that it take E =mV^2 to move a mass on a 90° circular arc. Amazing, don't you think? We have going from believing that not work has been done by the centripetal force to demonstrate that the work done is double the initial energy.

Have you found any error in this equations? Please let me know. If not, try develop this to your physics teacher and tell me what happends. Warning: lots of resistance is expected. After all, we have been teached from our childhood to believe that not work has been done in this case. Changing that will take some time.

A perpendicular force do work too!

Saying that a centripetal force do no work over a mass moving on a circle is an error. However is an error easy to understand, once you see what is involved.

If you like physics then you know that the centripetal force do no work on a circle because it is always perpendicular to its trajectory. Search on google and you’ll note that everybody think the same. After all, if you check the work formula, it sounds right. Sorry, I am here to demonstrate that we were plain wrong.

Prior to any demonstration I should say that I know something about physics, I am aware of the w=Fd \cos(\Theta) formula and I am (supossedly) in full possession of my mental faculties.

First let’s consider an “everyday” situation: It’s wendsday, you are on a spaceship traveling through the deep space when you find a 1000 Ton asteroid with a 2000km/h speed, and you feel creative today, so you decide to deflect that asteroid in a perfect circular 90° arc. From your physics knowledge you know that it will take energy to deflect it from its trajectory, but you know that if you deflect it always perpendicular to its trajectory then you will make no work over the asteroid and therefore deflecting it will cost you no energy. So, by using a strong cable, you start deflecting the asteroid in a big 90° arc. However, when you fynally stops, you find that the energy used is considerable. How can it be? You check the available data and double check the trajectory of the asteroid, just to find the trajectory was a perfect circle! Something somewhere must be wrong .

You call to your physics teacher, explain him the extrange results and he says everything is OK. You spent energy creating a force, and was that force what push the asteroid on a circle. Everybody knows that a force is required to make a circle, so was the centripetal force what created the circle, not the energy applied. That explains both where the energy was used, and the fact that a perpendicular force can’t do any work at all when the circle is created. Everybody is happy now! Everybody but you.

So you decide to check this “fact” by checking again the main computer data. You find that the force applied follows the equations

F_{cx} = -F_c Sin \Theta
F_{cy} = -F_c Cos \Theta

You also get the equations of displacement

\Delta x = R\Delta\Theta Cos \Theta
\Delta y = -R\Delta\Theta Sin \Theta

By the work formula, you find the work done is

\Delta E_x =-V^2 m Cos\Theta Sin\Theta \Delta\Theta
\Delta E_y =V^2 m Sin\Theta Cos\Theta \Delta\Theta

Wait a minute! That is clearly non zero work. There is positive work done in the y direction, and negative work done  in the x direction. So energy must be applied to make such work! Even negative work requires energy! This is nuts! We have been wrong about the work done over a circle for hundred of years!

Now, this clearly explains why took energy to deflect that asteroid. Even if the spaceship has low efficiency, is clear that some of the energy applied was used directly to do work over the asteroid. It is also clear why we don’t see any change in the asteroid energy: some additional energy was applied by the horizontal force to reduce the energy at the same time we were increasing it by applying the vertical force. Mistery solved!

Hey wait! If it takes energy to make a circle, why we see so many examples of circles that happend spontaneously? In fact, you can in theory attach the same asteroid to a big planet and no energy will be spent. How do you explain that?