Stuff that can't be right in relativity

[swansont]

It’s interesting to consider how much energy is required in the moving frame.

This is given by 1/2mv^2 where m is the mass being accelerated and v is the velocity attained

If the velocity required is c then the energy required is 1/2mc^2. The amount of mass that must be converted to generate that energy is given by E=mc^2. Ie the energy required is equivalent to half the rest mass of the object that’s accelerated. Note I may have missed out a factor due to the units but I think its ok if consistent units are used.

Ok so far we have managed to accelerate to the velocity of light and only used a finite amount of fuel equal to half the mass of the object we are accelerating. Admittedly where using a 100% efficient mass to energy converter but this is a thought experiment so that’s ok. However the mass must be carried on board and accelerated which will require additional energy (mass) to accelerate it. I have not done the maths but I suspect it would show for a finite amount of fuel you can get arbitrarily close to c but never c unless you start with an infinite amount of fuel..

 

Note We don’t have any 100% mass to energy converters and the usual method of propulsion, by ejecting the products of an energetic chemical reaction out of a nozzle probably have efficiencies in nano percents.

 

As a side note I am new here so hi all you science freaks.

 

 

The formula 1/2 mv^2 is an approximation, and increasingly fails to accurately give the proper kinetic energy as you approach c.

[alwynj48]

The formula 1/2 mv^2 is an approximation, and increasingly fails to accurately give the proper kinetic energy as you approach c.

 

Don't forget I am refering to the accelerating frame. If the formula fails as c is approached I can determine my absolute velocity from the error.

 

What ever velocity I happen to be moving wrt to some other frame will not change my phyics assuming relativity is right.

[D H]

Don't forget I am refering to the accelerating frame. If the formula fails as c is approached I can determine my absolute velocity from the error.

 

What ever velocity I happen to be moving wrt to some other frame will not change my phyics assuming relativity is right.

There are at least two problems here:

• There is no such thing as absolute velocity.
  
• You velocity are applying relativity incorrectly.

Your error is here:

It’s interesting to consider how much energy is required in the moving frame.

This is given by 1/2mv^2 where m is the mass being accelerated and v is the velocity attained

That equation is valid only for small velocities. The relativistic expression for kinetic energy is

 

[math]E_k = \biggl(\frac 1 {\sqrt{1-\frac {v^2}{c^2}}}-1\biggr)m_0c^2[/math]

 

When the velocity is much less than the speed of light this reduces to the familiar [math]1/2\,m_0v^2[/math]. On the other hand, the energy grows without bounds as the velocity approaches the speed of light.

Edited July 6, 2008 by swansont
correct tag

[swansont]

Don't forget I am refering to the accelerating frame. If the formula fails as c is approached I can determine my absolute velocity from the error.

 

What ever velocity I happen to be moving wrt to some other frame will not change my phyics assuming relativity is right.

 

No, you can't find your speed by applying an incorrect equation. Physics is the same in all inertial frames, but it has to be correct physics.

[alwynj48]

No, you can't find your speed by applying an incorrect equation. Physics is the same in all inertial frames, but it has to be correct physics.

 

You are probably not taking in to account we a referring to a craft's frame of reference in which the craft accelerates by throwing momentum out of the back to accelerate. I was assuming that the energy consumed (100% eff) was the usual kinetic energy E=1/2mv^2 to reach a given v. I believe your where making reference to the higher order terms of that equation . I suggested such terms can not be correct or physics would not be the same in all inertial frame. Well I was right about the higher order terms. In fact it must linear for the same reason. The equation is E=cmv where the terms have the usual meaning. The regular ke equation is simply not the correct one to use with or with out the high order terms. So we where both had the physics wrong LOL.

 

The result given above is suprising. But again if you think about the context it is seems reasonable. For given impulse the craft must accelerate by the same amount no matter what velocity it has already achieved. An other surprise is only a finite amount of energy is required to get to c ie mc^2 the equivalent energy of the mass being accelerated. This result is a quick fiddle with the equations and although I have checked them there may still be a mistake. If anyone is interested I can clean up the derivation and send them a copy in mcad and possibly an other format that mcad can produce.

 

So getting back to the original question. Can we accelerate to the velocity of light using on board fuel. I previously would have said no and even now I believe you can not jsut by throwing momentum out of the back because even with 100% efficiency you would must convert the total mass of the craft get to c. Speculating: having got very close to c a perhaps different thrust mechanism could propel a sub craft to c and beyond. Wow previously I did not think this was even theoretical possible. Though of cause there are huge practical difficulties.

[ajb]

A quick remark, please forget about the idea of mass being dependant on velocity. It is not. By mass we mean the "[math]m[/math]" in the mass-shell condition (in units c =1)

 

[math] E^{2}-p^{2} = m^{2}[/math]

 

(or indeed the "[math]m[/math]" in the 4-momenta and proper force.)

 

This is sometimes called the "rest mass" in the older literature. The so called "relativistic mass" is seldom used.

[swansont]

You are probably not taking in to account we a referring to a craft's frame of reference in which the craft accelerates by throwing momentum out of the back to accelerate. I was assuming that the energy consumed (100% eff) was the usual kinetic energy E=1/2mv^2 to reach a given v. I believe your where making reference to the higher order terms of that equation . I suggested such terms can not be correct or physics would not be the same in all inertial frame. Well I was right about the higher order terms. In fact it must linear for the same reason. The equation is E=cmv where the terms have the usual meaning. The regular ke equation is simply not the correct one to use with or with out the high order terms. So we where both had the physics wrong LOL.

 

Where did you find E=cmv? That's not right either.

 

The KE equation is the same in all frames. We just use 1/2 mv^2 for small speeds because it's a convenient and reasonable approximation.

 

The result given above is suprising. But again if you think about the context it is seems reasonable. For given impulse the craft must accelerate by the same amount no matter what velocity it has already achieved. An other surprise is only a finite amount of energy is required to get to c ie mc^2 the equivalent energy of the mass being accelerated. This result is a quick fiddle with the equations and although I have checked them there may still be a mistake. If anyone is interested I can clean up the derivation and send them a copy in mcad and possibly an other format that mcad can produce.

 

No, this is still wrong. There's no point in speculating further until you get the basics right. D H gave the equation a few posts back.

[alwynj48]

Where did you find E=cmv? That's not right either.

 

The KE equation is the same in all frames. We just use 1/2 mv^2 for small speeds because it's a convenient and reasonable approximation.

 

 

 

No, this is still wrong. There's no point in speculating further until you get the basics right. D H gave the equation a few posts back.

 

 

I will try one more time then rest my case.

 

I thing we can agree that KE is the work done in accelerating a mass to a given velocity ie the product of force times distance. Apparently you do not understand that is not the work done in throwing momentum out of the back of a spacecraft. That’s the equation I have given with the energy in units of mass. Specifically its the mass you must turn in to photons and direct out of the back of the spacecraft to generate the thrust to get to the given velocity. Previously I did think I could use KE by simple equating it to the energy of the fuel used. But that is incorrect for the reasons I have already given.

 

I have already explained I did not find the equation I derived it. I claim no originality it has probably been derived a million time before and been named after someone.

 

This is not rocket science oh wait a moment it is and what more its relativistic rocket science LOL Ok so it is rocket science but it only involves elementary physics, momentum and energy of a photon, laws of motion and momentum plus a bit e=mc^2.

 

I have just thought of this, perhaps it will help you: If you want to use the relativistic KE equation you must do so from the perspective of the stationary frame. So let consider how the space craft will appear from that frame. Ok so in the frame of space craft its converting mass in to photons and projecting them out of the back of the craft. It accelerated and eventual approaches c let suppose that some how it acquires some extra mass and hence accelerates past c.

From the perceptive of the stationary frame an observer might calculate the KE energy of the space craft as it approaches c and conclude that its KE is approaching infinite.

He might wonder how the spacecraft is obtaining this almost infinite amount of energy. So he looks very closely at the space craft and notice that the craft is powered by a mass to energy convertor.

He also notices that the space craft appears to him to be converting almost an infinite amounts mass to energy and this is where the almost infinite KE comes from.

 

However this curious observer not only knows about the relativistic KE equation he also knows about Lorentz transforms. He’s smart dude is this observer. So he transforms the mass to the frame of the space craft. Everything is now ok. The infinite mass has disappeared and in this transformed perspective the space craft can accelerate up to c and beyond with out requiring an infinite amount of anything. Note I have previously explained the craft will run out of fuel before it gets to c.

 

One more point suppose the statuary observer transformed his value of the KE of the space craft to the frame of the space craft???

Do you still thing KE is not dependent on your inertial frame? Rhetorical questions.

[D H]

This is not rocket science oh wait a moment it is and what more its relativistic rocket science LOL

No, it's not. It is just Newtonian mechanics applied erroneously. The Usenet Physics FAQ page on the relativistic rocket is relativistic rocket science.

[Pete]

A quick remark, please forget about the idea of mass being dependant on velocity. It is not. By mass we mean the "[math]m[/math]" in the mass-shell condition (in units c =1)

That is not true for all physicists. In fact if one were to look in the textual literature they'd see that m is often used to mean relativistic mass. And it would be unwise to forget about it because no other definition works in all generality, especially not the "mass = proper mass" definition. That only works for particles and closed systems.

[math] E^{2}-p^{2} = m^{2}[/math]

 

(or indeed the "[math]m[/math]" in the 4-momenta and proper force.)

 

This is sometimes called the "rest mass" in the older literature. The so called "relativistic mass" is seldom used.

In fact its used in over 50% of the textual literature (more like 70% in fact). Gary Oas did a study on this. His results are at http://arxiv.org/abs/physics/0504110

 

A bibliography of his work is at

http://arxiv.org/abs/physics/0504111

 

Pete

[ydoaPs]

Pete, it seems odd that you would be correct when every physicist here disagrees with you about physicists' usage of relativistic mass.

[Pete]

Pete, it seems odd that you would be correct when every physicist here disagrees with you about physicists' usage of relativistic mass.

I've done more research into this concept than most people so I'm sure I'm diffferent in that respect. I've come to learn more general applications of the concept of mass that most people. In fact I've come to learn that the concept of mass as rest mass cannot be made to apply in general and therefore cannot serve as a valid definition of mass in general.

 

Also, I don't understand what it is that every physicist here disagrees with. What is it that they are disagreeing with? The number of texts/books etc. in which the concept is used? The number of journal articles which employ the notion? If so then take a look at figure 1 in the article I quoted. It shows that there are more relativity textbooks utilize the concept of relativistic mass than those that don't (with the exclusion of the years 1980-1984)

 

First off let us recall exactly what I said. I.e. in response to the comment By mass we mean .. I wrote

That is not true for all physicists.

Do you beliece that is untrue? Now recall the statement

This is sometimes called the "rest mass" in the older literature. The so called "relativistic mass" is seldom used.

I was questioning the use of the term seldom when I wrote

In fact its used in over 50% of the textual literature (more like 70% in fact).

Do you think the source I provided for this data is wrong? If so then why? If you have a source which gives other data I would like to read it. Thanks.

 

Pete

Edited July 11, 2008 by Pete

[ydoaPs]

How current is the material that uses relativistic mass?

[Pete]

How current is the material that uses relativistic mass?

That information is available in the article I referenced above. If you read the article you'd see that in the years 2000-2005 is the most recent data in Fig. 1 and it shows that 8 out of 12 relativity texts use it. The article

 

The inertia of stress, Rodrigo Medina, Am. J. Phys. 74(11), November 2006

 

is more recent. The concept of the inertia of stress is one that comes from the concept or relativistic mass.

 

The current definition of mass is given at Argonne National Laboratory

http://www.neutron.anl.gov/hyper-physics/inertia.html and refers to relativistic mass rather than rest mass.

 

Same thing with Lawrence Berkeley National Laboratory

http://aether.lbl.gov/www/classes/p139/animation/sr.html

 

and CERN

http://teachers.web.cern.ch/teachers/archiv/HST2001/accelerators/teachers%20notes/cyclotron.htm

The real limiting factor is the basic design – all particles must orbit at the same frequency, whatever their speed. As particles approach the speed of light, however, they behave as if their mass is increasing. Accelerating them becomes more difficult and they start to lag behind the oscillating electric field. As cyclotrons approached 20 MeV they began to reach their limits and a new design had to be produced.

 

Then there are articles such as this

http://www2.fpm.wisc.edu/safety/Radiation/2000%20Manual/chapter12.pdf

 

Also for the sake of example, I constructed a list of online university lecture notes on the use of rel-mass. Its on my web site at - http://www.geocities.com/physics_world/ref/relativistic_mass.htm. See University Lecture Notes - Online Examples

 

Pete

Edited July 11, 2008 by Pete

[swansont]

I put a mass on a scale with a large digital readout that gives 1000N as the weight, i.e. the gravitational attraction between the two.

 

Someone comes by moving at a large speed such that gamma = 2. Of course, to the observer, it looks like the planet and mass on the scale are moving at that speed. What does the scale reading appear to be to the observer?

[Pete]

I put a mass on a scale with a large digital readout that gives 1000N as the weight, i.e. the gravitational attraction between the two.

 

Someone comes by moving at a large speed such that gamma = 2. Of course, to the observer, it looks like the planet and mass on the scale are moving at that speed.

I disagree. The observer can tell if he is at rest in a gravitational field. If the observer is moving at constant speed perpedicular to the field lines (i.e. sliding along the ground) then there will be a frame-dragging effect in the frame which is moving which is not there in the rest frame.

What does the scale reading appear to be to the observer?

If the scale is at rest in the gravitational field then the scale will read 2000N because W = mg. W_0 = m_0*g, W = gamma*m_0*g = 2m_0*g = 2W_0 = 2000N. This means that a moving body weighs more than the same body at rest.

 

Pete

[swansont]

If the scale is at rest in the gravitational field then the scale will read 2000N because W = mg. W_0 = m_0*g, W = gamma*m_0*g = 2m_0*g = 2W_0 = 2000N. This means that a moving body weighs more than the same body at rest.

 

Pete

 

Wouldn't g also double, from the mass of the planet doubling?

 

But, more importantly, how does the scale readout change for different observers?

[Pete]

Wouldn't g also double, from the mass of the planet doubling?

No. Its the body that is moving, not the planet. However, if the planet was moving then g would change.

 

Pete

[insane_alien]

No. Its the body that is moving, not the planet. However, if the planet was moving then g would change.

 

Pete

 

that would require a preffered reference fframe of which there are none.

 

i think you should study relativity a bit more closely.

[Mr Skeptic]

Wouldn't g also double, from the mass of the planet doubling?

 

But, more importantly, how does the scale readout change for different observers?

 

Wouldn't the electromagnetic forces between the atoms in the spring look distorted as well?

[swansont]

No. Its the body that is moving, not the planet. However, if the planet was moving then g would change.

 

Pete

 

Why does the mass of the object change but not the mass of the planet? How can you tell if the planet is moving?

[Pete]

that would require a preffered reference fframe of which there are none.

In SR there are no preferred frames. And I didn't say there were in GR. What I did said is that in GR one can at times choose a frame which is at rest with the source of gravity and that frame will have unique properties. I.e. that although there is no meaning for "the" rest frame one can at times (as in the example I gave) define such a frame which is unique, e.g. one in which the source of gravity is not moving. The analogy in EM would be a frame of reference at rest with respect to a static charge distribution. One can then say that they are at rest with respect to a unique frame of refererence. Unique does not mean absolute.

 

As a concrete example consider a uniform gravitiational field in frame S'. In that frame there is no g_01 cross term in the metric and thus no frame-dragging. If the observer transforms to a frame of reference S' which is moving at constant speed in a direction which is perpendicular to the field then in that frame there will be a cross term g_0'1' in the metric and thus in that frame there will be a frame-dragging effect. That means that if you launched a particle straight up it would be inertially deflected by the field. This wouldn't happen in the frame S. Thus one can determine a rest frame which is different than a rest frame in S'.

Why does the mass of the object change but not the mass of the planet? How can you tell if the planet is moving?

I didn't say that the mass of the planet didn't change. Sorry if you got that impression. I was using an example in which the observer was at rest with respect to the source of gravity. The mass does indeed change if the planet moves.

 

Pete

Edited July 13, 2008 by Pete

[swansont]

I didn't say that the mass of the planet didn't change. Sorry if you got that impression. I was using an example in which the observer was at rest with respect to the source of gravity. The mass does indeed change if the planet moves.

 

Pete

 

So how can the readout give 1000N for the planet-bound observer, and 4000N for the moving one?

[Pete]

So how can the readout give 1000N for the planet-bound observer, and 4000N for the moving one?

Good question. I guess I'd have to say that its what Einstein's theory of relativity predicts. But it was 2000N, not 4000N. And you have to keep in mind that the scale is at rest in the planet-bound observer's frame.

 

Consider this from another standpoint. Let us use the Equivalence Principle to analyze this. Consider a scale which is uniformly accelerating in the inertial frame S where the weighing surface has its normal vector in the direction of the scales acceleration. A body is sliding across the scales weighing surface. According to an observer in S who is momentarily at rest with respect to the rest frame of the scale the force on the body due to the scale will have the value F = dp/dt. Since the acceleration is perpendicular to motion it can be shown that F = m_ta where m_t = the transverse mass of the particle. It so happens that m_t = m = relativistic mass of particle. Therefore the force on the bodydue to the scale, which we call the weight of the body, is W = F = ma. Using the equivalence principle we then get W = mg where g = a = acceleration due to gravity and the field is a uniform gravitational field.

 

Pete

[swansont]

Good question. I guess I'd have to say that its what Einstein's theory of relativity predicts. But it was 2000N, not 4000N. And you have to keep in mind that the scale is at rest in the planet-bound observer's frame.

 

Why is the mass moving but the planet not moving? They are at rest with respect to each other.

 

Consider this from another standpoint. Let us use the Equivalence Principle to analyze this. Consider a scale which is uniformly accelerating in the inertial frame S where the weighing surface has its normal vector in the direction of the scales acceleration. A body is sliding across the scales weighing surface. According to an observer in S who is momentarily at rest with respect to the rest frame of the scale the force on the body due to the scale will have the value F = dp/dt. Since the acceleration is perpendicular to motion it can be shown that F = m_ta where m_t = the transverse mass of the particle. It so happens that m_t = m = relativistic mass of particle. Therefore the force on the bodydue to the scale, which we call the weight of the body, is W = F = ma. Using the equivalence principle we then get W = mg where g = a = acceleration due to gravity and the field is a uniform gravitational field.

 

Pete

 

 

Mass being a vector is one of the issues folks have with this definition of mass.